3.90.60 \(\int \frac {-8 e^{\frac {1}{4} (4 x^3+x^3 \log (5))} x+e^{\frac {1}{2} (4 x^3+x^3 \log (5))} (60 x^2+12 x^4+(15 x^2+3 x^4) \log (5))+(8 x+e^{\frac {1}{4} (4 x^3+x^3 \log (5))} (-60 x^2-12 x^4+(-15 x^2-3 x^4) \log (5))) \log (5+x^2)}{10+2 x^2} \, dx\)

Optimal. Leaf size=24 \[ \left (-e^{\frac {1}{4} x^3 (4+\log (5))}+\log \left (5+x^2\right )\right )^2 \]

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Rubi [B]  time = 1.96, antiderivative size = 72, normalized size of antiderivative = 3.00, number of steps used = 11, number of rules used = 9, integrand size = 131, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6688, 12, 6725, 2287, 2209, 2475, 2390, 2301, 2288} \begin {gather*} 5^{\frac {x^3}{2}} e^{2 x^3}+\log ^2\left (x^2+5\right )-\frac {2\ 5^{\frac {x^3}{4}} e^{x^3} \left (5 x \log \left (x^2+5\right )+x^3 \log \left (x^2+5\right )\right )}{x \left (x^2+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*E^((4*x^3 + x^3*Log[5])/4)*x + E^((4*x^3 + x^3*Log[5])/2)*(60*x^2 + 12*x^4 + (15*x^2 + 3*x^4)*Log[5])
+ (8*x + E^((4*x^3 + x^3*Log[5])/4)*(-60*x^2 - 12*x^4 + (-15*x^2 - 3*x^4)*Log[5]))*Log[5 + x^2])/(10 + 2*x^2),
x]

[Out]

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

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

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

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 6688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x \left (8-3\ 5^{\frac {x^3}{4}} e^{x^3} x \left (5+x^2\right ) (4+\log (5))\right ) \left (-5^{\frac {x^3}{4}} e^{x^3}+\log \left (5+x^2\right )\right )}{2 \left (5+x^2\right )} \, dx\\ &=\frac {1}{2} \int \frac {x \left (8-3\ 5^{\frac {x^3}{4}} e^{x^3} x \left (5+x^2\right ) (4+\log (5))\right ) \left (-5^{\frac {x^3}{4}} e^{x^3}+\log \left (5+x^2\right )\right )}{5+x^2} \, dx\\ &=\frac {1}{2} \int \left (3\ 5^{\frac {x^3}{2}} e^{2 x^3} x^2 (4+\log (5))+\frac {8 x \log \left (5+x^2\right )}{5+x^2}+\frac {5^{\frac {x^3}{4}} e^{x^3} x \left (-8-60 x \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )-12 x^3 \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )\right )}{5+x^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {5^{\frac {x^3}{4}} e^{x^3} x \left (-8-60 x \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )-12 x^3 \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )\right )}{5+x^2} \, dx+4 \int \frac {x \log \left (5+x^2\right )}{5+x^2} \, dx+\frac {1}{2} (3 (4+\log (5))) \int 5^{\frac {x^3}{2}} e^{2 x^3} x^2 \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {1}{4} x^3 (4+\log (5))} x \left (-8-60 x \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )-12 x^3 \left (1+\frac {\log (5)}{4}\right ) \log \left (5+x^2\right )\right )}{5+x^2} \, dx+2 \operatorname {Subst}\left (\int \frac {\log (5+x)}{5+x} \, dx,x,x^2\right )+\frac {1}{2} (3 (4+\log (5))) \int e^{\frac {1}{2} x^3 (4+\log (5))} x^2 \, dx\\ &=5^{\frac {x^3}{2}} e^{2 x^3}-\frac {2\ 5^{\frac {x^3}{4}} e^{x^3} \left (5 x \log \left (5+x^2\right )+x^3 \log \left (5+x^2\right )\right )}{x \left (5+x^2\right )}+2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x^2\right )\\ &=5^{\frac {x^3}{2}} e^{2 x^3}+\log ^2\left (5+x^2\right )-\frac {2\ 5^{\frac {x^3}{4}} e^{x^3} \left (5 x \log \left (5+x^2\right )+x^3 \log \left (5+x^2\right )\right )}{x \left (5+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 25, normalized size = 1.04 \begin {gather*} \left (-5^{\frac {x^3}{4}} e^{x^3}+\log \left (5+x^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^((4*x^3 + x^3*Log[5])/4)*x + E^((4*x^3 + x^3*Log[5])/2)*(60*x^2 + 12*x^4 + (15*x^2 + 3*x^4)*Lo
g[5]) + (8*x + E^((4*x^3 + x^3*Log[5])/4)*(-60*x^2 - 12*x^4 + (-15*x^2 - 3*x^4)*Log[5]))*Log[5 + x^2])/(10 + 2
*x^2),x]

[Out]

(-(5^(x^3/4)*E^x^3) + Log[5 + x^2])^2

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fricas [B]  time = 0.66, size = 43, normalized size = 1.79 \begin {gather*} -2 \, e^{\left (\frac {1}{4} \, x^{3} \log \relax (5) + x^{3}\right )} \log \left (x^{2} + 5\right ) + \log \left (x^{2} + 5\right )^{2} + e^{\left (\frac {1}{2} \, x^{3} \log \relax (5) + 2 \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3)+8*x)*log(x^2+5)+((3*x^4+15*x^2)*log
(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="fricas")

[Out]

-2*e^(1/4*x^3*log(5) + x^3)*log(x^2 + 5) + log(x^2 + 5)^2 + e^(1/2*x^3*log(5) + 2*x^3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3)+8*x)*log(x^2+5)+((3*x^4+15*x^2)*log
(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.45, size = 45, normalized size = 1.88

method	result	size
risch	\(5^{\frac {x^{3}}{2}} {\mathrm e}^{2 x^{3}}-2 \,5^{\frac {x^{3}}{4}} {\mathrm e}^{x^{3}} \ln \left (x^{2}+5\right )+\ln \left (x^{2}+5\right )^{2}\)	\(45\)
default	\(-2 \,{\mathrm e}^{\frac {x^{3} \ln \relax (5)}{4}+x^{3}} \ln \left (x^{2}+5\right )+\ln \left (x^{2}+5\right )^{2}+\frac {4 \,{\mathrm e}^{\frac {x^{3} \ln \relax (5)}{2}+2 x^{3}}}{\ln \relax (5)+4}+\frac {\ln \relax (5) {\mathrm e}^{\frac {x^{3} \ln \relax (5)}{2}+2 x^{3}}}{\ln \relax (5)+4}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-3*x^4-15*x^2)*ln(5)-12*x^4-60*x^2)*exp(1/4*x^3*ln(5)+x^3)+8*x)*ln(x^2+5)+((3*x^4+15*x^2)*ln(5)+12*x^4
+60*x^2)*exp(1/4*x^3*ln(5)+x^3)^2-8*x*exp(1/4*x^3*ln(5)+x^3))/(2*x^2+10),x,method=_RETURNVERBOSE)

[Out]

(5^(1/4*x^3))^2*exp(2*x^3)-2*5^(1/4*x^3)*exp(x^3)*ln(x^2+5)+ln(x^2+5)^2

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maxima [B]  time = 0.49, size = 43, normalized size = 1.79 \begin {gather*} -2 \, e^{\left (\frac {1}{4} \, x^{3} \log \relax (5) + x^{3}\right )} \log \left (x^{2} + 5\right ) + \log \left (x^{2} + 5\right )^{2} + e^{\left (\frac {1}{2} \, x^{3} \log \relax (5) + 2 \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3)+8*x)*log(x^2+5)+((3*x^4+15*x^2)*log
(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="maxima")

[Out]

-2*e^(1/4*x^3*log(5) + x^3)*log(x^2 + 5) + log(x^2 + 5)^2 + e^(1/2*x^3*log(5) + 2*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (x^2+5\right )\,\left (8\,x-{\mathrm {e}}^{\frac {x^3\,\ln \relax (5)}{4}+x^3}\,\left (\ln \relax (5)\,\left (3\,x^4+15\,x^2\right )+60\,x^2+12\,x^4\right )\right )-8\,x\,{\mathrm {e}}^{\frac {x^3\,\ln \relax (5)}{4}+x^3}+{\mathrm {e}}^{\frac {x^3\,\ln \relax (5)}{2}+2\,x^3}\,\left (\ln \relax (5)\,\left (3\,x^4+15\,x^2\right )+60\,x^2+12\,x^4\right )}{2\,x^2+10} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2 + 5)*(8*x - exp((x^3*log(5))/4 + x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4)) - 8*x*exp((x^3
*log(5))/4 + x^3) + exp((x^3*log(5))/2 + 2*x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4))/(2*x^2 + 10),x)

[Out]

int((log(x^2 + 5)*(8*x - exp((x^3*log(5))/4 + x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4)) - 8*x*exp((x^3
*log(5))/4 + x^3) + exp((x^3*log(5))/2 + 2*x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4))/(2*x^2 + 10), x)

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sympy [B]  time = 0.83, size = 44, normalized size = 1.83 \begin {gather*} - 2 e^{\frac {x^{3} \log {\relax (5 )}}{4} + x^{3}} \log {\left (x^{2} + 5 \right )} + e^{\frac {x^{3} \log {\relax (5 )}}{2} + 2 x^{3}} + \log {\left (x^{2} + 5 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-3*x**4-15*x**2)*ln(5)-12*x**4-60*x**2)*exp(1/4*x**3*ln(5)+x**3)+8*x)*ln(x**2+5)+((3*x**4+15*x**
2)*ln(5)+12*x**4+60*x**2)*exp(1/4*x**3*ln(5)+x**3)**2-8*x*exp(1/4*x**3*ln(5)+x**3))/(2*x**2+10),x)

[Out]

-2*exp(x**3*log(5)/4 + x**3)*log(x**2 + 5) + exp(x**3*log(5)/2 + 2*x**3) + log(x**2 + 5)**2

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