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

Optimal. Leaf size=220 \[ -\frac{2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt{c+d \sin (e+f x)}}-\frac{2 d \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f \left (c^2-d^2\right ) (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 b \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f (a+b) (b c-a d) \sqrt{c+d \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.630355, antiderivative size = 220, 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 = {2802, 3059, 2655, 2653, 12, 2807, 2805} \[ -\frac{2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt{c+d \sin (e+f x)}}-\frac{2 d \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f \left (c^2-d^2\right ) (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 b \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f (a+b) (b c-a d) \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2807

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

Rule 2805

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

Rubi steps

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

Mathematica [C]  time = 6.95446, size = 617, normalized size = 2.8 \[ -\frac{\frac{4 d^2 \cos (e+f x)}{\left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{-\frac{2 i \sec (e+f x) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{\frac{d (\sin (e+f x)+1)}{d-c}} \left (d \left (d \left (b^2-2 a^2\right ) \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+2 (a+b) (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )-2 b (c-d) (b c-a d) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )}{b \sqrt{-\frac{1}{c+d}} (b c-a d)}+\frac{2 \left (-2 a c d+2 b c^2-3 b d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )}{(a+b) \sqrt{c+d \sin (e+f x)}}+\frac{4 i (a d+b c) \sec (e+f x) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{\frac{d (\sin (e+f x)+1)}{d-c}} \left ((a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )-a d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )}{b \sqrt{-\frac{1}{c+d}} (b c-a d)}}{(c-d) (c+d)}}{2 f (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*d^2*Cos[e + f*x])/((c^2 - d^2)*Sqrt[c + d*Sin[e + f*x]]) + (((4*I)*(b*c + a*d)*((-(b*c) + a*d)*EllipticF[
I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] - a*d*EllipticPi[(b*(c + d))/(b*c -
a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)])*Sec[e + f*x]*Sqrt[-((d*(-1 +
Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d)])/(b*Sqrt[-(c + d)^(-1)]*(b*c - a*d)) - ((2*I)*(
-2*b*(c - d)*(b*c - a*d)*EllipticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] +
d*(2*(a + b)*(-(b*c) + a*d)*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d
)] + (-2*a^2 + b^2)*d*EllipticPi[(b*(c + d))/(b*c - a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x
]]], (c + d)/(c - d)]))*Sec[e + f*x]*Sqrt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c
+ d)])/(b*Sqrt[-(c + d)^(-1)]*(b*c - a*d)) + (2*(2*b*c^2 - 2*a*c*d - 3*b*d^2)*EllipticPi[(2*b)/(a + b), (-2*e
+ Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((a + b)*Sqrt[c + d*Sin[e + f*x]]))/((c -
d)*(c + d)))/(2*(b*c - a*d)*f)

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Maple [B]  time = 3.242, size = 610, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [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+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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+b*sin(f*x+e))/(c+d*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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

[Out]

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