\(\int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx\) [569]

Optimal result

Integrand size = 27, antiderivative size = 218 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=-\frac {\left (3 c^2-14 c d+43 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} (c-d)^3 f}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt {c+d} f}-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}} \] Output:

-1/32*(3*c^2-14*c*d+43*d^2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*si 
n(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/(c-d)^3/f+2*d^(5/2)*arctanh(a^(1/2)*d^(1/ 
2)*cos(f*x+e)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/(c-d)^3/(c+d)^(1 
/2)/f-1/4*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*c-11*d)*cos(f* 
x+e)/a/(c-d)^2/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 = 5.59 (sec) , antiderivative size = 863, normalized size of antiderivative = 3.96 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((8*Sin[(e + f*x)/2])/(c - d) - (4* 
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(c - d) + (2*(3*c - 11*d)*Sin[(e + 
f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(c - d)^2 + ((-3*c + 11*d 
)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/(c - d)^2 + ((1 + I)*(-1)^(3/4) 
*(3*c^2 - 14*c*d + 43*d^2)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f 
*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)/(c - d)^3 + (8*d^(5/2)*( 
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])^4)/((c - d)^3*Sqrt[c + d]) + (8*d^(5/2)*(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]*Sqr 
t[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])^4)/((-c + d)^3*Sqrt[c + d])))/(16*f*(a*(1 + Si...
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3245, 27, 3042, 3457, 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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]
 

Output:

-1/4*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^(5/2)) + ((-((Sqrt[2]*a^ 
(3/2)*(3*c^2 - 14*c*d + 43*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sq 
rt[a + a*Sin[e + f*x]])])/((c - d)*f)) + (64*a^(3/2)*d^(5/2)*ArcTanh[(Sqrt 
[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/((c - d 
)*Sqrt[c + d]*f))/(4*a^2*(c - d)) - (a*(3*c - 11*d)*Cos[e + f*x])/(2*(c - 
d)*f*(a + a*Sin[e + f*x])^(3/2)))/(8*a^2*(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 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 3457

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[b*(A*b - a*B)*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*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], 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[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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. \(731\) vs. \(2(185)=370\).

Time = 1.01 (sec) , antiderivative size = 732, normalized size of antiderivative = 3.36

method	result	size
default	\(\frac {\left (\left (-64 d^{3} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {5}{2}}+3 \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 ) \sqrt {2}\, a^{2} c^{2}-14 \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 ) \sqrt {2}\, a^{2} c d +43 \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 ) \sqrt {2}\, a^{2} d^{2}\right ) \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (128 d^{3} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {5}{2}}-6 \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 ) \sqrt {2}\, a^{2} c^{2}+28 \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 ) \sqrt {2}\, a^{2} c d -86 \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 ) \sqrt {2}\, a^{2} d^{2}\right )+128 d^{3} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {5}{2}}+6 \sqrt {a \left (c +d \right ) d}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c^{2}-28 \sqrt {a \left (c +d \right ) d}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c d +22 \sqrt {a \left (c +d \right ) d}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, d^{2}-20 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c^{2}+72 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c d -52 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} d^{2}-6 \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 ) \sqrt {2}\, a^{2} c^{2}+28 \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 ) \sqrt {2}\, a^{2} c d -86 \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 ) \sqrt {2}\, a^{2} d^{2}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \left (c -d \right )^{3} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(732\)

Input:

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

Output:

1/32/a^(9/2)*((-64*d^3*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2 
))*a^(5/2)+3*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/ 
a^(1/2))*2^(1/2)*a^2*c^2-14*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a) 
^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*c*d+43*(a*(c+d)*d)^(1/2)*arctanh(1/2*( 
a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*d^2)*cos(f*x+e)^2+sin(f 
*x+e)*(128*d^3*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/ 
2)-6*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2)) 
*2^(1/2)*a^2*c^2+28*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2 
^(1/2)/a^(1/2))*2^(1/2)*a^2*c*d-86*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f* 
x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*d^2)+128*d^3*arctanh((a-sin(f*x 
+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)+6*(a*(c+d)*d)^(1/2)*(a-sin(f*x 
+e)*a)^(3/2)*a^(1/2)*c^2-28*(a*(c+d)*d)^(1/2)*(a-sin(f*x+e)*a)^(3/2)*a^(1/ 
2)*c*d+22*(a*(c+d)*d)^(1/2)*(a-sin(f*x+e)*a)^(3/2)*a^(1/2)*d^2-20*(a*(c+d) 
*d)^(1/2)*(a-sin(f*x+e)*a)^(1/2)*a^(3/2)*c^2+72*(a*(c+d)*d)^(1/2)*(a-sin(f 
*x+e)*a)^(1/2)*a^(3/2)*c*d-52*(a*(c+d)*d)^(1/2)*(a-sin(f*x+e)*a)^(1/2)*a^( 
3/2)*d^2-6*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^ 
(1/2))*2^(1/2)*a^2*c^2+28*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^( 
1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*c*d-86*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a- 
sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*d^2)*(-a*(-1+sin(f*x+e))) 
^(1/2)/(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c-d)^3/cos(f*x+e)/(a+sin(f*x+e...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 865 vs. \(2 (185) = 370\).

Time = 0.45 (sec) , antiderivative size = 2015, normalized size of antiderivative = 9.24 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="fricas")
 

Output:

[-1/64*(sqrt(2)*((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 14* 
c*d + 43*d^2)*cos(f*x + e)^2 - 12*c^2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c 
*d + 43*d^2)*cos(f*x + e) + ((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^2 - 12 
*c^2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e))*sin(f* 
x + e))*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)) + 32*(a*d^2*cos(f*x + e)^3 + 3*a*d^2*cos(f* 
x + e)^2 - 2*a*d^2*cos(f*x + e) - 4*a*d^2 + (a*d^2*cos(f*x + e)^2 - 2*a*d^ 
2*cos(f*x + e) - 4*a*d^2)*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))*sqr 
t(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*((3*c^2 - 14*c*d + 1 
1*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 22*c*d + 15*d^2)* 
cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 - 14*c*d + 11*d^2)*cos(f...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4} d +\sin \left (f x +e \right )^{3} c +3 \sin \left (f x +e \right )^{3} d +3 \sin \left (f x +e \right )^{2} c +3 \sin \left (f x +e \right )^{2} d +3 \sin \left (f x +e \right ) c +\sin \left (f x +e \right ) d +c}d x \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**4*d + sin(e + f*x)**3*c 
 + 3*sin(e + f*x)**3*d + 3*sin(e + f*x)**2*c + 3*sin(e + f*x)**2*d + 3*sin 
(e + f*x)*c + sin(e + f*x)*d + c),x))/a**3