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

Optimal. Leaf size=110 \[ \frac {a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {d f-e f x}}\right )}{d e}+\frac {2 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e}+\frac {b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {2291, 2183, 2178} \[ \frac {a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {d f-e f x}}\right )}{d e}+\frac {2 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e}+\frac {b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

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 {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {d f-e f x}}}\right )^2}{d^2-e^2 x^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b F^{c x}\right )^2}{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^2}{x}+\frac {2 a b F^{c x}}{x}+\frac {b^2 F^{2 c x}}{x}\right ) \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {d f-e f x}}\right )}{d e}\\ &=\frac {a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {d f-e f x}}\right )}{d e}+\frac {(2 a 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^2 \operatorname {Subst}\left (\int \frac {F^{2 c x}}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {d f-e f x}}\right )}{d e}\\ &=\frac {2 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e}+\frac {b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {d f-e f x}}\right )}{d e}+\frac {a^2 \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 = 1.30, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {d f-e f x}}}\right )^2}{d^2-e^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 3.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a^{2} + \frac {2 \, a b}{F^{\frac {\sqrt {-e f x + d f} \sqrt {e x + d} c}{e f x - d f}}} + \frac {b^{2}}{F^{\frac {2 \, \sqrt {-e f x + d f} \sqrt {e x + d} c}{e f x - d f}}}}{e^{2} x^{2} - d^{2}}, x\right ) \]

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

[Out]

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

[Out]

Timed out

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,F^{\frac {\sqrt {e x +d}\, c}{\sqrt {-e f x +d f}}}+a \right )^{2}}{-e^{2} x^{2}+d^{2}}\, dx \]

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

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

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

[Out]

1/2*a^2*(log(e*x + d)/(d*e) - log(e*x - d)/(d*e)) - b^2*integrate(F^(2*sqrt(e*x + d)*c/(sqrt(-e*x + d)*sqrt(f)
))/(e^2*x^2 - d^2), x) - 2*a*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 \[ \int \frac {{\left (a+b\,{\mathrm {e}}^{\frac {c\,\ln \relax (F)\,\sqrt {d+e\,x}}{\sqrt {d\,f-e\,f\,x}}}\right )}^2}{d^2-e^2\,x^2} \,d x \]

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)^2/(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)))^2/(d^2 - e^2*x^2), x)

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

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)))**2/(-e**2*x**2+d**2),x)

[Out]

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

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