3.109 \(\int \frac {1}{(a+b \sin ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=144 \[ \frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{8 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} d (a+b)^{5/2}}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3184, 3173, 12, 3181, 205} \[ \frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} d (a+b)^{5/2}}+\frac {3 b (2 a+b) \sin (c+d x) \cos (c+d x)}{8 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}+\frac {b \sin (c+d x) \cos (c+d x)}{4 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^2)^(-3),x]

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*a^(5/2)*(a + b)^(5/2)*d) + (b*Cos[c +
d*x]*Sin[c + d*x])/(4*a*(a + b)*d*(a + b*Sin[c + d*x]^2)^2) + (3*b*(2*a + b)*Cos[c + d*x]*Sin[c + d*x])/(8*a^2
*(a + b)^2*d*(a + b*Sin[c + d*x]^2))

Rule 12

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

Rule 205

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

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 1.27, size = 125, normalized size = 0.87 \[ \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {\sqrt {a} b \sin (2 (c+d x)) \left (16 a^2-3 b (2 a+b) \cos (2 (c+d x))+16 a b+3 b^2\right )}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}}{8 a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(5/2) + (Sqrt[a]*b*(16*a^2 + 16*
a*b + 3*b^2 - 3*b*(2*a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/((a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2))/
(8*a^(5/2)*d)

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fricas [B]  time = 0.54, size = 843, normalized size = 5.85 \[ \left [-\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}, -\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 24 \, a^{3} b + 27 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 16 \, a^{2} b^{2} + 11 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{4} b + 19 \, a^{3} b^{2} + 14 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b + 4 \, a^{6} b^{2} + 6 \, a^{5} b^{3} + 4 \, a^{4} b^{4} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(d*x + c)^4 + 8*a^4 + 24*a^3*b + 27*a^2*b^2 + 14*a*b^3 + 3*b^4 - 2*(
8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*b^4)*cos(d*x + c)^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x +
c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2
- a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b
^2)) + 4*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (8*a^4*b + 19*a^3*b^2 + 14*a^2*b^3 + 3*a*b^4)*cos
(d*x + c))*sin(d*x + c))/((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^4 - 2*(a^7*b + 4*a^6*b^2
+ 6*a^5*b^3 + 4*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a
^3*b^5)*d), -1/16*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(d*x + c)^4 + 8*a^4 + 24*a^3*b + 27*a^2*b^2 + 14*a*b^3 +
3*b^4 - 2*(8*a^3*b + 16*a^2*b^2 + 11*a*b^3 + 3*b^4)*cos(d*x + c)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(
d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c))) + 2*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(d*
x + c)^3 - (8*a^4*b + 19*a^3*b^2 + 14*a^2*b^3 + 3*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 + 3*a^5*b^3 + 3
*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^4 - 2*(a^7*b + 4*a^6*b^2 + 6*a^5*b^3 + 4*a^4*b^4 + a^3*b^5)*d*cos(d*x + c)^
2 + (a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5)*d)]

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giac [A]  time = 0.14, size = 211, normalized size = 1.47 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {8 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, a b^{2} \tan \left (d x + c\right )^{3} + 3 \, b^{3} \tan \left (d x + c\right )^{3} + 8 \, a^{2} b \tan \left (d x + c\right ) + 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/8*((pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))
*(8*a^2 + 8*a*b + 3*b^2)/((a^4 + 2*a^3*b + a^2*b^2)*sqrt(a^2 + a*b)) + (8*a^2*b*tan(d*x + c)^3 + 11*a*b^2*tan(
d*x + c)^3 + 3*b^3*tan(d*x + c)^3 + 8*a^2*b*tan(d*x + c) + 5*a*b^2*tan(d*x + c))/((a^4 + 2*a^3*b + a^2*b^2)*(a
*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)^2))/d

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maple [B]  time = 0.40, size = 334, normalized size = 2.32 \[ \frac {\left (\tan ^{3}\left (d x +c \right )\right ) b}{d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{2} a \left (a +b \right )}+\frac {3 b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{8 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{2} a^{2} \left (a +b \right )}+\frac {\tan \left (d x +c \right ) b}{d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 b^{2} \tan \left (d x +c \right )}{8 d \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )^{2} a \left (a^{2}+2 a b +b^{2}\right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b}{d \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {a \left (a +b \right )}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b^{2}}{8 d \,a^{2} \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c)^2)^3,x)

[Out]

1/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/a/(a+b)*tan(d*x+c)^3*b+3/8/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/a^2*b
^2/(a+b)*tan(d*x+c)^3+1/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a^2+2*a*b+b^2)*tan(d*x+c)*b+5/8/d/(a*tan(d*x+c)
^2+tan(d*x+c)^2*b+a)^2*b^2/a/(a^2+2*a*b+b^2)*tan(d*x+c)+1/d/(a^2+2*a*b+b^2)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d
*x+c)/(a*(a+b))^(1/2))+1/d/(a^2+2*a*b+b^2)/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b+3/8/d/
a^2/(a^2+2*a*b+b^2)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b^2

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maxima [A]  time = 0.55, size = 211, normalized size = 1.47 \[ \frac {\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/8*((8*a^2 + 8*a*b + 3*b^2)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt((a +
b)*a)) + ((8*a^2*b + 11*a*b^2 + 3*b^3)*tan(d*x + c)^3 + (8*a^2*b + 5*a*b^2)*tan(d*x + c))/(a^6 + 2*a^5*b + a^
4*b^2 + (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*tan(d*x + c)^4 + 2*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*
b^3)*tan(d*x + c)^2))/d

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mupad [B]  time = 13.82, size = 176, normalized size = 1.22 \[ \frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,d\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d*x)^2)^3,x)

[Out]

((tan(c + d*x)^3*(8*a*b + 3*b^2))/(8*a^2*(a + b)) + (tan(c + d*x)*(8*a*b + 5*b^2))/(8*a*(2*a*b + a^2 + b^2)))/
(d*(tan(c + d*x)^4*(2*a*b + a^2 + b^2) + a^2 + tan(c + d*x)^2*(2*a*b + 2*a^2))) + (atan((tan(c + d*x)*(2*a + 2
*b)*(2*a*b + a^2 + b^2))/(2*a^(1/2)*(a + b)^(5/2)))*(8*a*b + 8*a^2 + 3*b^2))/(8*a^(5/2)*d*(a + b)^(5/2))

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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(1/(a+b*sin(d*x+c)**2)**3,x)

[Out]

Timed out

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