\(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [584]

Optimal result

Integrand size = 29, antiderivative size = 115 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^2 (c+5 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.80 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (5 c+d+(c+5 d) \sin (e+f x))}{3 (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{3/2}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

(-2*a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(5* 
c + d + (c + 5*d)*Sin[e + f*x]))/(3*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e 
 + f*x)/2])*(c + d*Sin[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3241, 27, 2011, 3042, 3250}

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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3241

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac {2 a \int -\frac {a (c+5 d)+a \sin (e+f x) (c+5 d)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \frac {a (c+5 d)+a \sin (e+f x) (c+5 d)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {a (c+5 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (c+5 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3250

\(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^2 (c+5 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\)

Input:

Int[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^(5/2),x]
 

Output:

(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + 
d*Sin[e + f*x])^(3/2)) - (2*a^2*(c + 5*d)*Cos[e + f*x])/(3*d*(c + d)^2*f*S 
qrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
 && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

rule 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(103)=206\).

Time = 14.34 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.67

method	result	size
default	\(-\frac {2 \left (\left (\left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2}+\left (\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+24 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-19 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d c +\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (20 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-20 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a}{3 f \left (c^{2}+2 d c +d^{2}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{\frac {5}{2}}}\)	\(307\)

Input:

int((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3/f/(c^2+2*c*d+d^2)*(((2*cos(1/2*f*x+1/2*e)^2-5)*sin(1/2*f*x+1/2*e)+2*c 
os(1/2*f*x+1/2*e)^3+3*cos(1/2*f*x+1/2*e))*c^2+((4*cos(1/2*f*x+1/2*e)^4+16* 
cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)-4*cos(1/2*f*x+1/2*e)^5+24*cos(1 
/2*f*x+1/2*e)^3-19*cos(1/2*f*x+1/2*e))*d*c+cos(1/2*f*x+1/2*e)*((20*cos(1/2 
*f*x+1/2*e)^2-20*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-2)*sin(1/2*f*x+1/2* 
e)^2+2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))*d^2)*((2*sin(1/2*f*x+1/2*e)* 
cos(1/2*f*x+1/2*e)+1)*a)^(1/2)*a/(cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1/2*e))/( 
c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (103) = 206\).

Time = 0.11 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.81 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {2 \, {\left ({\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )^{2} + 4 \, a c - 4 \, a d + {\left (5 \, a c + a d\right )} \cos \left (f x + e\right ) - {\left (4 \, a c - 4 \, a d - {\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{3 \, {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{3} + {\left (2 \, c^{3} d + 5 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{4} + 2 \, c^{3} d + 2 \, c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f + {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d + 2 \, c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

2/3*((a*c + 5*a*d)*cos(f*x + e)^2 + 4*a*c - 4*a*d + (5*a*c + a*d)*cos(f*x 
+ e) - (4*a*c - 4*a*d - (a*c + 5*a*d)*cos(f*x + e))*sin(f*x + e))*sqrt(a*s 
in(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^2*d^2 + 2*c*d^3 + d^4)*f*cos 
(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^ 
4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 
 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^ 
2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2* 
d^2 + 4*c*d^3 + d^4)*f)*sin(f*x + e))
 

Sympy [F]

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

integrate((a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**(5/2),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**(3/2)/(c + d*sin(e + f*x))**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (103) = 206\).

Time = 0.20 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.67 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left ({\left (5 \, c^{2} + c d\right )} a^{\frac {3}{2}} - \frac {{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, {\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, {\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {{\left (5 \, c^{2} + c d\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{3 \, {\left (c^{2} + 2 \, c d + d^{2} + \frac {{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (c + \frac {2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac {5}{2}} f} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

-2/3*((5*c^2 + c*d)*a^(3/2) - (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e 
)/(cos(f*x + e) + 1) + 2*(4*c^2 - 7*c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^2/(c 
os(f*x + e) + 1)^2 - 2*(4*c^2 - 7*c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^3/(cos 
(f*x + e) + 1)^3 + (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e)^4/(cos(f* 
x + e) + 1)^4 - (5*c^2 + c*d)*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) 
*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/((c^2 + 2*c*d + d^2 + (c^2 + 2* 
c*d + d^2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(c + 2*d*sin(f*x + e)/(cos 
(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)^(5/2)*f)
                                                                                    
                                                                                    
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 23.30 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.37 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}\,\left (\frac {a\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (c+5\,d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{3\,d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}+\frac {a\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (3\,c-d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (c\,3{}\mathrm {i}-d\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (c\,1{}\mathrm {i}+d\,5{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{3\,d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}\right )}{{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-\frac {{\left (c+d\right )}^2\,1{}\mathrm {i}}{{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {2\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (2\,c^2+2\,c\,d+d^2\right )}{d^2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (4\,c+d\right )}{d}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (2\,c^2+2\,c\,d+d^2\right )\,2{}\mathrm {i}}{d^2\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (4\,c+d\right )\,1{}\mathrm {i}}{d\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}} \] Input:

int((a + a*sin(e + f*x))^(3/2)/(c + d*sin(e + f*x))^(5/2),x)
 

Output:

-((c + d*sin(e + f*x))^(1/2)*((a*exp(e*1i + f*x*1i)*(c + 5*d)*(a + a*sin(e 
 + f*x))^(1/2)*4i)/(3*d^2*f*(c*1i + d*1i)^2) + (a*exp(e*3i + f*x*3i)*(3*c 
- d)*(a + a*sin(e + f*x))^(1/2)*4i)/(d^2*f*(c*1i + d*1i)^2) - (a*exp(e*2i 
+ f*x*2i)*(c*3i - d*1i)*(a + a*sin(e + f*x))^(1/2)*4i)/(d^2*f*(c*1i + d*1i 
)^2) - (a*exp(e*4i + f*x*4i)*(c*1i + d*5i)*(a + a*sin(e + f*x))^(1/2)*4i)/ 
(3*d^2*f*(c*1i + d*1i)^2)))/(exp(e*5i + f*x*5i) - ((c + d)^2*1i)/(c*1i + d 
*1i)^2 - (2*exp(e*3i + f*x*3i)*(2*c*d + 2*c^2 + d^2))/d^2 + (exp(e*1i + f* 
x*1i)*(4*c + d))/d + (exp(e*2i + f*x*2i)*(c + d)^2*(2*c*d + 2*c^2 + d^2)*2 
i)/(d^2*(c*1i + d*1i)^2) - (exp(e*4i + f*x*4i)*(c + d)^2*(4*c + d)*1i)/(d* 
(c*1i + d*1i)^2))
 

Reduce [F]

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

int((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)
 

Output:

sqrt(a)*a*(int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f* 
x))/(sin(e + f*x)**3*d**3 + 3*sin(e + f*x)**2*c*d**2 + 3*sin(e + f*x)*c**2 
*d + c**3),x) + int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1))/(sin 
(e + f*x)**3*d**3 + 3*sin(e + f*x)**2*c*d**2 + 3*sin(e + f*x)*c**2*d + c** 
3),x))