3.4.94 \(\int \frac {1}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}} \, dx\) [394]

Optimal. Leaf size=143 \[ -\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{2 a c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

-1/2*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)+1/2*cos(f*x+e)/a/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)/a/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 = 0.19, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 4, 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 a f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{2 a 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/((a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

-1/2*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2)) + Cos[e + f*x]/(2*a*f*Sqrt[a + a*S
in[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*a*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}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}} \, dx &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a 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 a c}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a 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 a c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a 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 a 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 = 0.40, size = 170, normalized size = 1.19 \begin {gather*} \frac {\cos (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 )+\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 (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-2 \sin (e+f x)\right )}{4 c f (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 11.61, size = 114, normalized size = 0.80

method	result	size
default	\(\frac {\left (\ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-\ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right )}{2 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^(3/2)/(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))*cos(f*x+e)^2-ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*
x+e)^2+sin(f*x+e))*cos(f*x+e)/(a*(1+sin(f*x+e)))^(3/2)/(-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))^(3/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 282, normalized size = 1.97 \begin {gather*} \left [\frac {\sqrt {a c} \cos \left (f x + e\right )^{3} \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} \sin \left (f x + e\right )}{4 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}}, -\frac {\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 ) \cos \left (f x + e\right )^{3} - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{2 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(a*c)*cos(f*x + e)^3*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))/(a^2*c^2*f*cos(f*x + e)^3), -1/2*(sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*s
qrt(-c*sin(f*x + e) + c)/(a*c*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^3 - sqrt(a*sin(f*x + e) + a)*sqrt(-c*si
n(f*x + e) + c)*sin(f*x + e))/(a^2*c^2*f*cos(f*x + e)^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \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))**(3/2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\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))^(3/2)*(c - c*sin(e + f*x))^(3/2)),x)

[Out]

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

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