\(\int \frac {e^{-x^2 \log (3)-(x-x^2) \log (3) \log (25)} (-3+((-6 x-6 x^2) \log (3)+(-3+3 x+6 x^2) \log (3) \log (25)) \log (1+x))}{(1+x) \log ^2(1+x)} \, dx\) [5635]

Optimal result

Integrand size = 70, antiderivative size = 24 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3^{1-x (x+(1-x) \log (25))}}{\log (1+x)} \]

[Out]

3/exp(ln(3)*(x+2*(1-x)*ln(5))*x)/ln(1+x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(24)=48\).

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.29, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2326} \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3^{1-x^2} e^{-\left (\left (x-x^2\right ) \log (3) \log (25)\right )} \left (\left (-2 x^2-x+1\right ) \log (3) \log (25)+2 \left (x^2+x\right ) \log (3)\right )}{(x+1) ((1-2 x) \log (3) \log (25)+2 x \log (3)) \log (x+1)} \]

[In]

Int[(E^(-(x^2*Log[3]) - (x - x^2)*Log[3]*Log[25])*(-3 + ((-6*x - 6*x^2)*Log[3] + (-3 + 3*x + 6*x^2)*Log[3]*Log
[25])*Log[1 + x]))/((1 + x)*Log[1 + x]^2),x]

[Out]

(3^(1 - x^2)*(2*(x + x^2)*Log[3] + (1 - x - 2*x^2)*Log[3]*Log[25]))/(E^((x - x^2)*Log[3]*Log[25])*(1 + x)*(2*x
*Log[3] + (1 - 2*x)*Log[3]*Log[25])*Log[1 + x])

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {3^{1-x^2} e^{-\left (\left (x-x^2\right ) \log (3) \log (25)\right )} \left (2 \left (x+x^2\right ) \log (3)+\left (1-x-2 x^2\right ) \log (3) \log (25)\right )}{(1+x) (2 x \log (3)+(1-2 x) \log (3) \log (25)) \log (1+x)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(24)=48\).

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.04 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3^{(-1+x) (-1+x (-1+\log (25)))} \left (x^2 \log (9) (-1+\log (25))-\log (3) \log (25)+x (-\log (9)+\log (3) \log (25))\right )}{(1+x) \log (3) (2 x (-1+\log (25))-\log (25)) \log (1+x)} \]

[In]

Integrate[(E^(-(x^2*Log[3]) - (x - x^2)*Log[3]*Log[25])*(-3 + ((-6*x - 6*x^2)*Log[3] + (-3 + 3*x + 6*x^2)*Log[
3]*Log[25])*Log[1 + x]))/((1 + x)*Log[1 + x]^2),x]

[Out]

(3^((-1 + x)*(-1 + x*(-1 + Log[25])))*(x^2*Log[9]*(-1 + Log[25]) - Log[3]*Log[25] + x*(-Log[9] + Log[3]*Log[25
])))/((1 + x)*Log[3]*(2*x*(-1 + Log[25]) - Log[25])*Log[1 + x])

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {3 \,9^{x \left (-1+x \right ) \ln \left (5\right )} 3^{-x^{2}}}{\ln \left (1+x \right )}\)	\(27\)
parallelrisch	\(\frac {3 \,{\mathrm e}^{\ln \left (3\right ) x \left (2 x \ln \left (5\right )+\ln \left (\frac {1}{25}\right )-x \right )}}{\ln \left (1+x \right )}\)	\(30\)
norman	\(\frac {3 \,{\mathrm e}^{-2 \left (-x^{2}+x \right ) \ln \left (3\right ) \ln \left (5\right )-x^{2} \ln \left (3\right )}}{\ln \left (1+x \right )}\)	\(32\)

[In]

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

[Out]

3/((1/9)^(x*(-1+x)*ln(5)))/(3^(x^2))/ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3 \, e^{\left (-x^{2} \log \left (3\right ) + 2 \, {\left (x^{2} - x\right )} \log \left (5\right ) \log \left (3\right )\right )}}{\log \left (x + 1\right )} \]

[In]

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

[Out]

3*e^(-x^2*log(3) + 2*(x^2 - x)*log(5)*log(3))/log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3 e^{- x^{2} \log {\left (3 \right )} - \left (- 2 x^{2} + 2 x\right ) \log {\left (3 \right )} \log {\left (5 \right )}}}{\log {\left (x + 1 \right )}} \]

[In]

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

[Out]

3*exp(-x**2*log(3) - (-2*x**2 + 2*x)*log(3)*log(5))/log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3 \, e^{\left (2 \, x^{2} \log \left (5\right ) \log \left (3\right ) - x^{2} \log \left (3\right ) - 2 \, x \log \left (5\right ) \log \left (3\right )\right )}}{\log \left (x + 1\right )} \]

[In]

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

[Out]

3*e^(2*x^2*log(5)*log(3) - x^2*log(3) - 2*x*log(5)*log(3))/log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3 \, e^{\left (2 \, x^{2} \log \left (5\right ) \log \left (3\right ) - x^{2} \log \left (3\right ) - 2 \, x \log \left (5\right ) \log \left (3\right )\right )}}{\log \left (x + 1\right )} \]

[In]

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

[Out]

3*e^(2*x^2*log(5)*log(3) - x^2*log(3) - 2*x*log(5)*log(3))/log(x + 1)

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-x^2 \log (3)-\left (x-x^2\right ) \log (3) \log (25)} \left (-3+\left (\left (-6 x-6 x^2\right ) \log (3)+\left (-3+3 x+6 x^2\right ) \log (3) \log (25)\right ) \log (1+x)\right )}{(1+x) \log ^2(1+x)} \, dx=\frac {3\,3^{2\,x^2\,\ln \left (5\right )}}{3^{x^2}\,3^{2\,x\,\ln \left (5\right )}\,\ln \left (x+1\right )} \]

[In]

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

[Out]

(3*3^(2*x^2*log(5)))/(3^(x^2)*3^(2*x*log(5))*log(x + 1))