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

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Rubi [A]  time = 0.17, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 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 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)
])

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 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]

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*}

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Mathematica [C]  time = 0.16, size = 218, normalized size = 2.53 \[ \frac {2 i (c+d x) \sqrt {\frac {d \left (f^2 x-1\right )}{f^2 (c+d x)}} \sqrt {\frac {d \left (f^2 x+1\right )}{f^2 (c+d x)}} \left (\operatorname {EllipticF}\left (i \sinh ^{-1}\left (\frac {\sqrt {-c-\frac {d}{f^2}}}{\sqrt {c+d x}}\right ),\frac {c f^2-d}{c f^2+d}\right )-\Pi \left (\frac {(b c-a d) f^2}{b \left (c f^2+d\right )};i \sinh ^{-1}\left (\frac {\sqrt {-c-\frac {d}{f^2}}}{\sqrt {c+d x}}\right )|\frac {c f^2-d}{c f^2+d}\right )\right )}{\sqrt {1-f^4 x^2} \sqrt {-c-\frac {d}{f^2}} (a d-b c)} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-f^{4} x^{2} + 1} {\left (b x + a\right )} \sqrt {d x + c}}\,{d x} \]

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)

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maple [B]  time = 0.03, size = 205, normalized size = 2.38 \[ -\frac {2 \left (c \,f^{2}-d \right ) \sqrt {-\frac {\left (f^{2} x +1\right ) d}{c \,f^{2}-d}}\, \sqrt {-\frac {\left (f^{2} x -1\right ) d}{c \,f^{2}+d}}\, \sqrt {\frac {\left (d x +c \right ) f^{2}}{c \,f^{2}-d}}\, \sqrt {-f^{4} x^{2}+1}\, \sqrt {d x +c}\, \EllipticPi \left (\sqrt {\frac {\left (d x +c \right ) f^{2}}{c \,f^{2}-d}}, -\frac {\left (c \,f^{2}-d \right ) b}{\left (a d -b c \right ) f^{2}}, \sqrt {\frac {c \,f^{2}-d}{c \,f^{2}+d}}\right )}{\left (a d -b c \right ) \left (d \,f^{4} x^{3}+c \,f^{4} x^{2}-d x -c \right ) f^{2}} \]

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)/(c*f^2-d)*f^2)^(1/2),-(c*f^2-d)/(a*d-b*c)*b/f^2,((c*f^2-d)/(c*f^2+d))^(1/2))*
(-(f^2*x+1)/(c*f^2-d)*d)^(1/2)*(-(f^2*x-1)/(c*f^2+d)*d)^(1/2)*((d*x+c)/(c*f^2-d)*f^2)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-f^{4} x^{2} + 1} {\left (b x + a\right )} \sqrt {d x + c}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {1-f^4\,x^2}\,\left (a+b\,x\right )\,\sqrt {c+d\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \left (f^{2} x - 1\right ) \left (f^{2} x + 1\right )} \left (a + b x\right ) \sqrt {c + d x}}\, dx \]

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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