3.46.68 \(\int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} (18+15 x+22 x^3+10 x^4+(-44 x^2-20 x^3) \log (11+5 x)+(22 x+10 x^2) \log ^2(11+5 x))}{11 x^2+5 x^3+(-22 x-10 x^2) \log (11+5 x)+(11+5 x) \log ^2(11+5 x)} \, dx\)

Optimal. Leaf size=24 \[ 1+e^{x^2+\frac {3}{-x+\log (1+5 (2+x))}} \]

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Rubi [F]  time = 3.37, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} \left (18+15 x+22 x^3+10 x^4+\left (-44 x^2-20 x^3\right ) \log (11+5 x)+\left (22 x+10 x^2\right ) \log ^2(11+5 x)\right )}{11 x^2+5 x^3+\left (-22 x-10 x^2\right ) \log (11+5 x)+(11+5 x) \log ^2(11+5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3 - x^3 + x^2*Log[11 + 5*x])/(-x + Log[11 + 5*x]))*(18 + 15*x + 22*x^3 + 10*x^4 + (-44*x^2 - 20*x^3)*
Log[11 + 5*x] + (22*x + 10*x^2)*Log[11 + 5*x]^2))/(11*x^2 + 5*x^3 + (-22*x - 10*x^2)*Log[11 + 5*x] + (11 + 5*x
)*Log[11 + 5*x]^2),x]

[Out]

2*Defer[Int][E^((3 - x^3 + x^2*Log[11 + 5*x])/(-x + Log[11 + 5*x]))*x, x] + 3*Defer[Int][E^((3 - x^3 + x^2*Log
[11 + 5*x])/(-x + Log[11 + 5*x]))/(x - Log[11 + 5*x])^2, x] - 15*Defer[Int][E^((3 - x^3 + x^2*Log[11 + 5*x])/(
-x + Log[11 + 5*x]))/((11 + 5*x)*(x - Log[11 + 5*x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} \left (18+15 x+22 x^3+10 x^4+\left (-44 x^2-20 x^3\right ) \log (11+5 x)+\left (22 x+10 x^2\right ) \log ^2(11+5 x)\right )}{(11+5 x) (x-\log (11+5 x))^2} \, dx\\ &=\int \left (2 e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} x+\frac {3 e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} (6+5 x)}{(11+5 x) (x-\log (11+5 x))^2}\right ) \, dx\\ &=2 \int e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} x \, dx+3 \int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} (6+5 x)}{(11+5 x) (x-\log (11+5 x))^2} \, dx\\ &=2 \int e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} x \, dx+3 \int \left (\frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}}}{(x-\log (11+5 x))^2}-\frac {5 e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}}}{(11+5 x) (x-\log (11+5 x))^2}\right ) \, dx\\ &=2 \int e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}} x \, dx+3 \int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}}}{(x-\log (11+5 x))^2} \, dx-15 \int \frac {e^{\frac {3-x^3+x^2 \log (11+5 x)}{-x+\log (11+5 x)}}}{(11+5 x) (x-\log (11+5 x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 44, normalized size = 1.83 \begin {gather*} e^{\frac {-3+x^3}{x-\log (11+5 x)}} (11+5 x)^{-\frac {x^2}{x-\log (11+5 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 - x^3 + x^2*Log[11 + 5*x])/(-x + Log[11 + 5*x]))*(18 + 15*x + 22*x^3 + 10*x^4 + (-44*x^2 - 20
*x^3)*Log[11 + 5*x] + (22*x + 10*x^2)*Log[11 + 5*x]^2))/(11*x^2 + 5*x^3 + (-22*x - 10*x^2)*Log[11 + 5*x] + (11
+ 5*x)*Log[11 + 5*x]^2),x]

[Out]

E^((-3 + x^3)/(x - Log[11 + 5*x]))/(11 + 5*x)^(x^2/(x - Log[11 + 5*x]))

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fricas [A]  time = 0.57, size = 30, normalized size = 1.25 \begin {gather*} e^{\left (\frac {x^{3} - x^{2} \log \left (5 \, x + 11\right ) - 3}{x - \log \left (5 \, x + 11\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+22*x)*log(5*x+11)^2+(-20*x^3-44*x^2)*log(5*x+11)+10*x^4+22*x^3+15*x+18)*exp((x^2*log(5*x+11
)-x^3+3)/(log(5*x+11)-x))/((5*x+11)*log(5*x+11)^2+(-10*x^2-22*x)*log(5*x+11)+5*x^3+11*x^2),x, algorithm="frica
s")

[Out]

e^((x^3 - x^2*log(5*x + 11) - 3)/(x - log(5*x + 11)))

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giac [B]  time = 0.91, size = 55, normalized size = 2.29 \begin {gather*} e^{\left (\frac {x^{3}}{x - \log \left (5 \, x + 11\right )} - \frac {x^{2} \log \left (5 \, x + 11\right )}{x - \log \left (5 \, x + 11\right )} - \frac {3}{x - \log \left (5 \, x + 11\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+22*x)*log(5*x+11)^2+(-20*x^3-44*x^2)*log(5*x+11)+10*x^4+22*x^3+15*x+18)*exp((x^2*log(5*x+11
)-x^3+3)/(log(5*x+11)-x))/((5*x+11)*log(5*x+11)^2+(-10*x^2-22*x)*log(5*x+11)+5*x^3+11*x^2),x, algorithm="giac"
)

[Out]

e^(x^3/(x - log(5*x + 11)) - x^2*log(5*x + 11)/(x - log(5*x + 11)) - 3/(x - log(5*x + 11)))

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maple [A]  time = 0.03, size = 31, normalized size = 1.29

method	result	size
risch	\({\mathrm e}^{\frac {-x^{2} \ln \left (5 x +11\right )+x^{3}-3}{-\ln \left (5 x +11\right )+x}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+22*x)*ln(5*x+11)^2+(-20*x^3-44*x^2)*ln(5*x+11)+10*x^4+22*x^3+15*x+18)*exp((x^2*ln(5*x+11)-x^3+3)/
(ln(5*x+11)-x))/((5*x+11)*ln(5*x+11)^2+(-10*x^2-22*x)*ln(5*x+11)+5*x^3+11*x^2),x,method=_RETURNVERBOSE)

[Out]

exp((-x^2*ln(5*x+11)+x^3-3)/(-ln(5*x+11)+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (10 \, x^{4} + 22 \, x^{3} + 2 \, {\left (5 \, x^{2} + 11 \, x\right )} \log \left (5 \, x + 11\right )^{2} - 4 \, {\left (5 \, x^{3} + 11 \, x^{2}\right )} \log \left (5 \, x + 11\right ) + 15 \, x + 18\right )} e^{\left (\frac {x^{3} - x^{2} \log \left (5 \, x + 11\right ) - 3}{x - \log \left (5 \, x + 11\right )}\right )}}{5 \, x^{3} + {\left (5 \, x + 11\right )} \log \left (5 \, x + 11\right )^{2} + 11 \, x^{2} - 2 \, {\left (5 \, x^{2} + 11 \, x\right )} \log \left (5 \, x + 11\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+22*x)*log(5*x+11)^2+(-20*x^3-44*x^2)*log(5*x+11)+10*x^4+22*x^3+15*x+18)*exp((x^2*log(5*x+11
)-x^3+3)/(log(5*x+11)-x))/((5*x+11)*log(5*x+11)^2+(-10*x^2-22*x)*log(5*x+11)+5*x^3+11*x^2),x, algorithm="maxim
a")

[Out]

integrate((10*x^4 + 22*x^3 + 2*(5*x^2 + 11*x)*log(5*x + 11)^2 - 4*(5*x^3 + 11*x^2)*log(5*x + 11) + 15*x + 18)*
e^((x^3 - x^2*log(5*x + 11) - 3)/(x - log(5*x + 11)))/(5*x^3 + (5*x + 11)*log(5*x + 11)^2 + 11*x^2 - 2*(5*x^2
+ 11*x)*log(5*x + 11)), x)

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mupad [B]  time = 3.64, size = 44, normalized size = 1.83 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^3-3}{x-\ln \left (5\,x+11\right )}}}{{\left (5\,x+11\right )}^{\frac {x^2}{x-\ln \left (5\,x+11\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(x^2*log(5*x + 11) - x^3 + 3)/(x - log(5*x + 11)))*(15*x + log(5*x + 11)^2*(22*x + 10*x^2) - log(5*x
+ 11)*(44*x^2 + 20*x^3) + 22*x^3 + 10*x^4 + 18))/(log(5*x + 11)^2*(5*x + 11) - log(5*x + 11)*(22*x + 10*x^2)
+ 11*x^2 + 5*x^3),x)

[Out]

exp((x^3 - 3)/(x - log(5*x + 11)))/(5*x + 11)^(x^2/(x - log(5*x + 11)))

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sympy [A]  time = 0.60, size = 24, normalized size = 1.00 \begin {gather*} e^{\frac {- x^{3} + x^{2} \log {\left (5 x + 11 \right )} + 3}{- x + \log {\left (5 x + 11 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+22*x)*ln(5*x+11)**2+(-20*x**3-44*x**2)*ln(5*x+11)+10*x**4+22*x**3+15*x+18)*exp((x**2*ln(5*
x+11)-x**3+3)/(ln(5*x+11)-x))/((5*x+11)*ln(5*x+11)**2+(-10*x**2-22*x)*ln(5*x+11)+5*x**3+11*x**2),x)

[Out]

exp((-x**3 + x**2*log(5*x + 11) + 3)/(-x + log(5*x + 11)))

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