3.12.15 \(\int \frac {e^2 (5 x^2+4 x^3)+e^{e^6 x^2-2 e^3 x^3+x^4} (e^8 (50 x^2+80 x^3+32 x^4)+e^5 (-150 x^3-240 x^4-96 x^5)+e^2 (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6))+e^2 (15 x^2+8 x^3) \log (x)}{25+40 x+16 x^2} \, dx\)

Optimal. Leaf size=34 \[ e^2 x \left (e^{\left (e^3-x\right )^2 x^2}+\frac {x^2 \log (x)}{5+4 x}\right ) \]

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Rubi [B]  time = 1.36, antiderivative size = 105, normalized size of antiderivative = 3.09, number of steps used = 16, number of rules used = 11, integrand size = 138, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 6688, 12, 6742, 2288, 43, 2357, 2295, 2304, 2314, 31} \begin {gather*} \frac {1}{4} e^2 x^2 \log (x)+\frac {e^{\left (e^3-x\right )^2 x^2+2} \left (2 x^4-3 e^3 x^3+e^6 x^2\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+\frac {25 e^2 x \log (x)}{16 (4 x+5)}-\frac {5}{16} e^2 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(5*x^2 + 4*x^3) + E^(E^6*x^2 - 2*E^3*x^3 + x^4)*(E^8*(50*x^2 + 80*x^3 + 32*x^4) + E^5*(-150*x^3 - 240
*x^4 - 96*x^5) + E^2*(25 + 40*x + 16*x^2 + 100*x^4 + 160*x^5 + 64*x^6)) + E^2*(15*x^2 + 8*x^3)*Log[x])/(25 + 4
0*x + 16*x^2),x]

[Out]

(E^(2 + (E^3 - x)^2*x^2)*(E^6*x^2 - 3*E^3*x^3 + 2*x^4))/((E^3 - x)^2*x - (E^3 - x)*x^2) - (5*E^2*x*Log[x])/16
+ (E^2*x^2*Log[x])/4 + (25*E^2*x*Log[x])/(16*(5 + 4*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{(5+4 x)^2} \, dx\\ &=\int \frac {e^2 \left (x^2 (5+4 x)+e^{\left (e^3-x\right )^2 x^2} (5+4 x)^2 \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+x^2 (15+8 x) \log (x)\right )}{(5+4 x)^2} \, dx\\ &=e^2 \int \frac {x^2 (5+4 x)+e^{\left (e^3-x\right )^2 x^2} (5+4 x)^2 \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+x^2 (15+8 x) \log (x)}{(5+4 x)^2} \, dx\\ &=e^2 \int \left (e^{\left (e^3-x\right )^2 x^2} \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+\frac {x^2 (5+4 x+15 \log (x)+8 x \log (x))}{(5+4 x)^2}\right ) \, dx\\ &=e^2 \int e^{\left (e^3-x\right )^2 x^2} \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right ) \, dx+e^2 \int \frac {x^2 (5+4 x+15 \log (x)+8 x \log (x))}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \left (\frac {x^2}{5+4 x}+\frac {x^2 (15+8 x) \log (x)}{(5+4 x)^2}\right ) \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \frac {x^2}{5+4 x} \, dx+e^2 \int \frac {x^2 (15+8 x) \log (x)}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \left (-\frac {5}{16}+\frac {x}{4}+\frac {25}{16 (5+4 x)}\right ) \, dx+e^2 \int \left (-\frac {5 \log (x)}{16}+\frac {1}{2} x \log (x)+\frac {125 \log (x)}{16 (5+4 x)^2}\right ) \, dx\\ &=-\frac {5 e^2 x}{16}+\frac {e^2 x^2}{8}+\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+\frac {25}{64} e^2 \log (5+4 x)-\frac {1}{16} \left (5 e^2\right ) \int \log (x) \, dx+\frac {1}{2} e^2 \int x \log (x) \, dx+\frac {1}{16} \left (125 e^2\right ) \int \frac {\log (x)}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}-\frac {5}{16} e^2 x \log (x)+\frac {1}{4} e^2 x^2 \log (x)+\frac {25 e^2 x \log (x)}{16 (5+4 x)}+\frac {25}{64} e^2 \log (5+4 x)-\frac {1}{16} \left (25 e^2\right ) \int \frac {1}{5+4 x} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}-\frac {5}{16} e^2 x \log (x)+\frac {1}{4} e^2 x^2 \log (x)+\frac {25 e^2 x \log (x)}{16 (5+4 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 35, normalized size = 1.03 \begin {gather*} e^2 \left (e^{\left (e^3-x\right )^2 x^2} x+\frac {x^3 \log (x)}{5+4 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(5*x^2 + 4*x^3) + E^(E^6*x^2 - 2*E^3*x^3 + x^4)*(E^8*(50*x^2 + 80*x^3 + 32*x^4) + E^5*(-150*x^3
- 240*x^4 - 96*x^5) + E^2*(25 + 40*x + 16*x^2 + 100*x^4 + 160*x^5 + 64*x^6)) + E^2*(15*x^2 + 8*x^3)*Log[x])/(
25 + 40*x + 16*x^2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 46, normalized size = 1.35 \begin {gather*} \frac {x^{3} e^{2} \log \relax (x) + {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6} + 2\right )}}{4 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+(-96*x^5-240*x^4-150*x^3)*exp(
2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp
(2))/(16*x^2+40*x+25),x, algorithm="fricas")

[Out]

(x^3*e^2*log(x) + (4*x^2 + 5*x)*e^(x^4 - 2*x^3*e^3 + x^2*e^6 + 2))/(4*x + 5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (8 \, x^{3} + 15 \, x^{2}\right )} e^{2} \log \relax (x) + {\left (4 \, x^{3} + 5 \, x^{2}\right )} e^{2} + {\left (2 \, {\left (16 \, x^{4} + 40 \, x^{3} + 25 \, x^{2}\right )} e^{8} - 6 \, {\left (16 \, x^{5} + 40 \, x^{4} + 25 \, x^{3}\right )} e^{5} + {\left (64 \, x^{6} + 160 \, x^{5} + 100 \, x^{4} + 16 \, x^{2} + 40 \, x + 25\right )} e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )}}{16 \, x^{2} + 40 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+(-96*x^5-240*x^4-150*x^3)*exp(
2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp
(2))/(16*x^2+40*x+25),x, algorithm="giac")

[Out]

integrate(((8*x^3 + 15*x^2)*e^2*log(x) + (4*x^3 + 5*x^2)*e^2 + (2*(16*x^4 + 40*x^3 + 25*x^2)*e^8 - 6*(16*x^5 +
40*x^4 + 25*x^3)*e^5 + (64*x^6 + 160*x^5 + 100*x^4 + 16*x^2 + 40*x + 25)*e^2)*e^(x^4 - 2*x^3*e^3 + x^2*e^6))/
(16*x^2 + 40*x + 25), x)

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maple [A]  time = 0.25, size = 52, normalized size = 1.53

method	result	size
risch	\(\frac {{\mathrm e}^{2} \left (64 x^{3}-100 x -125\right ) \ln \relax (x )}{256 x +320}+\frac {25 \,{\mathrm e}^{2} \ln \relax (x )}{64}+x \,{\mathrm e}^{-2 x^{3} {\mathrm e}^{3}+x^{4}+x^{2} {\mathrm e}^{6}+2}\)	\(52\)
default	\({\mathrm e}^{2} x \,{\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}+\frac {x^{2} {\mathrm e}^{2} \ln \relax (x )}{4}-\frac {5 x \,{\mathrm e}^{2} \ln \relax (x )}{16}+\frac {25 \,{\mathrm e}^{2} \ln \relax (x ) x}{16 \left (4 x +5\right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^3+15*x^2)*exp(2)*ln(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+(-96*x^5-240*x^4-150*x^3)*exp(2)*exp(
3)+(64*x^6+160*x^5+100*x^4+16*x^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2))/(1
6*x^2+40*x+25),x,method=_RETURNVERBOSE)

[Out]

1/64*exp(2)*(64*x^3-100*x-125)/(4*x+5)*ln(x)+25/64*exp(2)*ln(x)+x*exp(-2*x^3*exp(3)+x^4+x^2*exp(6)+2)

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maxima [B]  time = 0.60, size = 155, normalized size = 4.56 \begin {gather*} \frac {1}{64} \, {\left (8 \, x^{2} - 40 \, x + \frac {125}{4 \, x + 5} + 75 \, \log \left (4 \, x + 5\right )\right )} e^{2} + \frac {5}{64} \, {\left (4 \, x - \frac {25}{4 \, x + 5} - 10 \, \log \left (4 \, x + 5\right )\right )} e^{2} - \frac {25}{64} \, e^{2} \log \left (4 \, x + 5\right ) + \frac {25}{64} \, e^{2} \log \relax (x) - \frac {32 \, x^{3} e^{2} - 40 \, x^{2} e^{2} - 100 \, x e^{2} - 64 \, {\left (4 \, x^{2} e^{2} + 5 \, x e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )} - {\left (64 \, x^{3} e^{2} - 100 \, x e^{2} - 125 \, e^{2}\right )} \log \relax (x)}{64 \, {\left (4 \, x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+(-96*x^5-240*x^4-150*x^3)*exp(
2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp
(2))/(16*x^2+40*x+25),x, algorithm="maxima")

[Out]

1/64*(8*x^2 - 40*x + 125/(4*x + 5) + 75*log(4*x + 5))*e^2 + 5/64*(4*x - 25/(4*x + 5) - 10*log(4*x + 5))*e^2 -
25/64*e^2*log(4*x + 5) + 25/64*e^2*log(x) - 1/64*(32*x^3*e^2 - 40*x^2*e^2 - 100*x*e^2 - 64*(4*x^2*e^2 + 5*x*e^
2)*e^(x^4 - 2*x^3*e^3 + x^2*e^6) - (64*x^3*e^2 - 100*x*e^2 - 125*e^2)*log(x))/(4*x + 5)

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mupad [B]  time = 1.37, size = 91, normalized size = 2.68 \begin {gather*} x\,{\mathrm {e}}^{x^4-2\,{\mathrm {e}}^3\,x^3+{\mathrm {e}}^6\,x^2+2}-\frac {25\,\ln \left (x+\frac {5}{4}\right )\,{\mathrm {e}}^2}{64}+\frac {{\mathrm {e}}^2\,\left (25\,\ln \left (x+\frac {5}{4}\right )-20\,x+8\,x^2\right )}{64}+\frac {\frac {25\,x^2\,{\mathrm {e}}^2}{16}+\frac {5\,x^3\,{\mathrm {e}}^2}{8}-\frac {x^4\,{\mathrm {e}}^2}{2}+x^4\,{\mathrm {e}}^2\,\ln \relax (x)}{4\,x^2+5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*(5*x^2 + 4*x^3) + exp(x^2*exp(6) - 2*x^3*exp(3) + x^4)*(exp(8)*(50*x^2 + 80*x^3 + 32*x^4) - exp(5)
*(150*x^3 + 240*x^4 + 96*x^5) + exp(2)*(40*x + 16*x^2 + 100*x^4 + 160*x^5 + 64*x^6 + 25)) + exp(2)*log(x)*(15*
x^2 + 8*x^3))/(40*x + 16*x^2 + 25),x)

[Out]

x*exp(x^2*exp(6) - 2*x^3*exp(3) + x^4 + 2) - (25*log(x + 5/4)*exp(2))/64 + (exp(2)*(25*log(x + 5/4) - 20*x + 8
*x^2))/64 + ((25*x^2*exp(2))/16 + (5*x^3*exp(2))/8 - (x^4*exp(2))/2 + x^4*exp(2)*log(x))/(5*x + 4*x^2)

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sympy [B]  time = 0.56, size = 63, normalized size = 1.85 \begin {gather*} x e^{2} e^{x^{4} - 2 x^{3} e^{3} + x^{2} e^{6}} + \frac {25 e^{2} \log {\relax (x )}}{64} + \frac {\left (64 x^{3} e^{2} - 100 x e^{2} - 125 e^{2}\right ) \log {\relax (x )}}{256 x + 320} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**3+15*x**2)*exp(2)*ln(x)+((32*x**4+80*x**3+50*x**2)*exp(2)*exp(3)**2+(-96*x**5-240*x**4-150*x*
*3)*exp(2)*exp(3)+(64*x**6+160*x**5+100*x**4+16*x**2+40*x+25)*exp(2))*exp(x**2*exp(3)**2-2*x**3*exp(3)+x**4)+(
4*x**3+5*x**2)*exp(2))/(16*x**2+40*x+25),x)

[Out]

x*exp(2)*exp(x**4 - 2*x**3*exp(3) + x**2*exp(6)) + 25*exp(2)*log(x)/64 + (64*x**3*exp(2) - 100*x*exp(2) - 125*
exp(2))*log(x)/(256*x + 320)

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