\(\int \frac {1}{5} (e^{x/5} (20+14 x)+e^{x/5} (10 x+x^2) \log (2 x^2)) \, dx\) [2060]

Optimal result

Integrand size = 39, antiderivative size = 19 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x \left (4+x \log \left (2 x^2\right )\right ) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(19)=38\).

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.68, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2207, 2225, 1607, 2227, 2634} \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x^2 \log \left (2 x^2\right )-10 e^{x/5} x-20 e^{x/5}+2 e^{x/5} (7 x+10) \]

[In]

Int[(E^(x/5)*(20 + 14*x) + E^(x/5)*(10*x + x^2)*Log[2*x^2])/5,x]

[Out]

-20*E^(x/5) - 10*E^(x/5)*x + 2*E^(x/5)*(10 + 7*x) + E^(x/5)*x^2*Log[2*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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

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

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int e^{x/5} (20+14 x) \, dx+\frac {1}{5} \int e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right ) \, dx \\ & = 2 e^{x/5} (10+7 x)+\frac {1}{5} \int e^{x/5} x (10+x) \log \left (2 x^2\right ) \, dx-14 \int e^{x/5} \, dx \\ & = -70 e^{x/5}+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )-\frac {1}{5} \int 10 e^{x/5} x \, dx \\ & = -70 e^{x/5}+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )-2 \int e^{x/5} x \, dx \\ & = -70 e^{x/5}-10 e^{x/5} x+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )+10 \int e^{x/5} \, dx \\ & = -20 e^{x/5}-10 e^{x/5} x+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x \left (4+x \log \left (2 x^2\right )\right ) \]

[In]

Integrate[(E^(x/5)*(20 + 14*x) + E^(x/5)*(10*x + x^2)*Log[2*x^2])/5,x]

[Out]

E^(x/5)*x*(4 + x*Log[2*x^2])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21

method	result	size
norman	\({\mathrm e}^{\frac {x}{5}} \ln \left (2 x^{2}\right ) x^{2}+4 x \,{\mathrm e}^{\frac {x}{5}}\)	\(23\)
parallelrisch	\({\mathrm e}^{\frac {x}{5}} \ln \left (2 x^{2}\right ) x^{2}+4 x \,{\mathrm e}^{\frac {x}{5}}\)	\(23\)
default	\(4 x \,{\mathrm e}^{\frac {x}{5}}+\left (\ln \left (2 x^{2}\right )-2 \ln \left (x \right )\right ) x^{2} {\mathrm e}^{\frac {x}{5}}+2 \ln \left (x \right ) x^{2} {\mathrm e}^{\frac {x}{5}}\)	\(39\)
risch	\(2 \ln \left (x \right ) x^{2} {\mathrm e}^{\frac {x}{5}}+\frac {\left (100-20 x -i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x^{2} \ln \left (2\right )\right ) {\mathrm e}^{\frac {x}{5}}}{2}+\frac {\left (-250+70 x \right ) {\mathrm e}^{\frac {x}{5}}}{5}\)	\(100\)

[In]

int(1/5*(x^2+10*x)*exp(1/5*x)*ln(2*x^2)+1/5*(14*x+20)*exp(1/5*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 4 \, x e^{\left (\frac {1}{5} \, x\right )} \]

[In]

integrate(1/5*(x^2+10*x)*exp(1/5*x)*log(2*x^2)+1/5*(14*x+20)*exp(1/5*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=\left (x^{2} \log {\left (2 x^{2} \right )} + 4 x\right ) e^{\frac {x}{5}} \]

[In]

integrate(1/5*(x**2+10*x)*exp(1/5*x)*ln(2*x**2)+1/5*(14*x+20)*exp(1/5*x),x)

[Out]

(x**2*log(2*x**2) + 4*x)*exp(x/5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 4 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} + 20 \, e^{\left (\frac {1}{5} \, x\right )} \]

[In]

integrate(1/5*(x^2+10*x)*exp(1/5*x)*log(2*x^2)+1/5*(14*x+20)*exp(1/5*x),x, algorithm="maxima")

[Out]

x^2*e^(1/5*x)*log(2*x^2) + 4*(x - 5)*e^(1/5*x) + 20*e^(1/5*x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 2 \, {\left (7 \, x - 25\right )} e^{\left (\frac {1}{5} \, x\right )} - 10 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} \]

[In]

integrate(1/5*(x^2+10*x)*exp(1/5*x)*log(2*x^2)+1/5*(14*x+20)*exp(1/5*x),x, algorithm="giac")

[Out]

x^2*e^(1/5*x)*log(2*x^2) + 2*(7*x - 25)*e^(1/5*x) - 10*(x - 5)*e^(1/5*x)

Mupad [B] (verification not implemented)

Time = 8.83 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x\,{\mathrm {e}}^{x/5}\,\left (x\,\ln \left (2\,x^2\right )+4\right ) \]

[In]

int((exp(x/5)*(14*x + 20))/5 + (exp(x/5)*log(2*x^2)*(10*x + x^2))/5,x)

[Out]

x*exp(x/5)*(x*log(2*x^2) + 4)