\(\int \frac {25+9 x^3-21 x^4+e^{2 x} (x^2-2 x^3-2 x^4)+(-25+14 x^3+e^{2 x} (x^2+2 x^3)) \log (x)}{x^2} \, dx\) [5288]

Optimal result

Integrand size = 61, antiderivative size = 26 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=x \left (x-\left (e^{2 x}+\frac {25}{x^2}+7 x\right ) (x-\log (x))\right ) \]

[Out]

x*(x-(x-ln(x))*(7*x+exp(x)^2+25/x^2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50, number of steps used = 9, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {14, 2326, 2372} \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=-7 x^3+x^2+7 x^2 \log (x)-e^{2 x} \left (x^2-x \log (x)\right )+\frac {25 \log (x)}{x} \]

[In]

Int[(25 + 9*x^3 - 21*x^4 + E^(2*x)*(x^2 - 2*x^3 - 2*x^4) + (-25 + 14*x^3 + E^(2*x)*(x^2 + 2*x^3))*Log[x])/x^2,
x]

[Out]

x^2 - 7*x^3 + (25*Log[x])/x + 7*x^2*Log[x] - E^(2*x)*(x^2 - x*Log[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]

Rule 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{2 x} \left (-1+2 x+2 x^2-\log (x)-2 x \log (x)\right )+\frac {25+9 x^3-21 x^4-25 \log (x)+14 x^3 \log (x)}{x^2}\right ) \, dx \\ & = -\int e^{2 x} \left (-1+2 x+2 x^2-\log (x)-2 x \log (x)\right ) \, dx+\int \frac {25+9 x^3-21 x^4-25 \log (x)+14 x^3 \log (x)}{x^2} \, dx \\ & = -e^{2 x} \left (x^2-x \log (x)\right )+\int \left (\frac {25+9 x^3-21 x^4}{x^2}+\frac {\left (-25+14 x^3\right ) \log (x)}{x^2}\right ) \, dx \\ & = -e^{2 x} \left (x^2-x \log (x)\right )+\int \frac {25+9 x^3-21 x^4}{x^2} \, dx+\int \frac {\left (-25+14 x^3\right ) \log (x)}{x^2} \, dx \\ & = \frac {25 \log (x)}{x}+7 x^2 \log (x)-e^{2 x} \left (x^2-x \log (x)\right )-\int \left (\frac {25}{x^2}+7 x\right ) \, dx+\int \left (\frac {25}{x^2}+9 x-21 x^2\right ) \, dx \\ & = x^2-7 x^3+\frac {25 \log (x)}{x}+7 x^2 \log (x)-e^{2 x} \left (x^2-x \log (x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=-x^2 \left (-1+e^{2 x}+7 x\right )+\left (\frac {25}{x}+e^{2 x} x+7 x^2\right ) \log (x) \]

[In]

Integrate[(25 + 9*x^3 - 21*x^4 + E^(2*x)*(x^2 - 2*x^3 - 2*x^4) + (-25 + 14*x^3 + E^(2*x)*(x^2 + 2*x^3))*Log[x]
)/x^2,x]

[Out]

-(x^2*(-1 + E^(2*x) + 7*x)) + (25/x + E^(2*x)*x + 7*x^2)*Log[x]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54

method	result	size
risch	\(\frac {\left ({\mathrm e}^{2 x} x^{2}+7 x^{3}+25\right ) \ln \left (x \right )}{x}-{\mathrm e}^{2 x} x^{2}-7 x^{3}+x^{2}\)	\(40\)
default	\(x \,{\mathrm e}^{2 x} \ln \left (x \right )-{\mathrm e}^{2 x} x^{2}+7 x^{2} \ln \left (x \right )+x^{2}+\frac {25 \ln \left (x \right )}{x}-7 x^{3}\)	\(41\)
parts	\(x \,{\mathrm e}^{2 x} \ln \left (x \right )-{\mathrm e}^{2 x} x^{2}+7 x^{2} \ln \left (x \right )+x^{2}+\frac {25 \ln \left (x \right )}{x}-7 x^{3}\)	\(41\)
parallelrisch	\(\frac {{\mathrm e}^{2 x} \ln \left (x \right ) x^{2}-{\mathrm e}^{2 x} x^{3}+7 x^{3} \ln \left (x \right )-7 x^{4}+x^{3}+25 \ln \left (x \right )}{x}\)	\(44\)

[In]

int((((2*x^3+x^2)*exp(x)^2+14*x^3-25)*ln(x)+(-2*x^4-2*x^3+x^2)*exp(x)^2-21*x^4+9*x^3+25)/x^2,x,method=_RETURNV
ERBOSE)

[Out]

(exp(2*x)*x^2+7*x^3+25)/x*ln(x)-exp(2*x)*x^2-7*x^3+x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=-\frac {7 \, x^{4} + x^{3} e^{\left (2 \, x\right )} - x^{3} - {\left (7 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 25\right )} \log \left (x\right )}{x} \]

[In]

integrate((((2*x^3+x^2)*exp(x)^2+14*x^3-25)*log(x)+(-2*x^4-2*x^3+x^2)*exp(x)^2-21*x^4+9*x^3+25)/x^2,x, algorit
hm="fricas")

[Out]

-(7*x^4 + x^3*e^(2*x) - x^3 - (7*x^3 + x^2*e^(2*x) + 25)*log(x))/x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=- 7 x^{3} + x^{2} + \left (- x^{2} + x \log {\left (x \right )}\right ) e^{2 x} + \frac {\left (7 x^{3} + 25\right ) \log {\left (x \right )}}{x} \]

[In]

integrate((((2*x**3+x**2)*exp(x)**2+14*x**3-25)*ln(x)+(-2*x**4-2*x**3+x**2)*exp(x)**2-21*x**4+9*x**3+25)/x**2,
x)

[Out]

-7*x**3 + x**2 + (-x**2 + x*log(x))*exp(2*x) + (7*x**3 + 25)*log(x)/x

Maxima [F]

\[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=\int { -\frac {21 \, x^{4} - 9 \, x^{3} + {\left (2 \, x^{4} + 2 \, x^{3} - x^{2}\right )} e^{\left (2 \, x\right )} - {\left (14 \, x^{3} + {\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} - 25\right )} \log \left (x\right ) - 25}{x^{2}} \,d x } \]

[In]

integrate((((2*x^3+x^2)*exp(x)^2+14*x^3-25)*log(x)+(-2*x^4-2*x^3+x^2)*exp(x)^2-21*x^4+9*x^3+25)/x^2,x, algorit
hm="maxima")

[Out]

-7*x^3 + 7*x^2*log(x) + 1/2*(2*x - 1)*e^(2*x)*log(x) + x^2 - 1/2*(2*x^2 - 2*x + 1)*e^(2*x) - 1/2*(2*x - 1)*e^(
2*x) + 1/2*e^(2*x)*log(x) + 25*log(x)/x - 1/2*Ei(2*x) + 1/2*e^(2*x) - 1/2*integrate((2*x - 1)*e^(2*x)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=-\frac {7 \, x^{4} + x^{3} e^{\left (2 \, x\right )} - 7 \, x^{3} \log \left (x\right ) - x^{2} e^{\left (2 \, x\right )} \log \left (x\right ) - x^{3} - 25 \, \log \left (x\right )}{x} \]

[In]

integrate((((2*x^3+x^2)*exp(x)^2+14*x^3-25)*log(x)+(-2*x^4-2*x^3+x^2)*exp(x)^2-21*x^4+9*x^3+25)/x^2,x, algorit
hm="giac")

[Out]

-(7*x^4 + x^3*e^(2*x) - 7*x^3*log(x) - x^2*e^(2*x)*log(x) - x^3 - 25*log(x))/x

Mupad [B] (verification not implemented)

Time = 14.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {25+9 x^3-21 x^4+e^{2 x} \left (x^2-2 x^3-2 x^4\right )+\left (-25+14 x^3+e^{2 x} \left (x^2+2 x^3\right )\right ) \log (x)}{x^2} \, dx=\frac {25\,\ln \left (x\right )}{x}+x^2\,\left (7\,\ln \left (x\right )-{\mathrm {e}}^{2\,x}+1\right )-7\,x^3+x\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right ) \]

[In]

int((9*x^3 - exp(2*x)*(2*x^3 - x^2 + 2*x^4) - 21*x^4 + log(x)*(exp(2*x)*(x^2 + 2*x^3) + 14*x^3 - 25) + 25)/x^2
,x)

[Out]

(25*log(x))/x + x^2*(7*log(x) - exp(2*x) + 1) - 7*x^3 + x*exp(2*x)*log(x)