3.31.78 \(\int \frac {e^{1+x} (26 x-8 x^3)+e^{1+x} (-10+8 x^2) \log (-5+4 x^2)}{-5+4 x^2} \, dx\)

Optimal. Leaf size=20 \[ 2 e^{1+x} \left (1-x+\log \left (-5+4 x^2\right )\right ) \]

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Rubi [A]  time = 2.49, antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 36, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6725, 2271, 2178, 2176, 2194, 2269, 2554, 12} \begin {gather*} 2 e^{x+1} \log \left (4 x^2-5\right )-2 e^{x+1} x+2 e^{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + x)*(26*x - 8*x^3) + E^(1 + x)*(-10 + 8*x^2)*Log[-5 + 4*x^2])/(-5 + 4*x^2),x]

[Out]

2*E^(1 + x) - 2*E^(1 + x)*x + 2*E^(1 + x)*Log[-5 + 4*x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2269

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d +
e*x)^n), 1/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x]

Rule 2271

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_) + (c_)*(x_)^2), x_Symbol] :> Int[ExpandIntegran
d[F^(g*(d + e*x)^n), u^m/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x] && PolynomialQ[u, x] && Intege
rQ[m]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {26 e^{1+x} x}{-5+4 x^2}-\frac {8 e^{1+x} x^3}{-5+4 x^2}-\frac {10 e^{1+x} \log \left (-5+4 x^2\right )}{-5+4 x^2}+\frac {8 e^{1+x} x^2 \log \left (-5+4 x^2\right )}{-5+4 x^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{1+x} x^3}{-5+4 x^2} \, dx\right )+8 \int \frac {e^{1+x} x^2 \log \left (-5+4 x^2\right )}{-5+4 x^2} \, dx-10 \int \frac {e^{1+x} \log \left (-5+4 x^2\right )}{-5+4 x^2} \, dx+26 \int \frac {e^{1+x} x}{-5+4 x^2} \, dx\\ &=2 e^{1+x} \log \left (-5+4 x^2\right )-8 \int \left (\frac {1}{4} e^{1+x} x+\frac {5 e^{1+x} x}{4 \left (-5+4 x^2\right )}\right ) \, dx-8 \int \frac {e x \left (-4 e^x-\sqrt {5} e^{\frac {\sqrt {5}}{2}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\sqrt {5} e^{-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{10-8 x^2} \, dx+10 \int \frac {2 e^{1-\frac {\sqrt {5}}{2}} x \left (-e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{\sqrt {5} \left (5-4 x^2\right )} \, dx+26 \int \left (-\frac {e^{1+x}}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^{1+x}}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x} \log \left (-5+4 x^2\right )-2 \int e^{1+x} x \, dx-\frac {13}{2} \int \frac {e^{1+x}}{\sqrt {5}-2 x} \, dx+\frac {13}{2} \int \frac {e^{1+x}}{\sqrt {5}+2 x} \, dx-10 \int \frac {e^{1+x} x}{-5+4 x^2} \, dx-(8 e) \int \frac {x \left (-4 e^x-\sqrt {5} e^{\frac {\sqrt {5}}{2}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\sqrt {5} e^{-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{10-8 x^2} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \left (-e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{5-4 x^2} \, dx\\ &=-2 e^{1+x} x+\frac {13}{4} e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+\frac {13}{4} e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+2 \int e^{1+x} \, dx-10 \int \left (-\frac {e^{1+x}}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^{1+x}}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-(8 e) \int \left (\frac {2 e^x x}{-5+4 x^2}+\frac {\sqrt {5} e^{-\frac {\sqrt {5}}{2}} x \left (e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )-\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{2 \left (-5+4 x^2\right )}\right ) \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (\frac {e^{\sqrt {5}} x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}-\frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+\frac {13}{4} e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+\frac {13}{4} e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+\frac {5}{2} \int \frac {e^{1+x}}{\sqrt {5}-2 x} \, dx-\frac {5}{2} \int \frac {e^{1+x}}{\sqrt {5}+2 x} \, dx-(16 e) \int \frac {e^x x}{-5+4 x^2} \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \left (e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )-\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{-5+4 x^2} \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx+\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+2 e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )-(16 e) \int \left (-\frac {e^x}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^x}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (\frac {e^{\sqrt {5}} x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}-\frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}\right ) \, dx+\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+2 e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+(4 e) \int \frac {e^x}{\sqrt {5}-2 x} \, dx-(4 e) \int \frac {e^x}{\sqrt {5}+2 x} \, dx+\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx-\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx-\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx+\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx-\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+x} \log \left (-5+4 x^2\right )+\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx-\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx+\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx-\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+x} \log \left (-5+4 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 20, normalized size = 1.00 \begin {gather*} -2 e^{1+x} \left (-1+x-\log \left (-5+4 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + x)*(26*x - 8*x^3) + E^(1 + x)*(-10 + 8*x^2)*Log[-5 + 4*x^2])/(-5 + 4*x^2),x]

[Out]

-2*E^(1 + x)*(-1 + x - Log[-5 + 4*x^2])

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fricas [A]  time = 0.57, size = 24, normalized size = 1.20 \begin {gather*} -2 \, {\left (x - 1\right )} e^{\left (x + 1\right )} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-10)*exp(1)*exp(x)*log(4*x^2-5)+(-8*x^3+26*x)*exp(1)*exp(x))/(4*x^2-5),x, algorithm="fricas")

[Out]

-2*(x - 1)*e^(x + 1) + 2*e^(x + 1)*log(4*x^2 - 5)

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giac [A]  time = 0.37, size = 28, normalized size = 1.40 \begin {gather*} -2 \, x e^{\left (x + 1\right )} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) + 2 \, e^{\left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-10)*exp(1)*exp(x)*log(4*x^2-5)+(-8*x^3+26*x)*exp(1)*exp(x))/(4*x^2-5),x, algorithm="giac")

[Out]

-2*x*e^(x + 1) + 2*e^(x + 1)*log(4*x^2 - 5) + 2*e^(x + 1)

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maple [A]  time = 0.47, size = 25, normalized size = 1.25

method	result	size
risch	\(2 \ln \left (4 x^{2}-5\right ) {\mathrm e}^{x +1}-2 \left (x -1\right ) {\mathrm e}^{x +1}\)	\(25\)
norman	\(2 \,{\mathrm e} \,{\mathrm e}^{x}-2 x \,{\mathrm e} \,{\mathrm e}^{x}+2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (4 x^{2}-5\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2-10)*exp(1)*exp(x)*ln(4*x^2-5)+(-8*x^3+26*x)*exp(1)*exp(x))/(4*x^2-5),x,method=_RETURNVERBOSE)

[Out]

2*ln(4*x^2-5)*exp(x+1)-2*(x-1)*exp(x+1)

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maxima [A]  time = 0.45, size = 28, normalized size = 1.40 \begin {gather*} -2 \, {\left (x e - e\right )} e^{x} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-10)*exp(1)*exp(x)*log(4*x^2-5)+(-8*x^3+26*x)*exp(1)*exp(x))/(4*x^2-5),x, algorithm="maxima")

[Out]

-2*(x*e - e)*e^x + 2*e^(x + 1)*log(4*x^2 - 5)

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mupad [B]  time = 0.27, size = 28, normalized size = 1.40 \begin {gather*} 2\,\mathrm {e}\,{\mathrm {e}}^x-2\,x\,\mathrm {e}\,{\mathrm {e}}^x+2\,\ln \left (4\,x^2-5\right )\,\mathrm {e}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*exp(x)*(26*x - 8*x^3) + log(4*x^2 - 5)*exp(1)*exp(x)*(8*x^2 - 10))/(4*x^2 - 5),x)

[Out]

2*exp(1)*exp(x) - 2*x*exp(1)*exp(x) + 2*log(4*x^2 - 5)*exp(1)*exp(x)

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sympy [A]  time = 0.52, size = 27, normalized size = 1.35 \begin {gather*} \left (- 2 e x + 2 e \log {\left (4 x^{2} - 5 \right )} + 2 e\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2-10)*exp(1)*exp(x)*ln(4*x**2-5)+(-8*x**3+26*x)*exp(1)*exp(x))/(4*x**2-5),x)

[Out]

(-2*E*x + 2*E*log(4*x**2 - 5) + 2*E)*exp(x)

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