\(\int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx\) [562]

Optimal result

Integrand size = 27, antiderivative size = 164 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=-\frac {(c-5 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^2 f}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^2 \sqrt {c+d} f}-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 2.74 (sec) , antiderivative size = 747, normalized size of antiderivative = 4.55 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2] - (c - 
d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (1 + I)*(-1)^(3/4)*(c - 5*d)*Ar 
cTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + 
Sin[(e + f*x)/2])^2 - (d^(3/2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + Root 
Sum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + 
f*x)/4]]) + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + 
Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*# 
1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + T 
an[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3 
*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d 
] + (d^(3/2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + RootSum[c + 4*d*#1 + 2 
*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d 
]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]* 
#1 - 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + 
Tan[(e + f*x)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*# 
1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) 
& ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d]))/(2*(c - d)^2*f 
*(a*(1 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3245, 27, 3042, 3464, 3042, 3128, 219, 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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}dx\)

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

\(\displaystyle \frac {-\frac {8 a d^2 \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {\sqrt {2} \sqrt {a} (c-5 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {8 \sqrt {a} d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d) \sqrt {c+d}}-\frac {\sqrt {2} \sqrt {a} (c-5 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

Input:

Int[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])),x]
 

Output:

(-((Sqrt[2]*Sqrt[a]*(c - 5*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt 
[a + a*Sin[e + f*x]])])/((c - d)*f)) - (8*Sqrt[a]*d^(3/2)*ArcTanh[(Sqrt[a] 
*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/((c - d)*S 
qrt[c + d]*f))/(4*a^2*(c - d)) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + 
f*x])^(3/2))
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3245

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (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 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(347\) vs. \(2(135)=270\).

Time = 1.00 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.12

method	result	size
default	\(-\frac {\left (8 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {3}{2}} d^{2} \sin \left (f x +e \right )+\sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (f x +e \right ) a c -5 \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (f x +e \right ) a d +8 d^{2} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {3}{2}}+\sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c -5 \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a d +2 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, c -2 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a}\, d \right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{4 a^{\frac {5}{2}} \sqrt {a \left (c +d \right ) d}\, \left (c -d \right )^{2} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(348\)

Input:

int(1/(a+sin(f*x+e)*a)^(3/2)/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

-1/4/a^(5/2)*(8*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3 
/2)*d^2*sin(f*x+e)+2^(1/2)*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^ 
(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a*c-5*2^(1/2)*(a*(c+d)*d)^(1/2)*arctanh( 
1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a*d+8*d^2*arctanh(( 
a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)+2^(1/2)*(a*(c+d)*d)^( 
1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a*c-5*2^(1/2)*(a* 
(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a*d+2*( 
a*(c+d)*d)^(1/2)*(a-sin(f*x+e)*a)^(1/2)*a^(1/2)*c-2*(a*(c+d)*d)^(1/2)*(a-s 
in(f*x+e)*a)^(1/2)*a^(1/2)*d)*(-a*(-1+sin(f*x+e)))^(1/2)/(a*(c+d)*d)^(1/2) 
/(c-d)^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. 507 vs. \(2 (135) = 270\).

Time = 0.28 (sec) , antiderivative size = 1299, normalized size of antiderivative = 7.92 \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorithm="fricas")
 

Output:

[-1/8*(sqrt(2)*((c - 5*d)*cos(f*x + e)^2 - (c - 5*d)*cos(f*x + e) - ((c - 
5*d)*cos(f*x + e) + 2*c - 10*d)*sin(f*x + e) - 2*c + 10*d)*sqrt(a)*log(-(a 
*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) 
 - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + 
 e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e 
) - 2)) - 4*(a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x 
+ e) + 2*a*d)*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - 
(6*c*d + 7*d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - 4*((c*d + d^2)*cos(f* 
x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) + (c^2 
 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x 
+ e) + a)*sqrt(d/(a*c + a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2* 
cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*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*((c - d)*cos(f*x + e) - (c - d)* 
sin(f*x + e) + c - d)*sqrt(a*sin(f*x + e) + a))/((a^2*c^2 - 2*a^2*c*d + a^ 
2*d^2)*f*cos(f*x + e)^2 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) - 
 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f* 
cos(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f)*sin(f*x + e)), -1/8*(s 
qrt(2)*((c - 5*d)*cos(f*x + e)^2 - (c - 5*d)*cos(f*x + e) - ((c - 5*d)*...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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