3.50.18 \(\int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x (-6 x^2-10 x^3-10 x^4)+(500 x^3-40 e^x x^3+200 x^4+20 x^5) \log (25 x-2 e^x x+10 x^2+x^3)}{-125+10 e^x-50 x-5 x^2} \, dx\)

Optimal. Leaf size=28 \[ x^4 \left (-\frac {1}{5 x}-\log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right ) \]

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Rubi [A]  time = 2.13, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 7, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6741, 12, 6688, 6742, 14, 2551, 43} \begin {gather*} x^4 \left (-\log \left (x (x+5)^2-2 e^x x\right )\right )-\frac {x^3}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*x^2 + 155*x^3 + 103*x^4 + 15*x^5 + E^x*(-6*x^2 - 10*x^3 - 10*x^4) + (500*x^3 - 40*E^x*x^3 + 200*x^4 +
20*x^5)*Log[25*x - 2*E^x*x + 10*x^2 + x^3])/(-125 + 10*E^x - 50*x - 5*x^2),x]

[Out]

-1/5*x^3 - x^4*Log[-2*E^x*x + 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 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 6688

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

Rule 6741

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

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 \frac {-75 x^2-155 x^3-103 x^4-15 x^5-e^x \left (-6 x^2-10 x^3-10 x^4\right )-\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{5 \left (25-2 e^x+10 x+x^2\right )} \, dx\\ &=\frac {1}{5} \int \frac {-75 x^2-155 x^3-103 x^4-15 x^5-e^x \left (-6 x^2-10 x^3-10 x^4\right )-\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{25-2 e^x+10 x+x^2} \, dx\\ &=\frac {1}{5} \int \frac {x^2 \left (75+155 x+103 x^2+15 x^3-2 e^x \left (3+5 x+5 x^2\right )+20 x \left (-2 e^x+(5+x)^2\right ) \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right )}{2 e^x-(5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2}-x^2 \left (3+5 x+5 x^2+20 x \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right )\right ) \, dx\\ &=-\left (\frac {1}{5} \int x^2 \left (3+5 x+5 x^2+20 x \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right ) \, dx\right )+\int \frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2} \, dx\\ &=-\left (\frac {1}{5} \int \left (x^2 \left (3+5 x+5 x^2\right )+20 x^3 \log \left (-2 e^x x+x (5+x)^2\right )\right ) \, dx\right )+\int \left (\frac {15 x^4}{25-2 e^x+10 x+x^2}+\frac {8 x^5}{25-2 e^x+10 x+x^2}+\frac {x^6}{25-2 e^x+10 x+x^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int x^2 \left (3+5 x+5 x^2\right ) \, dx\right )-4 \int x^3 \log \left (-2 e^x x+x (5+x)^2\right ) \, dx+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx\\ &=-x^4 \log \left (-2 e^x x+x (5+x)^2\right )-\frac {1}{5} \int \left (3 x^2+5 x^3+5 x^4\right ) \, dx+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^3 \left (-25-20 x-3 x^2+2 e^x (1+x)\right )}{2 e^x-(5+x)^2} \, dx\\ &=-\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \left (x^3 (1+x)-\frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2}\right ) \, dx\\ &=-\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int x^3 (1+x) \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx-\int \frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2} \, dx\\ &=-\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \left (x^3+x^4\right ) \, dx-\int \left (\frac {15 x^4}{25-2 e^x+10 x+x^2}+\frac {8 x^5}{25-2 e^x+10 x+x^2}+\frac {x^6}{25-2 e^x+10 x+x^2}\right ) \, dx\\ &=-\frac {x^3}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{5} \left (-x^3-5 x^4 \log \left (-2 e^x x+x (5+x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*x^2 + 155*x^3 + 103*x^4 + 15*x^5 + E^x*(-6*x^2 - 10*x^3 - 10*x^4) + (500*x^3 - 40*E^x*x^3 + 200*
x^4 + 20*x^5)*Log[25*x - 2*E^x*x + 10*x^2 + x^3])/(-125 + 10*E^x - 50*x - 5*x^2),x]

[Out]

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

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fricas [A]  time = 0.79, size = 29, normalized size = 1.04 \begin {gather*} -x^{4} \log \left (x^{3} + 10 \, x^{2} - 2 \, x e^{x} + 25 \, x\right ) - \frac {1}{5} \, x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*exp(x)*x^3+20*x^5+200*x^4+500*x^3)*log(-2*exp(x)*x+x^3+10*x^2+25*x)+(-10*x^4-10*x^3-6*x^2)*exp
(x)+15*x^5+103*x^4+155*x^3+75*x^2)/(10*exp(x)-5*x^2-50*x-125),x, algorithm="fricas")

[Out]

-x^4*log(x^3 + 10*x^2 - 2*x*e^x + 25*x) - 1/5*x^3

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giac [A]  time = 0.25, size = 29, normalized size = 1.04 \begin {gather*} -x^{4} \log \left (x^{3} + 10 \, x^{2} - 2 \, x e^{x} + 25 \, x\right ) - \frac {1}{5} \, x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*exp(x)*x^3+20*x^5+200*x^4+500*x^3)*log(-2*exp(x)*x+x^3+10*x^2+25*x)+(-10*x^4-10*x^3-6*x^2)*exp
(x)+15*x^5+103*x^4+155*x^3+75*x^2)/(10*exp(x)-5*x^2-50*x-125),x, algorithm="giac")

[Out]

-x^4*log(x^3 + 10*x^2 - 2*x*e^x + 25*x) - 1/5*x^3

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maple [C]  time = 0.13, size = 188, normalized size = 6.71

method	result	size
risch	\(-x^{4} \ln \left (x^{2}-2 \,{\mathrm e}^{x}+10 x +25\right )-x^{4} \ln \relax (x )+\frac {i \pi \,x^{4} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right ) \mathrm {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )}{2}-\frac {i \pi \,x^{4} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )^{2}}{2}+\frac {i \pi \,x^{4} \mathrm {csgn}\left (i \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right ) \mathrm {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )^{2}}{2}-\frac {i \pi \,x^{4} \mathrm {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )^{3}}{2}-\frac {x^{3}}{5}\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*exp(x)*x^3+20*x^5+200*x^4+500*x^3)*ln(-2*exp(x)*x+x^3+10*x^2+25*x)+(-10*x^4-10*x^3-6*x^2)*exp(x)+15*
x^5+103*x^4+155*x^3+75*x^2)/(10*exp(x)-5*x^2-50*x-125),x,method=_RETURNVERBOSE)

[Out]

-x^4*ln(x^2-2*exp(x)+10*x+25)-x^4*ln(x)+1/2*I*Pi*x^4*csgn(I*x)*csgn(I*(-x^2+2*exp(x)-10*x-25))*csgn(I*x*(-x^2+
2*exp(x)-10*x-25))-1/2*I*Pi*x^4*csgn(I*x)*csgn(I*x*(-x^2+2*exp(x)-10*x-25))^2+1/2*I*Pi*x^4*csgn(I*(-x^2+2*exp(
x)-10*x-25))*csgn(I*x*(-x^2+2*exp(x)-10*x-25))^2-1/2*I*Pi*x^4*csgn(I*x*(-x^2+2*exp(x)-10*x-25))^3-1/5*x^3

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maxima [A]  time = 0.39, size = 31, normalized size = 1.11 \begin {gather*} -x^{4} \log \left (x^{2} + 10 \, x - 2 \, e^{x} + 25\right ) - x^{4} \log \relax (x) - \frac {1}{5} \, x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*exp(x)*x^3+20*x^5+200*x^4+500*x^3)*log(-2*exp(x)*x+x^3+10*x^2+25*x)+(-10*x^4-10*x^3-6*x^2)*exp
(x)+15*x^5+103*x^4+155*x^3+75*x^2)/(10*exp(x)-5*x^2-50*x-125),x, algorithm="maxima")

[Out]

-x^4*log(x^2 + 10*x - 2*e^x + 25) - x^4*log(x) - 1/5*x^3

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mupad [B]  time = 4.44, size = 29, normalized size = 1.04 \begin {gather*} -x^4\,\ln \left (25\,x-2\,x\,{\mathrm {e}}^x+10\,x^2+x^3\right )-\frac {x^3}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(75*x^2 - exp(x)*(6*x^2 + 10*x^3 + 10*x^4) + 155*x^3 + 103*x^4 + 15*x^5 + log(25*x - 2*x*exp(x) + 10*x^2
+ x^3)*(500*x^3 - 40*x^3*exp(x) + 200*x^4 + 20*x^5))/(50*x - 10*exp(x) + 5*x^2 + 125),x)

[Out]

- x^4*log(25*x - 2*x*exp(x) + 10*x^2 + x^3) - x^3/5

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sympy [A]  time = 0.53, size = 29, normalized size = 1.04 \begin {gather*} - x^{4} \log {\left (x^{3} + 10 x^{2} - 2 x e^{x} + 25 x \right )} - \frac {x^{3}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*exp(x)*x**3+20*x**5+200*x**4+500*x**3)*ln(-2*exp(x)*x+x**3+10*x**2+25*x)+(-10*x**4-10*x**3-6*x
**2)*exp(x)+15*x**5+103*x**4+155*x**3+75*x**2)/(10*exp(x)-5*x**2-50*x-125),x)

[Out]

-x**4*log(x**3 + 10*x**2 - 2*x*exp(x) + 25*x) - x**3/5

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