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

Optimal. Leaf size=140 \[ \frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{\cos (e+f x)}{4 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{\cos (e+f x)}{4 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[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]])

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Rubi [A]  time = 0.268799, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2743, 2741, 3770} \[ \frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{\cos (e+f x)}{4 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{\cos (e+f x)}{4 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[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 2743

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 2741

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 3770

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx &=\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{\tanh ^{-1}(\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]  time = 0.617281, size = 224, normalized size = 1.6 \[ \frac{\cos (e+f x) \left (-3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\sin (e+f x) \left (4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-2\right )+4\right )}{8 c^2 f (\sin (e+f x)-1)^2 \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(4 - 3*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Cos[2*(e + f*x)]*(Log[Cos[(e + f*x)/2] - Sin[(
e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 3*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (-2 +
4*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 4*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x]))/(8*c^2
*f*(-1 + Sin[e + f*x])^2*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.188, size = 252, normalized size = 1.8 \begin{align*}{\frac{\cos \left ( fx+e \right ) }{4\,f} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\,\sin \left ( fx+e \right ) -2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2 \right ){\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.35324, size = 990, normalized size = 7.07 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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