3.34.5 \(\int \frac {e^{\frac {e^3 x+e^{2 x} x^2 \log (e^4+64 x)}{e^3}} (e^7+64 e^3 x+64 e^{2 x} x^2+e^{2 x} (128 x^2+128 x^3+e^4 (2 x+2 x^2)) \log (e^4+64 x))}{e^7+64 e^3 x} \, dx\)

Optimal. Leaf size=23 \[ e^{x+e^{-3+2 x} x^2 \log \left (e^4+64 x\right )} \]

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Rubi [A]  time = 1.17, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 2288} \begin {gather*} e^x \left (64 x+e^4\right )^{e^{2 x-3} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((E^3*x + E^(2*x)*x^2*Log[E^4 + 64*x])/E^3)*(E^7 + 64*E^3*x + 64*E^(2*x)*x^2 + E^(2*x)*(128*x^2 + 128*x
^3 + E^4*(2*x + 2*x^2))*Log[E^4 + 64*x]))/(E^7 + 64*E^3*x),x]

[Out]

E^x*(E^4 + 64*x)^(E^(-3 + 2*x)*x^2)

Rule 2288

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{-3+x} \left (e^4+64 x\right )^{-1+e^{-3+2 x} x^2} \left (e^7+64 e^3 x+64 e^{2 x} x^2+2 e^{2 x} x (1+x) \left (e^4+64 x\right ) \log \left (e^4+64 x\right )\right ) \, dx\\ &=e^x \left (e^4+64 x\right )^{e^{-3+2 x} x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.00 \begin {gather*} e^x \left (e^4+64 x\right )^{e^{-3+2 x} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^3*x + E^(2*x)*x^2*Log[E^4 + 64*x])/E^3)*(E^7 + 64*E^3*x + 64*E^(2*x)*x^2 + E^(2*x)*(128*x^2 +
128*x^3 + E^4*(2*x + 2*x^2))*Log[E^4 + 64*x]))/(E^7 + 64*E^3*x),x]

[Out]

E^x*(E^4 + 64*x)^(E^(-3 + 2*x)*x^2)

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fricas [A]  time = 0.56, size = 24, normalized size = 1.04 \begin {gather*} e^{\left ({\left (x^{2} e^{\left (2 \, x\right )} \log \left (64 \, x + e^{4}\right ) + x e^{3}\right )} e^{\left (-3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*exp(4)+128*x^3+128*x^2)*exp(2*x)*log(exp(4)+64*x)+64*exp(2*x)*x^2+exp(3)*exp(4)+64*x*e
xp(3))*exp((x^2*exp(2*x)*log(exp(4)+64*x)+x*exp(3))/exp(3))/(exp(3)*exp(4)+64*x*exp(3)),x, algorithm="fricas")

[Out]

e^((x^2*e^(2*x)*log(64*x + e^4) + x*e^3)*e^(-3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (64 \, x^{2} e^{\left (2 \, x\right )} + 2 \, {\left (64 \, x^{3} + 64 \, x^{2} + {\left (x^{2} + x\right )} e^{4}\right )} e^{\left (2 \, x\right )} \log \left (64 \, x + e^{4}\right ) + 64 \, x e^{3} + e^{7}\right )} e^{\left ({\left (x^{2} e^{\left (2 \, x\right )} \log \left (64 \, x + e^{4}\right ) + x e^{3}\right )} e^{\left (-3\right )}\right )}}{64 \, x e^{3} + e^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*exp(4)+128*x^3+128*x^2)*exp(2*x)*log(exp(4)+64*x)+64*exp(2*x)*x^2+exp(3)*exp(4)+64*x*e
xp(3))*exp((x^2*exp(2*x)*log(exp(4)+64*x)+x*exp(3))/exp(3))/(exp(3)*exp(4)+64*x*exp(3)),x, algorithm="giac")

[Out]

integrate((64*x^2*e^(2*x) + 2*(64*x^3 + 64*x^2 + (x^2 + x)*e^4)*e^(2*x)*log(64*x + e^4) + 64*x*e^3 + e^7)*e^((
x^2*e^(2*x)*log(64*x + e^4) + x*e^3)*e^(-3))/(64*x*e^3 + e^7), x)

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maple [A]  time = 0.49, size = 21, normalized size = 0.91

method	result	size
risch	\(\left ({\mathrm e}^{4}+64 x \right )^{x^{2} {\mathrm e}^{2 x -3}} {\mathrm e}^{x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+2*x)*exp(4)+128*x^3+128*x^2)*exp(2*x)*ln(exp(4)+64*x)+64*exp(2*x)*x^2+exp(3)*exp(4)+64*x*exp(3))*
exp((x^2*exp(2*x)*ln(exp(4)+64*x)+x*exp(3))/exp(3))/(exp(3)*exp(4)+64*x*exp(3)),x,method=_RETURNVERBOSE)

[Out]

(exp(4)+64*x)^(x^2*exp(2*x-3))*exp(x)

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maxima [A]  time = 0.62, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (x^{2} e^{\left (2 \, x - 3\right )} \log \left (64 \, x + e^{4}\right ) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*exp(4)+128*x^3+128*x^2)*exp(2*x)*log(exp(4)+64*x)+64*exp(2*x)*x^2+exp(3)*exp(4)+64*x*e
xp(3))*exp((x^2*exp(2*x)*log(exp(4)+64*x)+x*exp(3))/exp(3))/(exp(3)*exp(4)+64*x*exp(3)),x, algorithm="maxima")

[Out]

e^(x^2*e^(2*x - 3)*log(64*x + e^4) + x)

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mupad [B]  time = 0.42, size = 20, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^x\,{\left (64\,x+{\mathrm {e}}^4\right )}^{x^2\,{\mathrm {e}}^{2\,x-3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(-3)*(x*exp(3) + x^2*log(64*x + exp(4))*exp(2*x)))*(exp(7) + 64*x*exp(3) + 64*x^2*exp(2*x) + log(6
4*x + exp(4))*exp(2*x)*(exp(4)*(2*x + 2*x^2) + 128*x^2 + 128*x^3)))/(exp(7) + 64*x*exp(3)),x)

[Out]

exp(x)*(64*x + exp(4))^(x^2*exp(2*x - 3))

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sympy [A]  time = 1.04, size = 26, normalized size = 1.13 \begin {gather*} e^{\frac {x^{2} e^{2 x} \log {\left (64 x + e^{4} \right )} + x e^{3}}{e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+2*x)*exp(4)+128*x**3+128*x**2)*exp(2*x)*ln(exp(4)+64*x)+64*exp(2*x)*x**2+exp(3)*exp(4)+64*
x*exp(3))*exp((x**2*exp(2*x)*ln(exp(4)+64*x)+x*exp(3))/exp(3))/(exp(3)*exp(4)+64*x*exp(3)),x)

[Out]

exp((x**2*exp(2*x)*log(64*x + exp(4)) + x*exp(3))*exp(-3))

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