3.6.57 \(\int \frac {(F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}})^n}{1-a^2 x^2} \, dx\) [557]

Optimal. Leaf size=77 \[ -\frac {F^{-\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n \text {Ei}\left (\frac {n \sqrt {1-a x} \log (F)}{\sqrt {1+a x}}\right )}{a} \]

[Out]

-(F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n*Ei(n*ln(F)*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a/(F^(n*(-a*x+1)^(1/2)/(a*x+1)^
(1/2)))

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Rubi [A]
time = 0.17, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {2319, 2329, 2209} \begin {gather*} -\frac {F^{-\frac {n \sqrt {1-a x}}{\sqrt {a x+1}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {a x+1}}}\right )^n \text {Ei}\left (\frac {n \sqrt {1-a x} \log (F)}{\sqrt {a x+1}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(F^(Sqrt[1 - a*x]/Sqrt[1 + a*x]))^n/(1 - a^2*x^2),x]

[Out]

-(((F^(Sqrt[1 - a*x]/Sqrt[1 + a*x]))^n*ExpIntegralEi[(n*Sqrt[1 - a*x]*Log[F])/Sqrt[1 + a*x]])/(a*F^((n*Sqrt[1
- a*x])/Sqrt[1 + a*x])))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2319

Int[(u_.)*((a_.)*(F_)^(v_))^(n_), x_Symbol] :> Dist[(a*F^v)^n/F^(n*v), Int[u*F^(n*v), x], x] /; FreeQ[{F, a, n
}, x] &&  !IntegerQ[n]

Rule 2329

Int[((a_.) + (b_.)*(F_)^(((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]))^(n_.)/((A_) + (C_.)*(x_)^
2), x_Symbol] :> Dist[2*e*(g/(C*(e*f - d*g))), Subst[Int[(a + b*F^(c*x))^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*
x]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0
]

Rubi steps

\begin {align*} \int \frac {\left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n}{1-a^2 x^2} \, dx &=\left (F^{-\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n\right ) \int \frac {F^{\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}}}{1-a^2 x^2} \, dx\\ &=-\frac {\left (F^{-\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n\right ) \text {Subst}\left (\int \frac {F^{n x}}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a}\\ &=-\frac {F^{-\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n \text {Ei}\left (\frac {n \sqrt {1-a x} \log (F)}{\sqrt {1+a x}}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 77, normalized size = 1.00 \begin {gather*} -\frac {F^{-\frac {n \sqrt {1-a x}}{\sqrt {1+a x}}} \left (F^{\frac {\sqrt {1-a x}}{\sqrt {1+a x}}}\right )^n \text {Ei}\left (\frac {n \sqrt {1-a x} \log (F)}{\sqrt {1+a x}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(F^(Sqrt[1 - a*x]/Sqrt[1 + a*x]))^n/(1 - a^2*x^2),x]

[Out]

-(((F^(Sqrt[1 - a*x]/Sqrt[1 + a*x]))^n*ExpIntegralEi[(n*Sqrt[1 - a*x]*Log[F])/Sqrt[1 + a*x]])/(a*F^((n*Sqrt[1
- a*x])/Sqrt[1 + a*x])))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (F^{\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}}\right )^{n}}{-a^{2} x^{2}+1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n/(-a^2*x^2+1),x)

[Out]

int((F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n/(-a^2*x^2+1),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate(F^(sqrt(-a*x + 1)*n/sqrt(a*x + 1))/(a^2*x^2 - 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(F^(sqrt(-a*x + 1)/sqrt(a*x + 1)))^n/(a^2*x^2 - 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\left (F^{\frac {\sqrt {- a x + 1}}{\sqrt {a x + 1}}}\right )^{n}}{a^{2} x^{2} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F**((-a*x+1)**(1/2)/(a*x+1)**(1/2)))**n/(-a**2*x**2+1),x)

[Out]

-Integral((F**(sqrt(-a*x + 1)/sqrt(a*x + 1)))**n/(a**2*x**2 - 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((F^((-a*x+1)^(1/2)/(a*x+1)^(1/2)))^n/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(F^(sqrt(-a*x + 1)/sqrt(a*x + 1)))^n/(a^2*x^2 - 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {{\left (F^{\frac {\sqrt {1-a\,x}}{\sqrt {a\,x+1}}}\right )}^n}{a^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(F^((1 - a*x)^(1/2)/(a*x + 1)^(1/2)))^n/(a^2*x^2 - 1),x)

[Out]

int(-(F^((1 - a*x)^(1/2)/(a*x + 1)^(1/2)))^n/(a^2*x^2 - 1), x)

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