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

Optimal. Leaf size=244 \[ -\frac{d (c+3 d) \cos (e+f x)}{a f (c-d)^2 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

-((d*(c + 3*d)*Cos[e + f*x])/(a*(c - d)^2*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])) - Cos[e + f*x]/((c - d)*f*(a +
a*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - ((c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d
*Sin[e + f*x]])/(a*(c - d)^2*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (EllipticF[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(a*(c - d)*f*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.326915, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2768, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{d (c+3 d) \cos (e+f x)}{a f (c-d)^2 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(c + 3*d)*Cos[e + f*x])/(a*(c - d)^2*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])) - Cos[e + f*x]/((c - d)*f*(a +
a*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - ((c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d
*Sin[e + f*x]])/(a*(c - d)^2*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (EllipticF[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(a*(c - d)*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2768

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

Rule 2754

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

Rule 2752

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

Rule 2663

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

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}+\frac{d \int \frac{-\frac{3 a}{2}+\frac{1}{2} a \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{a^2 (c-d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{(2 d) \int \frac{\frac{1}{4} a (3 c+d)+\frac{1}{4} a (c+3 d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{a^2 (c-d)^2 (c+d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{2 a (c-d)}-\frac{(c+3 d) \int \sqrt{c+d \sin (e+f x)} \, dx}{2 a (c-d)^2 (c+d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\left ((c+3 d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{2 a (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{2 a (c-d) \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{a (c-d)^2 (c+d) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{a (c-d) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.99742, size = 264, normalized size = 1.08 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (-\left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+\left (c^2+4 c d+3 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+(c+3 d) (c+d \sin (e+f x))-\frac{2 \left ((c+d)^2 \cos \left (\frac{1}{2} (e+f x)\right )+d \sin \left (\frac{1}{2} (e+f x)\right ) ((c+3 d) \cos (e+f x)+2 (c+d))\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}\right )}{a f (c-d)^2 (c+d) (\sin (e+f x)+1) \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*((-2*((c + d)^2*Cos[(e + f*x)/2] + d*(2*(c + d) + (c + 3*d)*Cos[e + f
*x])*Sin[(e + f*x)/2]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (c + 3*d)*(c + d*Sin[e + f*x]) + (c^2 + 4*c*d
+ 3*d^2)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - (c^2 - d^2)*Elli
pticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(a*(c - d)^2*(c + d)*f*(1 + S
in[e + f*x])*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 1.509, size = 925, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(f*x+e)^2*sin(f*x+e)*d+c*cos(f*x+e)^2)^(1/2)*((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/
(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^
(1/2))*c^3+3*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(
c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-(d/(c-d)*sin(f*x+e)+1/(c
-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*
x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d^2-3*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d
/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))
^(1/2))*d^3-4*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/
(c-d))^(1/2)*EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d+4*(d/(c-d)*sin(f*x+e)+1
/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/(c-d)*sin
(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3-c*cos(f*x+e)^2*d^2-3*cos(f*x+e)^2*d^3+c^2*d*sin(f*x+e)-d^3*s
in(f*x+e)-c^2*d+d^3)/d/(-(c+d*sin(f*x+e))*(-1+sin(f*x+e))*(1+sin(f*x+e)))^(1/2)/(c^2-d^2)/(c-d)/a/cos(f*x+e)/(
c+d*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 (d \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+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sin \left (f x + e\right ) + c}}{a c^{2} + 2 \, a c d + a d^{2} -{\left (2 \, a c d + a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*sin(f*x + e) + c)/(a*c^2 + 2*a*c*d + a*d^2 - (2*a*c*d + a*d^2)*cos(f*x + e)^2 - (a*d^2*cos(f*x
+ e)^2 - a*c^2 - 2*a*c*d - a*d^2)*sin(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}{\left (d \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+d*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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