3.15.15 \(\int \frac {24 x+24 x^2+e^8 (x+x^2)+e^4 (10 x+10 x^2)+e^{\frac {5 x^2-\log (1+x)}{x}} (6 x^2+5 x^3+(1+x) \log (1+x))}{x+x^2} \, dx\)

Optimal. Leaf size=31 \[ -x+\left (e^{5 x-\frac {\log (1+x)}{x}}+\left (5+e^4\right )^2\right ) x+\log (2) \]

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Rubi [F]  time = 1.32, 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 {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24*x + 24*x^2 + E^8*(x + x^2) + E^4*(10*x + 10*x^2) + E^((5*x^2 - Log[1 + x])/x)*(6*x^2 + 5*x^3 + (1 + x)
*Log[1 + x]))/(x + x^2),x]

[Out]

(24 + 10*E^4 + E^8)*x + 6*Defer[Int][E^(5*x)*x*(1 + x)^(-1 - x^(-1)), x] + 5*Defer[Int][E^(5*x)*x^2*(1 + x)^(-
1 - x^(-1)), x] + Log[1 + x]*Defer[Int][E^(5*x)/(x*(1 + x)^x^(-1)), x] - Defer[Int][Defer[Int][E^(5*x)/(x*(1 +
x)^x^(-1)), x]/(1 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x (1+x)} \, dx\\ &=\int \left (24+10 e^4+e^8+e^{5 x} x (1+x)^{-1-\frac {1}{x}} (6+5 x)+\frac {e^{5 x} (1+x)^{-1/x} \log (1+x)}{x}\right ) \, dx\\ &=\left (24+10 e^4+e^8\right ) x+\int e^{5 x} x (1+x)^{-1-\frac {1}{x}} (6+5 x) \, dx+\int \frac {e^{5 x} (1+x)^{-1/x} \log (1+x)}{x} \, dx\\ &=\left (24+10 e^4+e^8\right ) x+\log (1+x) \int \frac {e^{5 x} (1+x)^{-1/x}}{x} \, dx+\int \left (6 e^{5 x} x (1+x)^{-1-\frac {1}{x}}+5 e^{5 x} x^2 (1+x)^{-1-\frac {1}{x}}\right ) \, dx-\int \frac {\int \frac {e^{5 x} (1+x)^{-1/x}}{x} \, dx}{1+x} \, dx\\ &=\left (24+10 e^4+e^8\right ) x+5 \int e^{5 x} x^2 (1+x)^{-1-\frac {1}{x}} \, dx+6 \int e^{5 x} x (1+x)^{-1-\frac {1}{x}} \, dx+\log (1+x) \int \frac {e^{5 x} (1+x)^{-1/x}}{x} \, dx-\int \frac {\int \frac {e^{5 x} (1+x)^{-1/x}}{x} \, dx}{1+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 29, normalized size = 0.94 \begin {gather*} \left (24+10 e^4+e^8\right ) x+e^{5 x} x (1+x)^{-1/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*x + 24*x^2 + E^8*(x + x^2) + E^4*(10*x + 10*x^2) + E^((5*x^2 - Log[1 + x])/x)*(6*x^2 + 5*x^3 + (
1 + x)*Log[1 + x]))/(x + x^2),x]

[Out]

(24 + 10*E^4 + E^8)*x + (E^(5*x)*x)/(1 + x)^x^(-1)

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fricas [A]  time = 0.94, size = 32, normalized size = 1.03 \begin {gather*} x e^{8} + 10 \, x e^{4} + x e^{\left (\frac {5 \, x^{2} - \log \left (x + 1\right )}{x}\right )} + 24 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(x+1)+5*x^3+6*x^2)*exp((-log(x+1)+5*x^2)/x)+(x^2+x)*exp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+
24*x)/(x^2+x),x, algorithm="fricas")

[Out]

x*e^8 + 10*x*e^4 + x*e^((5*x^2 - log(x + 1))/x) + 24*x

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giac [A]  time = 0.41, size = 32, normalized size = 1.03 \begin {gather*} x e^{8} + 10 \, x e^{4} + x e^{\left (\frac {5 \, x^{2} - \log \left (x + 1\right )}{x}\right )} + 24 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(x+1)+5*x^3+6*x^2)*exp((-log(x+1)+5*x^2)/x)+(x^2+x)*exp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+
24*x)/(x^2+x),x, algorithm="giac")

[Out]

x*e^8 + 10*x*e^4 + x*e^((5*x^2 - log(x + 1))/x) + 24*x

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maple [A]  time = 0.24, size = 29, normalized size = 0.94

method	result	size
risch	\(x \left (x +1\right )^{-\frac {1}{x}} {\mathrm e}^{5 x}+10 x \,{\mathrm e}^{4}+x \,{\mathrm e}^{8}+24 x\)	\(29\)
norman	\(x \,{\mathrm e}^{\frac {-\ln \left (x +1\right )+5 x^{2}}{x}}+\left ({\mathrm e}^{8}+10 \,{\mathrm e}^{4}+24\right ) x\)	\(33\)
default	\(24 x +x \,{\mathrm e}^{8}+x \,{\mathrm e}^{\frac {-\ln \left (x +1\right )+5 x^{2}}{x}}+10 x \,{\mathrm e}^{4}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*ln(x+1)+5*x^3+6*x^2)*exp((-ln(x+1)+5*x^2)/x)+(x^2+x)*exp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x
^2+x),x,method=_RETURNVERBOSE)

[Out]

x*(x+1)^(-1/x)*exp(5*x)+10*x*exp(4)+x*exp(8)+24*x

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maxima [A]  time = 0.58, size = 58, normalized size = 1.87 \begin {gather*} {\left (x - \log \left (x + 1\right )\right )} e^{8} + 10 \, {\left (x - \log \left (x + 1\right )\right )} e^{4} + x e^{\left (5 \, x - \frac {\log \left (x + 1\right )}{x}\right )} + e^{8} \log \left (x + 1\right ) + 10 \, e^{4} \log \left (x + 1\right ) + 24 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(x+1)+5*x^3+6*x^2)*exp((-log(x+1)+5*x^2)/x)+(x^2+x)*exp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+
24*x)/(x^2+x),x, algorithm="maxima")

[Out]

(x - log(x + 1))*e^8 + 10*(x - log(x + 1))*e^4 + x*e^(5*x - log(x + 1)/x) + e^8*log(x + 1) + 10*e^4*log(x + 1)
+ 24*x

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mupad [B]  time = 1.29, size = 28, normalized size = 0.90 \begin {gather*} 24\,x+10\,x\,{\mathrm {e}}^4+x\,{\mathrm {e}}^8+\frac {x\,{\mathrm {e}}^{5\,x}}{{\left (x+1\right )}^{1/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*x + exp(-(log(x + 1) - 5*x^2)/x)*(log(x + 1)*(x + 1) + 6*x^2 + 5*x^3) + exp(4)*(10*x + 10*x^2) + exp(8
)*(x + x^2) + 24*x^2)/(x + x^2),x)

[Out]

24*x + 10*x*exp(4) + x*exp(8) + (x*exp(5*x))/(x + 1)^(1/x)

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sympy [A]  time = 11.68, size = 26, normalized size = 0.84 \begin {gather*} x e^{\frac {5 x^{2} - \log {\left (x + 1 \right )}}{x}} + x \left (24 + 10 e^{4} + e^{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*ln(x+1)+5*x**3+6*x**2)*exp((-ln(x+1)+5*x**2)/x)+(x**2+x)*exp(4)**2+(10*x**2+10*x)*exp(4)+24*
x**2+24*x)/(x**2+x),x)

[Out]

x*exp((5*x**2 - log(x + 1))/x) + x*(24 + 10*exp(4) + exp(8))

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