\(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx\) [543]

Optimal result

Integrand size = 27, antiderivative size = 179 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx=-\frac {a^{3/2} (c+7 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{3/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-1/4*a^(3/2)*(c+7*d)*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/2)/(a+a*s 
in(f*x+e))^(1/2))/d^(3/2)/(c+d)^(5/2)/f+1/2*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))^2-1/4*a^2*(c+7*d)*cos(f*x+e)/d/(c 
+d)^2/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.85 (sec) , antiderivative size = 888, normalized size of antiderivative = 4.96 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*(((c + 7*d)*((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]*L 
og[-#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)*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)^(7/2) + ((c + 7 
*d)*(-((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2])) + Sqrt[c + d]*RootSu 
m[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)*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)^(7/2) - (8*Sqrt[d]*(-c + d)*(Cos[(e + f*x)/2] - Si 
n[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x])^2) - (4*Sqrt[d]*(c + 7*d)*( 
Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)^2*(c + d*Sin[e + f*x])))...
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3241, 27, 2011, 3042, 3251, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}-\frac {a \int -\frac {a (c+7 d)+a \sin (e+f x) (c+7 d)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2}dx}{2 d (c+d)}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3251

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

\(\displaystyle \frac {a (c+7 d) \left (-\frac {a \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c+d)}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

(a^2*(c - d)*Cos[e + f*x])/(2*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d* 
Sin[e + f*x])^2) + (a*(c + 7*d)*(-((Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e 
 + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[d]*(c + d)^(3/2)*f 
)) - (a*Cos[e + f*x])/((c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f 
*x]))))/(4*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 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 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(428\) vs. \(2(155)=310\).

Time = 0.66 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.40

method	result	size
default	\(\frac {\left (-\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right )^{2} a^{2} c \,d^{2}-7 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right )^{2} a^{2} d^{3}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c^{2} d -14 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c \,d^{2}+\left (-a \left (-1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c d +7 \left (-a \left (-1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{2}-\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{3}-7 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{2} d +\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2}-8 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, a c d -9 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{2}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 \sqrt {a \left (c +d \right ) d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(429\)

Input:

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

Output:

1/4*(-arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2 
*a^2*c*d^2-7*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f 
*x+e)^2*a^2*d^3-2*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))* 
sin(f*x+e)*a^2*c^2*d-14*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^( 
1/2))*sin(f*x+e)*a^2*c*d^2+(-a*(-1+sin(f*x+e)))^(3/2)*(a*(c+d)*d)^(1/2)*c* 
d+7*(-a*(-1+sin(f*x+e)))^(3/2)*(a*(c+d)*d)^(1/2)*d^2-arctanh((-a*(-1+sin(f 
*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^3-7*arctanh((-a*(-1+sin(f*x+e)))^ 
(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^2*d+(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d 
)^(1/2)*a*c^2-8*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-9*(-a*( 
-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2)*(-a*(-1+sin(f*x+e)))^(1/2)* 
(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/d/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. 621 vs. \(2 (155) = 310\).

Time = 0.23 (sec) , antiderivative size = 1558, normalized size of antiderivative = 8.70 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/16*((a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*co 
s(f*x + e)^3 - (2*a*c^2*d + 15*a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + (a*c^3 
+ 7*a*c^2*d + a*c*d^2 + 7*a*d^3)*cos(f*x + e) + (a*c^3 + 9*a*c^2*d + 15*a* 
c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + 2*(a*c^2*d + 7*a*c* 
d^2)*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*(a*c^2 - 6*a*c*d + 5*a*d^2 - (a*c*d + 7*a*d^2)*cos(f*x 
+ e)^2 + (a*c^2 - 7*a*c*d - 2*a*d^2)*cos(f*x + e) - (a*c^2 - 6*a*c*d + 5*a 
*d^2 + (a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + 
 a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 
+ 4*c*d^4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d^3 + 2*c*d 
^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5) 
*f + ((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (155) = 310\).

Time = 0.16 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.74 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c^{2} d + 2 \, c d^{2} + d^{3}\right )} \sqrt {-c d - d^{2}}} + \frac {2 \, {\left (2 \, a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 14 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c^{2} d + 2 \, c d^{2} + d^{3}\right )} {\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}\right )}}{8 \, f} \] Input:

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

Output:

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 7 
*a*d*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^2*d + 2*c*d^2 + d^3)*sqrt(-c*d - d^2 
)) + 2*(2*a*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)^3 + 14*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/ 
2*f*x + 1/2*e)^3 + a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 
 1/2*f*x + 1/2*e) - 8*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*p 
i + 1/2*f*x + 1/2*e) - 9*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/ 
4*pi + 1/2*f*x + 1/2*e))/((c^2*d + 2*c*d^2 + d^3)*(2*d*sin(-1/4*pi + 1/2*f 
*x + 1/2*e)^2 - c - d)^2))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(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) + int((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))