\(\int \frac {e^{2 x^2+4 e^{4 x} x^2} (e^5 (-18+36 x^2)+e^{5+4 x} (72 x^2+144 x^3))}{x^3} \, dx\) [10250]

Optimal result

Integrand size = 53, antiderivative size = 22 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 e^{5+2 x \left (x+2 e^{4 x} x\right )}}{x^2} \]

[Out]

9*exp(5)*exp((2*x*exp(4*x)+x)*x)^2/x^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(22)=44\).

Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.23, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6873, 12, 2326} \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 e^{2 \left (2 e^{4 x}+1\right ) x^2+5} \left (4 e^{4 x} x^3+2 e^{4 x} x^2+x^2\right )}{x^3 \left (4 e^{4 x} x^2+\left (2 e^{4 x}+1\right ) x\right )} \]

[In]

Int[(E^(2*x^2 + 4*E^(4*x)*x^2)*(E^5*(-18 + 36*x^2) + E^(5 + 4*x)*(72*x^2 + 144*x^3)))/x^3,x]

[Out]

(9*E^(5 + 2*(1 + 2*E^(4*x))*x^2)*(x^2 + 2*E^(4*x)*x^2 + 4*E^(4*x)*x^3))/(x^3*((1 + 2*E^(4*x))*x + 4*E^(4*x)*x^
2))

Rule 12

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

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {18 e^{5+2 \left (1+2 e^{4 x}\right ) x^2} \left (-1+2 x^2+4 e^{4 x} x^2+8 e^{4 x} x^3\right )}{x^3} \, dx \\ & = 18 \int \frac {e^{5+2 \left (1+2 e^{4 x}\right ) x^2} \left (-1+2 x^2+4 e^{4 x} x^2+8 e^{4 x} x^3\right )}{x^3} \, dx \\ & = \frac {9 e^{5+2 \left (1+2 e^{4 x}\right ) x^2} \left (x^2+2 e^{4 x} x^2+4 e^{4 x} x^3\right )}{x^3 \left (\left (1+2 e^{4 x}\right ) x+4 e^{4 x} x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 e^{5+\left (2+4 e^{4 x}\right ) x^2}}{x^2} \]

[In]

Integrate[(E^(2*x^2 + 4*E^(4*x)*x^2)*(E^5*(-18 + 36*x^2) + E^(5 + 4*x)*(72*x^2 + 144*x^3)))/x^3,x]

[Out]

(9*E^(5 + (2 + 4*E^(4*x))*x^2))/x^2

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {9 \,{\mathrm e}^{4 x^{2} {\mathrm e}^{4 x}+2 x^{2}+5}}{x^{2}}\)	\(23\)
parallelrisch	\(\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{2 x^{2} \left (2 \,{\mathrm e}^{4 x}+1\right )}}{x^{2}}\)	\(23\)
norman	\(\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{4 x^{2} {\mathrm e}^{4 x}+2 x^{2}}}{x^{2}}\)	\(24\)

[In]

int(((144*x^3+72*x^2)*exp(5)*exp(4*x)+(36*x^2-18)*exp(5))*exp(2*x^2*exp(4*x)+x^2)^2/x^3,x,method=_RETURNVERBOS
E)

[Out]

9/x^2*exp(4*x^2*exp(4*x)+2*x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 \, e^{\left (2 \, {\left (x^{2} e^{5} + 2 \, x^{2} e^{\left (4 \, x + 5\right )}\right )} e^{\left (-5\right )} + 5\right )}}{x^{2}} \]

[In]

integrate(((144*x^3+72*x^2)*exp(5)*exp(4*x)+(36*x^2-18)*exp(5))*exp(2*x^2*exp(4*x)+x^2)^2/x^3,x, algorithm="fr
icas")

[Out]

9*e^(2*(x^2*e^5 + 2*x^2*e^(4*x + 5))*e^(-5) + 5)/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 e^{5} e^{4 x^{2} e^{4 x} + 2 x^{2}}}{x^{2}} \]

[In]

integrate(((144*x**3+72*x**2)*exp(5)*exp(4*x)+(36*x**2-18)*exp(5))*exp(2*x**2*exp(4*x)+x**2)**2/x**3,x)

[Out]

9*exp(5)*exp(4*x**2*exp(4*x) + 2*x**2)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9 \, e^{\left (4 \, x^{2} e^{\left (4 \, x\right )} + 2 \, x^{2} + 5\right )}}{x^{2}} \]

[In]

integrate(((144*x^3+72*x^2)*exp(5)*exp(4*x)+(36*x^2-18)*exp(5))*exp(2*x^2*exp(4*x)+x^2)^2/x^3,x, algorithm="ma
xima")

[Out]

9*e^(4*x^2*e^(4*x) + 2*x^2 + 5)/x^2

Giac [F]

\[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\int { \frac {18 \, {\left ({\left (2 \, x^{2} - 1\right )} e^{5} + 4 \, {\left (2 \, x^{3} + x^{2}\right )} e^{\left (4 \, x + 5\right )}\right )} e^{\left (4 \, x^{2} e^{\left (4 \, x\right )} + 2 \, x^{2}\right )}}{x^{3}} \,d x } \]

[In]

integrate(((144*x^3+72*x^2)*exp(5)*exp(4*x)+(36*x^2-18)*exp(5))*exp(2*x^2*exp(4*x)+x^2)^2/x^3,x, algorithm="gi
ac")

[Out]

integrate(18*((2*x^2 - 1)*e^5 + 4*(2*x^3 + x^2)*e^(4*x + 5))*e^(4*x^2*e^(4*x) + 2*x^2)/x^3, x)

Mupad [B] (verification not implemented)

Time = 14.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2 x^2+4 e^{4 x} x^2} \left (e^5 \left (-18+36 x^2\right )+e^{5+4 x} \left (72 x^2+144 x^3\right )\right )}{x^3} \, dx=\frac {9\,{\mathrm {e}}^5\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{4\,x}}}{x^2} \]

[In]

int((exp(4*x^2*exp(4*x) + 2*x^2)*(exp(5)*(36*x^2 - 18) + exp(4*x)*exp(5)*(72*x^2 + 144*x^3)))/x^3,x)

[Out]

(9*exp(5)*exp(2*x^2)*exp(4*x^2*exp(4*x)))/x^2