3.4.33 \(\int \frac {e^{2 \log ^2(x)} (64 x+32 x^2-96 x^3+(64 x+32 x^2-96 x^3) \log (15)+(16 x+8 x^2-24 x^3) \log ^2(15)+(128 x-64 x^2+64 x^3+(128 x-64 x^2+64 x^3) \log (15)+(32 x-16 x^2+16 x^3) \log ^2(15)) \log (x))}{32-80 x+160 x^2-200 x^3+210 x^4-161 x^5+105 x^6-50 x^7+20 x^8-5 x^9+x^{10}} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{2 \log ^2(x)} x^2 (4+2 \log (15))^2}{\left (2-x+x^2\right )^4} \]

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Rubi [B]  time = 0.27, antiderivative size = 108, normalized size of antiderivative = 3.60, number of steps used = 1, number of rules used = 1, integrand size = 162, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {2288} \begin {gather*} \frac {4 x e^{2 \log ^2(x)} \left (4 x^3-4 x^2+\left (x^3-x^2+2 x\right ) \log ^2(15)+4 \left (x^3-x^2+2 x\right ) \log (15)+8 x\right )}{x^{10}-5 x^9+20 x^8-50 x^7+105 x^6-161 x^5+210 x^4-200 x^3+160 x^2-80 x+32} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*Log[x]^2)*(64*x + 32*x^2 - 96*x^3 + (64*x + 32*x^2 - 96*x^3)*Log[15] + (16*x + 8*x^2 - 24*x^3)*Log[1
5]^2 + (128*x - 64*x^2 + 64*x^3 + (128*x - 64*x^2 + 64*x^3)*Log[15] + (32*x - 16*x^2 + 16*x^3)*Log[15]^2)*Log[
x]))/(32 - 80*x + 160*x^2 - 200*x^3 + 210*x^4 - 161*x^5 + 105*x^6 - 50*x^7 + 20*x^8 - 5*x^9 + x^10),x]

[Out]

(4*E^(2*Log[x]^2)*x*(8*x - 4*x^2 + 4*x^3 + 4*(2*x - x^2 + x^3)*Log[15] + (2*x - x^2 + x^3)*Log[15]^2))/(32 - 8
0*x + 160*x^2 - 200*x^3 + 210*x^4 - 161*x^5 + 105*x^6 - 50*x^7 + 20*x^8 - 5*x^9 + x^10)

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {4 e^{2 \log ^2(x)} x \left (8 x-4 x^2+4 x^3+4 \left (2 x-x^2+x^3\right ) \log (15)+\left (2 x-x^2+x^3\right ) \log ^2(15)\right )}{32-80 x+160 x^2-200 x^3+210 x^4-161 x^5+105 x^6-50 x^7+20 x^8-5 x^9+x^{10}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.97 \begin {gather*} \frac {4 e^{2 \log ^2(x)} x^2 (2+\log (15))^2}{\left (2-x+x^2\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*Log[x]^2)*(64*x + 32*x^2 - 96*x^3 + (64*x + 32*x^2 - 96*x^3)*Log[15] + (16*x + 8*x^2 - 24*x^3)
*Log[15]^2 + (128*x - 64*x^2 + 64*x^3 + (128*x - 64*x^2 + 64*x^3)*Log[15] + (32*x - 16*x^2 + 16*x^3)*Log[15]^2
)*Log[x]))/(32 - 80*x + 160*x^2 - 200*x^3 + 210*x^4 - 161*x^5 + 105*x^6 - 50*x^7 + 20*x^8 - 5*x^9 + x^10),x]

[Out]

(4*E^(2*Log[x]^2)*x^2*(2 + Log[15])^2)/(2 - x + x^2)^4

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fricas [B]  time = 0.97, size = 70, normalized size = 2.33 \begin {gather*} \frac {4 \, {\left (x^{2} \log \left (15\right )^{2} + 4 \, x^{2} \log \left (15\right ) + 4 \, x^{2}\right )} e^{\left (2 \, \log \relax (x)^{2}\right )}}{x^{8} - 4 \, x^{7} + 14 \, x^{6} - 28 \, x^{5} + 49 \, x^{4} - 56 \, x^{3} + 56 \, x^{2} - 32 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2+32*x)*log(15)^2+(64*x^3-64*x^2+128*x)*log(15)+64*x^3-64*x^2+128*x)*log(x)+(-24*x^3+
8*x^2+16*x)*log(15)^2+(-96*x^3+32*x^2+64*x)*log(15)-96*x^3+32*x^2+64*x)*exp(log(x)^2)^2/(x^10-5*x^9+20*x^8-50*
x^7+105*x^6-161*x^5+210*x^4-200*x^3+160*x^2-80*x+32),x, algorithm="fricas")

[Out]

4*(x^2*log(15)^2 + 4*x^2*log(15) + 4*x^2)*e^(2*log(x)^2)/(x^8 - 4*x^7 + 14*x^6 - 28*x^5 + 49*x^4 - 56*x^3 + 56
*x^2 - 32*x + 16)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {8 \, {\left (12 \, x^{3} + {\left (3 \, x^{3} - x^{2} - 2 \, x\right )} \log \left (15\right )^{2} - 4 \, x^{2} + 4 \, {\left (3 \, x^{3} - x^{2} - 2 \, x\right )} \log \left (15\right ) - 2 \, {\left (4 \, x^{3} + {\left (x^{3} - x^{2} + 2 \, x\right )} \log \left (15\right )^{2} - 4 \, x^{2} + 4 \, {\left (x^{3} - x^{2} + 2 \, x\right )} \log \left (15\right ) + 8 \, x\right )} \log \relax (x) - 8 \, x\right )} e^{\left (2 \, \log \relax (x)^{2}\right )}}{x^{10} - 5 \, x^{9} + 20 \, x^{8} - 50 \, x^{7} + 105 \, x^{6} - 161 \, x^{5} + 210 \, x^{4} - 200 \, x^{3} + 160 \, x^{2} - 80 \, x + 32}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2+32*x)*log(15)^2+(64*x^3-64*x^2+128*x)*log(15)+64*x^3-64*x^2+128*x)*log(x)+(-24*x^3+
8*x^2+16*x)*log(15)^2+(-96*x^3+32*x^2+64*x)*log(15)-96*x^3+32*x^2+64*x)*exp(log(x)^2)^2/(x^10-5*x^9+20*x^8-50*
x^7+105*x^6-161*x^5+210*x^4-200*x^3+160*x^2-80*x+32),x, algorithm="giac")

[Out]

integrate(-8*(12*x^3 + (3*x^3 - x^2 - 2*x)*log(15)^2 - 4*x^2 + 4*(3*x^3 - x^2 - 2*x)*log(15) - 2*(4*x^3 + (x^3
- x^2 + 2*x)*log(15)^2 - 4*x^2 + 4*(x^3 - x^2 + 2*x)*log(15) + 8*x)*log(x) - 8*x)*e^(2*log(x)^2)/(x^10 - 5*x^
9 + 20*x^8 - 50*x^7 + 105*x^6 - 161*x^5 + 210*x^4 - 200*x^3 + 160*x^2 - 80*x + 32), x)

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maple [B]  time = 0.07, size = 77, normalized size = 2.57

method	result	size
risch	\(\frac {4 \left (\ln \relax (5)^{2}+2 \ln \relax (3) \ln \relax (5)+\ln \relax (3)^{2}+4 \ln \relax (5)+4 \ln \relax (3)+4\right ) x^{2} {\mathrm e}^{2 \ln \relax (x )^{2}}}{x^{8}-4 x^{7}+14 x^{6}-28 x^{5}+49 x^{4}-56 x^{3}+56 x^{2}-32 x +16}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3-16*x^2+32*x)*ln(15)^2+(64*x^3-64*x^2+128*x)*ln(15)+64*x^3-64*x^2+128*x)*ln(x)+(-24*x^3+8*x^2+16*
x)*ln(15)^2+(-96*x^3+32*x^2+64*x)*ln(15)-96*x^3+32*x^2+64*x)*exp(ln(x)^2)^2/(x^10-5*x^9+20*x^8-50*x^7+105*x^6-
161*x^5+210*x^4-200*x^3+160*x^2-80*x+32),x,method=_RETURNVERBOSE)

[Out]

4*(ln(5)^2+2*ln(3)*ln(5)+ln(3)^2+4*ln(5)+4*ln(3)+4)*x^2/(x^8-4*x^7+14*x^6-28*x^5+49*x^4-56*x^3+56*x^2-32*x+16)
*exp(2*ln(x)^2)

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maxima [B]  time = 0.69, size = 74, normalized size = 2.47 \begin {gather*} \frac {4 \, {\left (\log \relax (5)^{2} + 2 \, {\left (\log \relax (5) + 2\right )} \log \relax (3) + \log \relax (3)^{2} + 4 \, \log \relax (5) + 4\right )} x^{2} e^{\left (2 \, \log \relax (x)^{2}\right )}}{x^{8} - 4 \, x^{7} + 14 \, x^{6} - 28 \, x^{5} + 49 \, x^{4} - 56 \, x^{3} + 56 \, x^{2} - 32 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2+32*x)*log(15)^2+(64*x^3-64*x^2+128*x)*log(15)+64*x^3-64*x^2+128*x)*log(x)+(-24*x^3+
8*x^2+16*x)*log(15)^2+(-96*x^3+32*x^2+64*x)*log(15)-96*x^3+32*x^2+64*x)*exp(log(x)^2)^2/(x^10-5*x^9+20*x^8-50*
x^7+105*x^6-161*x^5+210*x^4-200*x^3+160*x^2-80*x+32),x, algorithm="maxima")

[Out]

4*(log(5)^2 + 2*(log(5) + 2)*log(3) + log(3)^2 + 4*log(5) + 4)*x^2*e^(2*log(x)^2)/(x^8 - 4*x^7 + 14*x^6 - 28*x
^5 + 49*x^4 - 56*x^3 + 56*x^2 - 32*x + 16)

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mupad [B]  time = 0.83, size = 54, normalized size = 1.80 \begin {gather*} \frac {4\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2}\,{\left (\ln \left (15\right )+2\right )}^2}{{\left (x^2-x+2\right )}^3}+\frac {4\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2}\,{\left (\ln \left (15\right )+2\right )}^2\,\left (x-2\right )}{{\left (x^2-x+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*log(x)^2)*(64*x + log(x)*(128*x + log(15)*(128*x - 64*x^2 + 64*x^3) + log(15)^2*(32*x - 16*x^2 + 16
*x^3) - 64*x^2 + 64*x^3) + log(15)*(64*x + 32*x^2 - 96*x^3) + log(15)^2*(16*x + 8*x^2 - 24*x^3) + 32*x^2 - 96*
x^3))/(160*x^2 - 80*x - 200*x^3 + 210*x^4 - 161*x^5 + 105*x^6 - 50*x^7 + 20*x^8 - 5*x^9 + x^10 + 32),x)

[Out]

(4*exp(2*log(x)^2)*(log(15) + 2)^2)/(x^2 - x + 2)^3 + (4*exp(2*log(x)^2)*(log(15) + 2)^2*(x - 2))/(x^2 - x + 2
)^4

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sympy [B]  time = 0.63, size = 70, normalized size = 2.33 \begin {gather*} \frac {\left (16 x^{2} + 4 x^{2} \log {\left (15 \right )}^{2} + 16 x^{2} \log {\left (15 \right )}\right ) e^{2 \log {\relax (x )}^{2}}}{x^{8} - 4 x^{7} + 14 x^{6} - 28 x^{5} + 49 x^{4} - 56 x^{3} + 56 x^{2} - 32 x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3-16*x**2+32*x)*ln(15)**2+(64*x**3-64*x**2+128*x)*ln(15)+64*x**3-64*x**2+128*x)*ln(x)+(-24*
x**3+8*x**2+16*x)*ln(15)**2+(-96*x**3+32*x**2+64*x)*ln(15)-96*x**3+32*x**2+64*x)*exp(ln(x)**2)**2/(x**10-5*x**
9+20*x**8-50*x**7+105*x**6-161*x**5+210*x**4-200*x**3+160*x**2-80*x+32),x)

[Out]

(16*x**2 + 4*x**2*log(15)**2 + 16*x**2*log(15))*exp(2*log(x)**2)/(x**8 - 4*x**7 + 14*x**6 - 28*x**5 + 49*x**4
- 56*x**3 + 56*x**2 - 32*x + 16)

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