3.49.87 \(\int \frac {-32 x-4 e^x x-4 x^2+(16+12 x+2 x^2+e^x (2+3 x+x^2)) \log (16+16 x+4 x^2)+e^{x^2} (4 x+2 x^2) \log ^3(16+16 x+4 x^2)}{(2+x) \log ^3(16+16 x+4 x^2)} \, dx\)

Optimal. Leaf size=25 \[ 2+e^{x^2}+\frac {x \left (8+e^x+x\right )}{\log ^2\left ((4+2 x)^2\right )} \]

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Rubi [F]  time = 1.46, 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 {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32*x - 4*E^x*x - 4*x^2 + (16 + 12*x + 2*x^2 + E^x*(2 + 3*x + x^2))*Log[16 + 16*x + 4*x^2] + E^x^2*(4*x +
2*x^2)*Log[16 + 16*x + 4*x^2]^3)/((2 + x)*Log[16 + 16*x + 4*x^2]^3),x]

[Out]

E^x^2 - 12/Log[4*(2 + x)^2]^2 + (6*(2 + x))/Log[4*(2 + x)^2]^2 + (x*(2 + x))/Log[4*(2 + x)^2]^2 + (4*(2 + x))/
Log[4*(2 + x)^2] + (x*(2 + x))/Log[4*(2 + x)^2] - ((2 + x)*(4 + x))/Log[4*(2 + x)^2] - 4*Defer[Int][E^x/Log[4*
(2 + x)^2]^3, x] + 8*Defer[Int][E^x/((2 + x)*Log[4*(2 + x)^2]^3), x] - Defer[Int][E^x/Log[4*(2 + x)^2]^2, x] +
Defer[Int][(E^x*(2 + x))/Log[4*(2 + x)^2]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{x^2} x-\frac {4 x \left (8+e^x+x\right )}{(2+x) \log ^3\left (4 (2+x)^2\right )}+\frac {e^x (1+x)+2 (4+x)}{\log ^2\left (4 (2+x)^2\right )}\right ) \, dx\\ &=2 \int e^{x^2} x \, dx-4 \int \frac {x \left (8+e^x+x\right )}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (1+x)+2 (4+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-4 \int \left (\frac {e^x x}{(2+x) \log ^3\left (4 (2+x)^2\right )}+\frac {x (8+x)}{(2+x) \log ^3\left (4 (2+x)^2\right )}\right ) \, dx+\int \left (\frac {e^x (1+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {2 (4+x)}{\log ^2\left (4 (2+x)^2\right )}\right ) \, dx\\ &=e^{x^2}+2 \int \frac {4+x}{\log ^2\left (4 (2+x)^2\right )} \, dx-4 \int \frac {e^x x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-4 \int \frac {x (8+x)}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (1+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}-2 \int \frac {1}{\log \left (4 (2+x)^2\right )} \, dx+2 \int \frac {4+x}{\log \left (4 (2+x)^2\right )} \, dx-4 \int \left (\frac {6}{\log ^3\left (4 (2+x)^2\right )}+\frac {x}{\log ^3\left (4 (2+x)^2\right )}-\frac {12}{(2+x) \log ^3\left (4 (2+x)^2\right )}\right ) \, dx-4 \int \left (\frac {e^x}{\log ^3\left (4 (2+x)^2\right )}-\frac {2 e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )}\right ) \, dx+\int \left (-\frac {e^x}{\log ^2\left (4 (2+x)^2\right )}+\frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )}\right ) \, dx\\ &=e^{x^2}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}+2 \int \left (\frac {2}{\log \left (4 (2+x)^2\right )}+\frac {2+x}{\log \left (4 (2+x)^2\right )}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx-4 \int \frac {x}{\log ^3\left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-24 \int \frac {1}{\log ^3\left (4 (2+x)^2\right )} \, dx+48 \int \frac {1}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}-2 \int \frac {1}{\log ^2\left (4 (2+x)^2\right )} \, dx-2 \int \frac {x}{\log ^2\left (4 (2+x)^2\right )} \, dx+2 \int \frac {2+x}{\log \left (4 (2+x)^2\right )} \, dx-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+4 \int \frac {1}{\log \left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-24 \operatorname {Subst}\left (\int \frac {1}{\log ^3\left (4 x^2\right )} \, dx,x,2+x\right )+48 \operatorname {Subst}\left (\int \frac {1}{x \log ^3\left (4 x^2\right )} \, dx,x,2+x\right )-\frac {(2+x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{2 \sqrt {(2+x)^2}}-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-\frac {(2+x) \text {Ei}\left (\frac {1}{2} \log \left (4 (2+x)^2\right )\right )}{2 \sqrt {(2+x)^2}}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}-2 \int \frac {1}{\log \left (4 (2+x)^2\right )} \, dx-2 \int \frac {x}{\log \left (4 (2+x)^2\right )} \, dx-2 \operatorname {Subst}\left (\int \frac {1}{\log ^2\left (4 x^2\right )} \, dx,x,2+x\right )+2 \operatorname {Subst}\left (\int \frac {x}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+4 \operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-6 \operatorname {Subst}\left (\int \frac {1}{\log ^2\left (4 x^2\right )} \, dx,x,2+x\right )+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx+24 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (4 (2+x)^2\right )\right )-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-\frac {(2+x) \text {Ei}\left (\frac {1}{2} \log \left (4 (2+x)^2\right )\right )}{2 \sqrt {(2+x)^2}}-\frac {12}{\log ^2\left (4 (2+x)^2\right )}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {4 (2+x)}{\log \left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}-2 \int \left (-\frac {2}{\log \left (4 (2+x)^2\right )}+\frac {2+x}{\log \left (4 (2+x)^2\right )}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-3 \operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx+\frac {(2+x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{\sqrt {(2+x)^2}}-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx+\operatorname {Subst}\left (\int \frac {1}{\log (4 x)} \, dx,x,(2+x)^2\right )-\operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )\\ &=e^{x^2}+\frac {(2+x) \text {Ei}\left (\frac {1}{2} \log \left (4 (2+x)^2\right )\right )}{2 \sqrt {(2+x)^2}}-\frac {12}{\log ^2\left (4 (2+x)^2\right )}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {4 (2+x)}{\log \left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}+\frac {1}{4} \text {li}\left (4 (2+x)^2\right )-2 \int \frac {2+x}{\log \left (4 (2+x)^2\right )} \, dx-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+4 \int \frac {1}{\log \left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-\frac {(2+x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{4 \sqrt {(2+x)^2}}-\frac {(2+x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{2 \sqrt {(2+x)^2}}-\frac {(3 (2+x)) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{4 \sqrt {(2+x)^2}}-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-\frac {(2+x) \text {Ei}\left (\frac {1}{2} \log \left (4 (2+x)^2\right )\right )}{\sqrt {(2+x)^2}}-\frac {12}{\log ^2\left (4 (2+x)^2\right )}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {4 (2+x)}{\log \left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}+\frac {1}{4} \text {li}\left (4 (2+x)^2\right )-2 \operatorname {Subst}\left (\int \frac {x}{\log \left (4 x^2\right )} \, dx,x,2+x\right )-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+4 \operatorname {Subst}\left (\int \frac {1}{\log \left (4 x^2\right )} \, dx,x,2+x\right )+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ &=e^{x^2}-\frac {(2+x) \text {Ei}\left (\frac {1}{2} \log \left (4 (2+x)^2\right )\right )}{\sqrt {(2+x)^2}}-\frac {12}{\log ^2\left (4 (2+x)^2\right )}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {4 (2+x)}{\log \left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}+\frac {1}{4} \text {li}\left (4 (2+x)^2\right )-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx+\frac {(2+x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 (2+x)^2\right )\right )}{\sqrt {(2+x)^2}}-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx-\operatorname {Subst}\left (\int \frac {1}{\log (4 x)} \, dx,x,(2+x)^2\right )\\ &=e^{x^2}-\frac {12}{\log ^2\left (4 (2+x)^2\right )}+\frac {6 (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log ^2\left (4 (2+x)^2\right )}+\frac {4 (2+x)}{\log \left (4 (2+x)^2\right )}+\frac {x (2+x)}{\log \left (4 (2+x)^2\right )}-\frac {(2+x) (4+x)}{\log \left (4 (2+x)^2\right )}-4 \int \frac {e^x}{\log ^3\left (4 (2+x)^2\right )} \, dx+8 \int \frac {e^x}{(2+x) \log ^3\left (4 (2+x)^2\right )} \, dx-\int \frac {e^x}{\log ^2\left (4 (2+x)^2\right )} \, dx+\int \frac {e^x (2+x)}{\log ^2\left (4 (2+x)^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 24, normalized size = 0.96 \begin {gather*} e^{x^2}+\frac {x \left (8+e^x+x\right )}{\log ^2\left (4 (2+x)^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x - 4*E^x*x - 4*x^2 + (16 + 12*x + 2*x^2 + E^x*(2 + 3*x + x^2))*Log[16 + 16*x + 4*x^2] + E^x^2*
(4*x + 2*x^2)*Log[16 + 16*x + 4*x^2]^3)/((2 + x)*Log[16 + 16*x + 4*x^2]^3),x]

[Out]

E^x^2 + (x*(8 + E^x + x))/Log[4*(2 + x)^2]^2

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fricas [A]  time = 0.98, size = 43, normalized size = 1.72 \begin {gather*} \frac {e^{\left (x^{2}\right )} \log \left (4 \, x^{2} + 16 \, x + 16\right )^{2} + x^{2} + x e^{x} + 8 \, x}{\log \left (4 \, x^{2} + 16 \, x + 16\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+12*x+16)*log(4*x^2+16*x+16)-4*e
xp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16*x+16)^3,x, algorithm="fricas")

[Out]

(e^(x^2)*log(4*x^2 + 16*x + 16)^2 + x^2 + x*e^x + 8*x)/log(4*x^2 + 16*x + 16)^2

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giac [A]  time = 0.35, size = 43, normalized size = 1.72 \begin {gather*} \frac {e^{\left (x^{2}\right )} \log \left (4 \, x^{2} + 16 \, x + 16\right )^{2} + x^{2} + x e^{x} + 8 \, x}{\log \left (4 \, x^{2} + 16 \, x + 16\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+12*x+16)*log(4*x^2+16*x+16)-4*e
xp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16*x+16)^3,x, algorithm="giac")

[Out]

(e^(x^2)*log(4*x^2 + 16*x + 16)^2 + x^2 + x*e^x + 8*x)/log(4*x^2 + 16*x + 16)^2

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maple [C]  time = 0.88, size = 83, normalized size = 3.32

method	result	size
risch	\({\mathrm e}^{x^{2}}-\frac {4 x \left (x +{\mathrm e}^{x}+8\right )}{\left (\pi \mathrm {csgn}\left (i \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \left (2+x \right )\right )^{2}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+4*x)*exp(x^2)*ln(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+12*x+16)*ln(4*x^2+16*x+16)-4*exp(x)*x-
4*x^2-32*x)/(2+x)/ln(4*x^2+16*x+16)^3,x,method=_RETURNVERBOSE)

[Out]

exp(x^2)-4*x*(x+exp(x)+8)/(Pi*csgn(I*(2+x))^2*csgn(I*(2+x)^2)-2*Pi*csgn(I*(2+x))*csgn(I*(2+x)^2)^2+Pi*csgn(I*(
2+x)^2)^3+4*I*ln(2)+4*I*ln(2+x))^2

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maxima [B]  time = 0.48, size = 59, normalized size = 2.36 \begin {gather*} \frac {x^{2} + 4 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )} e^{\left (x^{2}\right )} + x e^{x} + 8 \, x}{4 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+12*x+16)*log(4*x^2+16*x+16)-4*e
xp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16*x+16)^3,x, algorithm="maxima")

[Out]

1/4*(x^2 + 4*(log(2)^2 + 2*log(2)*log(x + 2) + log(x + 2)^2)*e^(x^2) + x*e^x + 8*x)/(log(2)^2 + 2*log(2)*log(x
+ 2) + log(x + 2)^2)

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mupad [B]  time = 0.21, size = 55, normalized size = 2.20 \begin {gather*} {\mathrm {e}}^{x^2}+\frac {8\,x}{{\ln \left (4\,x^2+16\,x+16\right )}^2}+\frac {x^2}{{\ln \left (4\,x^2+16\,x+16\right )}^2}+\frac {x\,{\mathrm {e}}^x}{{\ln \left (4\,x^2+16\,x+16\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x + 4*x*exp(x) + 4*x^2 - log(16*x + 4*x^2 + 16)*(12*x + exp(x)*(3*x + x^2 + 2) + 2*x^2 + 16) - exp(x^
2)*log(16*x + 4*x^2 + 16)^3*(4*x + 2*x^2))/(log(16*x + 4*x^2 + 16)^3*(x + 2)),x)

[Out]

exp(x^2) + (8*x)/log(16*x + 4*x^2 + 16)^2 + x^2/log(16*x + 4*x^2 + 16)^2 + (x*exp(x))/log(16*x + 4*x^2 + 16)^2

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sympy [A]  time = 0.51, size = 42, normalized size = 1.68 \begin {gather*} \frac {x e^{x}}{\log {\left (4 x^{2} + 16 x + 16 \right )}^{2}} + \frac {x^{2} + 8 x}{\log {\left (4 x^{2} + 16 x + 16 \right )}^{2}} + e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+4*x)*exp(x**2)*ln(4*x**2+16*x+16)**3+((x**2+3*x+2)*exp(x)+2*x**2+12*x+16)*ln(4*x**2+16*x+16
)-4*exp(x)*x-4*x**2-32*x)/(2+x)/ln(4*x**2+16*x+16)**3,x)

[Out]

x*exp(x)/log(4*x**2 + 16*x + 16)**2 + (x**2 + 8*x)/log(4*x**2 + 16*x + 16)**2 + exp(x**2)

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