\(\int \frac {\sin ^5(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [112]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a^2 \cos (e+f x)}{3 b^2 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}+\frac {2 a (2 a+3 b) \cos (e+f x)}{3 b^2 (a+b)^2 f \sqrt {a+b-b \cos ^2(e+f x)}} \] Output:

-arctan(b^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*a^2*c 
os(f*x+e)/b^2/(a+b)/f/(a+b-b*cos(f*x+e)^2)^(3/2)+2/3*a*(2*a+3*b)*cos(f*x+e 
)/b^2/(a+b)^2/f/(a+b-b*cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {2 \sqrt {2} a \cos (e+f x) \left (3 a^2+7 a b+3 b^2-b (2 a+3 b) \cos (2 (e+f x))\right )}{(a+b)^2 (2 a+b-b \cos (2 (e+f x)))^{3/2}}-\frac {3 \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{\sqrt {-b}}}{3 b^2 f} \] Input:

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

Output:

((2*Sqrt[2]*a*Cos[e + f*x]*(3*a^2 + 7*a*b + 3*b^2 - b*(2*a + 3*b)*Cos[2*(e 
 + f*x)]))/((a + b)^2*(2*a + b - b*Cos[2*(e + f*x)])^(3/2)) - (3*Log[Sqrt[ 
2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b])/ 
(3*b^2*f)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3665, 315, 25, 298, 224, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

\(\displaystyle -\frac {\frac {\frac {3 (a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{b}-\frac {a (3 a+5 b) \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \cos (e+f x) \left (1-\cos ^2(e+f x)\right )}{3 b (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {\frac {\frac {3 (a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{b}-\frac {a (3 a+5 b) \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \cos (e+f x) \left (1-\cos ^2(e+f x)\right )}{3 b (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {\frac {3 (a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{b^{3/2}}-\frac {a (3 a+5 b) \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \cos (e+f x) \left (1-\cos ^2(e+f x)\right )}{3 b (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

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

Output:

-((-1/3*(a*Cos[e + f*x]*(1 - Cos[e + f*x]^2))/(b*(a + b)*(a + b - b*Cos[e 
+ f*x]^2)^(3/2)) + ((3*(a + b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - 
b*Cos[e + f*x]^2]])/b^(3/2) - (a*(3*a + 5*b)*Cos[e + f*x])/(b*(a + b)*Sqrt 
[a + b - b*Cos[e + f*x]^2]))/(3*b*(a + b)))/f)
 

Defintions of rubi rules used

rule 25

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

rule 216

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 3042

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

rule 3665

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(242\) vs. \(2(117)=234\).

Time = 0.63 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.85

method	result	size
default	\(\frac {\sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}\, \left (\frac {\arctan \left (\frac {\sqrt {b}\, \left (\sin \left (f x +e \right )^{2}-\frac {-a +b}{2 b}\right )}{\sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}\right )}{2 b^{\frac {5}{2}}}-\frac {a^{2} \left (2 b \sin \left (f x +e \right )^{2}+3 a +b \right ) \cos \left (f x +e \right )^{2}}{3 b^{2} \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}\, \left (a +b \sin \left (f x +e \right )^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a \cos \left (f x +e \right )^{2}}{b^{2} \left (a +b \right ) \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(243\)

Input:

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

Output:

(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2)*(1/2/b^(5/2)*arctan(b^(1/2)*(sin 
(f*x+e)^2-1/2*(-a+b)/b)/(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2))-1/3*a^2 
/b^2*(2*b*sin(f*x+e)^2+3*a+b)*cos(f*x+e)^2/(-(-a-b*sin(f*x+e)^2)*cos(f*x+e 
)^2)^(1/2)/(a+b*sin(f*x+e)^2)/(a^2+2*a*b+b^2)+2*a/b^2*cos(f*x+e)^2/(a+b)/( 
-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1 
/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (117) = 234\).

Time = 0.56 (sec) , antiderivative size = 885, normalized size of antiderivative = 6.76 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/24*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^ 
2*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e) 
^2)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 
 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^ 
2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 
+ 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 
 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + 
 e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)) + 8*(2*(2*a^2*b^2 + 3*a*b^3 
)*cos(f*x + e)^3 - 3*(a^3*b + 3*a^2*b^2 + 2*a*b^3)*cos(f*x + e))*sqrt(-b*c 
os(f*x + e)^2 + a + b))/((a^2*b^5 + 2*a*b^6 + b^7)*f*cos(f*x + e)^4 - 2*(a 
^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^4*b^3 + 4*a^3*b^ 
4 + 6*a^2*b^5 + 4*a*b^6 + b^7)*f), 1/12*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos( 
f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3*a^2* 
b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2)*sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e 
)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + 
e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e) 
^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e))) - 4*(2*(2*a^2*b^2 + 3*a*b^3)*c 
os(f*x + e)^3 - 3*(a^3*b + 3*a^2*b^2 + 2*a*b^3)*cos(f*x + e))*sqrt(-b*cos( 
f*x + e)^2 + a + b))/((a^2*b^5 + 2*a*b^6 + b^7)*f*cos(f*x + e)^4 - 2*(a^3* 
b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^4*b^3 + 4*a^3*b^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (117) = 234\).

Time = 0.12 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.16 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \cos \left (f x + e\right )^{2}}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} - \frac {2 \, a}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b^{2}} - \frac {2}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}\right )} \cos \left (f x + e\right ) + \frac {3 \, \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {2 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} - \frac {3 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b^{2}} + \frac {2 \, a \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b^{2}} - \frac {2 \, \cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {4 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \] Input:

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

Output:

-1/3*((3*cos(f*x + e)^2/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b) - 2*a/((-b*c 
os(f*x + e)^2 + a + b)^(3/2)*b^2) - 2/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b 
))*cos(f*x + e) + 3*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 2*cos 
(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)^2) + cos(f*x + e)/((-b* 
cos(f*x + e)^2 + a + b)^(3/2)*(a + b)) - 3*cos(f*x + e)/(sqrt(-b*cos(f*x + 
 e)^2 + a + b)*b^2) + 2*a*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*(a 
 + b)*b^2) - 2*cos(f*x + e)/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b) + 4*cos( 
f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*b))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (117) = 234\).

Time = 0.79 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.50 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {{\left (3 \, a^{3} b^{8} + 5 \, a^{2} b^{9}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}} + \frac {3 \, {\left (a^{3} b^{8} + 7 \, a^{2} b^{9} + 8 \, a b^{10}\right )}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, {\left (a^{3} b^{8} + 7 \, a^{2} b^{9} + 8 \, a b^{10}\right )}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, a^{3} b^{8} + 5 \, a^{2} b^{9}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} + \sqrt {a}}{2 \, \sqrt {b}}\right )}{b^{\frac {5}{2}}}}{3 \, f} \] Input:

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

Output:

-1/3*(((((3*a^3*b^8 + 5*a^2*b^9)*tan(1/2*f*x + 1/2*e)^2/(a^2*b^10 + 2*a*b^ 
11 + b^12) + 3*(a^3*b^8 + 7*a^2*b^9 + 8*a*b^10)/(a^2*b^10 + 2*a*b^11 + b^1 
2))*tan(1/2*f*x + 1/2*e)^2 - 3*(a^3*b^8 + 7*a^2*b^9 + 8*a*b^10)/(a^2*b^10 
+ 2*a*b^11 + b^12))*tan(1/2*f*x + 1/2*e)^2 - (3*a^3*b^8 + 5*a^2*b^9)/(a^2* 
b^10 + 2*a*b^11 + b^12))/(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2 
*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2) - 6*arctan(-1/2*(sqrt(a)*tan 
(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/ 
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(a))/sqrt(b))/b^(5/2))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*sin(e + f*x)**5)/(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)