\(\int \sin ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [83]

Optimal result

Integrand size = 25, antiderivative size = 169 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {(a-5 b) (a+b)^2 \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{16 b f}+\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f} \] Output:

1/16*(a-5*b)*(a+b)^2*arctan(b^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2)) 
/b^(3/2)/f+1/16*(a-5*b)*(a+b)*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/b/f+1/ 
24*(a-5*b)*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(3/2)/b/f-1/6*cos(f*x+e)*(a+b-b 
*cos(f*x+e)^2)^(5/2)/b/f
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.90 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))} \left (3 a^2+29 a b+23 b^2-b (7 a+9 b) \cos (2 (e+f x))+b^2 \cos (4 (e+f x))\right )}{3 \sqrt {2} b}+\frac {(a+b)^2 (-a+5 b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{(-b)^{3/2}}}{16 f} \] Input:

Integrate[Sin[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(-1/3*(Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*(3*a^2 + 29*a*b + 2 
3*b^2 - b*(7*a + 9*b)*Cos[2*(e + f*x)] + b^2*Cos[4*(e + f*x)]))/(Sqrt[2]*b 
) + ((a + b)^2*(-a + 5*b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b 
 - b*Cos[2*(e + f*x)]]])/(-b)^(3/2))/(16*f)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3665, 299, 211, 211, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

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

\(\Big \downarrow \) 299

\(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \int \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}d\cos (e+f x)}{6 b}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \int \sqrt {-b \cos ^2(e+f x)+a+b}d\cos (e+f x)+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {(a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\)

Input:

Int[Sin[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

-(((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(5/2))/(6*b) - ((a - 5*b)*((Co 
s[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/4 + (3*(a + b)*(((a + b)*ArcT 
an[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*Sqrt[b]) + ( 
Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/2))/4))/(6*b))/f)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 299

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

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. \(445\) vs. \(2(149)=298\).

Time = 0.45 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.64

method	result	size
default	\(-\frac {\sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\, \left (16 \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b^{\frac {7}{2}} \cos \left (f x +e \right )^{4}-4 \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b^{\frac {5}{2}} \left (13 b +7 a \right ) \cos \left (f x +e \right )^{2}+66 b^{\frac {7}{2}} \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+72 a \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b^{\frac {5}{2}}+6 a^{2} \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b^{\frac {3}{2}}+3 \arctan \left (\frac {-2 b \cos \left (f x +e \right )^{2}+a +b}{2 \sqrt {b}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}}\right ) a^{3} b -9 \arctan \left (\frac {-2 b \cos \left (f x +e \right )^{2}+a +b}{2 \sqrt {b}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}}\right ) a^{2} b^{2}-27 b^{3} \arctan \left (\frac {-2 b \cos \left (f x +e \right )^{2}+a +b}{2 \sqrt {b}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}}\right ) a -15 b^{4} \arctan \left (\frac {-2 b \cos \left (f x +e \right )^{2}+a +b}{2 \sqrt {b}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}}\right )\right )}{96 b^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(446\)

Input:

int(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/96*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(16*(-b*cos(f*x+e)^4+(a+b)*c 
os(f*x+e)^2)^(1/2)*b^(7/2)*cos(f*x+e)^4-4*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e 
)^2)^(1/2)*b^(5/2)*(13*b+7*a)*cos(f*x+e)^2+66*b^(7/2)*(-b*cos(f*x+e)^4+(a+ 
b)*cos(f*x+e)^2)^(1/2)+72*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^( 
5/2)+6*a^2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(3/2)+3*arctan(1/2 
*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2 
))*a^3*b-9*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+ 
b)*cos(f*x+e)^2)^(1/2))*a^2*b^2-27*b^3*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/ 
b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))*a-15*b^4*arctan(1/2*(- 
2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))) 
/b^(5/2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.43 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right ) - 8 \, {\left (8 \, b^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (7 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b + 12 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{384 \, b^{2} f}, -\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) + 4 \, {\left (8 \, b^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (7 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b + 12 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{192 \, b^{2} f}\right ] \] Input:

integrate(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*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)*c 
os(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*(8*b^3*cos(f*x + e)^5 - 2*(7*a*b^2 + 13*b^3)*cos(f*x + e) 
^3 + 3*(a^2*b + 12*a*b^2 + 11*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + 
a + b))/(b^2*f), -1/192*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(b)*arcta 
n(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*(8 
*b^3*cos(f*x + e)^5 - 2*(7*a*b^2 + 13*b^3)*cos(f*x + e)^3 + 3*(a^2*b + 12* 
a*b^2 + 11*b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(b^2*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**3*(a+b*sin(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.47 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\frac {3 \, {\left (a + b\right )}^{2} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {18 \, {\left (a + b\right )} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 18 \, {\left (a + b\right )} \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 12 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) - 18 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \cos \left (f x + e\right ) - \frac {8 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \cos \left (f x + e\right )}{b} + \frac {2 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \cos \left (f x + e\right )}{b} + \frac {3 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \cos \left (f x + e\right )}{b}}{48 \, f} \] Input:

integrate(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

1/48*(3*(a + b)^2*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(3/2) + 3*(a 
+ b)^2*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*a*arcsi 
n(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*sqrt(b)*arcsin(b*co 
s(f*x + e)/sqrt((a + b)*b)) - 12*(-b*cos(f*x + e)^2 + a + b)^(3/2)*cos(f*x 
 + e) - 18*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*cos(f*x + e) - 8*(-b*co 
s(f*x + e)^2 + a + b)^(5/2)*cos(f*x + e)/b + 2*(-b*cos(f*x + e)^2 + a + b) 
^(3/2)*(a + b)*cos(f*x + e)/b + 3*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)^ 
2*cos(f*x + e)/b)/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5251 vs. \(2 (149) = 298\).

Time = 1.04 (sec) , antiderivative size = 5251, normalized size of antiderivative = 31.07 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

-1/24*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*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^(3/2) - 2*(3*(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))^11*a^3 - 9*(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))^11*a^2*b - 27*(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))^11*a*b^2 - 15*(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*t 
an(1/2*f*x + 1/2*e)^2 + a))^11*b^3 + 33*(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))^10*a^(7/2) + 93*(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))^10*a^(5/2)*b - 297*(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))^10*a^(3/2)*b^2 - 165*(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))^10*sqrt(a)*b^3 + 165*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqr 
t(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...
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \] Input:

int(sin(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2),x)
 

Output:

int(sin(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2), x)
 

Reduce [F]

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

int(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sin(e + f*x)**2*b + a)*sin(e + f*x)**5,x)*b + int(sqrt(sin(e + f* 
x)**2*b + a)*sin(e + f*x)**3,x)*a