\(\int \frac {1}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [471]

Optimal result

Integrand size = 16, antiderivative size = 225 \[ \int \frac {1}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {b \cos (e+f x) \sin (e+f x)}{3 a (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 b (2 a+b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {2 (2 a+b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 a^2 (a+b)^2 f \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}-\frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{3 a (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

1/3*b*cos(f*x+e)*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)+2/3*b*(2*a+ 
b)*cos(f*x+e)*sin(f*x+e)/a^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+2/3*(2*a+b 
)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin(f*x+e)^2)^(1/2)/a^2/(a+b)^2/ 
f/((a+b*sin(f*x+e)^2)/a)^(1/2)-1/3*InverseJacobiAM(f*x+e,(-b/a)^(1/2))*((a 
+b*sin(f*x+e)^2)/a)^(1/2)/a/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 a^2 (2 a+b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} E\left (e+f x\left |-\frac {b}{a}\right .\right )-a^2 (a+b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \left (-5 a^2-5 a b-b^2+b (2 a+b) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{3 a^2 (a+b)^2 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3663

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3652

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3651

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3657

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3656

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

\(\Big \downarrow \) 3662

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3661

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

Input:

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

Output:

(b*Cos[e + f*x]*Sin[e + f*x])/(3*a*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) 
 + ((2*b*(2*a + b)*Cos[e + f*x]*Sin[e + f*x])/(a*(a + b)*f*Sqrt[a + b*Sin[ 
e + f*x]^2]) + ((2*(2*a + b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*Sin[e + 
 f*x]^2])/(f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (a*(a + b)*EllipticF[e + f* 
x, -(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(f*Sqrt[a + b*Sin[e + f*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. \(546\) vs. \(2(210)=420\).

Time = 1.28 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.43

method	result	size
default	\(-\frac {\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \sin \left (f x +e \right )^{2}+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \sin \left (f x +e \right )^{2}-4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b \sin \left (f x +e \right )^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2} \sin \left (f x +e \right )^{2}+4 a \,b^{2} \sin \left (f x +e \right )^{5}+2 b^{3} \sin \left (f x +e \right )^{5}+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+a^{2} \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +5 a^{2} b \sin \left (f x +e \right )^{3}-a \,b^{2} \sin \left (f x +e \right )^{3}-2 b^{3} \sin \left (f x +e \right )^{3}-5 a^{2} b \sin \left (f x +e \right )-3 a \,b^{2} \sin \left (f x +e \right )}{3 \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}} a^{2} \left (a +b \right )^{2} \cos \left (f x +e \right ) f}\)	\(547\)

Input:

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

Output:

-1/3*((cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+ 
e),(-b/a)^(1/2))*a^2*b*sin(f*x+e)^2+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^ 
2)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*a*b^2*sin(f*x+e)^2-4*(cos(f 
*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-b/a)^(1 
/2))*a^2*b*sin(f*x+e)^2-2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2 
)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*a*b^2*sin(f*x+e)^2+4*a*b^2*sin(f*x+e) 
^5+2*b^3*sin(f*x+e)^5+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*El 
lipticF(sin(f*x+e),(-b/a)^(1/2))*a^3+a^2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f* 
x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*b-4*(cos(f*x+e)^2)^(1/ 
2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*a^3-2*( 
cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-b/ 
a)^(1/2))*a^2*b+5*a^2*b*sin(f*x+e)^3-a*b^2*sin(f*x+e)^3-2*b^3*sin(f*x+e)^3 
-5*a^2*b*sin(f*x+e)-3*a*b^2*sin(f*x+e))/(a+b*sin(f*x+e)^2)^(3/2)/a^2/(a+b) 
^2/cos(f*x+e)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 1531, normalized size of antiderivative = 6.80 \[ \int \frac {1}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sin(f*x+e)^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(f*x + e)^4 - 2*(2*I*a^2*b^3 + 3*I*a*b^4 + I*b^5)*cos(f*x + e)^2)*sq 
rt(-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(f*x + e)^4 + 2 
*(4*I*a^3*b^2 + 8*I*a^2*b^3 + 5*I*a*b^4 + I*b^5)*cos(f*x + e)^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(f*x + e) + I*sin(f*x + e))), (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(f*x 
+ e)^4 - 2*(-2*I*a^2*b^3 - 3*I*a*b^4 - I*b^5)*cos(f*x + e)^2)*sqrt(-b)*sqr 
t((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(f*x + e)^4 + 2*(-4*I*a^3*b 
^2 - 8*I*a^2*b^3 - 5*I*a*b^4 - I*b^5)*cos(f*x + e)^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(f*x + e) - I*sin(f*x + e))), (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(f*x + e)^4 - 2*(-3*I*a^3*b^2 - 8*I*a^2*b^3 - 7*I*a*b^4 - 
2*I*b^5)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - (-6*I*a^5 - 17*I 
*a^4*b - 17*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...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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