\(\int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [608]

Optimal result

Integrand size = 29, antiderivative size = 260 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2 d^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{a^{5/2} f}-\frac {\sqrt {c-d} \left (3 c^2+14 c d+43 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a+a \sin (e+f x))^{5/2}} \] Output:

-2*d^(5/2)*arctan(a^(1/2)*d^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c+d*s 
in(f*x+e))^(1/2))/a^(5/2)/f-1/32*(c-d)^(1/2)*(3*c^2+14*c*d+43*d^2)*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*si 
n(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/f-1/16*(c-d)*(3*c+11*d)*cos(f*x+e)*(c+d*s 
in(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)*cos(f*x+e)*(c+d*sin( 
f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(5/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.90 (sec) , antiderivative size = 1845, normalized size of antiderivative = 7.10 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-1/4*(c - d)^2/(Cos[(e + f*x)/2] 
 + Sin[(e + f*x)/2])^3 - (3*(c - d)*(c + 5*d))/(16*(Cos[(e + f*x)/2] + Sin 
[(e + f*x)/2])) + (3*(c^2*Sin[(e + f*x)/2] + 4*c*d*Sin[(e + f*x)/2] - 5*d^ 
2*Sin[(e + f*x)/2]))/(8*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2) + (c^2*Si 
n[(e + f*x)/2] - 2*c*d*Sin[(e + f*x)/2] + d^2*Sin[(e + f*x)/2])/(2*(Cos[(e 
 + f*x)/2] + Sin[(e + f*x)/2])^4))*Sqrt[c + d*Sin[e + f*x]])/(f*(a*(1 + Si 
n[e + f*x]))^(5/2)) + ((Sqrt[2]*(3*c^3 + 11*c^2*d + 29*c*d^2 - 43*d^3)*Log 
[1 + Tan[(e + f*x)/2]] - Sqrt[2]*(3*c^3 + 11*c^2*d + 29*c*d^2 - 43*d^3)*Lo 
g[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]] - (32*I)*Sqrt[c - d]*d^(5/2)*(Log[(c - I 
*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin 
[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2])/(16*d^(7/2)*(I + Tan[(e + f*x 
)/2]))] - Log[(c + I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^( 
-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2])/(16*d^(7/2)*(- 
I + Tan[(e + f*x)/2]))]))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*((3*c^3) 
/(32*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (11 
*c^2*d)/(32*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]] 
) + (29*c*d^2)/(32*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e 
+ f*x]]) - (11*d^3)/(32*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*S 
in[e + f*x]]) + (d^3*Sin[e + f*x])/((Cos[(e + f*x)/2] + Sin[(e + f*x)/2...
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3244, 27, 3042, 3456, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {32 a^2 \sin (e+f x) d^3+a^2 (3 c-d) \left (c^2+4 d c+11 d^2\right )}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {32 a^2 \sin (e+f x) d^3+a^2 (3 c-d) \left (c^2+4 d c+11 d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {32 a^2 \sin (e+f x) d^3+a^2 (3 c-d) \left (c^2+4 d c+11 d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {\frac {a^2 (c-d) \left (3 c^2+14 c d+43 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+32 a d^3 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (c-d) \left (3 c^2+14 c d+43 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+32 a d^3 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {\frac {a^2 (c-d) \left (3 c^2+14 c d+43 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {64 a^2 d^3 \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {a^2 (c-d) \left (3 c^2+14 c d+43 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {64 a^{3/2} d^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {-\frac {2 a^3 (c-d) \left (3 c^2+14 c d+43 d^2\right ) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {64 a^{3/2} d^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {64 a^{3/2} d^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {\sqrt {2} a^{3/2} \sqrt {c-d} \left (3 c^2+14 c d+43 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {a (c-d) (3 c+11 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f 
*x])^(5/2)) + (((-64*a^(3/2)*d^(5/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x]) 
/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (Sqrt[2]*a^(3/2 
)*Sqrt[c - d]*(3*c^2 + 14*c*d + 43*d^2)*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)/ 
(4*a^2) - (a*(c - d)*(3*c + 11*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/( 
2*f*(a + a*Sin[e + f*x])^(3/2)))/(8*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 3456

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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
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 - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{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] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(5127\) vs. \(2(219)=438\).

Time = 3409.48 (sec) , antiderivative size = 5128, normalized size of antiderivative = 19.72

method	result	size
default	\(\text {Expression too large to display}\)	\(5128\)

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 759 vs. \(2 (219) = 438\).

Time = 0.64 (sec) , antiderivative size = 3949, normalized size of antiderivative = 15.19 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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