3.93.70 \(\int \frac {-e^8-20 e^4 x-75 x^2+e^5 (-e^8-10 e^4 x-25 x^2)+e^{x^2} (1+e^5+2 x^2)+(e^8-e^{x^2}+10 e^4 x+25 x^2) \log (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3)}{-e^8 x^2+e^{x^2} x^2-10 e^4 x^3-25 x^4} \, dx\)

Optimal. Leaf size=33 \[ 3+\frac {-e^5+x+\log \left (x \left (-e^{x^2}+\left (e^4+5 x\right )^2\right )\right )}{x} \]

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Rubi [A]  time = 2.09, antiderivative size = 46, normalized size of antiderivative = 1.39, number of steps used = 15, number of rules used = 3, integrand size = 139, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6742, 14, 2551} \begin {gather*} \frac {\log \left (25 x^3+10 e^4 x^2-e^{x^2} x+e^8 x\right )}{x}-\frac {1+e^5}{x}+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^8 - 20*E^4*x - 75*x^2 + E^5*(-E^8 - 10*E^4*x - 25*x^2) + E^x^2*(1 + E^5 + 2*x^2) + (E^8 - E^x^2 + 10*E
^4*x + 25*x^2)*Log[E^8*x - E^x^2*x + 10*E^4*x^2 + 25*x^3])/(-(E^8*x^2) + E^x^2*x^2 - 10*E^4*x^3 - 25*x^4),x]

[Out]

x^(-1) - (1 + E^5)/x + Log[E^8*x - E^x^2*x + 10*E^4*x^2 + 25*x^3]/x

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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3\right )}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}+\frac {1+e^5+2 x^2-\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx+\int \frac {1+e^5+2 x^2-\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {25 \left (1-\frac {e^8}{25}\right )}{-e^8+e^{x^2}-10 e^4 x-25 x^2}+\frac {5 e^4}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}-\frac {10 e^4 x}{e^8-e^{x^2}+10 e^4 x+25 x^2}-\frac {25 x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2}\right ) \, dx+\int \left (\frac {1+e^5+2 x^2}{x^2}-\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x^2}\right ) \, dx\\ &=-\left (50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx\right )+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx+\int \frac {1+e^5+2 x^2}{x^2} \, dx-\int \frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x^2} \, dx\\ &=\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx+\int \left (2+\frac {1+e^5}{x^2}\right ) \, dx-\int \frac {e^8+20 e^4 x+75 x^2-e^{x^2} \left (1+2 x^2\right )}{x^2 \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \left (\frac {1+2 x^2}{x^2}+\frac {2 \left (5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3\right )}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}\right ) \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-2 \int \frac {5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \frac {1+2 x^2}{x^2} \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-2 \int \left (-\frac {25 \left (1-\frac {e^8}{25}\right )}{-e^8+e^{x^2}-10 e^4 x-25 x^2}+\frac {5 e^4}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}-\frac {10 e^4 x}{e^8-e^{x^2}+10 e^4 x+25 x^2}-\frac {25 x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2}\right ) \, dx-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \left (2+\frac {1}{x^2}\right ) \, dx\\ &=\frac {1}{x}-\frac {1+e^5}{x}+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 38, normalized size = 1.15 \begin {gather*} -\frac {e^5}{x}+\frac {\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^8 - 20*E^4*x - 75*x^2 + E^5*(-E^8 - 10*E^4*x - 25*x^2) + E^x^2*(1 + E^5 + 2*x^2) + (E^8 - E^x^2
+ 10*E^4*x + 25*x^2)*Log[E^8*x - E^x^2*x + 10*E^4*x^2 + 25*x^3])/(-(E^8*x^2) + E^x^2*x^2 - 10*E^4*x^3 - 25*x^4
),x]

[Out]

-(E^5/x) + Log[x*(E^8 - E^x^2 + 10*E^4*x + 25*x^2)]/x

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fricas [A]  time = 0.57, size = 35, normalized size = 1.06 \begin {gather*} -\frac {e^{5} - \log \left (25 \, x^{3} + 10 \, x^{2} e^{4} + x e^{8} - x e^{\left (x^{2}\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4)^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*
x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4)-25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-
10*x^3*exp(4)-25*x^4),x, algorithm="fricas")

[Out]

-(e^5 - log(25*x^3 + 10*x^2*e^4 + x*e^8 - x*e^(x^2)))/x

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4)^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*
x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4)-25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-
10*x^3*exp(4)-25*x^4),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.35, size = 37, normalized size = 1.12

method	result	size
norman	\(\frac {-{\mathrm e}^{5}+\ln \left (-{\mathrm e}^{x^{2}} x +x \,{\mathrm e}^{8}+10 x^{2} {\mathrm e}^{4}+25 x^{3}\right )}{x}\)	\(37\)
risch	\(\frac {\ln \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )}{x}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{2}+i \pi \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{3}+2 \,{\mathrm e}^{5}-2 \ln \relax (x )}{2 x}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*ln(-exp(x^2)*x+x*exp(4)^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*x^2+1)*
exp(x^2)+(-exp(4)^2-10*x*exp(4)-25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-10*x^3*
exp(4)-25*x^4),x,method=_RETURNVERBOSE)

[Out]

(-exp(5)+ln(-exp(x^2)*x+x*exp(4)^2+10*x^2*exp(4)+25*x^3))/x

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maxima [A]  time = 0.41, size = 34, normalized size = 1.03 \begin {gather*} -\frac {e^{5} - \log \left (25 \, x^{2} + 10 \, x e^{4} + e^{8} - e^{\left (x^{2}\right )}\right ) - \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x^2)+exp(4)^2+10*x*exp(4)+25*x^2)*log(-exp(x^2)*x+x*exp(4)^2+10*x^2*exp(4)+25*x^3)+(exp(5)+2*
x^2+1)*exp(x^2)+(-exp(4)^2-10*x*exp(4)-25*x^2)*exp(5)-exp(4)^2-20*x*exp(4)-75*x^2)/(x^2*exp(x^2)-x^2*exp(4)^2-
10*x^3*exp(4)-25*x^4),x, algorithm="maxima")

[Out]

-(e^5 - log(25*x^2 + 10*x*e^4 + e^8 - e^(x^2)) - log(x))/x

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mupad [B]  time = 9.13, size = 35, normalized size = 1.06 \begin {gather*} -\frac {{\mathrm {e}}^5-\ln \left (x\,{\mathrm {e}}^8-x\,{\mathrm {e}}^{x^2}+10\,x^2\,{\mathrm {e}}^4+25\,x^3\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(8) - log(x*exp(8) - x*exp(x^2) + 10*x^2*exp(4) + 25*x^3)*(exp(8) - exp(x^2) + 10*x*exp(4) + 25*x^2) +
20*x*exp(4) - exp(x^2)*(exp(5) + 2*x^2 + 1) + 75*x^2 + exp(5)*(exp(8) + 10*x*exp(4) + 25*x^2))/(10*x^3*exp(4)
- x^2*exp(x^2) + x^2*exp(8) + 25*x^4),x)

[Out]

-(exp(5) - log(x*exp(8) - x*exp(x^2) + 10*x^2*exp(4) + 25*x^3))/x

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sympy [A]  time = 0.57, size = 32, normalized size = 0.97 \begin {gather*} \frac {\log {\left (25 x^{3} + 10 x^{2} e^{4} - x e^{x^{2}} + x e^{8} \right )}}{x} - \frac {e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x**2)+exp(4)**2+10*x*exp(4)+25*x**2)*ln(-exp(x**2)*x+x*exp(4)**2+10*x**2*exp(4)+25*x**3)+(exp
(5)+2*x**2+1)*exp(x**2)+(-exp(4)**2-10*x*exp(4)-25*x**2)*exp(5)-exp(4)**2-20*x*exp(4)-75*x**2)/(x**2*exp(x**2)
-x**2*exp(4)**2-10*x**3*exp(4)-25*x**4),x)

[Out]

log(25*x**3 + 10*x**2*exp(4) - x*exp(x**2) + x*exp(8))/x - exp(5)/x

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