3.78.75 \(\int \frac {e^x (-15-50 x^2+10 e^{2 x} x^2+(15 x+5 e^{2 x} x^2-50 x^3) \log (\frac {-3-e^{2 x} x+10 x^2}{x}))}{3 x+e^{2 x} x^2-10 x^3} \, dx\)

Optimal. Leaf size=22 \[ 5 e^x \log \left (-e^{2 x}-\frac {3}{x}+10 x\right ) \]

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Rubi [A]  time = 0.17, antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 1, number of rules used = 1, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} 5 e^x \log \left (-\frac {-10 x^2+e^{2 x} x+3}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-15 - 50*x^2 + 10*E^(2*x)*x^2 + (15*x + 5*E^(2*x)*x^2 - 50*x^3)*Log[(-3 - E^(2*x)*x + 10*x^2)/x]))/(
3*x + E^(2*x)*x^2 - 10*x^3),x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=5 e^x \log \left (-\frac {3+e^{2 x} x-10 x^2}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 22, normalized size = 1.00 \begin {gather*} 5 e^x \log \left (-e^{2 x}-\frac {3}{x}+10 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-15 - 50*x^2 + 10*E^(2*x)*x^2 + (15*x + 5*E^(2*x)*x^2 - 50*x^3)*Log[(-3 - E^(2*x)*x + 10*x^2)/
x]))/(3*x + E^(2*x)*x^2 - 10*x^3),x]

[Out]

5*E^x*Log[-E^(2*x) - 3/x + 10*x]

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fricas [A]  time = 1.00, size = 23, normalized size = 1.05 \begin {gather*} 5 \, e^{x} \log \left (\frac {10 \, x^{2} - x e^{\left (2 \, x\right )} - 3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="fricas")

[Out]

5*e^x*log((10*x^2 - x*e^(2*x) - 3)/x)

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giac [A]  time = 0.21, size = 26, normalized size = 1.18 \begin {gather*} 5 \, e^{x} \log \left (10 \, x^{2} - x e^{\left (2 \, x\right )} - 3\right ) - 5 \, e^{x} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="giac")

[Out]

5*e^x*log(10*x^2 - x*e^(2*x) - 3) - 5*e^x*log(x)

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maple [C]  time = 0.23, size = 189, normalized size = 8.59

method	result	size
risch	\(5 \,{\mathrm e}^{x} \ln \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )-5 \,{\mathrm e}^{x} \ln \relax (x )+\frac {5 i \pi \,\mathrm {csgn}\left (i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )^{2} {\mathrm e}^{x}}{2}-\frac {5 i \pi \,\mathrm {csgn}\left (i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}+\frac {5 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}-\frac {5 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )^{3} {\mathrm e}^{x}}{2}+5 \,{\mathrm e}^{x} \ln \relax (2)+5 \,{\mathrm e}^{x} \ln \relax (5)\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(x)^2*x^2-50*x^3+15*x)*ln((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(ln(ln((-x*exp(x)
^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/ln((-x*exp(x)^2+10*x^2-3)/x),x,method=_RETURNVERBOSE)

[Out]

5*exp(x)*ln(-1/10*x*exp(2*x)+x^2-3/10)-5*exp(x)*ln(x)+5/2*I*Pi*csgn(I*(-1/10*x*exp(2*x)+x^2-3/10))*csgn(I/x*(-
1/10*x*exp(2*x)+x^2-3/10))^2*exp(x)-5/2*I*Pi*csgn(I*(-1/10*x*exp(2*x)+x^2-3/10))*csgn(I/x*(-1/10*x*exp(2*x)+x^
2-3/10))*csgn(I/x)*exp(x)+5/2*I*Pi*csgn(I/x*(-1/10*x*exp(2*x)+x^2-3/10))^2*csgn(I/x)*exp(x)-5/2*I*Pi*csgn(I/x*
(-1/10*x*exp(2*x)+x^2-3/10))^3*exp(x)+5*exp(x)*ln(2)+5*exp(x)*ln(5)

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maxima [A]  time = 0.41, size = 26, normalized size = 1.18 \begin {gather*} 5 \, e^{x} \log \left (10 \, x^{2} - x e^{\left (2 \, x\right )} - 3\right ) - 5 \, e^{x} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="maxima")

[Out]

5*e^x*log(10*x^2 - x*e^(2*x) - 3) - 5*e^x*log(x)

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mupad [B]  time = 5.60, size = 20, normalized size = 0.91 \begin {gather*} 5\,{\mathrm {e}}^x\,\ln \left (10\,x-{\mathrm {e}}^{2\,x}-\frac {3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(log(-(x*exp(2*x) - 10*x^2 + 3)/x)))*(10*x^2*exp(2*x) + log(-(x*exp(2*x) - 10*x^2 + 3)/x)*(15*
x + 5*x^2*exp(2*x) - 50*x^3) - 50*x^2 - 15))/(log(-(x*exp(2*x) - 10*x^2 + 3)/x)*(3*x + x^2*exp(2*x) - 10*x^3))
,x)

[Out]

5*exp(x)*log(10*x - exp(2*x) - 3/x)

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sympy [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(((5*exp(x)**2*x**2-50*x**3+15*x)*ln((-x*exp(x)**2+10*x**2-3)/x)+10*exp(x)**2*x**2-50*x**2-15)*exp(ln
(ln((-x*exp(x)**2+10*x**2-3)/x))+x)/(exp(x)**2*x**2-10*x**3+3*x)/ln((-x*exp(x)**2+10*x**2-3)/x),x)

[Out]

Timed out

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