3.103.67 \(\int \frac {10 x+2 x^2+e^3 (8 x+4 x^2+2 x^3)+(10+2 x+e^3 (8+4 x+2 x^2)) \log (4+x+e^3 (4+x^2))}{4+x+e^3 (4+x^2)} \, dx\)

Optimal. Leaf size=17 \[ \left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )^2 \]

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Rubi [A]  time = 0.17, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6741, 12, 6686} \begin {gather*} \left (\log \left (e^3 \left (x^2+4\right )+x+4\right )+x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + 2*x^2 + E^3*(8*x + 4*x^2 + 2*x^3) + (10 + 2*x + E^3*(8 + 4*x + 2*x^2))*Log[4 + x + E^3*(4 + x^2)])
/(4 + x + E^3*(4 + x^2)),x]

[Out]

(x + Log[4 + x + E^3*(4 + x^2)])^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (5+4 e^3+\left (1+2 e^3\right ) x+e^3 x^2\right ) \left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )}{4 \left (1+e^3\right )+x+e^3 x^2} \, dx\\ &=2 \int \frac {\left (5+4 e^3+\left (1+2 e^3\right ) x+e^3 x^2\right ) \left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )}{4 \left (1+e^3\right )+x+e^3 x^2} \, dx\\ &=\left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 1.00 \begin {gather*} \left (x+\log \left (4+x+e^3 \left (4+x^2\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 2*x^2 + E^3*(8*x + 4*x^2 + 2*x^3) + (10 + 2*x + E^3*(8 + 4*x + 2*x^2))*Log[4 + x + E^3*(4 +
x^2)])/(4 + x + E^3*(4 + x^2)),x]

[Out]

(x + Log[4 + x + E^3*(4 + x^2)])^2

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fricas [B]  time = 0.51, size = 33, normalized size = 1.94 \begin {gather*} x^{2} + 2 \, x \log \left ({\left (x^{2} + 4\right )} e^{3} + x + 4\right ) + \log \left ({\left (x^{2} + 4\right )} e^{3} + x + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4*x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)
*exp(3)+4+x),x, algorithm="fricas")

[Out]

x^2 + 2*x*log((x^2 + 4)*e^3 + x + 4) + log((x^2 + 4)*e^3 + x + 4)^2

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giac [B]  time = 0.20, size = 37, normalized size = 2.18 \begin {gather*} x^{2} + 2 \, x \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) + \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4*x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)
*exp(3)+4+x),x, algorithm="giac")

[Out]

x^2 + 2*x*log(x^2*e^3 + x + 4*e^3 + 4) + log(x^2*e^3 + x + 4*e^3 + 4)^2

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maple [B]  time = 0.38, size = 34, normalized size = 2.00

method	result	size
norman	\(x^{2}+\ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )^{2}+2 x \ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )\)	\(34\)
risch	\(x^{2}+\ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )^{2}+2 x \ln \left (\left (x^{2}+4\right ) {\mathrm e}^{3}+4+x \right )\)	\(34\)
default	error in gcdex: invalid arguments\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+4*x+8)*exp(3)+2*x+10)*ln((x^2+4)*exp(3)+4+x)+(2*x^3+4*x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)*exp(3)
+4+x),x,method=_RETURNVERBOSE)

[Out]

x^2+ln((x^2+4)*exp(3)+4+x)^2+2*x*ln((x^2+4)*exp(3)+4+x)

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maxima [B]  time = 0.69, size = 439, normalized size = 25.82 \begin {gather*} 2 \, \sqrt {16 \, e^{6} + 16 \, e^{3} - 1} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )} - \frac {2 \, {\left (8 \, e^{6} + 8 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-6\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - {\left ({\left (4 \, e^{6} + 4 \, e^{3} - 1\right )} e^{\left (-9\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {2 \, {\left (12 \, e^{6} + 12 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-9\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - {\left (x^{2} e^{3} - 2 \, x\right )} e^{\left (-6\right )}\right )} e^{3} - 2 \, {\left (\frac {2 \, {\left (8 \, e^{6} + 8 \, e^{3} - 1\right )} \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-6\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} - 2 \, x e^{\left (-3\right )} + e^{\left (-6\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )\right )} e^{3} + 4 \, {\left (e^{\left (-3\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {2 \, \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right )} e^{3} + {\left (e^{3} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )^{2} - 4 \, x e^{3} + {\left (2 \, x e^{3} + 1\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right )\right )} e^{\left (-3\right )} + 2 \, x e^{\left (-3\right )} + 5 \, e^{\left (-3\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - e^{\left (-6\right )} \log \left (x^{2} e^{3} + x + 4 \, e^{3} + 4\right ) - \frac {10 \, \arctan \left (\frac {2 \, x e^{3} + 1}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}}\right ) e^{\left (-3\right )}}{\sqrt {16 \, e^{6} + 16 \, e^{3} - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+4*x+8)*exp(3)+2*x+10)*log((x^2+4)*exp(3)+4+x)+(2*x^3+4*x^2+8*x)*exp(3)+2*x^2+10*x)/((x^2+4)
*exp(3)+4+x),x, algorithm="maxima")

[Out]

2*sqrt(16*e^6 + 16*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-3) - 2*(8*e^6 + 8*e^3 - 1)*arc
tan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-6)/sqrt(16*e^6 + 16*e^3 - 1) - ((4*e^6 + 4*e^3 - 1)*e^(-9)*lo
g(x^2*e^3 + x + 4*e^3 + 4) - 2*(12*e^6 + 12*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-9)/sq
rt(16*e^6 + 16*e^3 - 1) - (x^2*e^3 - 2*x)*e^(-6))*e^3 - 2*(2*(8*e^6 + 8*e^3 - 1)*arctan((2*x*e^3 + 1)/sqrt(16*
e^6 + 16*e^3 - 1))*e^(-6)/sqrt(16*e^6 + 16*e^3 - 1) - 2*x*e^(-3) + e^(-6)*log(x^2*e^3 + x + 4*e^3 + 4))*e^3 +
4*(e^(-3)*log(x^2*e^3 + x + 4*e^3 + 4) - 2*arctan((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-3)/sqrt(16*e^6
+ 16*e^3 - 1))*e^3 + (e^3*log(x^2*e^3 + x + 4*e^3 + 4)^2 - 4*x*e^3 + (2*x*e^3 + 1)*log(x^2*e^3 + x + 4*e^3 + 4
))*e^(-3) + 2*x*e^(-3) + 5*e^(-3)*log(x^2*e^3 + x + 4*e^3 + 4) - e^(-6)*log(x^2*e^3 + x + 4*e^3 + 4) - 10*arct
an((2*x*e^3 + 1)/sqrt(16*e^6 + 16*e^3 - 1))*e^(-3)/sqrt(16*e^6 + 16*e^3 - 1)

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mupad [B]  time = 6.99, size = 16, normalized size = 0.94 \begin {gather*} {\left (x+\ln \left (x+{\mathrm {e}}^3\,\left (x^2+4\right )+4\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log(x + exp(3)*(x^2 + 4) + 4)*(2*x + exp(3)*(4*x + 2*x^2 + 8) + 10) + exp(3)*(8*x + 4*x^2 + 2*x^3)
+ 2*x^2)/(x + exp(3)*(x^2 + 4) + 4),x)

[Out]

(x + log(x + exp(3)*(x^2 + 4) + 4))^2

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sympy [B]  time = 0.22, size = 34, normalized size = 2.00 \begin {gather*} x^{2} + 2 x \log {\left (x + \left (x^{2} + 4\right ) e^{3} + 4 \right )} + \log {\left (x + \left (x^{2} + 4\right ) e^{3} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+4*x+8)*exp(3)+2*x+10)*ln((x**2+4)*exp(3)+4+x)+(2*x**3+4*x**2+8*x)*exp(3)+2*x**2+10*x)/((x*
*2+4)*exp(3)+4+x),x)

[Out]

x**2 + 2*x*log(x + (x**2 + 4)*exp(3) + 4) + log(x + (x**2 + 4)*exp(3) + 4)**2

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