\(\int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx\) [388]

Optimal result

Integrand size = 30, antiderivative size = 140 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x)}{4 f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{4 c f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{4 c^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, 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} \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{4 c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{4 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

Cos[e + f*x]/(4*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + Cos[e + f*x]/(4*c*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])/(4*c^2*f*Sqrt[a + a*Sin[e + f*x]]*S
qrt[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 \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{2 c} \\ & = \frac {\cos (e+f x)}{4 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{4 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{4 c^2} \\ & = \frac {\cos (e+f x)}{4 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{4 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x) \int \sec (e+f x) \, dx}{4 c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {\cos (e+f x)}{4 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{4 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{4 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (10-3 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (-2+\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+3 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (-5+2 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{8 \sqrt {3} c^2 f (-1+\sin (e+f x))^2 \sqrt {1+\sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[1/(Sqrt[3 + 3*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)),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]))/(8*Sqrt[3]*c^2*f*(-1 + Si
n[e + f*x])^2*Sqrt[1 + Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(423\) vs. \(2(122)=244\).

Time = 2.97 (sec) , antiderivative size = 424, normalized size of antiderivative = 3.03

method	result	size
default	\(-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+2 \left (\cos ^{3}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \cos \left (f x +e \right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+3 \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right )+\sin \left (f x +e \right )+1}{4 f \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )-\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, c^{2}}\)	\(424\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.69 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {{\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) - 2\right )}}{8 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}, -\frac {{\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) - 2\right )}}{4 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}\right ] \]

[In]

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),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*c^3*f*cos(f*x + e)^3 + 2*a*c^3*
f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*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*c^3*f*cos(f*x +
e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (-256 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 256\right )}{\sqrt {a} c^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{\sqrt {a} c^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1}{\sqrt {a} c^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}}{16 \, f} \]

[In]

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

[Out]

-1/16*(2*log(-256*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 256)/(sqrt(a)*c^(5/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*log(abs(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(sqrt(a)*c^(5/2)*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*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(sqr
t(a)*c^(5/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1
/2*e)^4))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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