3.89.3 $\int \frac{e^x(1-x-2\log (4x)+{\log }^2(4x))(6+3x^2+(-6+6x)\log (4x)-3x{\log }^2(4x))}{x-x^2-2x\log (4x)+x{\log }^2(4x)}dx$

Optimal. Leaf size=19 $$-3e^x(-x+(1-\log (4x){)}^2)$$

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Rubi [F]  time = 1.17, 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 = {} $$\begin{array}{c}\int \frac{e^x(1-x-2\log (4x)+{\log }^2(4x))(6+3x^2+(-6+6x)\log (4x)-3x{\log }^2(4x))}{x-x^2-2x\log (4x)+x{\log }^2(4x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

-3*E^x + 3*E^x*x + 6*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, x] + 3*Log[-x]^2 + 6*EulerGamma*Log[x] + 6*(Exp
IntegralE[1, -x] + ExpIntegralEi[x])*Log[x] + 6*E^x*Log[4*x] - 6*ExpIntegralEi[x]*Log[4*x] - 3*Defer[Int][E^x*
Log[4*x]^2, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{3e^x(2+x^2+2(-1+x)\log (4x)-x{\log }^2(4x))}{x}dx\\ & =3\int \frac{e^x(2+x^2+2(-1+x)\log (4x)-x{\log }^2(4x))}{x}dx\\ & =3\int (\frac{e^x(2+x^2)}{x}+\frac{2e^x(-1+x)\log (4x)}{x}-e^x{\log }^2(4x))dx\\ & =3\int \frac{e^x(2+x^2)}{x}dx-3\int e^x{\log }^2(4x)dx+6\int \frac{e^x(-1+x)\log (4x)}{x}dx\\ & =6e^x\log (4x)-6\text{Ei}(x)\log (4x)+3\int (\frac{2e^x}{x}+e^{x}x)dx-3\int e^x{\log }^2(4x)dx-6\int \frac{e^x-\text{Ei}(x)}{x}dx\\ & =6e^x\log (4x)-6\text{Ei}(x)\log (4x)+3\int e^{x}xdx-3\int e^x{\log }^2(4x)dx+6\int \frac{e^x}{x}dx-6\int (\frac{e^x}{x}-\frac{\text{Ei}(x)}{x})dx\\ & =3e^{x}x+6\text{Ei}(x)+6e^x\log (4x)-6\text{Ei}(x)\log (4x)-3\int e^{x}dx-3\int e^x{\log }^2(4x)dx-6\int \frac{e^x}{x}dx+6\int \frac{\text{Ei}(x)}{x}dx\\ & =-3e^x+3e^{x}x+6(E_1(-x)+\text{Ei}(x))\log (x)+6e^x\log (4x)-6\text{Ei}(x)\log (4x)-3\int e^x{\log }^2(4x)dx-6\int \frac{E_1(-x)}{x}dx\\ & =-3e^x+3e^{x}x+6x{}_3{F}_3(1,1,1;2,2,2;x)+3{\log }^2(-x)+6\gamma \log (x)+6(E_1(-x)+\text{Ei}(x))\log (x)+6e^x\log (4x)-6\text{Ei}(x)\log (4x)-3\int e^x{\log }^2(4x)dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.10, size = 22, normalized size = 1.16 $$\begin{array}{c}3e^x(-1+x+2\log (4x)-{\log }^2(4x))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

3*E^x*(-1 + x + 2*Log[4*x] - Log[4*x]^2)

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fricas [A]  time = 0.60, size = 23, normalized size = 1.21 $$\begin{array}{c}-3e^{(x+\log (\log {\left(4x\right)}^2-x-2\log \left(4x\right)+1))}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*e^(x + log(log(4*x)^2 - x - 2*log(4*x) + 1))

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giac [B]  time = 0.18, size = 46, normalized size = 2.42 $$\begin{array}{c}-12e^x\log (2{)}^2-12e^x\log (2)\log (x)-3e^x\log (x{)}^2+3x{e}^x+12e^x\log (2)+6e^x\log (x)-3e^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*e^x*log(2)^2 - 12*e^x*log(2)*log(x) - 3*e^x*log(x)^2 + 3*x*e^x + 12*e^x*log(2) + 6*e^x*log(x) - 3*e^x

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maple [A]  time = 0.04, size = 22, normalized size = 1.16

method	result	size
risch	$-3(\ln {\left(4x\right)}^2-2\ln \left(4x\right)-x+1){\mathrm{e}}^x$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(ln(4*x)^2-2*ln(4*x)-x+1)*exp(x)

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maxima [A]  time = 0.49, size = 33, normalized size = 1.74 $$\begin{array}{c}-3(4\log (2{)}^2+2(2\log (2)-1)\log (x)+\log (x{)}^2-x-4\log (2)+1)e^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(4*log(2)^2 + 2*(2*log(2) - 1)*log(x) + log(x)^2 - x - 4*log(2) + 1)*e^x

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mupad [B]  time = 7.07, size = 46, normalized size = 2.42 $$\begin{array}{c}6{\mathrm{e}}^x\ln (x)-12{\mathrm{e}}^x{\ln (2)}^2-3{\mathrm{e}}^x-3{\mathrm{e}}^x{\ln (x)}^2+12{\mathrm{e}}^x\ln (2)+3x{\mathrm{e}}^x-12{\mathrm{e}}^x\ln (2)\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*exp(x)*log(x) - 12*exp(x)*log(2)^2 - 3*exp(x) - 3*exp(x)*log(x)^2 + 12*exp(x)*log(2) + 3*x*exp(x) - 12*exp(x
)*log(2)*log(x)

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sympy [A]  time = 0.40, size = 22, normalized size = 1.16 $$\begin{array}{c}(3x-3\log {\left(4x\right)}^2+6\log \left(4x\right)-3)e^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x - 3*log(4*x)**2 + 6*log(4*x) - 3)*exp(x)

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