\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx\) [549]

Optimal result

Integrand size = 27, antiderivative size = 166 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx=\frac {a^{5/2} (c-d) (3 c+5 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{5/2} (c+d)^{3/2} f}-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 8.47 (sec) , antiderivative size = 892, normalized size of antiderivative = 5.37 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(5/2)*(-8*Sqrt[d]*Cos[(e + f*x)/2] + ((-3*c^2 - 2* 
c*d + 5*d^2)*((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2]) + Sqrt[c + d]* 
RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 
 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] - d*Sqrt[c + d 
]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 
 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt[c + d]*Log[-#1 + Tan[ 
(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Lo 
g[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4 
]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d 
*#1^2 - c*#1^3) & ]))/(c + d)^(5/2) + ((3*c^2 + 2*c*d - 5*d^2)*((c + d)*(e 
 + f*x - 2*Log[Sec[(e + f*x)/4]^2]) - Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c 
*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - 
d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x 
)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + T 
an[(e + f*x)/4]]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqr 
t[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4] 
]*#1^2 - 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]* 
Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c 
 + d)^(5/2) + 8*Sqrt[d]*Sin[(e + f*x)/2] - (4*(c - d)^2*Sqrt[d]*(Cos[(e + 
f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(4*d^(5/2...
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3241, 27, 3042, 3460, 3042, 3252, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 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 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*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 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(392\) vs. \(2(148)=296\).

Time = 10.71 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.37

method	result	size
default	\(-\frac {a^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (-\sin \left (f x +e \right ) d \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2}+2 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c d -5 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,d^{2}-2 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, c -2 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, d \right )-3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{3}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2} d +5 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c \,d^{2}+3 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, c^{2}+\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, d^{2}\right )}{d^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(393\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (148) = 296\).

Time = 0.19 (sec) , antiderivative size = 1322, normalized size of antiderivative = 7.96 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*((3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2*c^2*d + 
2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 2*a^2*c^2*d - 5*a^2 
*c*d^2)*cos(f*x + e) + (3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^3 
+ (3*a^2*c^2*d + 2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt 
(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a 
*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3 
)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d 
^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) 
 + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a* 
d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f 
*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 
- c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c 
*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(3*a^2*c^2 - 2*a^2 
*c*d - a^2*d^2 + 2*(a^2*c*d + a^2*d^2)*cos(f*x + e)^2 + (3*a^2*c^2 + a^2*d 
^2)*cos(f*x + e) - (3*a^2*c^2 - 2*a^2*c*d - a^2*d^2 - 2*(a^2*c*d + a^2*d^2 
)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^3 + d^4)*f*c 
os(f*x + e)^2 - (c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^ 
4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*c*d^3 + d^4)*f)*sin(f* 
x + e)), -1/2*((3*a^2*c^3 + 5*a^2*c^2*d - 3*a^2*c*d^2 - 5*a^2*d^3 - (3*a^2 
*c^2*d + 2*a^2*c*d^2 - 5*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 2*a^2*c...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (148) = 296\).

Time = 0.16 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.81 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {2} {\left (\frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{d^{2}} + \frac {\sqrt {2} {\left (3 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c d^{2} + d^{3}\right )} \sqrt {-c d - d^{2}}} - \frac {2 \, {\left (a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c d^{2} + d^{3}\right )} {\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}}\right )} \sqrt {a}}{2 \, f} \] Input:

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

Output:

1/2*sqrt(2)*(4*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f 
*x + 1/2*e)/d^2 + sqrt(2)*(3*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 
 2*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*a^2*d^2*sgn(cos(-1/4*pi 
 + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt 
(-c*d - d^2))/((c*d^2 + d^3)*sqrt(-c*d - d^2)) - 2*(a^2*c^2*sgn(cos(-1/4*p 
i + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 2*a^2*c*d*sgn(cos(- 
1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + a^2*d^2*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((c*d^2 + d^ 
3)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)))*sqrt(a)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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