3.2.98 \(\int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [198]

Optimal. Leaf size=474 \[ \frac {8 a^4 b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))} \]

[Out]

8*a^4*b^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(9/2)/d+12*a^2*b^2*(a^2+b^2)*arctan((b+a*
tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(9/2)/d+a^4*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^
2)^(1/2))/(a^2-b^2)^(9/2)/d+1/12*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))^2-3/4*a*cos(d*x+c)/(a+b)^4/d/(1-sin(d*x+c
))+1/12*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/12*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))^2+3/4*a*cos(d*x+c)/(a-b)^
4/d/(1+sin(d*x+c))-1/12*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))+1/2*a^4*b*cos(d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c)
)^2+3/2*a^5*b*cos(d*x+c)/(a^2-b^2)^4/d/(a+b*sin(d*x+c))+4*a^3*b^3*cos(d*x+c)/(a^2-b^2)^4/d/(a+b*sin(d*x+c))

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Rubi [A]
time = 0.66, antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2810, 2729, 2727, 2743, 2833, 12, 2739, 632, 210} \begin {gather*} \frac {12 a^2 b^2 \left (a^2+b^2\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {a^4 \left (2 a^2+b^2\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {8 a^4 b^2 \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac {3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac {3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^3,x]

[Out]

(8*a^4*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (12*a^2*b^2*(a^2 + b^2)*A
rcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (a^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])^2) - (
3*a*Cos[c + d*x])/(4*(a + b)^4*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[
c + d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])^2) + (3*a*Cos[c + d*x])/(4*(a - b)^4*d*(1 + Sin[c + d*x])) - Cos[c
+ d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2)
+ (3*a^5*b*Cos[c + d*x])/(2*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (4*a^3*b^3*Cos[c + d*x])/((a^2 - b^2)^4*d*
(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2810

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Int[ExpandIntegrand
[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^m/(1 - Sin[e + f*x]^2)^(p/2)), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a
^2 - b^2, 0] && IntegersQ[m, p/2]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac {1}{4 (a+b)^3 (-1+\sin (c+d x))^2}+\frac {3 a}{4 (a+b)^4 (-1+\sin (c+d x))}+\frac {1}{4 (a-b)^3 (1+\sin (c+d x))^2}-\frac {3 a}{4 (a-b)^4 (1+\sin (c+d x))}+\frac {a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3}+\frac {4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {6 a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac {(3 a) \int \frac {1}{1+\sin (c+d x)} \, dx}{4 (a-b)^4}+\frac {\int \frac {1}{(1+\sin (c+d x))^2} \, dx}{4 (a-b)^3}+\frac {(3 a) \int \frac {1}{-1+\sin (c+d x)} \, dx}{4 (a+b)^4}+\frac {\int \frac {1}{(-1+\sin (c+d x))^2} \, dx}{4 (a+b)^3}+\frac {\left (4 a^3 b^2\right ) \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{\left (a^2-b^2\right )^3}+\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^3} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (6 a^2 b^2 \left (a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}\\ &=\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{12 (a-b)^3}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{12 (a+b)^3}+\frac {\left (4 a^3 b^2\right ) \int \frac {a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac {a^4 \int \frac {-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )^3}+\frac {\left (12 a^2 b^2 \left (a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {a^4 \int \frac {2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac {\left (4 a^4 b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac {\left (24 a^2 b^2 \left (a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {\left (a^4 \left (2 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac {\left (8 a^4 b^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac {\left (16 a^4 b^2\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}+\frac {\left (a^4 \left (2 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=\frac {8 a^4 b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac {\left (2 a^4 \left (2 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=\frac {8 a^4 b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac {3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac {3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 351, normalized size = 0.74 \begin {gather*} \frac {\frac {96 a^2 \left (2 a^4+21 a^2 b^2+12 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2}}-\frac {\sec ^3(c+d x) \left (-264 a^6 b-358 a^4 b^3+8 a^2 b^5-16 b^7-8 \left (44 a^6 b+55 a^4 b^3+8 a^2 b^5-2 b^7\right ) \cos (2 (c+d x))-2 \left (28 a^6 b+89 a^4 b^3-12 a^2 b^5\right ) \cos (4 (c+d x))+22 a^5 b^2 \sin (c+d x)-264 a^3 b^4 \sin (c+d x)+32 a b^6 \sin (c+d x)+32 a^7 \sin (3 (c+d x))-91 a^5 b^2 \sin (3 (c+d x))-244 a^3 b^4 \sin (3 (c+d x))-12 a b^6 \sin (3 (c+d x))-17 a^5 b^2 \sin (5 (c+d x))-76 a^3 b^4 \sin (5 (c+d x))-12 a b^6 \sin (5 (c+d x))\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))^2}}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^3,x]

[Out]

((96*a^2*(2*a^4 + 21*a^2*b^2 + 12*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(9/2) - (
Sec[c + d*x]^3*(-264*a^6*b - 358*a^4*b^3 + 8*a^2*b^5 - 16*b^7 - 8*(44*a^6*b + 55*a^4*b^3 + 8*a^2*b^5 - 2*b^7)*
Cos[2*(c + d*x)] - 2*(28*a^6*b + 89*a^4*b^3 - 12*a^2*b^5)*Cos[4*(c + d*x)] + 22*a^5*b^2*Sin[c + d*x] - 264*a^3
*b^4*Sin[c + d*x] + 32*a*b^6*Sin[c + d*x] + 32*a^7*Sin[3*(c + d*x)] - 91*a^5*b^2*Sin[3*(c + d*x)] - 244*a^3*b^
4*Sin[3*(c + d*x)] - 12*a*b^6*Sin[3*(c + d*x)] - 17*a^5*b^2*Sin[5*(c + d*x)] - 76*a^3*b^4*Sin[5*(c + d*x)] - 1
2*a*b^6*Sin[5*(c + d*x)]))/((a^2 - b^2)^4*(a + b*Sin[c + d*x])^2))/(96*d)

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Maple [A]
time = 0.79, size = 354, normalized size = 0.75 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/3/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^3+1/2/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^2-1/2/(a-b)^4*(-2*a-b)/(tan(1/2*
d*x+1/2*c)+1)-1/3/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^2-1/2/(a+b)^4*(-2*a+b)/(
tan(1/2*d*x+1/2*c)-1)+2*a^2/(a-b)^4/(a+b)^4*((1/2*a*b^2*(5*a^2+6*b^2)*tan(1/2*d*x+1/2*c)^3+1/2*b*(4*a^4+15*a^2
*b^2+14*b^4)*tan(1/2*d*x+1/2*c)^2+11/2*b^2*a*(a^2+2*b^2)*tan(1/2*d*x+1/2*c)+2*b*a^4+7/2*a^2*b^3)/(a*tan(1/2*d*
x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*(2*a^4+21*a^2*b^2+12*b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d
*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.43, size = 1249, normalized size = 2.64 \begin {gather*} \left [\frac {4 \, a^{8} b - 16 \, a^{6} b^{3} + 24 \, a^{4} b^{5} - 16 \, a^{2} b^{7} + 4 \, b^{9} - 2 \, {\left (28 \, a^{8} b + 61 \, a^{6} b^{3} - 101 \, a^{4} b^{5} + 12 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (8 \, a^{8} b - 25 \, a^{6} b^{3} + 27 \, a^{4} b^{5} - 11 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (2 \, a^{6} b^{2} + 21 \, a^{4} b^{4} + 12 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (2 \, a^{7} b + 21 \, a^{5} b^{3} + 12 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - {\left (2 \, a^{8} + 23 \, a^{6} b^{2} + 33 \, a^{4} b^{4} + 12 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (2 \, a^{9} - 8 \, a^{7} b^{2} + 12 \, a^{5} b^{4} - 8 \, a^{3} b^{6} + 2 \, a b^{8} + {\left (17 \, a^{7} b^{2} + 59 \, a^{5} b^{4} - 64 \, a^{3} b^{6} - 12 \, a b^{8}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{9} - 9 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + 5 \, a^{3} b^{6} - 3 \, a b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{10} b^{2} - 5 \, a^{8} b^{4} + 10 \, a^{6} b^{6} - 10 \, a^{4} b^{8} + 5 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{11} b - 5 \, a^{9} b^{3} + 10 \, a^{7} b^{5} - 10 \, a^{5} b^{7} + 5 \, a^{3} b^{9} - a b^{11}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - {\left (a^{12} - 4 \, a^{10} b^{2} + 5 \, a^{8} b^{4} - 5 \, a^{4} b^{8} + 4 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, a^{8} b - 8 \, a^{6} b^{3} + 12 \, a^{4} b^{5} - 8 \, a^{2} b^{7} + 2 \, b^{9} - {\left (28 \, a^{8} b + 61 \, a^{6} b^{3} - 101 \, a^{4} b^{5} + 12 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a^{8} b - 25 \, a^{6} b^{3} + 27 \, a^{4} b^{5} - 11 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (2 \, a^{6} b^{2} + 21 \, a^{4} b^{4} + 12 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (2 \, a^{7} b + 21 \, a^{5} b^{3} + 12 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - {\left (2 \, a^{8} + 23 \, a^{6} b^{2} + 33 \, a^{4} b^{4} + 12 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (2 \, a^{9} - 8 \, a^{7} b^{2} + 12 \, a^{5} b^{4} - 8 \, a^{3} b^{6} + 2 \, a b^{8} + {\left (17 \, a^{7} b^{2} + 59 \, a^{5} b^{4} - 64 \, a^{3} b^{6} - 12 \, a b^{8}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{9} - 9 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + 5 \, a^{3} b^{6} - 3 \, a b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{10} b^{2} - 5 \, a^{8} b^{4} + 10 \, a^{6} b^{6} - 10 \, a^{4} b^{8} + 5 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{11} b - 5 \, a^{9} b^{3} + 10 \, a^{7} b^{5} - 10 \, a^{5} b^{7} + 5 \, a^{3} b^{9} - a b^{11}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - {\left (a^{12} - 4 \, a^{10} b^{2} + 5 \, a^{8} b^{4} - 5 \, a^{4} b^{8} + 4 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/12*(4*a^8*b - 16*a^6*b^3 + 24*a^4*b^5 - 16*a^2*b^7 + 4*b^9 - 2*(28*a^8*b + 61*a^6*b^3 - 101*a^4*b^5 + 12*a^
2*b^7)*cos(d*x + c)^4 - 4*(8*a^8*b - 25*a^6*b^3 + 27*a^4*b^5 - 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^
2 + 21*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a^3*b^5)*cos(d*x + c)^3*sin(d*x + c
) - (2*a^8 + 23*a^6*b^2 + 33*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x
+ c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/
(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b
^8 + (17*a^7*b^2 + 59*a^5*b^4 - 64*a^3*b^6 - 12*a*b^8)*cos(d*x + c)^4 - 2*(4*a^9 - 9*a^7*b^2 + 3*a^5*b^4 + 5*a
^3*b^6 - 3*a*b^8)*cos(d*x + c)^2)*sin(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10
- b^12)*d*cos(d*x + c)^5 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c
)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3), 1/6*(2*a
^8*b - 8*a^6*b^3 + 12*a^4*b^5 - 8*a^2*b^7 + 2*b^9 - (28*a^8*b + 61*a^6*b^3 - 101*a^4*b^5 + 12*a^2*b^7)*cos(d*x
+ c)^4 - 2*(8*a^8*b - 25*a^6*b^3 + 27*a^4*b^5 - 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^2 + 21*a^4*b^4
+ 12*a^2*b^6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a^3*b^5)*cos(d*x + c)^3*sin(d*x + c) - (2*a^8 + 2
3*a^6*b^2 + 33*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 -
b^2)*cos(d*x + c))) - (2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 + (17*a^7*b^2 + 59*a^5*b^4 - 64*a^
3*b^6 - 12*a*b^8)*cos(d*x + c)^4 - 2*(4*a^9 - 9*a^7*b^2 + 3*a^5*b^4 + 5*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^2)*sin
(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d*cos(d*x + c)^5 - 2*(a^11*b
- 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2
+ 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**3, x)

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Giac [A]
time = 12.29, size = 632, normalized size = 1.33 \begin {gather*} \frac {\frac {3 \, {\left (2 \, a^{6} + 21 \, a^{4} b^{2} + 12 \, a^{2} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (5 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{6} b + 7 \, a^{4} b^{3}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 56 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a^{4} b - 20 \, a^{2} b^{3} - b^{5}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(3*(2*a^6 + 21*a^4*b^2 + 12*a^2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/
2*c) + b)/sqrt(a^2 - b^2)))/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) + 3*(5*a^5*b^2*t
an(1/2*d*x + 1/2*c)^3 + 6*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^6*b*tan(1/2*d*x + 1/2*c)^2 + 15*a^4*b^3*tan(1/2
*d*x + 1/2*c)^2 + 14*a^2*b^5*tan(1/2*d*x + 1/2*c)^2 + 11*a^5*b^2*tan(1/2*d*x + 1/2*c) + 22*a^3*b^4*tan(1/2*d*x
+ 1/2*c) + 4*a^6*b + 7*a^4*b^3)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a*tan(1/2*d*x + 1/2*c)^2 +
2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 2*(3*a^5*tan(1/2*d*x + 1/2*c)^5 + 24*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*a*b
^4*tan(1/2*d*x + 1/2*c)^5 - 9*a^4*b*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 3*b^5*tan(1/2
*d*x + 1/2*c)^4 - 10*a^5*tan(1/2*d*x + 1/2*c)^3 - 56*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^4*tan(1/2*d*x + 1/
2*c)^3 + 36*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 36*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*tan(1/2*d*x + 1/2*c) + 24
*a^3*b^2*tan(1/2*d*x + 1/2*c) + 9*a*b^4*tan(1/2*d*x + 1/2*c) - 15*a^4*b - 20*a^2*b^3 - b^5)/((a^8 - 4*a^6*b^2
+ 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d

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Mupad [B]
time = 11.22, size = 1099, normalized size = 2.32 \begin {gather*} \frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (2\,a^6\,b+21\,a^4\,b^3+12\,a^2\,b^5\right )}{a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (14\,a^7+23\,a^5\,b^2+242\,a^3\,b^4+36\,a\,b^6\right )}{3\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^5+21\,a^3\,b^2+12\,a\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (30\,a^4\,b+71\,a^2\,b^3+4\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {a^2\,\left (42\,a^4\,b+61\,a^2\,b^3+2\,b^5\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^6\,b+16\,a^4\,b^3+43\,a^2\,b^5\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-2\,a^7+65\,a^5\,b^2+131\,a^3\,b^4+16\,a\,b^6\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (22\,a^6\,b+233\,a^4\,b^3+153\,a^2\,b^5+12\,b^7\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )}{a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-6\,a^6+105\,a^4\,b^2+208\,a^2\,b^4+8\,b^6\right )}{3\,\left (a^2-b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+12\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^2+12\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-a^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-4\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a^2-4\,b^2\right )+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {\frac {a^2\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )\,\left (2\,a^8\,b-8\,a^6\,b^3+12\,a^4\,b^5-8\,a^2\,b^7+2\,b^9\right )}{2\,{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}{{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}}}{2\,a^6+21\,a^4\,b^2+12\,a^2\,b^4}\right )\,\left (2\,a^4+21\,a^2\,b^2+12\,b^4\right )}{d\,{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + b*sin(c + d*x))^3,x)

[Out]

((3*tan(c/2 + (d*x)/2)^8*(2*a^6*b + 12*a^2*b^5 + 21*a^4*b^3))/(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)
- (2*tan(c/2 + (d*x)/2)^5*(36*a*b^6 + 14*a^7 + 242*a^3*b^4 + 23*a^5*b^2))/(3*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^
4 - 4*a^6*b^2)) - (4*tan(c/2 + (d*x)/2)^7*(12*a*b^4 + 2*a^5 + 21*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b
^2)) + (2*tan(c/2 + (d*x)/2)^2*(30*a^4*b + 4*b^5 + 71*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (a^2
*(42*a^4*b + 2*b^5 + 61*a^2*b^3))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (10*tan(c/2 + (d*x)/2)
^4*(4*a^6*b + 43*a^2*b^5 + 16*a^4*b^3))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (4*tan(c/2 + (d*
x)/2)^3*(16*a*b^6 - 2*a^7 + 131*a^3*b^4 + 65*a^5*b^2))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (
2*tan(c/2 + (d*x)/2)^6*(22*a^6*b + 12*b^7 + 153*a^2*b^5 + 233*a^4*b^3))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4
- 3*a^4*b^2)) + (a^3*tan(c/2 + (d*x)/2)^9*(2*a^4 + 12*b^4 + 21*a^2*b^2))/(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 -
4*a^6*b^2) - (a*tan(c/2 + (d*x)/2)*(8*b^6 - 6*a^6 + 208*a^2*b^4 + 105*a^4*b^2))/(3*(a^2 - b^2)*(a^6 - b^6 + 3*
a^2*b^4 - 3*a^4*b^2)))/(d*(tan(c/2 + (d*x)/2)^4*(2*a^2 + 12*b^2) - tan(c/2 + (d*x)/2)^6*(2*a^2 + 12*b^2) + a^2
*tan(c/2 + (d*x)/2)^10 - a^2 + tan(c/2 + (d*x)/2)^2*(a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^8*(a^2 - 4*b^2) + 8*a*b
*tan(c/2 + (d*x)/2)^3 - 8*a*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2)^9 - 4*a*b*tan(c/2 + (d*x)/2))) +
(a^2*atan(((a^2*(2*a^4 + 12*b^4 + 21*a^2*b^2)*(2*a^8*b + 2*b^9 - 8*a^2*b^7 + 12*a^4*b^5 - 8*a^6*b^3))/(2*(a +
b)^(9/2)*(a - b)^(9/2)) + (a^3*tan(c/2 + (d*x)/2)*(2*a^4 + 12*b^4 + 21*a^2*b^2)*(a^8 + b^8 - 4*a^2*b^6 + 6*a^
4*b^4 - 4*a^6*b^2))/((a + b)^(9/2)*(a - b)^(9/2)))/(2*a^6 + 12*a^2*b^4 + 21*a^4*b^2))*(2*a^4 + 12*b^4 + 21*a^2
*b^2))/(d*(a + b)^(9/2)*(a - b)^(9/2))

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