3.17.8 \(\int \frac {e^{8 x} (48 x^3+96 x^4) \log (16-24 e^4+9 e^8+e^{16 x} x^8+e^{8 x} (8 x^4-6 e^4 x^4))}{4-3 e^4+e^{8 x} x^4} \, dx\)

Optimal. Leaf size=23 \[ 3 \log ^2\left (\left (4-3 e^4+e^{8 x} x^4\right )^2\right ) \]

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Rubi [A]  time = 1.30, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {1593, 6684, 6741, 6686} \begin {gather*} 3 \log ^2\left (\left (-e^{8 x} x^4+3 e^4-4\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(8*x)*(48*x^3 + 96*x^4)*Log[16 - 24*E^4 + 9*E^8 + E^(16*x)*x^8 + E^(8*x)*(8*x^4 - 6*E^4*x^4)])/(4 - 3*E
^4 + E^(8*x)*x^4),x]

[Out]

3*Log[(-4 + 3*E^4 - E^(8*x)*x^4)^2]^2

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{8 x} x^3 (48+96 x) \log \left (16-24 e^4+9 e^8+e^{16 x} x^8+e^{8 x} \left (8 x^4-6 e^4 x^4\right )\right )}{4-3 e^4+e^{8 x} x^4} \, dx\\ &=\int \frac {e^{8 x} x^3 (48+96 x) \log \left (\left (-4 \left (1-\frac {3 e^4}{4}\right )-e^{8 x} x^4\right )^2\right )}{4 \left (1-\frac {3 e^4}{4}\right )+e^{8 x} x^4} \, dx\\ &=3 \log ^2\left (\left (-4+3 e^4-e^{8 x} x^4\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 23, normalized size = 1.00 \begin {gather*} 3 \log ^2\left (\left (4-3 e^4+e^{8 x} x^4\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(8*x)*(48*x^3 + 96*x^4)*Log[16 - 24*E^4 + 9*E^8 + E^(16*x)*x^8 + E^(8*x)*(8*x^4 - 6*E^4*x^4)])/(4
- 3*E^4 + E^(8*x)*x^4),x]

[Out]

3*Log[(4 - 3*E^4 + E^(8*x)*x^4)^2]^2

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fricas [A]  time = 0.67, size = 42, normalized size = 1.83 \begin {gather*} 3 \, \log \left (x^{8} e^{\left (16 \, x\right )} - 2 \, {\left (3 \, x^{4} e^{4} - 4 \, x^{4}\right )} e^{\left (8 \, x\right )} + 9 \, e^{8} - 24 \, e^{4} + 16\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x^4+48*x^3)*exp(2*x)^4*log(x^8*exp(2*x)^8+(-6*x^4*exp(4)+8*x^4)*exp(2*x)^4+9*exp(4)^2-24*exp(4)+
16)/(x^4*exp(2*x)^4-3*exp(4)+4),x, algorithm="fricas")

[Out]

3*log(x^8*e^(16*x) - 2*(3*x^4*e^4 - 4*x^4)*e^(8*x) + 9*e^8 - 24*e^4 + 16)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {48 \, {\left (2 \, x^{4} + x^{3}\right )} e^{\left (8 \, x\right )} \log \left (x^{8} e^{\left (16 \, x\right )} - 2 \, {\left (3 \, x^{4} e^{4} - 4 \, x^{4}\right )} e^{\left (8 \, x\right )} + 9 \, e^{8} - 24 \, e^{4} + 16\right )}{x^{4} e^{\left (8 \, x\right )} - 3 \, e^{4} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x^4+48*x^3)*exp(2*x)^4*log(x^8*exp(2*x)^8+(-6*x^4*exp(4)+8*x^4)*exp(2*x)^4+9*exp(4)^2-24*exp(4)+
16)/(x^4*exp(2*x)^4-3*exp(4)+4),x, algorithm="giac")

[Out]

integrate(48*(2*x^4 + x^3)*e^(8*x)*log(x^8*e^(16*x) - 2*(3*x^4*e^4 - 4*x^4)*e^(8*x) + 9*e^8 - 24*e^4 + 16)/(x^
4*e^(8*x) - 3*e^4 + 4), x)

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maple [B]  time = 0.43, size = 77, normalized size = 3.35

method	result	size
default	\(12 \left (\ln \left (\left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )^{2}\right )-2 \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )\right ) \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )+12 \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )^{2}\)	\(77\)
risch	\(-24 \ln \relax (3) \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )+12 \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right )^{2}-6 i \pi \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right ) \mathrm {csgn}\left (i \left (-\frac {x^{4} {\mathrm e}^{8 x}}{3}+{\mathrm e}^{4}-\frac {4}{3}\right )\right )^{2} \mathrm {csgn}\left (i \left (-\frac {x^{4} {\mathrm e}^{8 x}}{3}+{\mathrm e}^{4}-\frac {4}{3}\right )^{2}\right )+12 i \pi \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right ) \mathrm {csgn}\left (i \left (-\frac {x^{4} {\mathrm e}^{8 x}}{3}+{\mathrm e}^{4}-\frac {4}{3}\right )\right ) \mathrm {csgn}\left (i \left (-\frac {x^{4} {\mathrm e}^{8 x}}{3}+{\mathrm e}^{4}-\frac {4}{3}\right )^{2}\right )^{2}-6 i \pi \ln \left (-x^{4} {\mathrm e}^{8 x}+3 \,{\mathrm e}^{4}-4\right ) \mathrm {csgn}\left (i \left (-\frac {x^{4} {\mathrm e}^{8 x}}{3}+{\mathrm e}^{4}-\frac {4}{3}\right )^{2}\right )^{3}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((96*x^4+48*x^3)*exp(2*x)^4*ln(x^8*exp(2*x)^8+(-6*x^4*exp(4)+8*x^4)*exp(2*x)^4+9*exp(4)^2-24*exp(4)+16)/(x^
4*exp(2*x)^4-3*exp(4)+4),x,method=_RETURNVERBOSE)

[Out]

12*(ln((-x^4*exp(8*x)+3*exp(4)-4)^2)-2*ln(-x^4*exp(8*x)+3*exp(4)-4))*ln(-x^4*exp(8*x)+3*exp(4)-4)+12*ln(-x^4*e
xp(8*x)+3*exp(4)-4)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 48 \, \int \frac {{\left (2 \, x^{4} + x^{3}\right )} e^{\left (8 \, x\right )} \log \left (x^{8} e^{\left (16 \, x\right )} - 2 \, {\left (3 \, x^{4} e^{4} - 4 \, x^{4}\right )} e^{\left (8 \, x\right )} + 9 \, e^{8} - 24 \, e^{4} + 16\right )}{x^{4} e^{\left (8 \, x\right )} - 3 \, e^{4} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x^4+48*x^3)*exp(2*x)^4*log(x^8*exp(2*x)^8+(-6*x^4*exp(4)+8*x^4)*exp(2*x)^4+9*exp(4)^2-24*exp(4)+
16)/(x^4*exp(2*x)^4-3*exp(4)+4),x, algorithm="maxima")

[Out]

48*integrate((2*x^4 + x^3)*e^(8*x)*log(x^8*e^(16*x) - 2*(3*x^4*e^4 - 4*x^4)*e^(8*x) + 9*e^8 - 24*e^4 + 16)/(x^
4*e^(8*x) - 3*e^4 + 4), x)

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mupad [B]  time = 1.48, size = 42, normalized size = 1.83 \begin {gather*} 3\,{\ln \left (9\,{\mathrm {e}}^8-24\,{\mathrm {e}}^4+x^8\,{\mathrm {e}}^{16\,x}-{\mathrm {e}}^{8\,x}\,\left (6\,x^4\,{\mathrm {e}}^4-8\,x^4\right )+16\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(8*x)*log(9*exp(8) - 24*exp(4) + x^8*exp(16*x) - exp(8*x)*(6*x^4*exp(4) - 8*x^4) + 16)*(48*x^3 + 96*x^
4))/(x^4*exp(8*x) - 3*exp(4) + 4),x)

[Out]

3*log(9*exp(8) - 24*exp(4) + x^8*exp(16*x) - exp(8*x)*(6*x^4*exp(4) - 8*x^4) + 16)^2

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sympy [B]  time = 0.44, size = 42, normalized size = 1.83 \begin {gather*} 3 \log {\left (x^{8} e^{16 x} + \left (- 6 x^{4} e^{4} + 8 x^{4}\right ) e^{8 x} - 24 e^{4} + 16 + 9 e^{8} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((96*x**4+48*x**3)*exp(2*x)**4*ln(x**8*exp(2*x)**8+(-6*x**4*exp(4)+8*x**4)*exp(2*x)**4+9*exp(4)**2-24
*exp(4)+16)/(x**4*exp(2*x)**4-3*exp(4)+4),x)

[Out]

3*log(x**8*exp(16*x) + (-6*x**4*exp(4) + 8*x**4)*exp(8*x) - 24*exp(4) + 16 + 9*exp(8))**2

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