\(\int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx\) [401]

Optimal result

Integrand size = 30, antiderivative size = 134 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x)}{4 f (3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{12 f (3+3 \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{36 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

-1/4*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2)-1/4*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c
*sin(f*x+e))^(1/2)+1/4*arctanh(sin(f*x+e))*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2822, 2820, 3855} \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{4 a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

-1/4*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]) - Cos[e + f*x]/(4*a*f*(a + a*Sin[e +
f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(4*a^2*f*Sqrt[a + a*Sin[e + f*x]
]*Sqrt[c - c*Sin[e + f*x]])

Rule 2820

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Di
st[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 2822

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] + Dist[(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] /; FreeQ[{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 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}+\frac {\int \frac {1}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx}{2 a} \\ & = -\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{4 a^2} \\ & = -\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) \int \sec (e+f x) \, dx}{4 a^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-10-3 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (2+\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+3 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 \left (5+2 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{72 \sqrt {3} f (1+\sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[1/((3 + 3*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-10 - 3*Log[1 - Tan[(e + f*x)/2]
] + Cos[2*(e + f*x)]*(2 + Log[1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]]) + 3*Log[1 + Tan[(e + f*x)/2]]
- 2*(5 + 2*Log[1 - Tan[(e + f*x)/2]] - 2*Log[1 + Tan[(e + f*x)/2]])*Sin[e + f*x]))/(72*Sqrt[3]*f*(1 + Sin[e +
f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(421\) vs. \(2(122)=244\).

Time = 3.54 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.15

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \left (\cos ^{3}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+3 \sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos ^{2}\left (f x +e \right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \cos \left (f x +e \right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \cos \left (f x +e \right )+\sin \left (f x +e \right )+2 \cos \left (f x +e \right )-1}{4 f \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+2 \sin \left (f x +e \right )+\cos \left (f x +e \right )+2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2}}\)	\(422\)

[In]

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

[Out]

1/4/f*(sin(f*x+e)*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)-cos(f*x+e)^2*sin(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)-
1)-cos(f*x+e)^3*ln(-cot(f*x+e)+csc(f*x+e)+1)+cos(f*x+e)^3*ln(-cot(f*x+e)+csc(f*x+e)-1)+2*sin(f*x+e)*cos(f*x+e)
^2+2*ln(-cot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)*cos(f*x+e)-2*ln(-cot(f*x+e)+csc(f*x+e)-1)*sin(f*x+e)*cos(f*x+e)-2
*cos(f*x+e)^3+cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)-cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)-1)+3*sin(f*x+e)
*cos(f*x+e)+cos(f*x+e)^2+2*ln(-cot(f*x+e)+csc(f*x+e)+1)*cos(f*x+e)-2*ln(-cot(f*x+e)+csc(f*x+e)-1)*cos(f*x+e)+s
in(f*x+e)+2*cos(f*x+e)-1)/(sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2+2*sin(f*x+e)+cos(f*x+e)+2)/(-c*(sin(f*x+e)-1))^(
1/2)/(a*(sin(f*x+e)+1))^(1/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.81 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\left [\frac {{\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) + 2\right )}}{8 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}, -\frac {{\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) + 2\right )}}{4 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}\right ] \]

[In]

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

[Out]

[1/8*((cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f*x + e))*sqrt(a*c)*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)*(sin(f*x + e) + 2))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*
f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e)), -1/4*((cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2
*cos(f*x + e))*sqrt(-a*c)*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)*(sin(f*x + e) + 2))/(a^3*c*f*cos(f*x +
e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e))]

Sympy [F]

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

[In]

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

[Out]

Integral(1/((a*(sin(e + f*x) + 1))**(5/2)*sqrt(-c*(sin(e + f*x) - 1))), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {2 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{\frac {5}{2}} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{\frac {5}{2}} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1}{a^{\frac {5}{2}} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{16 \, f} \]

[In]

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

[Out]

1/16*(2*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^(5/2)*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si
n(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^(5/2)*sqrt(c)*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*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^(5/2)*sq
rt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*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

Mupad [F(-1)]

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

[In]

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

[Out]

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