\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx\) [550]

Optimal result

Integrand size = 27, antiderivative size = 194 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx=-\frac {a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac {3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-1/4*a^(5/2)*(3*c^2+10*c*d+19*d^2)*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)^(5/2)/f+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))^2+3/4*a^3*(c-d)*(c+ 
3*d)*cos(f*x+e)/d^2/(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 = 9.43 (sec) , antiderivative size = 920, normalized size of antiderivative = 4.74 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(5/2)*(((3*c^2 + 10*c*d + 19*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)*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) + ((3*c^2 + 10*c*d + 19*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)*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*(c - d) 
^2*Sqrt[d]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + 
f*x])^2) - (4*Sqrt[d]*(-5*c^2 - 6*c*d + 11*d^2)*(Cos[(e + f*x)/2] - Sin...
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.05, 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, 3459, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3459

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {a \left (-\frac {a \left (3 c^2+10 c d+19 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}-\frac {3 a^2 (c-d) (c+3 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {a \left (-\frac {a \left (3 c^2+10 c d+19 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 d (c+d)}-\frac {3 a^2 (c-d) (c+3 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 d (c+d)}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {a \left (\frac {a^2 \left (3 c^2+10 c d+19 d^2\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f (c+d)}-\frac {3 a^2 (c-d) (c+3 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 d (c+d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {a \left (\frac {a^{3/2} \left (3 c^2+10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f (c+d)^{3/2}}-\frac {3 a^2 (c-d) (c+3 d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 d (c+d)}\)

Input:

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

Output:

(a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*d*(c + d)*f*(c + d* 
Sin[e + f*x])^2) - (a*((a^(3/2)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a] 
*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(3/2)*( 
c + d)^(3/2)*f) - (3*a^2*(c - d)*(c + 3*d)*Cos[e + f*x])/(d*(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 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 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   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, A, B}, 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. \(566\) vs. \(2(170)=340\).

Time = 89.66 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.92

method	result	size
default	\(-\frac {a \left (-\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} d^{2} \left (3 c^{2}+10 c d +19 d^{2}\right ) \cos \left (f x +e \right )^{2}+2 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c d \left (3 c^{2}+10 c d +19 d^{2}\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^{2} c^{4}+10 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{3} d +22 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c^{2} d^{2}+10 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} c \,d^{3}+19 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{2} d^{4}+5 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c^{2} d +6 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c \,d^{2}-11 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{3}-3 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{3}-13 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2} d +3 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a c \,d^{2}+13 \sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{3}\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^{2} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(567\)

Input:

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

Output:

-1/4*a*(-arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(3* 
c^2+10*c*d+19*d^2)*cos(f*x+e)^2+2*sin(f*x+e)*arctanh((a-sin(f*x+e)*a)^(1/2 
)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(3*c^2+10*c*d+19*d^2)+3*arctanh((a-sin(f* 
x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^4+10*arctanh((a-sin(f*x+e)*a)^( 
1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^3*d+22*arctanh((a-sin(f*x+e)*a)^(1/2)*d/ 
(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2+10*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c* 
d+a*d^2)^(1/2))*a^2*c*d^3+19*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2 
)^(1/2))*a^2*d^4+5*(a-sin(f*x+e)*a)^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d+6*(a-sin 
(f*x+e)*a)^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2-11*(a-sin(f*x+e)*a)^(3/2)*(a*(c+d 
)*d)^(1/2)*d^3-3*(a-sin(f*x+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*a*c^3-13*(a-sin( 
f*x+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2*d+3*(a-sin(f*x+e)*a)^(1/2)*(a*(c+d 
)*d)^(1/2)*a*c*d^2+13*(a-sin(f*x+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*a*d^3)*(-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^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. 839 vs. \(2 (170) = 340\).

Time = 0.30 (sec) , antiderivative size = 1994, normalized size of antiderivative = 10.28 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/16*((3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2 
*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^3 - (6*a^2 
*c^3*d + 23*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + (3*a 
^2*c^4 + 10*a^2*c^3*d + 22*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f* 
x + e) + (3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^ 
2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + 2*(3* 
a^2*c^3*d + 10*a^2*c^2*d^2 + 19*a^2*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqr 
t(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^3 + 3*a^ 
2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 + (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d 
^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 8*a^2*c^2*d - 9*a^2*c*d^2 - 2*a^2*d^3)*c 
os(f*x + e) - (3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 - (5*a^2 
*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*s...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((a*sin(f*x + e) + a)^(5/2)/(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. 429 vs. \(2 (170) = 340\).

Time = 0.17 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.21 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(3*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
)) + 10*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 19*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^2*d^2 + 2*c*d^3 + d^4)*sqrt(-c*d - d^2)) - 2*(10*a 
^2*c^2*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 + 12*a^2*c*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 - 22*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi 
 + 1/2*f*x + 1/2*e)^3 - 3*a^2*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin( 
-1/4*pi + 1/2*f*x + 1/2*e) - 13*a^2*c^2*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*a^2*c*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) + 13*a^2*d^3*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 + 2*c*d^3 + d^4 
)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)^2))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(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)**2)/(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) + 2*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))