3.2.87 \(\int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [187]

Optimal. Leaf size=333 \[ \frac {2 a^5 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {8 a^3 b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \]

[Out]

2*a^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d+8*a^3*b^2*arctan((b+a*tan(1/2*d*x+1/2
*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d+1/12*cos(d*x+c)/(a+b)^2/d/(1-sin(d*x+c))^2+1/12*cos(d*x+c)/(a+b)^2/d/(
1-sin(d*x+c))-1/4*(3*a+b)*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/12*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))^2-1/12*
cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))+1/4*(3*a-b)*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))+a^4*b*cos(d*x+c)/(a^2-b^2)
^3/d/(a+b*sin(d*x+c))

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Rubi [A]
time = 0.47, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2810, 2729, 2727, 2743, 12, 2739, 632, 210} \begin {gather*} \frac {2 a^5 \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 )^{7/2}}+\frac {a^4 b \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {8 a^3 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 )^{7/2}}-\frac {(3 a+b) \cos (c+d x)}{4 d (a+b)^3 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 d (a-b)^3 (\sin (c+d x)+1)}-\frac {\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^2 (\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])^2,x]

[Out]

(2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (8*a^3*b^2*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + Cos[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])^2) + C
os[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])) - ((3*a + b)*Cos[c + d*x])/(4*(a + b)^3*d*(1 - Sin[c + d*x]))
- Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])^2) - Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])) + ((3*
a - b)*Cos[c + d*x])/(4*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/((a^2 - b^2)^3*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]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (\frac {1}{4 (a+b)^2 (-1+\sin (c+d x))^2}+\frac {3 a+b}{4 (a+b)^3 (-1+\sin (c+d x))}+\frac {1}{4 (a-b)^2 (1+\sin (c+d x))^2}+\frac {-3 a+b}{4 (a-b)^3 (1+\sin (c+d x))}+\frac {a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \frac {1}{(1+\sin (c+d x))^2} \, dx}{4 (a-b)^2}-\frac {(3 a-b) \int \frac {1}{1+\sin (c+d x)} \, dx}{4 (a-b)^3}+\frac {\int \frac {1}{(-1+\sin (c+d x))^2} \, dx}{4 (a+b)^2}+\frac {(3 a+b) \int \frac {1}{-1+\sin (c+d x)} \, dx}{4 (a+b)^3}+\frac {\left (4 a^3 b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{12 (a-b)^2}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{12 (a+b)^2}+\frac {a^4 \int \frac {a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac {\left (8 a^3 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 )^3 d}\\ &=\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {a^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}-\frac {\left (16 a^3 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 )^3 d}\\ &=\frac {8 a^3 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 )^{7/2} d}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 a^5\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 )^3 d}\\ &=\frac {8 a^3 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 )^{7/2} d}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (4 a^5\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 )^3 d}\\ &=\frac {2 a^5 \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 )^{7/2} d}+\frac {8 a^3 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 )^{7/2} d}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 1.24, size = 341, normalized size = 1.02 \begin {gather*} \frac {\frac {24 a^3 \left (a^2+4 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 )^{7/2}}+\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {4 (4 a+b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {1}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (-4 a+b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 a^4 b \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

((24*a^3*(a^2 + 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + 1/((a + b)^2*(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
) - (4*(4*a + b)*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])/((
a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4
*(-4*a + b)*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (12*a^4*b*Cos[c + d*x])/((a
- b)^3*(a + b)^3*(a + b*Sin[c + d*x])))/(12*d)

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Maple [A]
time = 0.52, size = 248, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {\frac {2 a^{3} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(248\)
default	\(\frac {\frac {2 a^{3} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(248\)
risch	\(\frac {\frac {14 i a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{3}+\frac {2 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{3}+8 a^{3} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+\frac {44 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {4 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {70 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+\frac {82 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}}{3}+6 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {22 a^{5} {\mathrm e}^{i \left (d x +c \right )}}{3}-\frac {4 b^{4} a \,{\mathrm e}^{i \left (d x +c \right )}}{3}+6 i a^{2} b^{3}-2 i b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+2 i b^{5} {\mathrm e}^{6 i \left (d x +c \right )}-6 i a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+14 i a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {2 i b^{5}}{3}+\frac {64 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{3}-\frac {16 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+4 b^{2} a^{3} {\mathrm e}^{i \left (d x +c \right )}+2 a^{5} {\mathrm e}^{7 i \left (d x +c \right )}+14 a^{5} {\mathrm e}^{5 i \left (d x +c \right )}+\frac {14 i b \,a^{4}}{3}+14 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}}{\left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left (a^{2}-b^{2}\right )^{3} d}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(729\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2*a^3/(a-b)^3/(a+b)^3*((b^2*tan(1/2*d*x+1/2*c)+a*b)/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+a)+(a^
2+4*b^2)/(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)))-1/3/(a-b)^2/(tan(1/2*d*x+1/
2*c)+1)^3+1/2/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^2+a/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/3/(a+b)^2/(tan(1/2*d*x+1/2*c
)-1)^3-1/2/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^2+a/(a+b)^3/(tan(1/2*d*x+1/2*c)-1))

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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))^2,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 = 0.45, size = 815, normalized size = 2.45 \begin {gather*} \left [-\frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} - 2 \, {\left (7 \, a^{6} b + 2 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (7 \, a^{6} b - 16 \, a^{4} b^{3} + 11 \, a^{2} b^{5} - 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{6} + 4 \, a^{4} b^{2}\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 (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} - {\left (4 \, a^{7} - 7 \, a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3}\right )}}, -\frac {a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} - {\left (7 \, a^{6} b + 2 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (7 \, a^{6} b - 16 \, a^{4} b^{3} + 11 \, a^{2} b^{5} - 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{6} + 4 \, a^{4} b^{2}\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 (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} - {\left (4 \, a^{7} - 7 \, a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\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))^2,x, algorithm="fricas")

[Out]

[-1/6*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - 2*(7*a^6*b + 2*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c)^4 - 2
*(7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 2*b^7)*cos(d*x + c)^2 - 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x +
c) + (a^6 + 4*a^4*b^2)*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*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (4*a^7 - 7*a^5*b^2 + 2*a^3*b^4 + a*b^6)
*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c
) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3), -1/3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 -
b^7 - (7*a^6*b + 2*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c)^4 - (7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 2*b^7)*c
os(d*x + c)^2 + 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x + c) + (a^6 + 4*a^4*b^2)*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))) - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 -
(4*a^7 - 7*a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2
*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*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 )^{2}}\, 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))**2,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**2, x)

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Giac [A]
time = 6.59, size = 406, normalized size = 1.22 \begin {gather*} \frac {2 \, {\left (\frac {3 \, {\left (a^{5} + 4 \, a^{3} b^{2}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4} b\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}} + \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a^{3} b - 2 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{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))^2,x, algorithm="giac")

[Out]

2/3*(3*(a^5 + 4*a^3*b^2)*(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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 3*(a^3*b^2*tan(1/2*d*x + 1/2*c) + a^4*b)/(
(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)) + (3*a^4*tan(1/
2*d*x + 1/2*c)^5 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 6*a*b^3*tan(1/2*d*x + 1
/2*c)^4 - 10*a^4*tan(1/2*d*x + 1/2*c)^3 - 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^4*tan(1/2*d*x + 1/2*c)^3 + 2
4*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) + 9*a^2*b^2*tan(1/2*d*x + 1/2*c) - 10*a^3*b - 2*a*
b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d

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Mupad [B]
time = 9.90, size = 722, normalized size = 2.17 \begin {gather*} \frac {\frac {2\,\left (13\,a^4\,b+2\,a^2\,b^3\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 )\,\left (-3\,a^5+14\,a^3\,b^2+4\,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 )}^6\,\left (a^4\,b+4\,a^2\,b^3\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (29\,a^4\,b+16\,a^2\,b^3\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 )}^5\,\left (7\,a^5+30\,a^3\,b^2+8\,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 )}^3\,\left (-7\,a^5+48\,a^3\,b^2+4\,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 )}^4\,\left (11\,a^4\,b+42\,a^2\,b^3-8\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {2\,a^3\,\mathrm {atan}\left (\frac {\frac {a^3\,\left (a^2+4\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{2\,a^5+8\,a^3\,b^2}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \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))^2,x)

[Out]

((2*(13*a^4*b + 2*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)*(4*a*b^4 - 3*a^5 +
14*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)^6*(a^4*b + 4*a^2*b^3))/(a^6 - b^
6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 + (d*x)/2)^2*(29*a^4*b + 16*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^
4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(8*a*b^4 + 7*a^5 + 30*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)^3*(4*a*b^4 - 7*a^5 + 48*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)^4*(11*a^4*b - 8*b^5 + 42*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*a^3*tan(c/2 + (d*x)/2)
^7*(a^2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - 2*a*tan(c/2 + (d*x)/2)
^2 + 2*a*tan(c/2 + (d*x)/2)^6 - a*tan(c/2 + (d*x)/2)^8 - 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 -
2*b*tan(c/2 + (d*x)/2)^7)) + (2*a^3*atan(((a^3*(a^2 + 4*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/((a +
b)^(7/2)*(a - b)^(7/2)) + (2*a^4*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a +
b)^(7/2)*(a - b)^(7/2)))/(2*a^5 + 8*a^3*b^2))*(a^2 + 4*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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