3.95.25 \(\int \frac {140+70 x-77 x^2+35 x^3+77 x^4+21 x^5+e^x (28 x+28 x^2+7 x^3)+(-56 x^2-56 x^3-14 x^4) \log (5 x)}{4 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=29 \[ 7 \left (e^x+\left (5-x^2\right ) \left (-x-\frac {2}{2+x}+\log (5 x)\right )\right ) \]

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Rubi [A]  time = 0.65, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 16, number of rules used = 7, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1594, 27, 6742, 2194, 44, 43, 2304} \begin {gather*} 7 x^3-7 x^2 \log (5 x)-21 x+7 e^x-\frac {14}{x+2}+35 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(140 + 70*x - 77*x^2 + 35*x^3 + 77*x^4 + 21*x^5 + E^x*(28*x + 28*x^2 + 7*x^3) + (-56*x^2 - 56*x^3 - 14*x^4
)*Log[5*x])/(4*x + 4*x^2 + x^3),x]

[Out]

7*E^x - 21*x + 7*x^3 - 14/(2 + x) + 35*Log[x] - 7*x^2*Log[5*x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1594

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

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, 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 \frac {140+70 x-77 x^2+35 x^3+77 x^4+21 x^5+e^x \left (28 x+28 x^2+7 x^3\right )+\left (-56 x^2-56 x^3-14 x^4\right ) \log (5 x)}{x \left (4+4 x+x^2\right )} \, dx\\ &=\int \frac {140+70 x-77 x^2+35 x^3+77 x^4+21 x^5+e^x \left (28 x+28 x^2+7 x^3\right )+\left (-56 x^2-56 x^3-14 x^4\right ) \log (5 x)}{x (2+x)^2} \, dx\\ &=\int \left (7 e^x+\frac {70}{(2+x)^2}+\frac {140}{x (2+x)^2}-\frac {77 x}{(2+x)^2}+\frac {35 x^2}{(2+x)^2}+\frac {77 x^3}{(2+x)^2}+\frac {21 x^4}{(2+x)^2}-14 x \log (5 x)\right ) \, dx\\ &=-\frac {70}{2+x}+7 \int e^x \, dx-14 \int x \log (5 x) \, dx+21 \int \frac {x^4}{(2+x)^2} \, dx+35 \int \frac {x^2}{(2+x)^2} \, dx-77 \int \frac {x}{(2+x)^2} \, dx+77 \int \frac {x^3}{(2+x)^2} \, dx+140 \int \frac {1}{x (2+x)^2} \, dx\\ &=7 e^x+\frac {7 x^2}{2}-\frac {70}{2+x}-7 x^2 \log (5 x)+21 \int \left (12-4 x+x^2+\frac {16}{(2+x)^2}-\frac {32}{2+x}\right ) \, dx+35 \int \left (1+\frac {4}{(2+x)^2}-\frac {4}{2+x}\right ) \, dx-77 \int \left (-\frac {2}{(2+x)^2}+\frac {1}{2+x}\right ) \, dx+77 \int \left (-4+x-\frac {8}{(2+x)^2}+\frac {12}{2+x}\right ) \, dx+140 \int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx\\ &=7 e^x-21 x+7 x^3-\frac {14}{2+x}+35 \log (x)-7 x^2 \log (5 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 32, normalized size = 1.10 \begin {gather*} 7 \left (e^x-3 x+x^3-\frac {2}{2+x}+5 \log (x)-x^2 \log (5 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(140 + 70*x - 77*x^2 + 35*x^3 + 77*x^4 + 21*x^5 + E^x*(28*x + 28*x^2 + 7*x^3) + (-56*x^2 - 56*x^3 -
14*x^4)*Log[5*x])/(4*x + 4*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 0.51, size = 50, normalized size = 1.72 \begin {gather*} \frac {7 \, {\left (x^{4} + 2 \, x^{3} - 3 \, x^{2} + {\left (x + 2\right )} e^{x} - {\left (x^{3} + 2 \, x^{2} - 5 \, x - 10\right )} \log \left (5 \, x\right ) - 6 \, x - 2\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-14*x^4-56*x^3-56*x^2)*log(5*x)+(7*x^3+28*x^2+28*x)*exp(x)+21*x^5+77*x^4+35*x^3-77*x^2+70*x+140)/(
x^3+4*x^2+4*x),x, algorithm="fricas")

[Out]

7*(x^4 + 2*x^3 - 3*x^2 + (x + 2)*e^x - (x^3 + 2*x^2 - 5*x - 10)*log(5*x) - 6*x - 2)/(x + 2)

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giac [B]  time = 0.23, size = 60, normalized size = 2.07 \begin {gather*} \frac {7 \, {\left (x^{4} - x^{3} \log \left (5 \, x\right ) + 2 \, x^{3} - 2 \, x^{2} \log \left (5 \, x\right ) - 3 \, x^{2} + x e^{x} + 5 \, x \log \relax (x) - 6 \, x + 2 \, e^{x} + 10 \, \log \relax (x) - 2\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-14*x^4-56*x^3-56*x^2)*log(5*x)+(7*x^3+28*x^2+28*x)*exp(x)+21*x^5+77*x^4+35*x^3-77*x^2+70*x+140)/(
x^3+4*x^2+4*x),x, algorithm="giac")

[Out]

7*(x^4 - x^3*log(5*x) + 2*x^3 - 2*x^2*log(5*x) - 3*x^2 + x*e^x + 5*x*log(x) - 6*x + 2*e^x + 10*log(x) - 2)/(x
+ 2)

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maple [A]  time = 0.05, size = 34, normalized size = 1.17

method	result	size
default	\(-7 x^{2} \ln \left (5 x \right )+7 x^{3}-21 x +35 \ln \relax (x )-\frac {14}{2+x}+7 \,{\mathrm e}^{x}\)	\(34\)
risch	\(-7 x^{2} \ln \left (5 x \right )+\frac {7 x^{4}+14 x^{3}+35 x \ln \relax (x )-21 x^{2}+7 \,{\mathrm e}^{x} x +70 \ln \relax (x )-42 x +14 \,{\mathrm e}^{x}-14}{2+x}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-14*x^4-56*x^3-56*x^2)*ln(5*x)+(7*x^3+28*x^2+28*x)*exp(x)+21*x^5+77*x^4+35*x^3-77*x^2+70*x+140)/(x^3+4*x
^2+4*x),x,method=_RETURNVERBOSE)

[Out]

-7*x^2*ln(5*x)+7*x^3-21*x+35*ln(x)-14/(2+x)+7*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 7 \, x^{3} - \frac {7}{2} \, x^{2} {\left (2 \, \log \relax (5) - 1\right )} - 7 \, x^{2} \log \relax (x) - \frac {7}{2} \, x^{2} - 21 \, x - \frac {28 \, e^{\left (-2\right )} E_{2}\left (-x - 2\right )}{x + 2} - \frac {14}{x + 2} + 7 \, \int \frac {{\left (x^{2} + 4 \, x\right )} e^{x}}{x^{2} + 4 \, x + 4}\,{d x} + 35 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-14*x^4-56*x^3-56*x^2)*log(5*x)+(7*x^3+28*x^2+28*x)*exp(x)+21*x^5+77*x^4+35*x^3-77*x^2+70*x+140)/(
x^3+4*x^2+4*x),x, algorithm="maxima")

[Out]

7*x^3 - 7/2*x^2*(2*log(5) - 1) - 7*x^2*log(x) - 7/2*x^2 - 21*x - 28*e^(-2)*exp_integral_e(2, -x - 2)/(x + 2) -
14/(x + 2) + 7*integrate((x^2 + 4*x)*e^x/(x^2 + 4*x + 4), x) + 35*log(x)

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mupad [B]  time = 6.90, size = 38, normalized size = 1.31 \begin {gather*} 7\,{\mathrm {e}}^x-21\,x+35\,\ln \relax (x)-7\,x^2\,\ln \relax (x)-\frac {14}{x+2}-7\,x^2\,\ln \relax (5)+7\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((70*x - log(5*x)*(56*x^2 + 56*x^3 + 14*x^4) - 77*x^2 + 35*x^3 + 77*x^4 + 21*x^5 + exp(x)*(28*x + 28*x^2 +
7*x^3) + 140)/(4*x + 4*x^2 + x^3),x)

[Out]

7*exp(x) - 21*x + 35*log(x) - 7*x^2*log(x) - 14/(x + 2) - 7*x^2*log(5) + 7*x^3

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sympy [A]  time = 0.40, size = 32, normalized size = 1.10 \begin {gather*} 7 x^{3} - 7 x^{2} \log {\left (5 x \right )} - 21 x + 7 e^{x} + 35 \log {\relax (x )} - \frac {14}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-14*x**4-56*x**3-56*x**2)*ln(5*x)+(7*x**3+28*x**2+28*x)*exp(x)+21*x**5+77*x**4+35*x**3-77*x**2+70*
x+140)/(x**3+4*x**2+4*x),x)

[Out]

7*x**3 - 7*x**2*log(5*x) - 21*x + 7*exp(x) + 35*log(x) - 14/(x + 2)

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