3.41.22 $\int \frac{3-18x+e^{25-11x+x^2}(11x-2x^2)-18x\log (x)}{x}dx$

Optimal. Leaf size=22 $$-e^{(-5+x{)}^2-x}+(3-18x)\log (x)$$

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 34, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.118, Rules used = {14, 2236, 43, 2295} $$\begin{array}{c}-e^{x^2-11x+25}-18x\log (x)+3\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(3 - 18*x + E^(25 - 11*x + x^2)*(11*x - 2*x^2) - 18*x*Log[x])/x,x]

[Out]

-E^(25 - 11*x + x^2) + 3*Log[x] - 18*x*Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-e^{25-11x+x^2}(-11+2x)-\frac{3(-1+6x+6x\log (x))}{x})dx\\ & =-(3\int \frac{-1+6x+6x\log (x)}{x}dx)-\int e^{25-11x+x^2}(-11+2x)dx\\ & =-e^{25-11x+x^2}-3\int (\frac{-1+6x}{x}+6\log (x))dx\\ & =-e^{25-11x+x^2}-3\int \frac{-1+6x}{x}dx-18\int \log (x)dx\\ & =-e^{25-11x+x^2}+18x-18x\log (x)-3\int (6-\frac{1}{x})dx\\ & =-e^{25-11x+x^2}+3\log (x)-18x\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 22, normalized size = 1.00 $$\begin{array}{c}-e^{25-11x+x^2}+3\log (x)-18x\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(3 - 18*x + E^(25 - 11*x + x^2)*(11*x - 2*x^2) - 18*x*Log[x])/x,x]

[Out]

-E^(25 - 11*x + x^2) + 3*Log[x] - 18*x*Log[x]

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fricas [A]  time = 0.88, size = 21, normalized size = 0.95 $$\begin{array}{c}-3(6x-1)\log (x)-e^{(x^2-11x+25)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="fricas")

[Out]

-3*(6*x - 1)*log(x) - e^(x^2 - 11*x + 25)

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giac [A]  time = 0.20, size = 21, normalized size = 0.95 $$\begin{array}{c}-18x\log (x)-e^{(x^2-11x+25)}+3\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="giac")

[Out]

-18*x*log(x) - e^(x^2 - 11*x + 25) + 3*log(x)

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maple [A]  time = 0.03, size = 22, normalized size = 1.00

method	result	size
default	$-{\mathrm{e}}^{x^2-11x+25}+3\ln (x)-18x\ln (x)$	$22$
norman	$-{\mathrm{e}}^{x^2-11x+25}+3\ln (x)-18x\ln (x)$	$22$
risch	$-{\mathrm{e}}^{x^2-11x+25}+3\ln (x)-18x\ln (x)$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-18*x*ln(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x,method=_RETURNVERBOSE)

[Out]

-exp(x^2-11*x+25)+3*ln(x)-18*x*ln(x)

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maxima [C]  time = 0.38, size = 77, normalized size = 3.50 $$\begin{array}{c}-\frac{11}{2}i\sqrt{\pi }\mathrm{erf}(ix-\frac{11}{2}i)e^{(-\frac{21}{4})}-\frac{1}{2}(\frac{11\sqrt{\pi }(2x-11)(\mathrm{erf}\left(\frac{1}{2}\sqrt{-{(2x-11)}^2}\right)-1)}{\sqrt{-{(2x-11)}^2}}+2e^{\left(\frac{1}{4}{(2x-11)}^2\right)})e^{(-\frac{21}{4})}-18x\log (x)+3\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="maxima")

[Out]

-11/2*I*sqrt(pi)*erf(I*x - 11/2*I)*e^(-21/4) - 1/2*(11*sqrt(pi)*(2*x - 11)*(erf(1/2*sqrt(-(2*x - 11)^2)) - 1)/
sqrt(-(2*x - 11)^2) + 2*e^(1/4*(2*x - 11)^2))*e^(-21/4) - 18*x*log(x) + 3*log(x)

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mupad [B]  time = 3.05, size = 22, normalized size = 1.00 $$\begin{array}{c}3\ln (x)-18x\ln (x)-{\mathrm{e}}^{-11x}{\mathrm{e}}^{x^2}{\mathrm{e}}^{25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(18*x - exp(x^2 - 11*x + 25)*(11*x - 2*x^2) + 18*x*log(x) - 3)/x,x)

[Out]

3*log(x) - 18*x*log(x) - exp(-11*x)*exp(x^2)*exp(25)

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sympy [A]  time = 0.29, size = 20, normalized size = 0.91 $$\begin{array}{c}-18x\log (x)-e^{x^2-11x+25}+3\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x*ln(x)+(-2*x**2+11*x)*exp(x**2-11*x+25)-18*x+3)/x,x)

[Out]

-18*x*log(x) - exp(x**2 - 11*x + 25) + 3*log(x)

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