3.553 \(\int \frac{a+b F^{\frac{c \sqrt{d+e x}}{\sqrt{d f-e f x}}}}{d^2-e^2 x^2} \, dx\)

Optimal. Leaf size=68 \[ \frac{a \log \left (\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}+\frac{b \text{Ei}\left (\frac{c \sqrt{d+e x} \log (F)}{\sqrt{d f-e f x}}\right )}{d e} \]

[Out]

(b*ExpIntegralEi[(c*Sqrt[d + e*x]*Log[F])/Sqrt[d*f - e*f*x]])/(d*e) + (a*Log[Sqrt[d + e*x]/Sqrt[d*f - e*f*x]])
/(d*e)

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Rubi [A]  time = 0.179581, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2291, 14, 2178} \[ \frac{a \log \left (\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}+\frac{b \text{Ei}\left (\frac{c \sqrt{d+e x} \log (F)}{\sqrt{d f-e f x}}\right )}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ExpIntegralEi[(c*Sqrt[d + e*x]*Log[F])/Sqrt[d*f - e*f*x]])/(d*e) + (a*Log[Sqrt[d + e*x]/Sqrt[d*f - e*f*x]])
/(d*e)

Rule 2291

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
]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2178

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

Rubi steps

\begin{align*} \int \frac{a+b F^{\frac{c \sqrt{d+e x}}{\sqrt{d f-e f x}}}}{d^2-e^2 x^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b F^{c x}}{x} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a}{x}+\frac{b F^{c x}}{x}\right ) \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}\\ &=\frac{a \log \left (\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}+\frac{b \operatorname{Subst}\left (\int \frac{F^{c x}}{x} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}\\ &=\frac{b \text{Ei}\left (\frac{c \sqrt{d+e x} \log (F)}{\sqrt{d f-e f x}}\right )}{d e}+\frac{a \log \left (\frac{\sqrt{d+e x}}{\sqrt{d f-e f x}}\right )}{d e}\\ \end{align*}

Mathematica [F]  time = 0.454371, size = 0, normalized size = 0. \[ \int \frac{a+b F^{\frac{c \sqrt{d+e x}}{\sqrt{d f-e f x}}}}{d^2-e^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-{e}^{2}{x}^{2}+{d}^{2}} \left ( a+b{F}^{{c\sqrt{ex+d}{\frac{1}{\sqrt{-efx+df}}}}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a + b/F^(sqrt(-e*f*x + d*f)*sqrt(e*x + d)*c/(e*f*x - d*f)))/(e^2*x^2 - d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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