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

Optimal. Leaf size=117 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}+\frac{3 \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.192404, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2736, 2687, 2650, 2649, 206} \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}+\frac{3 \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx &=\frac{\int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{a c}\\ &=-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}}+\frac{3 \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{2 a}\\ &=\frac{3 \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}}+\frac{3 \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 a c}\\ &=\frac{3 \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{4 a c f}\\ &=\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}+\frac{3 \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{a c f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.603284, size = 125, normalized size = 1.07 \[ -\frac{\sec (e+f x) \left (-3 \sin (e+f x)+(3+3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+1\right )}{4 a c f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.638, size = 134, normalized size = 1.2 \begin{align*} -{\frac{1}{8\,af\cos \left ( fx+e \right ) } \left ( 3\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ) c-3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }-6\,{c}^{3/2}\sin \left ( fx+e \right ) +2\,{c}^{3/2} \right ){c}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/c^(5/2)/a*(3*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f
*x+e)*c-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c*(c*(1+sin(f*x+e)))^(1/2)-6*c^(3/2)*s
in(f*x+e)+2*c^(3/2))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.13358, size = 563, normalized size = 4.81 \begin{align*} \frac{3 \, \sqrt{2}{\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) - \cos \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt{-c \sin \left (f x + e\right ) + c}{\left (3 \, \sin \left (f x + e\right ) - 1\right )}}{16 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*sqrt(2)*(cos(f*x + e)*sin(f*x + e) - cos(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*
sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*
x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(-c*sin(f*x + e)
+ c)*(3*sin(f*x + e) - 1))/(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., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{- c \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} + c \sqrt{- c \sin{\left (e + f x \right )} + c}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.99257, size = 628, normalized size = 5.37 \begin{align*} \frac{\frac{3 \, \sqrt{2} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{2 \, \sqrt{-c}}\right )}{a \sqrt{-c} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{4 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} + 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} \sqrt{c} - c\right )} a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{2 \,{\left (3 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{3} -{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} \sqrt{c} -{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} c - c^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} - 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} \sqrt{c} - c\right )}^{2} a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - sqrt(c
))/sqrt(-c))/(a*sqrt(-c)*c*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 4*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f
*x + 1/2*e)^2 + c) - sqrt(c))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqr
t(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1)
) + 2*(3*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3 - (sqrt(c)*tan(1/2*f*x + 1/2*e)
- sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*sqrt(c) - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)
^2 + c))*c - c^(3/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 - 2*(sqrt(c)*tan
(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^2*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f