3.76 \(\int \frac{1}{(a+b x) \sqrt{c+d x} \sqrt{1-f^4 x^2}} \, dx\)

Optimal. Leaf size=86 \[ -\frac{2 \sqrt{\frac{f^2 (c+d x)}{c f^2+d}} \Pi \left (\frac{2 b}{a f^2+b};\sin ^{-1}\left (\frac{\sqrt{1-f^2 x}}{\sqrt{2}}\right )|\frac{2 d}{c f^2+d}\right )}{\left (a f^2+b\right ) \sqrt{c+d x}} \]

[Out]

(-2*Sqrt[(f^2*(c + d*x))/(d + c*f^2)]*EllipticPi[(2*b)/(b + a*f^2), ArcSin[Sqrt[1 - f^2*x]/Sqrt[2]], (2*d)/(d
+ c*f^2)])/((b + a*f^2)*Sqrt[c + d*x])

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Rubi [A]  time = 0.174857, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {932, 168, 538, 537} \[ -\frac{2 \sqrt{\frac{f^2 (c+d x)}{c f^2+d}} \Pi \left (\frac{2 b}{a f^2+b};\sin ^{-1}\left (\frac{\sqrt{1-f^2 x}}{\sqrt{2}}\right )|\frac{2 d}{c f^2+d}\right )}{\left (a f^2+b\right ) \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*Sqrt[c + d*x]*Sqrt[1 - f^4*x^2]),x]

[Out]

(-2*Sqrt[(f^2*(c + d*x))/(d + c*f^2)]*EllipticPi[(2*b)/(b + a*f^2), ArcSin[Sqrt[1 - f^2*x]/Sqrt[2]], (2*d)/(d
+ c*f^2)])/((b + a*f^2)*Sqrt[c + d*x])

Rule 932

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[1/Sqrt[a], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]), x], x]] /; FreeQ[{a, c,
d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 538

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

Rule 537

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

Rubi steps

\begin{align*} \int \frac{1}{(a+b x) \sqrt{c+d x} \sqrt{1-f^4 x^2}} \, dx &=\int \frac{1}{(a+b x) \sqrt{c+d x} \sqrt{1-f^2 x} \sqrt{1+f^2 x}} \, dx\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^2} \left (b+a f^2-b x^2\right ) \sqrt{c+\frac{d}{f^2}-\frac{d x^2}{f^2}}} \, dx,x,\sqrt{1-f^2 x}\right )\right )\\ &=-\frac{\left (2 \sqrt{\frac{f^2 (c+d x)}{d+c f^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^2} \left (b+a f^2-b x^2\right ) \sqrt{1-\frac{d x^2}{\left (c+\frac{d}{f^2}\right ) f^2}}} \, dx,x,\sqrt{1-f^2 x}\right )}{\sqrt{c+d x}}\\ &=-\frac{2 \sqrt{\frac{f^2 (c+d x)}{d+c f^2}} \Pi \left (\frac{2 b}{b+a f^2};\sin ^{-1}\left (\frac{\sqrt{1-f^2 x}}{\sqrt{2}}\right )|\frac{2 d}{d+c f^2}\right )}{\left (b+a f^2\right ) \sqrt{c+d x}}\\ \end{align*}

Mathematica [C]  time = 0.107044, size = 218, normalized size = 2.53 \[ \frac{2 i (c+d x) \sqrt{\frac{d \left (f^2 x-1\right )}{f^2 (c+d x)}} \sqrt{\frac{d \left (f^2 x+1\right )}{f^2 (c+d x)}} \left (\text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-c-\frac{d}{f^2}}}{\sqrt{c+d x}}\right ),\frac{c f^2-d}{c f^2+d}\right )-\Pi \left (\frac{(b c-a d) f^2}{b \left (c f^2+d\right )};i \sinh ^{-1}\left (\frac{\sqrt{-c-\frac{d}{f^2}}}{\sqrt{c+d x}}\right )|\frac{c f^2-d}{c f^2+d}\right )\right )}{\sqrt{1-f^4 x^2} \sqrt{-c-\frac{d}{f^2}} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*Sqrt[c + d*x]*Sqrt[1 - f^4*x^2]),x]

[Out]

((2*I)*(c + d*x)*Sqrt[(d*(-1 + f^2*x))/(f^2*(c + d*x))]*Sqrt[(d*(1 + f^2*x))/(f^2*(c + d*x))]*(EllipticF[I*Arc
Sinh[Sqrt[-c - d/f^2]/Sqrt[c + d*x]], (-d + c*f^2)/(d + c*f^2)] - EllipticPi[((b*c - a*d)*f^2)/(b*(d + c*f^2))
, I*ArcSinh[Sqrt[-c - d/f^2]/Sqrt[c + d*x]], (-d + c*f^2)/(d + c*f^2)]))/((-(b*c) + a*d)*Sqrt[-c - d/f^2]*Sqrt
[1 - f^4*x^2])

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Maple [B]  time = 0.024, size = 205, normalized size = 2.4 \begin{align*} -2\,{\frac{ \left ( c{f}^{2}-d \right ) \sqrt{-{f}^{4}{x}^{2}+1}\sqrt{dx+c}}{ \left ( ad-bc \right ){f}^{2} \left ( d{f}^{4}{x}^{3}+c{f}^{4}{x}^{2}-dx-c \right ) }{\it EllipticPi} \left ( \sqrt{{\frac{{f}^{2} \left ( dx+c \right ) }{c{f}^{2}-d}}},-{\frac{ \left ( c{f}^{2}-d \right ) b}{ \left ( ad-bc \right ){f}^{2}}},\sqrt{{\frac{c{f}^{2}-d}{c{f}^{2}+d}}} \right ) \sqrt{-{\frac{ \left ({f}^{2}x+1 \right ) d}{c{f}^{2}-d}}}\sqrt{-{\frac{ \left ({f}^{2}x-1 \right ) d}{c{f}^{2}+d}}}\sqrt{{\frac{{f}^{2} \left ( dx+c \right ) }{c{f}^{2}-d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)^(1/2)/(-f^4*x^2+1)^(1/2),x)

[Out]

-2*(c*f^2-d)*EllipticPi(((d*x+c)*f^2/(c*f^2-d))^(1/2),-(c*f^2-d)*b/f^2/(a*d-b*c),((c*f^2-d)/(c*f^2+d))^(1/2))*
(-(f^2*x+1)*d/(c*f^2-d))^(1/2)*(-(f^2*x-1)*d/(c*f^2+d))^(1/2)*((d*x+c)*f^2/(c*f^2-d))^(1/2)*(-f^4*x^2+1)^(1/2)
*(d*x+c)^(1/2)/f^2/(a*d-b*c)/(d*f^4*x^3+c*f^4*x^2-d*x-c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-f^{4} x^{2} + 1}{\left (b x + a\right )} \sqrt{d x + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(1/2)/(-f^4*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-f^4*x^2 + 1)*(b*x + a)*sqrt(d*x + c)), 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/(b*x+a)/(d*x+c)^(1/2)/(-f^4*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)**(1/2)/(-f**4*x**2+1)**(1/2),x)

[Out]

Integral(1/(sqrt(-(f**2*x - 1)*(f**2*x + 1))*(a + b*x)*sqrt(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-f^{4} x^{2} + 1}{\left (b x + a\right )} \sqrt{d x + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(1/2)/(-f^4*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-f^4*x^2 + 1)*(b*x + a)*sqrt(d*x + c)), x)