\(\int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [602]

Optimal result

Integrand size = 29, antiderivative size = 251 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {(5 c-3 d) d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {(c-d)^{3/2} (c+9 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{3/2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.81 (sec) , antiderivative size = 1844, normalized size of antiderivative = 7.35 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-(d^2*Cos[(e + f*x)/2]) + d^2*Si 
n[(e + f*x)/2] - (c - d)^2/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (c^ 
2*Sin[(e + f*x)/2] - 2*c*d*Sin[(e + f*x)/2] + d^2*Sin[(e + f*x)/2])/(Cos[( 
e + f*x)/2] + Sin[(e + f*x)/2])^2)*Sqrt[c + d*Sin[e + f*x]])/(f*(a*(1 + Si 
n[e + f*x]))^(3/2)) + ((((c - d)^(3/2)*(c + 9*d)*Log[1 + Tan[(e + f*x)/2]] 
)/Sqrt[2] + I*(5*c - 3*d)*d^(3/2)*Log[((-I)*((-I)*c + d + (1 - I)*Sqrt[2]* 
Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c - I*d) 
*Tan[(e + f*x)/2]))/(d^(5/2)*(-5*c + 3*d)*(-I + Tan[(e + f*x)/2]))] + I*d^ 
(3/2)*(-5*c + 3*d)*Log[(I*(I*c + d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos 
[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2]))/( 
d^(5/2)*(-5*c + 3*d)*(I + Tan[(e + f*x)/2]))] - ((c - d)^(3/2)*(c + 9*d)*L 
og[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + 
f*x]] + (-c + d)*Tan[(e + f*x)/2]])/Sqrt[2])*(Cos[(e + f*x)/2] + Sin[(e + 
f*x)/2])^3*(c^3/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e 
+ f*x]]) + (7*c^2*d)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*S 
in[e + f*x]]) - (7*c*d^2)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c 
+ d*Sin[e + f*x]]) + (3*d^3)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt 
[c + d*Sin[e + f*x]]) + (5*c*d^2*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[ 
(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) - (3*d^3*Sin[e + f*x])/(2*(Cos[(e 
+ f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(a*(1 + Si...
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3244, 27, 3042, 3462, 3042, 3461, 3042, 3254, 218, 3261, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int -\frac {\sqrt {c+d \sin (e+f x)} \left (a \left (c^2+6 d c-3 d^2\right )-2 a (c-3 d) d \sin (e+f x)\right )}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {\int \frac {\left (c^3+7 d c^2-7 d^2 c+3 d^3\right ) a^2+2 (5 c-3 d) d^2 \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (c^3+7 d c^2-7 d^2 c+3 d^3\right ) a^2+2 (5 c-3 d) d^2 \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {\frac {a^2 (c+9 d) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+2 a d^2 (5 c-3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (c+9 d) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+2 a d^2 (5 c-3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {\frac {a^2 (c-d)^2 (c+9 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {4 a^2 d^2 (5 c-3 d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {a^2 (c-d)^2 (c+9 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {-\frac {2 a^3 (c-d)^2 (c+9 d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {\sqrt {2} a^{3/2} (c+9 d) (c-d)^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f 
*x])^(3/2)) + (((-4*a^(3/2)*(5*c - 3*d)*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Co 
s[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (Sqr 
t[2]*a^(3/2)*(c - d)^(3/2)*(c + 9*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + 
f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f)/a + 
 (2*a*(c - 3*d)*d*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin 
[e + f*x]]))/(4*a^2)
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3244

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

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[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]
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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]
 

rule 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] 
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3462

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

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 651 vs. \(2 (210) = 420\).

Time = 0.59 (sec) , antiderivative size = 3514, normalized size of antiderivative = 14.00 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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