3.1471 \(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=388 \[ -\frac {3 a^5 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {a^4 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

[Out]

4*a^4*(a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(7/2)/d-a^4*(2*a^2+b^2)*arcta
n((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(7/2)/d-2*a^2*(a^4-3*a^2*b^2+6*b^4)*arctan((b+a*tan(
1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(7/2)/d+1/2*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/2*cos(d*x+c)/
(a-b)^3/d/(1+sin(d*x+c))-1/2*a^4*cos(d*x+c)/b/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^2-3/2*a^5*cos(d*x+c)/b/(a^2-b^2)^
3/d/(a+b*sin(d*x+c))+2*a^3*(a^2-2*b^2)*cos(d*x+c)/b/(a^2-b^2)^3/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.58, antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2897, 2648, 2664, 2754, 12, 2660, 618, 204} \[ -\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {2 a^2 \left (-3 a^2 b^2+a^4+6 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {3 a^5 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(4*a^4*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(7/2)*d) - (a^4*(2*a^2
+ b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(7/2)*d) - (2*a^2*(a^4 - 3*a^2*b^2
+ 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(7/2)*d) + Cos[c + d*x]/(2*(a + b)
^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^3*d*(1 + Sin[c + d*x])) - (a^4*Cos[c + d*x])/(2*b*(a^2 - b^
2)^2*d*(a + b*Sin[c + d*x])^2) - (3*a^5*Cos[c + d*x])/(2*b*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (2*a^3*(a^2
- 2*b^2)*Cos[c + d*x])/(b*(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 204

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

Rule 618

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 2648

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 2660

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 2664

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] && Integer
Q[2*n]

Rule 2754

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]

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (-\frac {1}{2 (a+b)^3 (-1+\sin (c+d x))}+\frac {1}{2 (a-b)^3 (1+\sin (c+d x))}+\frac {a^4}{b^2 \left (-a^2+b^2\right ) (a+b \sin (c+d x))^3}-\frac {2 \left (-a^5+2 a^3 b^2\right )}{b^2 \left (-a^2+b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {a^6-3 a^4 b^2+6 a^2 b^4}{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)} \, dx}{2 (a-b)^3}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^3}+\frac {\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^3} \, dx}{b^2 \left (a^2-b^2\right )}-\frac {\left (a^2 \left (a^4-3 a^2 b^2+6 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}\\ &=\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac {a}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}+\frac {a^4 \int \frac {-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )^2}-\frac {\left (2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 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 b^2 \left (a^2-b^2\right )^3}+\frac {\left (2 a^4 \left (a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}+\frac {\left (4 a^2 \left (a^4-3 a^2 b^2+6 b^4\right )\right ) \operatorname {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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 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 b^2 \left (a^2-b^2\right )^3}+\frac {\left (4 a^4 \left (a^2-2 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (8 a^4 \left (a^2-2 b^2\right )\right ) \operatorname {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 )}{b^2 \left (a^2-b^2\right )^3 d}-\frac {\left (a^4 \left (2 a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 a^4 \left (2 a^2+b^2\right )\right ) \operatorname {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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} 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 )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 3.19, size = 195, normalized size = 0.50 \[ \frac {-\frac {6 a^2 \left (a^2+4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a^3 \cos (c+d x) \left (\left (a^2-8 b^2\right ) \sin (c+d x)-7 a b\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

((-6*a^2*(a^2 + 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]*
(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) +
(a^3*Cos[c + d*x]*(-7*a*b + (a^2 - 8*b^2)*Sin[c + d*x]))/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])^2))/(2*d)

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fricas [A]  time = 0.52, size = 906, normalized size = 2.34 \[ \left [\frac {4 \, a^{6} b - 12 \, a^{4} b^{3} + 12 \, a^{2} b^{5} - 4 \, b^{7} + 2 \, {\left (11 \, a^{6} b - 5 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{6} + 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\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^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}, \frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} + {\left (11 \, a^{6} b - 5 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{6} + 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\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^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(4*a^6*b - 12*a^4*b^3 + 12*a^2*b^5 - 4*b^7 + 2*(11*a^6*b - 5*a^4*b^3 - 8*a^2*b^5 + 2*b^7)*cos(d*x + c)^2
+ 3*((a^4*b^2 + 4*a^2*b^4)*cos(d*x + c)^3 - 2*(a^5*b + 4*a^3*b^3)*cos(d*x + c)*sin(d*x + c) - (a^6 + 5*a^4*b^2
+ 4*a^2*b^4)*cos(d*x + c))*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^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (a^7 - 11*a^5*b^2 + 4*a^3*b^4 + 6*a*b^6)*cos(d*
x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*
a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b
^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)), 1/2*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + (11*a^6*b - 5*a^4*b^3 -
8*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + 3*((a^4*b^2 + 4*a^2*b^4)*cos(d*x + c)^3 - 2*(a^5*b + 4*a^3*b^3)*cos(d*x +
c)*sin(d*x + c) - (a^6 + 5*a^4*b^2 + 4*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(s
qrt(a^2 - b^2)*cos(d*x + c))) - (2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (a^7 - 11*a^5*b^2 + 4*a^3*b^4 + 6*a
*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2
*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b
^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c))]

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giac [A]  time = 0.35, size = 351, normalized size = 0.90 \[ -\frac {\frac {3 \, {\left (a^{4} + 4 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{4} b}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-(3*(a^4 + 4*a^2*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)) + 2*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b^2*tan(1/2
*d*x + 1/2*c) - 3*a^2*b - 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)) + (a^5*tan(1
/2*d*x + 1/2*c)^3 + 6*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 7*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 14*a^2*b^3*tan(1/2*d*x
+ 1/2*c)^2 - a^5*tan(1/2*d*x + 1/2*c) + 22*a^3*b^2*tan(1/2*d*x + 1/2*c) + 7*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)^2))/d

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maple [A]  time = 0.54, size = 590, normalized size = 1.52 \[ -\frac {1}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a^{5} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {6 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {7 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {14 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {a^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {22 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {7 a^{4} b}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {3 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}}-\frac {12 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^3,x)

[Out]

-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)-1/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/d*a^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/
2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3-6/d*a^3/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*ta
n(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3*b^2-7/d*a^4/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2*b-14/d*a^2/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+
a)^2*tan(1/2*d*x+1/2*c)^2*b^3+1/d*a^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(
1/2*d*x+1/2*c)-22/d*a^3/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)
*b^2-7/d*a^4/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b-3/d*a^4/(a-b)^3/(a+b)^3/(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))-12/d*a^2/(a-b)^3/(a+b)^3/(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))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 18.23, size = 585, normalized size = 1.51 \[ \frac {\frac {13\,a^4\,b+2\,a^2\,b^3}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^5+6\,a^3\,b^2+8\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^4\,b+9\,a^2\,b^3+4\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-3\,a^4+40\,a^2\,b^2+8\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {9\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {\frac {3\,a^2\,\left (a^2+4\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{3\,a^4+12\,a^2\,b^2}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(cos(c + d*x)^2*(a + b*sin(c + d*x))^3),x)

[Out]

((13*a^4*b + 2*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)^3*(8*a*b^4 + a^5 + 6*a^3*b
^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)^2*(2*a^4*b + 4*b^5 + 9*a^2*b^3))/(a^6 - b^6 +
3*a^2*b^4 - 3*a^4*b^2) + (a*tan(c/2 + (d*x)/2)*(8*b^4 - 3*a^4 + 40*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b
^2) - (3*a^3*tan(c/2 + (d*x)/2)^5*(a^2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (9*a^2*b*tan(c/2 + (d*x
)/2)^4*(a^2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2 - tan(c/2 + (d*x
)/2)^2*(a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b*tan(c/2 + (d*x)/2)^5 - 4*a*b*tan(c/2 + (d*x)
/2))) - (3*a^2*atan(((3*a^2*(a^2 + 4*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/(2*(a + b)^(7/2)*(a - b)^
(7/2)) + (3*a^3*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)))/(3*a^4 + 12*a^2*b^2))*(a^2 + 4*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**4/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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