3.27.45 \(\int \frac {135 x+e^x (72+18 x)+(-18 e^x-27 x) \log (\frac {1}{2} (10 e^x x^4+15 x^5))}{2 e^x x^2+3 x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {9 \log \left (5 x^4 \left (e^x+\frac {3 x}{2}\right )\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6742, 14, 43, 2551} \begin {gather*} \frac {9 \log \left (\frac {5}{2} x^4 \left (3 x+2 e^x\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(135*x + E^x*(72 + 18*x) + (-18*E^x - 27*x)*Log[(10*E^x*x^4 + 15*x^5)/2])/(2*E^x*x^2 + 3*x^3),x]

[Out]

(9*Log[(5*x^4*(2*E^x + 3*x))/2])/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 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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

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 \left (-\frac {27 (-1+x)}{x \left (2 e^x+3 x\right )}+\frac {9 \left (4+x-\log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )\right )}{x^2}\right ) \, dx\\ &=9 \int \frac {4+x-\log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x^2} \, dx-27 \int \frac {-1+x}{x \left (2 e^x+3 x\right )} \, dx\\ &=9 \int \left (\frac {4+x}{x^2}-\frac {\log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x^2}\right ) \, dx-27 \int \left (\frac {1}{2 e^x+3 x}-\frac {1}{x \left (2 e^x+3 x\right )}\right ) \, dx\\ &=9 \int \frac {4+x}{x^2} \, dx-9 \int \frac {\log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x^2} \, dx-27 \int \frac {1}{2 e^x+3 x} \, dx+27 \int \frac {1}{x \left (2 e^x+3 x\right )} \, dx\\ &=\frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x}+9 \int \left (\frac {4}{x^2}+\frac {1}{x}\right ) \, dx-9 \int \frac {15 x+2 e^x (4+x)}{x^2 \left (2 e^x+3 x\right )} \, dx-27 \int \frac {1}{2 e^x+3 x} \, dx+27 \int \frac {1}{x \left (2 e^x+3 x\right )} \, dx\\ &=-\frac {36}{x}+9 \log (x)+\frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x}-9 \int \left (\frac {4+x}{x^2}-\frac {3 (-1+x)}{x \left (2 e^x+3 x\right )}\right ) \, dx-27 \int \frac {1}{2 e^x+3 x} \, dx+27 \int \frac {1}{x \left (2 e^x+3 x\right )} \, dx\\ &=-\frac {36}{x}+9 \log (x)+\frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x}-9 \int \frac {4+x}{x^2} \, dx-27 \int \frac {1}{2 e^x+3 x} \, dx+27 \int \frac {1}{x \left (2 e^x+3 x\right )} \, dx+27 \int \frac {-1+x}{x \left (2 e^x+3 x\right )} \, dx\\ &=-\frac {36}{x}+9 \log (x)+\frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x}-9 \int \left (\frac {4}{x^2}+\frac {1}{x}\right ) \, dx-27 \int \frac {1}{2 e^x+3 x} \, dx+27 \int \frac {1}{x \left (2 e^x+3 x\right )} \, dx+27 \int \left (\frac {1}{2 e^x+3 x}-\frac {1}{x \left (2 e^x+3 x\right )}\right ) \, dx\\ &=\frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 22, normalized size = 1.10 \begin {gather*} \frac {9 \log \left (\frac {5}{2} x^4 \left (2 e^x+3 x\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(135*x + E^x*(72 + 18*x) + (-18*E^x - 27*x)*Log[(10*E^x*x^4 + 15*x^5)/2])/(2*E^x*x^2 + 3*x^3),x]

[Out]

(9*Log[(5*x^4*(2*E^x + 3*x))/2])/x

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 19, normalized size = 0.95 \begin {gather*} \frac {9 \, \log \left (\frac {15}{2} \, x^{5} + 5 \, x^{4} e^{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(x)-27*x)*log(5*exp(x)*x^4+15/2*x^5)+(18*x+72)*exp(x)+135*x)/(2*exp(x)*x^2+3*x^3),x, algori
thm="fricas")

[Out]

9*log(15/2*x^5 + 5*x^4*e^x)/x

________________________________________________________________________________________

giac [A]  time = 0.22, size = 19, normalized size = 0.95 \begin {gather*} \frac {9 \, \log \left (\frac {15}{2} \, x^{5} + 5 \, x^{4} e^{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(x)-27*x)*log(5*exp(x)*x^4+15/2*x^5)+(18*x+72)*exp(x)+135*x)/(2*exp(x)*x^2+3*x^3),x, algori
thm="giac")

[Out]

9*log(15/2*x^5 + 5*x^4*e^x)/x

________________________________________________________________________________________

maple [A]  time = 0.13, size = 20, normalized size = 1.00

method	result	size
norman	\(\frac {9 \ln \left (5 \,{\mathrm e}^{x} x^{4}+\frac {15 x^{5}}{2}\right )}{x}\)	\(20\)
risch	\(\frac {9 \ln \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )}{x}-\frac {9 \left (i \pi \mathrm {csgn}\left (i x^{4} \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right )^{3}-i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x^{4} \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right )^{2}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right ) \mathrm {csgn}\left (i x^{4} \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right )^{2}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{4}\right )^{3}-i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right ) \mathrm {csgn}\left (i x^{4} \left (\frac {2 \,{\mathrm e}^{x}}{3}+x \right )\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+2 \ln \relax (2)-2 \ln \relax (5)-8 \ln \relax (x )\right )}{2 x}\)	\(337\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*exp(x)-27*x)*ln(5*exp(x)*x^4+15/2*x^5)+(18*x+72)*exp(x)+135*x)/(2*exp(x)*x^2+3*x^3),x,method=_RETURN
VERBOSE)

[Out]

9*ln(5*exp(x)*x^4+15/2*x^5)/x

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 25, normalized size = 1.25 \begin {gather*} \frac {9 \, {\left (\log \relax (5) - \log \relax (2) + \log \left (3 \, x + 2 \, e^{x}\right ) + 4 \, \log \relax (x)\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(x)-27*x)*log(5*exp(x)*x^4+15/2*x^5)+(18*x+72)*exp(x)+135*x)/(2*exp(x)*x^2+3*x^3),x, algori
thm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.64, size = 19, normalized size = 0.95 \begin {gather*} \frac {9\,\ln \left (5\,x^4\,{\mathrm {e}}^x+\frac {15\,x^5}{2}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((135*x + exp(x)*(18*x + 72) - log(5*x^4*exp(x) + (15*x^5)/2)*(27*x + 18*exp(x)))/(2*x^2*exp(x) + 3*x^3),x)

[Out]

(9*log(5*x^4*exp(x) + (15*x^5)/2))/x

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 19, normalized size = 0.95 \begin {gather*} \frac {9 \log {\left (\frac {15 x^{5}}{2} + 5 x^{4} e^{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*exp(x)-27*x)*ln(5*exp(x)*x**4+15/2*x**5)+(18*x+72)*exp(x)+135*x)/(2*exp(x)*x**2+3*x**3),x)

[Out]

9*log(15*x**5/2 + 5*x**4*exp(x))/x

________________________________________________________________________________________