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*Ei(c*ln(F)*(e*x+d)^(1/2)/(-e*f*x+d*f)^(1/2))/d/e+a*ln((e*x+d)^(1/2)/(-e*f*x+d*f)^(1/2))/d/e

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Rubi [A]  time = 0.18, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 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

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
]

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*}

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Mathematica [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {d f-e f x}}}}{d^2-e^2 x^2} \, dx \]

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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fricas [F]  time = 2.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a + \frac {b}{F^{\frac {\sqrt {-e f x + d f} \sqrt {e x + d} c}{e f x - d f}}}}{e^{2} x^{2} - d^{2}}, x\right ) \]

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

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)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {b \,F^{\frac {\sqrt {e x +d}\, c}{\sqrt {-e f x +d f}}}+a}{-e^{2} x^{2}+d^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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