\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx\) [557]

Optimal result

Integrand size = 27, antiderivative size = 247 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^3 f}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {a} (c-d)^3 (c+d)^{5/2} f}+\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-2^(1/2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^ 
(1/2)/(c-d)^3/f+1/4*d^(1/2)*(15*c^2+10*c*d+7*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))/a^(1/2)/(c-d)^3/(c+d)^(5/2) 
/f+1/2*d*cos(f*x+e)/(c^2-d^2)/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2+ 
1/4*d*(7*c+d)*cos(f*x+e)/(c^2-d^2)^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 = 4.91 (sec) , antiderivative size = 776, normalized size of antiderivative = 3.14 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(((32 + 32*I)*(-1)^(3/4)*ArcTanh[(1 
/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])])/(c - d)^3 + (Sqrt[d]*(15*c^ 
2 + 10*c*d + 7*d^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) & ]))/((c - d)^3*(c + d)^(5/2)) + (Sqrt[d]*(15*c^2 + 10*c*d + 7*d^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) & ]))/((-c 
+ d)^3*(c + d)^(5/2)) + (8*d*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c - 
d)*(c + d)*(c + d*Sin[e + f*x])^2) + (4*d*(7*c + d)*(Cos[(e + f*x)/2] - Si 
n[(e + f*x)/2]))/((c - d)^2*(c + d)^2*(c + d*Sin[e + f*x]))))/(16*f*Sqrt[a 
*(1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.16, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3258, 3042, 3463, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {\int -\frac {a^2 \left (8 c^2+9 d c+7 d^2\right )-a^2 d (7 c+d) \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a \left (c^2-d^2\right )}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (8 c^2+9 d c+7 d^2\right )-a^2 d (7 c+d) \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (8 c^2+9 d c+7 d^2\right )-a^2 d (7 c+d) \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {\frac {8 a^2 (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {a d \left (15 c^2+10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {8 a^2 (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {a d \left (15 c^2+10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {-\frac {16 a^2 (c+d)^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {a d \left (15 c^2+10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {a d \left (15 c^2+10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {8 \sqrt {2} a^{3/2} (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {\frac {2 a^2 d \left (15 c^2+10 c d+7 d^2\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {8 \sqrt {2} a^{3/2} (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 a^{3/2} \sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d) \sqrt {c+d}}-\frac {8 \sqrt {2} a^{3/2} (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{2 a \left (c^2-d^2\right )}+\frac {a d (7 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a \left (c^2-d^2\right )}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

Input:

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

Output:

(d*Cos[e + f*x])/(2*(c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + 
f*x])^2) + (((-8*Sqrt[2]*a^(3/2)*(c + d)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/ 
(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/((c - d)*f) + (2*a^(3/2)*Sqrt[d]*(15* 
c^2 + 10*c*d + 7*d^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))/(2*a*(c^2 - d^2)) + ( 
a*d*(7*c + d)*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d 
*Sin[e + f*x])))/(4*a*(c^2 - d^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 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 3258

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

rule 3463

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 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. \(1064\) vs. \(2(214)=428\).

Time = 1.12 (sec) , antiderivative size = 1065, normalized size of antiderivative = 4.31

method	result	size
default	\(\text {Expression too large to display}\)	\(1065\)

Input:

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

Output:

1/4*(-8*(a*(c+d)*d)^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a 
^(1/2))*2^(1/2)*a^5*c^3*d-4*(a*(c+d)*d)^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+ 
e)))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^5*c^2*d^2+10*arctanh((-a*(-1+sin(f*x 
+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*c^3*d^2+7*arctanh((-a*(-1+sin(f* 
x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*c^2*d^3+(-a*(-1+sin(f*x+e)))^(1 
/2)*(a*(c+d)*d)^(1/2)*a^(9/2)*d^4+(-a*(-1+sin(f*x+e)))^(3/2)*(a*(c+d)*d)^( 
1/2)*a^(7/2)*d^4+7*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2)) 
*a^(11/2)*sin(f*x+e)^2*d^5+15*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d 
)*d)^(1/2))*a^(11/2)*c^4*d+30*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d 
)*d)^(1/2))*a^(11/2)*sin(f*x+e)*c^3*d^2+20*arctanh((-a*(-1+sin(f*x+e)))^(1 
/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x+e)*c^2*d^3+14*arctanh((-a*(-1+si 
n(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x+e)*c*d^4-8*(a*(c+d) 
*d)^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)* 
sin(f*x+e)*a^5*c*d^3-4*(a*(c+d)*d)^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^ 
(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*sin(f*x+e)^2*a^5*c^2*d^2-8*(a*(c+d)*d)^(1/2 
)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*sin(f*x+ 
e)^2*a^5*c*d^3-8*(a*(c+d)*d)^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)* 
2^(1/2)/a^(1/2))*2^(1/2)*sin(f*x+e)*a^5*c^3*d-16*(a*(c+d)*d)^(1/2)*arctanh 
(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*sin(f*x+e)*a^5*c^ 
2*d^2+9*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a^(9/2)*c^3*d-7*(-...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1309 vs. \(2 (214) = 428\).

Time = 0.58 (sec) , antiderivative size = 2903, normalized size of antiderivative = 11.75 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*((15*a*c^4 + 40*a*c^3*d + 42*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4 - (15* 
a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^3 - (30*a*c^3*d + 35*a*c^2* 
d^2 + 24*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^2 + (15*a*c^4 + 10*a*c^3*d + 22*a 
*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e) + (15*a*c^4 + 40*a*c^3*d + 4 
2*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4 - (15*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4) 
*cos(f*x + e)^2 + 2*(15*a*c^3*d + 10*a*c^2*d^2 + 7*a*c*d^3)*cos(f*x + e))* 
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))) + 8*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a 
*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (2*a*c^3 
*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d 
+ 2*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a*c^3*d + 6*a 
*c^2*d^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e 
)^2 + 2*(a*c^3*d + 2*a*c^2*d^2 + a*c*d^3)*cos(f*x + e))*sin(f*x + e))*l...
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

integrate(1/(sqrt(a*sin(f*x + e) + a)*(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. 559 vs. \(2 (214) = 428\).

Time = 0.20 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.26 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*sqrt(2)*(sqrt(2)*(15*c^2*d + 10*c*d^2 + 7*d^3)*arctan(sqrt(2)*d*sin(-1 
/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((c^5*sgn(cos(-1/4*pi + 1/2*f*x 
 + 1/2*e)) - c^4*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c^3*d^2*sgn(cos 
(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*c^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
)) + c*d^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - d^5*sgn(cos(-1/4*pi + 1/2 
*f*x + 1/2*e)))*sqrt(-c*d - d^2)) + 4*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 
 1)/(c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*c^2*d*sgn(cos(-1/4*pi + 1 
/2*f*x + 1/2*e)) + 3*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - d^3*sgn(c 
os(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 
1)/(c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*c^2*d*sgn(cos(-1/4*pi + 1/ 
2*f*x + 1/2*e)) + 3*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - d^3*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*(14*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e 
)^3 + 2*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 9*c^2*d*sin(-1/4*pi + 1/2*f 
*x + 1/2*e) - 8*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) + d^3*sin(-1/4*pi + 1 
/2*f*x + 1/2*e))/((c^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c^2*d^2*sgn 
(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) 
)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)^2))/(sqrt(a)*f)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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