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

Optimal. Leaf size=95 \[ \frac {\cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{2 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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)

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Rubi [A]
time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2822, 2820, 3855} \begin {gather*} \frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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]])

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*} \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx &=\frac {\cos (e+f x)}{2 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}{2 c}\\ &=\frac {\cos (e+f x)}{2 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}{2 c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x)}{2 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)}{2 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 161, normalized size = 1.69 \begin {gather*} -\frac {\cos (e+f x) \left (1-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{2 c f (-1+\sin (e+f x)) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 18.80, size = 165, normalized size = 1.74

method	result	size
default	\(\frac {\left (\ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )-\ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )+\sin \left (f x +e \right )-\ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )\right ) \cos \left (f x +e \right )}{2 f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(165\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 337, normalized size = 3.55 \begin {gather*} \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 ] \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

[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*s
qrt(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))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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))**(3/2),x)

[Out]

Integral(1/(sqrt(a*(sin(e + f*x) + 1))*(-c*(sin(e + f*x) - 1))**(3/2)), x)

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Giac [A]
time = 0.52, size = 177, normalized size = 1.86 \begin {gather*} -\frac {\sqrt {c} {\left (\frac {\log \left (-64 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 64\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 | \sin \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}{\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 ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}\right )}}{4 \, f} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

-1/4*sqrt(c)*(log(-64*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 64)/(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(sin(-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/(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))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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