3.4.87 \(\int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [387]

3.4.87.1 Optimal result
3.4.87.2 Mathematica [A] (verified)
3.4.87.3 Rubi [A] (verified)
3.4.87.4 Maple [B] (verified)
3.4.87.5 Fricas [A] (verification not implemented)
3.4.87.6 Sympy [F]
3.4.87.7 Maxima [F]
3.4.87.8 Giac [B] (verification not implemented)
3.4.87.9 Mupad [F(-1)]

3.4.87.1 Optimal result

Integrand size = 30, antiderivative size = 95 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x)}{2 f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{2 c f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

output
1/2*cos(f*x+e)/f/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+1/2*arctanh 
(sin(f*x+e))*cos(f*x+e)/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
 
3.4.87.2 Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.66 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \left (-\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+\left (1+\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{2 \sqrt {3} c f (-1+\sin (e+f x)) \sqrt {1+\sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

input
Integrate[1/(Sqrt[3 + 3*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
 
output
(Cos[(e + f*x)/2]^2*(-Log[1 - Tan[(e + f*x)/2]] + Log[1 + Tan[(e + f*x)/2] 
] + (1 + Log[1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]])*Sin[e + f* 
x])*(-1 + Tan[(e + f*x)/2])*(1 + Tan[(e + f*x)/2]))/(2*Sqrt[3]*c*f*(-1 + S 
in[e + f*x])*Sqrt[1 + Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])
 
3.4.87.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3222, 3042, 3220, 3042, 4257}

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.

\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3220

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input
Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
 
output
Cos[e + f*x]/(2*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + ( 
ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c 
 - c*Sin[e + f*x]])
 

3.4.87.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3220
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ 
.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x] 
]*Sqrt[c + d*Sin[e + f*x]])   Int[1/Cos[e + f*x], x], x] /; FreeQ[{a, b, c, 
 d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
 

rule 3222
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[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[(m + n + 1)/(a*(2*m + 1) 
)   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free 
Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && 
 ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || 
!SumSimplerQ[n, 1])
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.4.87.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(246\) vs. \(2(83)=166\).

Time = 3.30 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.60

method result size
default \(-\frac {\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \cos \left (f x +e \right )-\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \cos \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )-\sin \left (f x +e \right )-1}{2 f \left (1+\cos \left (f x +e \right )-\sin \left (f x +e \right )\right ) c \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\) \(247\)

input
int(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 
output
-1/2/f*(cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)-1)-ln(-cot(f*x+e)+csc(f*x+e 
)-1)*sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)+ln(-c 
ot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)*cos(f*x+e)+ln(-cot(f*x+e)+csc(f*x+e)-1) 
*cos(f*x+e)-ln(-cot(f*x+e)+csc(f*x+e)+1)*cos(f*x+e)+cos(f*x+e)^2-sin(f*x+e 
)*cos(f*x+e)-sin(f*x+e)-1)/(1+cos(f*x+e)-sin(f*x+e))/c/(a*(sin(f*x+e)+1))^ 
(1/2)/(-c*(sin(f*x+e)-1))^(1/2)
 
3.4.87.5 Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.27 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\left [\frac {\sqrt {a c} {\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) - \cos \left (f x + e\right )\right )} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}}, -\frac {\sqrt {-a c} {\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) - \cos \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}}\right ] \]

input
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fr 
icas")
 
output
[1/4*(sqrt(a*c)*(cos(f*x + e)*sin(f*x + e) - cos(f*x + e))*log(-(a*c*cos(f 
*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + a)*sqrt 
(-c*sin(f*x + e) + c)*sin(f*x + e))/cos(f*x + e)^3) - 2*sqrt(a*sin(f*x + e 
) + a)*sqrt(-c*sin(f*x + e) + c))/(a*c^2*f*cos(f*x + e)*sin(f*x + e) - a*c 
^2*f*cos(f*x + e)), -1/2*(sqrt(-a*c)*(cos(f*x + e)*sin(f*x + e) - cos(f*x 
+ e))*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c) 
/(a*c*cos(f*x + e)*sin(f*x + e))) + sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f 
*x + e) + c))/(a*c^2*f*cos(f*x + e)*sin(f*x + e) - a*c^2*f*cos(f*x + e))]
 
3.4.87.6 Sympy [F]

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

input
integrate(1/(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(3/2),x)
 
output
Integral(1/(sqrt(a*(sin(e + f*x) + 1))*(-c*(sin(e + f*x) - 1))**(3/2)), x)
 
3.4.87.7 Maxima [F]

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

input
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="ma 
xima")
 
output
integrate(1/(sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(3/2)), x)
 
3.4.87.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (83) = 166\).

Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c} {\left (\frac {\log \left (-8 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8\right )}{\sqrt {a} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{\sqrt {a} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {1}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} \sqrt {a} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{4 \, f} \]

input
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="gi 
ac")
 
output
1/4*sqrt(c)*(log(-8*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 8)/(sqrt(a)*c^2*sgn 
(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 2* 
log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/(sqrt(a)*c^2*sgn(cos(-1/4*pi + 1/ 
2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 1/((cos(-1/4*pi + 1 
/2*f*x + 1/2*e)^2 - 1)*sqrt(a)*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn 
(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f
 
3.4.87.9 Mupad [F(-1)]

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

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