\(\int \frac {1}{(a+b \sin ^2(x))^{5/2}} \, dx\) [145]

Optimal result

Integrand size = 12, antiderivative size = 165 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\frac {b \cos (x) \sin (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}+\frac {2 b (2 a+b) \cos (x) \sin (x)}{3 a^2 (a+b)^2 \sqrt {a+b \sin ^2(x)}}+\frac {2 (2 a+b) E\left (x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(x)}}{3 a^2 (a+b)^2 \sqrt {\frac {a+b \sin ^2(x)}{a}}}-\frac {\operatorname {EllipticF}\left (x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(x)}{a}}}{3 a (a+b) \sqrt {a+b \sin ^2(x)}} \] Output:

1/3*b*cos(x)*sin(x)/a/(a+b)/(a+b*sin(x)^2)^(3/2)+2/3*b*(2*a+b)*cos(x)*sin( 
x)/a^2/(a+b)^2/(a+b*sin(x)^2)^(1/2)+2/3*(2*a+b)*EllipticE(sin(x),(-b/a)^(1 
/2))*(a+b*sin(x)^2)^(1/2)/a^2/(a+b)^2/((a+b*sin(x)^2)/a)^(1/2)-1/3*Inverse 
JacobiAM(x,(-b/a)^(1/2))*((a+b*sin(x)^2)/a)^(1/2)/a/(a+b)/(a+b*sin(x)^2)^( 
1/2)
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\frac {2 a^2 (2 a+b) \left (\frac {2 a+b-b \cos (2 x)}{a}\right )^{3/2} E\left (x\left |-\frac {b}{a}\right .\right )-a^2 (a+b) \left (\frac {2 a+b-b \cos (2 x)}{a}\right )^{3/2} \operatorname {EllipticF}\left (x,-\frac {b}{a}\right )-\sqrt {2} b \left (-5 a^2-5 a b-b^2+b (2 a+b) \cos (2 x)\right ) \sin (2 x)}{3 a^2 (a+b)^2 (2 a+b-b \cos (2 x))^{3/2}} \] Input:

Integrate[(a + b*Sin[x]^2)^(-5/2),x]
 

Output:

(2*a^2*(2*a + b)*((2*a + b - b*Cos[2*x])/a)^(3/2)*EllipticE[x, -(b/a)] - a 
^2*(a + b)*((2*a + b - b*Cos[2*x])/a)^(3/2)*EllipticF[x, -(b/a)] - Sqrt[2] 
*b*(-5*a^2 - 5*a*b - b^2 + b*(2*a + b)*Cos[2*x])*Sin[2*x])/(3*a^2*(a + b)^ 
2*(2*a + b - b*Cos[2*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {3042, 3663, 25, 3042, 3652, 3042, 3651, 3042, 3657, 3042, 3656, 3662, 3042, 3661}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sin (x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3663

\(\displaystyle \frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}-\frac {\int -\frac {-b \sin ^2(x)+3 a+2 b}{\left (b \sin ^2(x)+a\right )^{3/2}}dx}{3 a (a+b)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {-b \sin ^2(x)+3 a+2 b}{\left (b \sin ^2(x)+a\right )^{3/2}}dx}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-b \sin (x)^2+3 a+2 b}{\left (b \sin (x)^2+a\right )^{3/2}}dx}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3652

\(\displaystyle \frac {\frac {\int \frac {2 b (2 a+b) \sin ^2(x)+a (3 a+b)}{\sqrt {b \sin ^2(x)+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 b (2 a+b) \sin (x)^2+a (3 a+b)}{\sqrt {b \sin (x)^2+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3651

\(\displaystyle \frac {\frac {2 (2 a+b) \int \sqrt {b \sin ^2(x)+a}dx-a (a+b) \int \frac {1}{\sqrt {b \sin ^2(x)+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 (2 a+b) \int \sqrt {b \sin (x)^2+a}dx-a (a+b) \int \frac {1}{\sqrt {b \sin (x)^2+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3657

\(\displaystyle \frac {\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} \int \sqrt {\frac {b \sin ^2(x)}{a}+1}dx}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-a (a+b) \int \frac {1}{\sqrt {b \sin (x)^2+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} \int \sqrt {\frac {b \sin (x)^2}{a}+1}dx}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-a (a+b) \int \frac {1}{\sqrt {b \sin (x)^2+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3656

\(\displaystyle \frac {\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} E\left (x\left |-\frac {b}{a}\right .\right )}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-a (a+b) \int \frac {1}{\sqrt {b \sin (x)^2+a}}dx}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3662

\(\displaystyle \frac {\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} E\left (x\left |-\frac {b}{a}\right .\right )}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}dx}{\sqrt {a+b \sin ^2(x)}}}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} E\left (x\left |-\frac {b}{a}\right .\right )}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sin (x)^2}{a}+1}}dx}{\sqrt {a+b \sin ^2(x)}}}{a (a+b)}+\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}}{3 a (a+b)}+\frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3661

\(\displaystyle \frac {b \sin (x) \cos (x)}{3 a (a+b) \left (a+b \sin ^2(x)\right )^{3/2}}+\frac {\frac {2 b (2 a+b) \sin (x) \cos (x)}{a (a+b) \sqrt {a+b \sin ^2(x)}}+\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(x)} E\left (x\left |-\frac {b}{a}\right .\right )}{\sqrt {\frac {b \sin ^2(x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(x)}{a}+1} \operatorname {EllipticF}\left (x,-\frac {b}{a}\right )}{\sqrt {a+b \sin ^2(x)}}}{a (a+b)}}{3 a (a+b)}\)

Input:

Int[(a + b*Sin[x]^2)^(-5/2),x]
 

Output:

(b*Cos[x]*Sin[x])/(3*a*(a + b)*(a + b*Sin[x]^2)^(3/2)) + ((2*b*(2*a + b)*C 
os[x]*Sin[x])/(a*(a + b)*Sqrt[a + b*Sin[x]^2]) + ((2*(2*a + b)*EllipticE[x 
, -(b/a)]*Sqrt[a + b*Sin[x]^2])/Sqrt[1 + (b*Sin[x]^2)/a] - (a*(a + b)*Elli 
pticF[x, -(b/a)]*Sqrt[1 + (b*Sin[x]^2)/a])/Sqrt[a + b*Sin[x]^2])/(a*(a + b 
)))/(3*a*(a + b))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3651

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + 
 (f_.)*(x_)]^2], x_Symbol] :> Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]^2], x] 
, x] + Simp[(A*b - a*B)/b   Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /; Fre 
eQ[{a, b, e, f, A, B}, x]
 

rule 3652

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(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] - Simp[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 3656

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a 
]/f)*EllipticE[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
 

rule 3657

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[a 
+ b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin[e + f*x]^2/a)]   Int[Sqrt[1 + (b*Sin[e 
+ f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]
 

rule 3661

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S 
qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 
 0]
 

rule 3662

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[ 
1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a + b*Sin[e + f*x]^2]   Int[1/Sqrt[1 + (b*Si 
n[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]
 

rule 3663

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C 
os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a 
+ b))), x] + Simp[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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(395\) vs. \(2(150)=300\).

Time = 1.18 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.40

method	result	size
default	\(-\frac {\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \sin \left (x \right )^{2}+\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \sin \left (x \right )^{2}-4 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \sin \left (x \right )^{2}-2 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \sin \left (x \right )^{2}+4 a \,b^{2} \sin \left (x \right )^{5}+2 b^{3} \sin \left (x \right )^{5}+\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b -4 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-2 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (x \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +5 a^{2} b \sin \left (x \right )^{3}-a \,b^{2} \sin \left (x \right )^{3}-2 b^{3} \sin \left (x \right )^{3}-5 \sin \left (x \right ) b \,a^{2}-3 \sin \left (x \right ) b^{2} a}{3 \left (a +b \sin \left (x \right )^{2}\right )^{\frac {3}{2}} \left (a +b \right )^{2} a^{2} \cos \left (x \right )}\)	\(396\)

Input:

int(1/(a+b*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*((cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1/2)*EllipticF(sin(x),(-b/a)^(1 
/2))*a^2*b*sin(x)^2+(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1/2)*EllipticF(si 
n(x),(-b/a)^(1/2))*a*b^2*sin(x)^2-4*(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1 
/2)*EllipticE(sin(x),(-b/a)^(1/2))*a^2*b*sin(x)^2-2*(cos(x)^2)^(1/2)*((a+b 
*sin(x)^2)/a)^(1/2)*EllipticE(sin(x),(-b/a)^(1/2))*a*b^2*sin(x)^2+4*a*b^2* 
sin(x)^5+2*b^3*sin(x)^5+(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1/2)*Elliptic 
F(sin(x),(-b/a)^(1/2))*a^3+(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1/2)*Ellip 
ticF(sin(x),(-b/a)^(1/2))*a^2*b-4*(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a)^(1/2 
)*EllipticE(sin(x),(-b/a)^(1/2))*a^3-2*(cos(x)^2)^(1/2)*((a+b*sin(x)^2)/a) 
^(1/2)*EllipticE(sin(x),(-b/a)^(1/2))*a^2*b+5*a^2*b*sin(x)^3-a*b^2*sin(x)^ 
3-2*b^3*sin(x)^3-5*sin(x)*b*a^2-3*sin(x)*b^2*a)/(a+b*sin(x)^2)^(3/2)/(a+b) 
^2/a^2/cos(x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 1406, normalized size of antiderivative = 8.52 \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sin(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

1/3*((2*(2*I*a^3*b^2 + 5*I*a^2*b^3 + 4*I*a*b^4 + I*b^5 + (2*I*a*b^4 + I*b^ 
5)*cos(x)^4 - 2*(2*I*a^2*b^3 + 3*I*a*b^4 + I*b^5)*cos(x)^2)*sqrt(-b)*sqrt( 
(a^2 + a*b)/b^2) - (-4*I*a^4*b - 12*I*a^3*b^2 - 13*I*a^2*b^3 - 6*I*a*b^4 - 
 I*b^5 + (-4*I*a^2*b^3 - 4*I*a*b^4 - I*b^5)*cos(x)^4 + 2*(4*I*a^3*b^2 + 8* 
I*a^2*b^3 + 5*I*a*b^4 + I*b^5)*cos(x)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a 
*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 
 2*a + b)/b)*(cos(x) + I*sin(x))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)* 
sqrt((a^2 + a*b)/b^2))/b^2) + (2*(-2*I*a^3*b^2 - 5*I*a^2*b^3 - 4*I*a*b^4 - 
 I*b^5 + (-2*I*a*b^4 - I*b^5)*cos(x)^4 - 2*(-2*I*a^2*b^3 - 3*I*a*b^4 - I*b 
^5)*cos(x)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (4*I*a^4*b + 12*I*a^3*b^2 + 
 13*I*a^2*b^3 + 6*I*a*b^4 + I*b^5 + (4*I*a^2*b^3 + 4*I*a*b^4 + I*b^5)*cos( 
x)^4 + 2*(-4*I*a^3*b^2 - 8*I*a^2*b^3 - 5*I*a*b^4 - I*b^5)*cos(x)^2)*sqrt(- 
b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt(( 
2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(x) - I*sin(x))), (8*a^2 + 8*a 
*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*(-3*I*a^4*b - 
11*I*a^3*b^2 - 15*I*a^2*b^3 - 9*I*a*b^4 - 2*I*b^5 + (-3*I*a^2*b^3 - 5*I*a* 
b^4 - 2*I*b^5)*cos(x)^4 - 2*(-3*I*a^3*b^2 - 8*I*a^2*b^3 - 7*I*a*b^4 - 2*I* 
b^5)*cos(x)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (-6*I*a^5 - 17*I*a^4*b - 1 
7*I*a^3*b^2 - 7*I*a^2*b^3 - I*a*b^4 + (-6*I*a^3*b^2 - 5*I*a^2*b^3 - I*a*b^ 
4)*cos(x)^4 + 2*(6*I*a^4*b + 11*I*a^3*b^2 + 6*I*a^2*b^3 + I*a*b^4)*cos(...
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \sin ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+b*sin(x)**2)**(5/2),x)
 

Output:

Integral((a + b*sin(x)**2)**(-5/2), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*sin(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*sin(x)^2 + a)^(-5/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*sin(x)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*sin(x)^2 + a)^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\sin \left (x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

int(1/(a + b*sin(x)^2)^(5/2),x)
 

Output:

int(1/(a + b*sin(x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (x \right )^{2} b +a}}{\sin \left (x \right )^{6} b^{3}+3 \sin \left (x \right )^{4} a \,b^{2}+3 \sin \left (x \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(1/(a+b*sin(x)^2)^(5/2),x)
 

Output:

int(sqrt(sin(x)**2*b + a)/(sin(x)**6*b**3 + 3*sin(x)**4*a*b**2 + 3*sin(x)* 
*2*a**2*b + a**3),x)