\(\int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx\) [6641]

Optimal result

Integrand size = 40, antiderivative size = 29 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=\frac {x-\log (3) \left (5 \left (e^{(4+\log (x))^2}+x\right )-\frac {1}{\log (5)}\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 14, 2326} \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=\frac {\log (3)}{x \log (5)}-5 x^7 \log (3) e^{\log ^2(x)+16} \]

[In]

Int[(-Log[3] + E^(16 + Log[x]^2)*x^8*(-35*Log[3]*Log[5] - 10*Log[3]*Log[5]*Log[x]))/(x^2*Log[5]),x]

[Out]

-5*E^(16 + Log[x]^2)*x^7*Log[3] + Log[3]/(x*Log[5])

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2326

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{align*} \text {integral}& = \frac {\int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2} \, dx}{\log (5)} \\ & = \frac {\int \left (-\frac {\log (3)}{x^2}-5 e^{16+\log ^2(x)} x^6 \log (3) \log (5) (7+2 \log (x))\right ) \, dx}{\log (5)} \\ & = \frac {\log (3)}{x \log (5)}-(5 \log (3)) \int e^{16+\log ^2(x)} x^6 (7+2 \log (x)) \, dx \\ & = -5 e^{16+\log ^2(x)} x^7 \log (3)+\frac {\log (3)}{x \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=\frac {\log (3) \left (\frac {1}{x}-5 e^{16+\log ^2(x)} x^7 \log (5)\right )}{\log (5)} \]

[In]

Integrate[(-Log[3] + E^(16 + Log[x]^2)*x^8*(-35*Log[3]*Log[5] - 10*Log[3]*Log[5]*Log[x]))/(x^2*Log[5]),x]

[Out]

(Log[3]*(x^(-1) - 5*E^(16 + Log[x]^2)*x^7*Log[5]))/Log[5]

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
risch	\(-5 \ln \left (3\right ) x^{7} {\mathrm e}^{\ln \left (x \right )^{2}+16}+\frac {\ln \left (3\right )}{\ln \left (5\right ) x}\)	\(26\)
norman	\(\frac {\frac {\ln \left (3\right )}{\ln \left (5\right )}-5 \ln \left (3\right ) {\mathrm e}^{\ln \left (x \right )^{2}+8 \ln \left (x \right )+16}}{x}\)	\(28\)
parts	\(-\frac {5 \ln \left (3\right ) {\mathrm e}^{\ln \left (x \right )^{2}+8 \ln \left (x \right )+16}}{x}+\frac {\ln \left (3\right )}{\ln \left (5\right ) x}\)	\(30\)
parallelrisch	\(-\frac {5 \ln \left (5\right ) \ln \left (3\right ) {\mathrm e}^{\ln \left (x \right )^{2}+8 \ln \left (x \right )+16}-\ln \left (3\right )}{\ln \left (5\right ) x}\)	\(32\)
default	\(\frac {-\frac {5 \ln \left (3\right ) \ln \left (5\right ) {\mathrm e}^{\ln \left (x \right )^{2}+8 \ln \left (x \right )+16}}{x}+\frac {\ln \left (3\right )}{x}}{\ln \left (5\right )}\)	\(33\)

[In]

int(((-10*ln(3)*ln(5)*ln(x)-35*ln(3)*ln(5))*exp(ln(x)^2+8*ln(x)+16)-ln(3))/x^2/ln(5),x,method=_RETURNVERBOSE)

[Out]

-5*ln(3)*x^7*exp(ln(x)^2+16)+1/ln(5)*ln(3)/x

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=-\frac {5 \, e^{\left (\log \left (x\right )^{2} + 8 \, \log \left (x\right ) + 16\right )} \log \left (5\right ) \log \left (3\right ) - \log \left (3\right )}{x \log \left (5\right )} \]

[In]

integrate(((-10*log(3)*log(5)*log(x)-35*log(3)*log(5))*exp(log(x)^2+8*log(x)+16)-log(3))/x^2/log(5),x, algorit
hm="fricas")

[Out]

-(5*e^(log(x)^2 + 8*log(x) + 16)*log(5)*log(3) - log(3))/(x*log(5))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=- 5 x^{7} e^{\log {\left (x \right )}^{2} + 16} \log {\left (3 \right )} + \frac {\log {\left (3 \right )}}{x \log {\left (5 \right )}} \]

[In]

integrate(((-10*ln(3)*ln(5)*ln(x)-35*ln(3)*ln(5))*exp(ln(x)**2+8*ln(x)+16)-ln(3))/x**2/ln(5),x)

[Out]

-5*x**7*exp(log(x)**2 + 16)*log(3) + log(3)/(x*log(5))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.24 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=\frac {35 i \, \sqrt {\pi } \operatorname {erf}\left (i \, \log \left (x\right ) + \frac {7}{2} i\right ) e^{\frac {15}{4}} \log \left (5\right ) \log \left (3\right ) + 5 \, {\left (\frac {7 \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, \log \left (x\right ) + 7\right )}^{2}}\right ) - 1\right )} {\left (2 \, \log \left (x\right ) + 7\right )}}{\sqrt {-{\left (2 \, \log \left (x\right ) + 7\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, \log \left (x\right ) + 7\right )}^{2}\right )}\right )} e^{\frac {15}{4}} \log \left (5\right ) \log \left (3\right ) + \frac {2 \, \log \left (3\right )}{x}}{2 \, \log \left (5\right )} \]

[In]

integrate(((-10*log(3)*log(5)*log(x)-35*log(3)*log(5))*exp(log(x)^2+8*log(x)+16)-log(3))/x^2/log(5),x, algorit
hm="maxima")

[Out]

1/2*(35*I*sqrt(pi)*erf(I*log(x) + 7/2*I)*e^(15/4)*log(5)*log(3) + 5*(7*sqrt(pi)*(erf(1/2*sqrt(-(2*log(x) + 7)^
2)) - 1)*(2*log(x) + 7)/sqrt(-(2*log(x) + 7)^2) - 2*e^(1/4*(2*log(x) + 7)^2))*e^(15/4)*log(5)*log(3) + 2*log(3
)/x)/log(5)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=-\frac {5 \, x^{8} e^{\left (\log \left (x\right )^{2} + 16\right )} \log \left (5\right ) \log \left (3\right ) - \log \left (3\right )}{x \log \left (5\right )} \]

[In]

integrate(((-10*log(3)*log(5)*log(x)-35*log(3)*log(5))*exp(log(x)^2+8*log(x)+16)-log(3))/x^2/log(5),x, algorit
hm="giac")

[Out]

-(5*x^8*e^(log(x)^2 + 16)*log(5)*log(3) - log(3))/(x*log(5))

Mupad [B] (verification not implemented)

Time = 11.88 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-\log (3)+e^{16+\log ^2(x)} x^8 (-35 \log (3) \log (5)-10 \log (3) \log (5) \log (x))}{x^2 \log (5)} \, dx=\frac {\ln \left (3\right )}{x\,\ln \left (5\right )}-5\,x^7\,{\mathrm {e}}^{16}\,{\mathrm {e}}^{{\ln \left (x\right )}^2}\,\ln \left (3\right ) \]

[In]

int(-(log(3) + exp(8*log(x) + log(x)^2 + 16)*(35*log(3)*log(5) + 10*log(3)*log(5)*log(x)))/(x^2*log(5)),x)

[Out]

log(3)/(x*log(5)) - 5*x^7*exp(16)*exp(log(x)^2)*log(3)