3.66.55 \(\int \frac {-e^3 x+(64+32 x+4 x^2) \log (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}})}{(160+80 x+10 x^2) \log ^2(\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}})} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 x}{5 \log \left (\frac {1}{4} e^{9+\frac {e^3 x}{16 (4+x)}}\right )} \]

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Rubi [F]  time = 1.39, 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 {gather*} \int \frac {-e^3 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )}{\left (160+80 x+10 x^2\right ) \log ^2\left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^3*x) + (64 + 32*x + 4*x^2)*Log[E^((576 + 144*x + E^3*x)/(64 + 16*x))/4])/((160 + 80*x + 10*x^2)*Log[E
^((576 + 144*x + E^3*x)/(64 + 16*x))/4]^2),x]

[Out]

-8/(5*Log[E^((576 + (144 + E^3)*x)/(16*(4 + x)))/4]) - (E^3*Defer[Int][1/((4 + x)*Log[E^((576 + (144 + E^3)*x)
/(16*(4 + x)))/4]^2), x])/10 + (2*Defer[Int][Log[E^((576 + (144 + E^3)*x)/(16*(4 + x)))/4]^(-1), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^3 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )}{10 (4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )} \, dx\\ &=\frac {1}{10} \int \frac {-e^3 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )} \, dx\\ &=\frac {1}{10} \int \frac {-e^3 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{4} e^{\frac {576+144 x+e^3 x}{64+16 x}}\right )}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{64+16 x}}\right )} \, dx\\ &=\frac {1}{10} \int \left (-\frac {e^3 x}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {64}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {32 x}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {4 x^2}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}\right ) \, dx\\ &=\frac {2}{5} \int \frac {x^2}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx+\frac {16}{5} \int \frac {x}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx+\frac {32}{5} \int \frac {1}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx-\frac {1}{10} e^3 \int \frac {x}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx\\ &=\frac {128 \log \left (\log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )\right )}{5 e^3}+\frac {2}{5} \int \left (\frac {1}{\log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {16}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}-\frac {8}{(4+x) \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}\right ) \, dx+\frac {16}{5} \int \left (-\frac {4}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {1}{(4+x) \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}\right ) \, dx-\frac {1}{10} e^3 \int \left (-\frac {4}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {1}{(4+x) \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}\right ) \, dx\\ &=\frac {128 \log \left (\log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )\right )}{5 e^3}+\frac {2}{5} \int \frac {1}{\log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx+\frac {32}{5} \int \frac {1}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx-\frac {64}{5} \int \frac {1}{(4+x)^2 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx-\frac {1}{10} e^3 \int \frac {1}{(4+x) \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx+\frac {1}{5} \left (2 e^3\right ) \int \frac {1}{(4+x)^2 \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx\\ &=-\frac {8}{5 \log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )}+\frac {2}{5} \int \frac {1}{\log \left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx-\frac {1}{10} e^3 \int \frac {1}{(4+x) \log ^2\left (\frac {1}{4} e^{\frac {576+\left (144+e^3\right ) x}{16 (4+x)}}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.24, size = 127, normalized size = 4.38 \begin {gather*} \frac {2 \left (e^6 x+4 e^3 \left (-16+x^2\right ) \log \left (\frac {1}{4} e^{9+\frac {e^3 x}{64+16 x}}\right )+16 x (4+x)^2 \log ^2\left (\frac {1}{4} e^{9+\frac {e^3 x}{64+16 x}}\right )\right )}{5 \log \left (\frac {1}{4} e^{9+\frac {e^3 x}{64+16 x}}\right ) \left (e^3+4 (4+x) \log \left (\frac {1}{4} e^{9+\frac {e^3 x}{64+16 x}}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^3*x) + (64 + 32*x + 4*x^2)*Log[E^((576 + 144*x + E^3*x)/(64 + 16*x))/4])/((160 + 80*x + 10*x^2)
*Log[E^((576 + 144*x + E^3*x)/(64 + 16*x))/4]^2),x]

[Out]

(2*(E^6*x + 4*E^3*(-16 + x^2)*Log[E^(9 + (E^3*x)/(64 + 16*x))/4] + 16*x*(4 + x)^2*Log[E^(9 + (E^3*x)/(64 + 16*
x))/4]^2))/(5*Log[E^(9 + (E^3*x)/(64 + 16*x))/4]*(E^3 + 4*(4 + x)*Log[E^(9 + (E^3*x)/(64 + 16*x))/4])^2)

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fricas [B]  time = 0.60, size = 150, normalized size = 5.17 \begin {gather*} -\frac {32 \, {\left (x^{2} e^{6} + 1024 \, {\left (x^{2} + 4 \, x\right )} \log \relax (2)^{2} + 20736 \, x^{2} + 288 \, {\left (x^{2} + 2 \, x - 8\right )} e^{3} - 64 \, {\left (144 \, x^{2} + {\left (x^{2} + 2 \, x - 8\right )} e^{3} + 576 \, x\right )} \log \relax (2) + 82944 \, x\right )}}{5 \, {\left (32768 \, {\left (x + 4\right )} \log \relax (2)^{3} - 1024 \, {\left ({\left (3 \, x + 8\right )} e^{3} + 432 \, x + 1728\right )} \log \relax (2)^{2} - x e^{9} - 144 \, {\left (3 \, x + 4\right )} e^{6} - 20736 \, {\left (3 \, x + 8\right )} e^{3} + 32 \, {\left ({\left (3 \, x + 4\right )} e^{6} + 288 \, {\left (3 \, x + 8\right )} e^{3} + 62208 \, x + 248832\right )} \log \relax (2) - 2985984 \, x - 11943936\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+32*x+64)*log(1/4*exp((x*exp(3)+144*x+576)/(16*x+64)))-x*exp(3))/(10*x^2+80*x+160)/log(1/4*ex
p((x*exp(3)+144*x+576)/(16*x+64)))^2,x, algorithm="fricas")

[Out]

-32/5*(x^2*e^6 + 1024*(x^2 + 4*x)*log(2)^2 + 20736*x^2 + 288*(x^2 + 2*x - 8)*e^3 - 64*(144*x^2 + (x^2 + 2*x -
8)*e^3 + 576*x)*log(2) + 82944*x)/(32768*(x + 4)*log(2)^3 - 1024*((3*x + 8)*e^3 + 432*x + 1728)*log(2)^2 - x*e
^9 - 144*(3*x + 4)*e^6 - 20736*(3*x + 8)*e^3 + 32*((3*x + 4)*e^6 + 288*(3*x + 8)*e^3 + 62208*x + 248832)*log(2
) - 2985984*x - 11943936)

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giac [B]  time = 0.17, size = 105, normalized size = 3.62 \begin {gather*} -\frac {32 \, {\left (x e^{3} - 32 \, x \log \relax (2) + 144 \, x\right )}}{5 \, {\left (64 \, e^{3} \log \relax (2) - 1024 \, \log \relax (2)^{2} - e^{6} - 288 \, e^{3} + 9216 \, \log \relax (2) - 20736\right )}} - \frac {8192 \, {\left (2 \, e^{3} \log \relax (2) - 9 \, e^{3}\right )}}{5 \, {\left (x e^{3} - 32 \, x \log \relax (2) + 144 \, x - 128 \, \log \relax (2) + 576\right )} {\left (64 \, e^{3} \log \relax (2) - 1024 \, \log \relax (2)^{2} - e^{6} - 288 \, e^{3} + 9216 \, \log \relax (2) - 20736\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+32*x+64)*log(1/4*exp((x*exp(3)+144*x+576)/(16*x+64)))-x*exp(3))/(10*x^2+80*x+160)/log(1/4*ex
p((x*exp(3)+144*x+576)/(16*x+64)))^2,x, algorithm="giac")

[Out]

-32/5*(x*e^3 - 32*x*log(2) + 144*x)/(64*e^3*log(2) - 1024*log(2)^2 - e^6 - 288*e^3 + 9216*log(2) - 20736) - 81
92/5*(2*e^3*log(2) - 9*e^3)/((x*e^3 - 32*x*log(2) + 144*x - 128*log(2) + 576)*(64*e^3*log(2) - 1024*log(2)^2 -
 e^6 - 288*e^3 + 9216*log(2) - 20736))

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maple [C]  time = 0.58, size = 34, normalized size = 1.17




method result size



risch \(-\frac {4 i x}{5 \left (4 i \ln \relax (2)-2 i \ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}\right )\right )}\) \(34\)
norman \(\frac {\frac {8}{5} x +\frac {2}{5} x^{2}}{\left (4+x \right ) \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )}\) \(39\)
default \(\frac {32 x}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \left (x \,{\mathrm e}^{3}+144 x +576\right )}{16 x +64}+144\right )}+\frac {128 \,{\mathrm e}^{3} \ln \left (x \,{\mathrm e}^{3}+16 \left (\ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}\right ) x +144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \left (x \,{\mathrm e}^{3}+144 x +576\right )}{16 x +64}+576\right )}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \left (x \,{\mathrm e}^{3}+144 x +576\right )}{16 x +64}+144\right )^{2}}-\frac {128 \,{\mathrm e}^{3} \ln \left (x \,{\mathrm e}^{3}+16 \left (\ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}\right ) x +144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \left (x \,{\mathrm e}^{3}+144 x +576\right )}{16 x +64}+576\right )}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2}}-\frac {8192 \,{\mathrm e}^{3} \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2} \left (x \,{\mathrm e}^{3}+16 x \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 x^{2} {\mathrm e}^{3}}{16 x +64}-\frac {2304 x^{2}}{16 x +64}-\frac {18432 x}{16 x +64}+144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \,{\mathrm e}^{3} x}{16 x +64}-\frac {36864}{16 x +64}+576\right )}+\frac {8192 x \,{\mathrm e}^{6}}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2} \left (x \,{\mathrm e}^{3}+16 x \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 x^{2} {\mathrm e}^{3}}{16 x +64}-\frac {2304 x^{2}}{16 x +64}-\frac {18432 x}{16 x +64}+144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \,{\mathrm e}^{3} x}{16 x +64}-\frac {36864}{16 x +64}+576\right ) \left (16 x +64\right )}+\frac {1179648 x \,{\mathrm e}^{3}}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2} \left (x \,{\mathrm e}^{3}+16 x \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 x^{2} {\mathrm e}^{3}}{16 x +64}-\frac {2304 x^{2}}{16 x +64}-\frac {18432 x}{16 x +64}+144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \,{\mathrm e}^{3} x}{16 x +64}-\frac {36864}{16 x +64}+576\right ) \left (16 x +64\right )}+\frac {4718592 \,{\mathrm e}^{3}}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2} \left (x \,{\mathrm e}^{3}+16 x \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 x^{2} {\mathrm e}^{3}}{16 x +64}-\frac {2304 x^{2}}{16 x +64}-\frac {18432 x}{16 x +64}+144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \,{\mathrm e}^{3} x}{16 x +64}-\frac {36864}{16 x +64}+576\right ) \left (16 x +64\right )}-\frac {73728 \,{\mathrm e}^{3}}{5 \left ({\mathrm e}^{3}+16 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 \,{\mathrm e}^{3} x}{16 x +64}-\frac {2304 x}{16 x +64}-\frac {9216}{16 x +64}+144\right )^{2} \left (x \,{\mathrm e}^{3}+16 x \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {16 x^{2} {\mathrm e}^{3}}{16 x +64}-\frac {2304 x^{2}}{16 x +64}-\frac {18432 x}{16 x +64}+144 x +64 \ln \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{3}+144 x +576}{16 x +64}}}{4}\right )-\frac {64 \,{\mathrm e}^{3} x}{16 x +64}-\frac {36864}{16 x +64}+576\right )}\) \(1296\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+32*x+64)*ln(1/4*exp((x*exp(3)+144*x+576)/(16*x+64)))-x*exp(3))/(10*x^2+80*x+160)/ln(1/4*exp((x*exp
(3)+144*x+576)/(16*x+64)))^2,x,method=_RETURNVERBOSE)

[Out]

-4/5*I*x/(4*I*ln(2)-2*I*ln(exp(1/16*(x*exp(3)+144*x+576)/(4+x))))

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maxima [B]  time = 0.57, size = 1184, normalized size = 40.83 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+32*x+64)*log(1/4*exp((x*exp(3)+144*x+576)/(16*x+64)))-x*exp(3))/(10*x^2+80*x+160)/log(1/4*ex
p((x*exp(3)+144*x+576)/(16*x+64)))^2,x, algorithm="maxima")

[Out]

-128/5*(64*(2*log(2) - 9)/(131072*log(2)^3 + (32768*log(2)^3 + 48*(2*log(2) - 9)*e^6 - 768*(4*log(2)^2 - 36*lo
g(2) + 81)*e^3 - 442368*log(2)^2 - e^9 + 1990656*log(2) - 2985984)*x + 64*(2*log(2) - 9)*e^6 - 2048*(4*log(2)^
2 - 36*log(2) + 81)*e^3 - 1769472*log(2)^2 + 7962624*log(2) - 11943936) - log(x*(e^3 - 32*log(2) + 144) - 128*
log(2) + 576)/(32*(2*log(2) - 9)*e^3 - 1024*log(2)^2 - e^6 + 9216*log(2) - 20736))*e^3 + 128/5*e^(-3)*log(-x*(
e^3 - 32*log(2) + 144) + 128*log(2) - 576) - 512/5*(128*(2*log(2) - 9)*log(x*(e^3 - 32*log(2) + 144) - 128*log
(2) + 576)/(32768*log(2)^3 + 48*(2*log(2) - 9)*e^6 - 768*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 442368*log(2)^2 -
 e^9 + 1990656*log(2) - 2985984) + ((32*(2*log(2) - 9)*e^3 - 1024*log(2)^2 - e^6 + 9216*log(2) - 20736)*x^2 +
64*((2*log(2) - 9)*e^3 - 64*log(2)^2 + 576*log(2) - 1296)*x + 16384*log(2)^2 - 147456*log(2) + 331776)/(419430
4*log(2)^4 - 75497472*log(2)^3 + (1048576*log(2)^4 - 18874368*log(2)^3 - 64*(2*log(2) - 9)*e^9 + 1536*(4*log(2
)^2 - 36*log(2) + 81)*e^6 - 16384*(8*log(2)^3 - 108*log(2)^2 + 486*log(2) - 729)*e^3 + 127401984*log(2)^2 + e^
12 - 382205952*log(2) + 429981696)*x - 64*(2*log(2) - 9)*e^9 + 3072*(4*log(2)^2 - 36*log(2) + 81)*e^6 - 49152*
(8*log(2)^3 - 108*log(2)^2 + 486*log(2) - 729)*e^3 + 509607936*log(2)^2 - 1528823808*log(2) + 1719926784))*log
(1/4*e^(1/16*x*e^3/(x + 4) + 9*x/(x + 4) + 36/(x + 4))) + 4096/5*(64*(2*log(2) - 9)/(131072*log(2)^3 + (32768*
log(2)^3 + 48*(2*log(2) - 9)*e^6 - 768*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 442368*log(2)^2 - e^9 + 1990656*log
(2) - 2985984)*x + 64*(2*log(2) - 9)*e^6 - 2048*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 1769472*log(2)^2 + 7962624
*log(2) - 11943936) - log(x*(e^3 - 32*log(2) + 144) - 128*log(2) + 576)/(32*(2*log(2) - 9)*e^3 - 1024*log(2)^2
 - e^6 + 9216*log(2) - 20736))*log(1/4*e^(1/16*x*e^3/(x + 4) + 9*x/(x + 4) + 36/(x + 4))) - 512/5*(16*(2*log(2
) - 9)*e^6 + 256*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 8*(4096*log(2)^3 + (1024*log(2)^3 + (2*log(2) - 9)*e^6 -
24*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 13824*log(2)^2 + 62208*log(2) - 93312)*x - 96*(4*log(2)^2 - 36*log(2) +
 81)*e^3 - 55296*log(2)^2 + 248832*log(2) - 373248)*log(x*(e^3 - 32*log(2) + 144) - 128*log(2) + 576) - e^9)/(
(48*(2*log(2) - 9)*e^9 - 768*(4*log(2)^2 - 36*log(2) + 81)*e^6 + 4096*(8*log(2)^3 - 108*log(2)^2 + 486*log(2)
- 729)*e^3 - e^12)*x + 192*(2*log(2) - 9)*e^9 - 3072*(4*log(2)^2 - 36*log(2) + 81)*e^6 + 16384*(8*log(2)^3 - 1
08*log(2)^2 + 486*log(2) - 729)*e^3 - 4*e^12) - 256/5*(64*(2*log(2) - 9)*e^3 + ((32*(2*log(2) - 9)*e^3 - 1024*
log(2)^2 - e^6 + 9216*log(2) - 20736)*x + 128*(2*log(2) - 9)*e^3 - 4096*log(2)^2 + 36864*log(2) - 82944)*log(x
*(e^3 - 32*log(2) + 144) - 128*log(2) + 576))/((32*(2*log(2) - 9)*e^6 - 256*(4*log(2)^2 - 36*log(2) + 81)*e^3
- e^9)*x + 128*(2*log(2) - 9)*e^6 - 1024*(4*log(2)^2 - 36*log(2) + 81)*e^3 - 4*e^9) + 8192/5*log(1/4*e^(1/16*x
*e^3/(x + 4) + 9*x/(x + 4) + 36/(x + 4)))/((32*(2*log(2) - 9)*e^3 - 1024*log(2)^2 - e^6 + 9216*log(2) - 20736)
*x + 64*(2*log(2) - 9)*e^3 - 4096*log(2)^2 + 36864*log(2) - 82944) + 512/5/(x*(e^3 - 32*log(2) + 144) + 4*e^3
- 128*log(2) + 576)

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mupad [B]  time = 4.82, size = 113, normalized size = 3.90 \begin {gather*} \frac {32\,\left (82944\,x-2304\,{\mathrm {e}}^3+1024\,x^2\,{\ln \relax (2)}^2+512\,{\mathrm {e}}^3\,\ln \relax (2)+576\,x\,{\mathrm {e}}^3-36864\,x\,\ln \relax (2)+288\,x^2\,{\mathrm {e}}^3+x^2\,{\mathrm {e}}^6+4096\,x\,{\ln \relax (2)}^2-9216\,x^2\,\ln \relax (2)+20736\,x^2-64\,x^2\,{\mathrm {e}}^3\,\ln \relax (2)-128\,x\,{\mathrm {e}}^3\,\ln \relax (2)\right )}{5\,{\left ({\mathrm {e}}^3-32\,\ln \relax (2)+144\right )}^2\,\left (144\,x-128\,\ln \relax (2)+x\,{\mathrm {e}}^3-32\,x\,\ln \relax (2)+576\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*exp(3) - log(exp((144*x + x*exp(3) + 576)/(16*x + 64))/4)*(32*x + 4*x^2 + 64))/(log(exp((144*x + x*exp
(3) + 576)/(16*x + 64))/4)^2*(80*x + 10*x^2 + 160)),x)

[Out]

(32*(82944*x - 2304*exp(3) + 1024*x^2*log(2)^2 + 512*exp(3)*log(2) + 576*x*exp(3) - 36864*x*log(2) + 288*x^2*e
xp(3) + x^2*exp(6) + 4096*x*log(2)^2 - 9216*x^2*log(2) + 20736*x^2 - 64*x^2*exp(3)*log(2) - 128*x*exp(3)*log(2
)))/(5*(exp(3) - 32*log(2) + 144)^2*(144*x - 128*log(2) + x*exp(3) - 32*x*log(2) + 576))

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sympy [B]  time = 1.11, size = 150, normalized size = 5.17 \begin {gather*} \frac {32 x}{- 160 \log {\relax (2 )} + 5 e^{3} + 720} + \frac {- 73728 e^{3} + 16384 e^{3} \log {\relax (2 )}}{x \left (- 9953280 \log {\relax (2 )} - 138240 e^{3} \log {\relax (2 )} - 480 e^{6} \log {\relax (2 )} - 163840 \log {\relax (2 )}^{3} + 5 e^{9} + 15360 e^{3} \log {\relax (2 )}^{2} + 2160 e^{6} + 2211840 \log {\relax (2 )}^{2} + 311040 e^{3} + 14929920\right ) - 39813120 \log {\relax (2 )} - 368640 e^{3} \log {\relax (2 )} - 655360 \log {\relax (2 )}^{3} - 640 e^{6} \log {\relax (2 )} + 40960 e^{3} \log {\relax (2 )}^{2} + 2880 e^{6} + 8847360 \log {\relax (2 )}^{2} + 829440 e^{3} + 59719680} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+32*x+64)*ln(1/4*exp((x*exp(3)+144*x+576)/(16*x+64)))-x*exp(3))/(10*x**2+80*x+160)/ln(1/4*ex
p((x*exp(3)+144*x+576)/(16*x+64)))**2,x)

[Out]

32*x/(-160*log(2) + 5*exp(3) + 720) + (-73728*exp(3) + 16384*exp(3)*log(2))/(x*(-9953280*log(2) - 138240*exp(3
)*log(2) - 480*exp(6)*log(2) - 163840*log(2)**3 + 5*exp(9) + 15360*exp(3)*log(2)**2 + 2160*exp(6) + 2211840*lo
g(2)**2 + 311040*exp(3) + 14929920) - 39813120*log(2) - 368640*exp(3)*log(2) - 655360*log(2)**3 - 640*exp(6)*l
og(2) + 40960*exp(3)*log(2)**2 + 2880*exp(6) + 8847360*log(2)**2 + 829440*exp(3) + 59719680)

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