3.70.3 \(\int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+(-10 x-4 x^2) \log (e^x (5 x+2 x^2))}{5+2 x} \, dx\)

Optimal. Leaf size=31 \[ -e^x-x+x^2 \left (1-\log \left (e^x (x+x (4+2 x))\right )\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 28, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6742, 2194, 1850, 2551, 771} \begin {gather*} x^2+x^2 \left (-\log \left (e^x x (2 x+5)\right )\right )-x-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + E^x*(-5 - 2*x) + 3*x - 5*x^2 - 2*x^3 + (-10*x - 4*x^2)*Log[E^x*(5*x + 2*x^2)])/(5 + 2*x),x]

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2194

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 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 (-e^x+\frac {-5+3 x-5 x^2-2 x^3-10 x \log \left (e^x x (5+2 x)\right )-4 x^2 \log \left (e^x x (5+2 x)\right )}{5+2 x}\right ) \, dx\\ &=-\int e^x \, dx+\int \frac {-5+3 x-5 x^2-2 x^3-10 x \log \left (e^x x (5+2 x)\right )-4 x^2 \log \left (e^x x (5+2 x)\right )}{5+2 x} \, dx\\ &=-e^x+\int \left (\frac {-5+3 x-5 x^2-2 x^3}{5+2 x}-2 x \log \left (e^x x (5+2 x)\right )\right ) \, dx\\ &=-e^x-2 \int x \log \left (e^x x (5+2 x)\right ) \, dx+\int \frac {-5+3 x-5 x^2-2 x^3}{5+2 x} \, dx\\ &=-e^x-x^2 \log \left (e^x x (5+2 x)\right )+\int \frac {x \left (5+9 x+2 x^2\right )}{5+2 x} \, dx+\int \left (\frac {3}{2}-x^2-\frac {25}{2 (5+2 x)}\right ) \, dx\\ &=-e^x+\frac {3 x}{2}-\frac {x^3}{3}-\frac {25}{4} \log (5+2 x)-x^2 \log \left (e^x x (5+2 x)\right )+\int \left (-\frac {5}{2}+2 x+x^2+\frac {25}{2 (5+2 x)}\right ) \, dx\\ &=-e^x-x+x^2-x^2 \log \left (e^x x (5+2 x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 28, normalized size = 0.90 \begin {gather*} -e^x-x+x^2-x^2 \log \left (e^x x (5+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + E^x*(-5 - 2*x) + 3*x - 5*x^2 - 2*x^3 + (-10*x - 4*x^2)*Log[E^x*(5*x + 2*x^2)])/(5 + 2*x),x]

[Out]

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

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fricas [A]  time = 0.75, size = 29, normalized size = 0.94 \begin {gather*} -x^{2} \log \left ({\left (2 \, x^{2} + 5 \, x\right )} e^{x}\right ) + x^{2} - x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x-5)/(5+2*x),x, algorithm="fric
as")

[Out]

-x^2*log((2*x^2 + 5*x)*e^x) + x^2 - x - e^x

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giac [A]  time = 0.27, size = 31, normalized size = 1.00 \begin {gather*} -x^{3} - x^{2} \log \left (2 \, x^{2} + 5 \, x\right ) + x^{2} - x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x-5)/(5+2*x),x, algorithm="giac
")

[Out]

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

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maple [A]  time = 0.15, size = 30, normalized size = 0.97

method	result	size
default	\(-x^{2} \ln \left (\left (2 x^{2}+5 x \right ) {\mathrm e}^{x}\right )-x +x^{2}-{\mathrm e}^{x}\)	\(30\)
norman	\(-x^{2} \ln \left (\left (2 x^{2}+5 x \right ) {\mathrm e}^{x}\right )-x +x^{2}-{\mathrm e}^{x}\)	\(30\)
risch	\(-x^{2} \ln \left ({\mathrm e}^{x}\right )-x^{2} \ln \left (\frac {5}{2}+x \right )-x^{2} \ln \relax (x )-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (\frac {5}{2}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{3}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{3}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right ) \mathrm {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right ) \mathrm {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i \left (\frac {5}{2}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{2}}{2}-x^{2} \ln \relax (2)+x^{2}-x -{\mathrm e}^{x}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2-10*x)*ln((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x-5)/(5+2*x),x,method=_RETURNVERBOSE)

[Out]

-x^2*ln((2*x^2+5*x)*exp(x))-x+x^2-exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x^{3} - x^{2} \log \relax (x) + x^{2} + \frac {5}{2} \, e^{\left (-\frac {5}{2}\right )} E_{1}\left (-x - \frac {5}{2}\right ) - \frac {1}{4} \, {\left (4 \, x^{2} - 25\right )} \log \left (2 \, x + 5\right ) - x - 2 \, \int \frac {x e^{x}}{2 \, x + 5}\,{d x} - \frac {25}{4} \, \log \left (2 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x-5)/(5+2*x),x, algorithm="maxi
ma")

[Out]

-x^3 - x^2*log(x) + x^2 + 5/2*e^(-5/2)*exp_integral_e(1, -x - 5/2) - 1/4*(4*x^2 - 25)*log(2*x + 5) - x - 2*int
egrate(x*e^x/(2*x + 5), x) - 25/4*log(2*x + 5)

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mupad [B]  time = 4.37, size = 29, normalized size = 0.94 \begin {gather*} x^2-{\mathrm {e}}^x-x^2\,\ln \left ({\mathrm {e}}^x\,\left (2\,x^2+5\,x\right )\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 - exp(x) - x^2*log(exp(x)*(5*x + 2*x^2)) - x

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sympy [A]  time = 0.63, size = 44, normalized size = 1.42 \begin {gather*} x^{2} - \frac {49 x}{24} + \left (\frac {25}{24} - x^{2}\right ) \log {\left (\left (2 x^{2} + 5 x\right ) e^{x} \right )} - e^{x} - \frac {25 \log {\left (2 x^{2} + 5 x \right )}}{24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2-10*x)*ln((2*x**2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x**3-5*x**2+3*x-5)/(5+2*x),x)

[Out]

x**2 - 49*x/24 + (25/24 - x**2)*log((2*x**2 + 5*x)*exp(x)) - exp(x) - 25*log(2*x**2 + 5*x)/24

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