3.5.90 \(\int \frac {2 x^2+(1-2 x+e^2 x) \log ^2(15+e)}{x \log ^2(15+e)} \, dx\)

Optimal. Leaf size=21 \[ -2 x+e^2 x+\frac {x^2}{\log ^2(15+e)}+\log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 14} \begin {gather*} \frac {x^2}{\log ^2(15+e)}-\left (2-e^2\right ) x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + (1 - 2*x + E^2*x)*Log[15 + E]^2)/(x*Log[15 + E]^2),x]

[Out]

-((2 - E^2)*x) + x^2/Log[15 + E]^2 + Log[x]

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {2 x^2+\left (1-2 x+e^2 x\right ) \log ^2(15+e)}{x} \, dx}{\log ^2(15+e)}\\ &=\frac {\int \left (2 x-\left (2-e^2\right ) \log ^2(15+e)+\frac {\log ^2(15+e)}{x}\right ) \, dx}{\log ^2(15+e)}\\ &=-\left (\left (2-e^2\right ) x\right )+\frac {x^2}{\log ^2(15+e)}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} -2 x+e^2 x+\frac {x^2}{\log ^2(15+e)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + (1 - 2*x + E^2*x)*Log[15 + E]^2)/(x*Log[15 + E]^2),x]

[Out]

-2*x + E^2*x + x^2/Log[15 + E]^2 + Log[x]

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fricas [A]  time = 0.88, size = 30, normalized size = 1.43 \begin {gather*} \frac {{\left (x e^{2} - 2 \, x + \log \relax (x)\right )} \log \left (e + 15\right )^{2} + x^{2}}{\log \left (e + 15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*x+1-2*x)*log(exp(1)+15)^2+2*x^2)/x/log(exp(1)+15)^2,x, algorithm="fricas")

[Out]

((x*e^2 - 2*x + log(x))*log(e + 15)^2 + x^2)/log(e + 15)^2

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giac [B]  time = 0.29, size = 44, normalized size = 2.10 \begin {gather*} \frac {x e^{2} \log \left (e + 15\right )^{2} - 2 \, x \log \left (e + 15\right )^{2} + \log \left (e + 15\right )^{2} \log \left ({\left | x \right |}\right ) + x^{2}}{\log \left (e + 15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*x+1-2*x)*log(exp(1)+15)^2+2*x^2)/x/log(exp(1)+15)^2,x, algorithm="giac")

[Out]

(x*e^2*log(e + 15)^2 - 2*x*log(e + 15)^2 + log(e + 15)^2*log(abs(x)) + x^2)/log(e + 15)^2

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maple [A]  time = 0.06, size = 22, normalized size = 1.05

method	result	size
risch	\(\ln \relax (x )+{\mathrm e}^{2} x -2 x +\frac {x^{2}}{\ln \left ({\mathrm e}+15\right )^{2}}\)	\(22\)
norman	\(\frac {\frac {x^{2}}{\ln \left ({\mathrm e}+15\right )}+\ln \left ({\mathrm e}+15\right ) \left ({\mathrm e}^{2}-2\right ) x}{\ln \left ({\mathrm e}+15\right )}+\ln \relax (x )\)	\(35\)
default	\(\frac {x^{2}+{\mathrm e}^{2} \ln \left ({\mathrm e}+15\right )^{2} x -2 x \ln \left ({\mathrm e}+15\right )^{2}+\ln \left ({\mathrm e}+15\right )^{2} \ln \relax (x )}{\ln \left ({\mathrm e}+15\right )^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)*x+1-2*x)*ln(exp(1)+15)^2+2*x^2)/x/ln(exp(1)+15)^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)+exp(2)*x-2*x+x^2/ln(exp(1)+15)^2

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maxima [A]  time = 0.37, size = 35, normalized size = 1.67 \begin {gather*} \frac {x {\left (e^{2} - 2\right )} \log \left (e + 15\right )^{2} + \log \relax (x) \log \left (e + 15\right )^{2} + x^{2}}{\log \left (e + 15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*x+1-2*x)*log(exp(1)+15)^2+2*x^2)/x/log(exp(1)+15)^2,x, algorithm="maxima")

[Out]

(x*(e^2 - 2)*log(e + 15)^2 + log(x)*log(e + 15)^2 + x^2)/log(e + 15)^2

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mupad [B]  time = 0.07, size = 20, normalized size = 0.95 \begin {gather*} \ln \relax (x)+x\,\left ({\mathrm {e}}^2-2\right )+\frac {x^2}{{\ln \left (\mathrm {e}+15\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(1) + 15)^2*(x*exp(2) - 2*x + 1) + 2*x^2)/(x*log(exp(1) + 15)^2),x)

[Out]

log(x) + x*(exp(2) - 2) + x^2/log(exp(1) + 15)^2

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sympy [B]  time = 0.13, size = 46, normalized size = 2.19 \begin {gather*} \frac {x^{2} + x \left (- 2 \log {\left (e + 15 \right )}^{2} + e^{2} \log {\left (e + 15 \right )}^{2}\right ) + \log {\relax (x )} \log {\left (e + 15 \right )}^{2}}{\log {\left (e + 15 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*x+1-2*x)*ln(exp(1)+15)**2+2*x**2)/x/ln(exp(1)+15)**2,x)

[Out]

(x**2 + x*(-2*log(E + 15)**2 + exp(2)*log(E + 15)**2) + log(x)*log(E + 15)**2)/log(E + 15)**2

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