3.958 \(\int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}} x^2 \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=89 \[ -\frac {2 \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}} \]

[Out]

-2*EllipticPi(1/2*(-c*x+1)^(1/2)*2^(1/2),2,2^(1/2)*(e/(c*d+e))^(1/2))*(c*(e*x+d)/(c*d+e))^(1/2)*(-c^2*x^2+1)^(
1/2)/x/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1574, 933, 168, 538, 537} \[ -\frac {2 \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - 1/(c^2*x^2)]*x^2*Sqrt[d + e*x]),x]

[Out]

(-2*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2, ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d +
e)])/(Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*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 933

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[Sqrt[1 + (c*x^2)/a]/Sqrt[a + c*x^2], 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 1574

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Dist[(x^(2*n*Fr
acPart[p])*(a + c/x^(2*n))^FracPart[p])/(c + a*x^(2*n))^FracPart[p], Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + a*x^
(2*n))^p, x], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[mn2, -2*n] &&  !IntegerQ[p] &&  !IntegerQ[q] &&
PosQ[n]

Rubi steps

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

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Mathematica [C]  time = 0.64, size = 188, normalized size = 2.11 \[ -\frac {2 i (d+e x) \sqrt {\frac {e (c x-1)}{c (d+e x)}} \sqrt {\frac {c e x+e}{c d+c e x}} \left (F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {c d+e}{c}}}{\sqrt {d+e x}}\right )|\frac {c d-e}{c d+e}\right )-\Pi \left (\frac {c d}{c d+e};i \sinh ^{-1}\left (\frac {\sqrt {-\frac {c d+e}{c}}}{\sqrt {d+e x}}\right )|\frac {c d-e}{c d+e}\right )\right )}{d x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {-\frac {c d+e}{c}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - 1/(c^2*x^2)]*x^2*Sqrt[d + e*x]),x]

[Out]

((-2*I)*Sqrt[(e*(-1 + c*x))/(c*(d + e*x))]*(d + e*x)*Sqrt[(e + c*e*x)/(c*d + c*e*x)]*(EllipticF[I*ArcSinh[Sqrt
[-((c*d + e)/c)]/Sqrt[d + e*x]], (c*d - e)/(c*d + e)] - EllipticPi[(c*d)/(c*d + e), I*ArcSinh[Sqrt[-((c*d + e)
/c)]/Sqrt[d + e*x]], (c*d - e)/(c*d + e)]))/(d*Sqrt[-((c*d + e)/c)]*Sqrt[1 - 1/(c^2*x^2)]*x)

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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/x^2/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e x + d} x^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(e*x + d)*x^2*sqrt(-1/(c^2*x^2) + 1)), x)

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maple [A]  time = 0.14, size = 148, normalized size = 1.66 \[ -\frac {2 \left (c d -e \right ) \sqrt {-\frac {\left (c x +1\right ) e}{c d -e}}\, \sqrt {-\frac {\left (c x -1\right ) e}{c d +e}}\, \sqrt {\frac {\left (e x +d \right ) c}{c d -e}}\, \EllipticPi \left (\sqrt {\frac {\left (e x +d \right ) c}{c d -e}}, \frac {c d -e}{c d}, \sqrt {\frac {c d -e}{c d +e}}\right )}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, \sqrt {e x +d}\, c d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2),x)

[Out]

-2/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*(c*d-e)*EllipticPi(((e*x+d)*c/(c*d-e))^(1/2),(c*d-e)/c/d,((c*d-e)/(c*d+e))^(1
/2))*(-(c*x+1)*e/(c*d-e))^(1/2)*(-(c*x-1)*e/(c*d+e))^(1/2)*((e*x+d)*c/(c*d-e))^(1/2)/(e*x+d)^(1/2)/c/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e x + d} x^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(e*x + d)*x^2*sqrt(-1/(c^2*x^2) + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,\sqrt {1-\frac {1}{c^2\,x^2}}\,\sqrt {d+e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(1 - 1/(c^2*x^2))^(1/2)*(d + e*x)^(1/2)),x)

[Out]

int(1/(x^2*(1 - 1/(c^2*x^2))^(1/2)*(d + e*x)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(1-1/c**2/x**2)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt(-(-1 + 1/(c*x))*(1 + 1/(c*x)))*sqrt(d + e*x)), x)

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