3.48.20 \(\int (16 x^3+e^{2 x+2 x^3} (2+6 x^2)-16 x^3 \log (5)+4 x^3 \log ^2(5)+e^{x+x^3} (8 x+4 x^2+12 x^4+(-4 x-2 x^2-6 x^4) \log (5))) \, dx\)

Optimal. Leaf size=21 \[ \left (e^{x+x^3}+x (2 x-x \log (5))\right )^2 \]

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Rubi [B]  time = 0.15, antiderivative size = 56, normalized size of antiderivative = 2.67, number of steps used = 7, number of rules used = 5, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6, 6706, 6688, 12, 2288} \begin {gather*} x^4 (2-\log (5))^2+e^{2 x^3+2 x}+\frac {2 e^{x^3+x} \left (3 x^3+x\right ) x (2-\log (5))}{3 x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16*x^3 + E^(2*x + 2*x^3)*(2 + 6*x^2) - 16*x^3*Log[5] + 4*x^3*Log[5]^2 + E^(x + x^3)*(8*x + 4*x^2 + 12*x^4
+ (-4*x - 2*x^2 - 6*x^4)*Log[5]),x]

[Out]

E^(2*x + 2*x^3) + (2*E^(x + x^3)*x*(x + 3*x^3)*(2 - Log[5]))/(1 + 3*x^2) + x^4*(2 - Log[5])^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{2 x+2 x^3} \left (2+6 x^2\right )+x^3 (16-16 \log (5))+4 x^3 \log ^2(5)+e^{x+x^3} \left (8 x+4 x^2+12 x^4+\left (-4 x-2 x^2-6 x^4\right ) \log (5)\right )\right ) \, dx\\ &=\int \left (e^{2 x+2 x^3} \left (2+6 x^2\right )+e^{x+x^3} \left (8 x+4 x^2+12 x^4+\left (-4 x-2 x^2-6 x^4\right ) \log (5)\right )+x^3 \left (16-16 \log (5)+4 \log ^2(5)\right )\right ) \, dx\\ &=x^4 (2-\log (5))^2+\int e^{2 x+2 x^3} \left (2+6 x^2\right ) \, dx+\int e^{x+x^3} \left (8 x+4 x^2+12 x^4+\left (-4 x-2 x^2-6 x^4\right ) \log (5)\right ) \, dx\\ &=e^{2 x+2 x^3}+x^4 (2-\log (5))^2+\int 2 e^{x+x^3} x \left (2+x+3 x^3\right ) (2-\log (5)) \, dx\\ &=e^{2 x+2 x^3}+x^4 (2-\log (5))^2+(2 (2-\log (5))) \int e^{x+x^3} x \left (2+x+3 x^3\right ) \, dx\\ &=e^{2 x+2 x^3}+\frac {2 e^{x+x^3} x \left (x+3 x^3\right ) (2-\log (5))}{1+3 x^2}+x^4 (2-\log (5))^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.90 \begin {gather*} \left (e^{x+x^3}-x^2 (-2+\log (5))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16*x^3 + E^(2*x + 2*x^3)*(2 + 6*x^2) - 16*x^3*Log[5] + 4*x^3*Log[5]^2 + E^(x + x^3)*(8*x + 4*x^2 + 1
2*x^4 + (-4*x - 2*x^2 - 6*x^4)*Log[5]),x]

[Out]

(E^(x + x^3) - x^2*(-2 + Log[5]))^2

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fricas [B]  time = 0.68, size = 51, normalized size = 2.43 \begin {gather*} x^{4} \log \relax (5)^{2} - 4 \, x^{4} \log \relax (5) + 4 \, x^{4} - 2 \, {\left (x^{2} \log \relax (5) - 2 \, x^{2}\right )} e^{\left (x^{3} + x\right )} + e^{\left (2 \, x^{3} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2+2)*exp(x^3+x)^2+((-6*x^4-2*x^2-4*x)*log(5)+12*x^4+4*x^2+8*x)*exp(x^3+x)+4*x^3*log(5)^2-16*x^3
*log(5)+16*x^3,x, algorithm="fricas")

[Out]

x^4*log(5)^2 - 4*x^4*log(5) + 4*x^4 - 2*(x^2*log(5) - 2*x^2)*e^(x^3 + x) + e^(2*x^3 + 2*x)

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giac [B]  time = 0.14, size = 55, normalized size = 2.62 \begin {gather*} x^{4} \log \relax (5)^{2} - 4 \, x^{4} \log \relax (5) + 4 \, x^{4} - 2 \, x^{2} e^{\left (x^{3} + x\right )} \log \relax (5) + 4 \, x^{2} e^{\left (x^{3} + x\right )} + e^{\left (2 \, x^{3} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2+2)*exp(x^3+x)^2+((-6*x^4-2*x^2-4*x)*log(5)+12*x^4+4*x^2+8*x)*exp(x^3+x)+4*x^3*log(5)^2-16*x^3
*log(5)+16*x^3,x, algorithm="giac")

[Out]

x^4*log(5)^2 - 4*x^4*log(5) + 4*x^4 - 2*x^2*e^(x^3 + x)*log(5) + 4*x^2*e^(x^3 + x) + e^(2*x^3 + 2*x)

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maple [A]  time = 0.09, size = 40, normalized size = 1.90

method	result	size
norman	\({\mathrm e}^{2 x^{3}+2 x}+\left (\ln \relax (5)^{2}-4 \ln \relax (5)+4\right ) x^{4}+{\mathrm e}^{x^{3}+x} \left (-2 \ln \relax (5)+4\right ) x^{2}\)	\(40\)
default	\({\mathrm e}^{x^{3}+x} \left (-2 \ln \relax (5)+4\right ) x^{2}+{\mathrm e}^{2 x^{3}+2 x}+4 x^{4}-4 x^{4} \ln \relax (5)+x^{4} \ln \relax (5)^{2}\)	\(46\)
risch	\({\mathrm e}^{2 x \left (x^{2}+1\right )}-2 \left (\ln \relax (5)-2\right ) x^{2} {\mathrm e}^{x \left (x^{2}+1\right )}+x^{4} \ln \relax (5)^{2}-4 x^{4} \ln \relax (5)+4 x^{4}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^2+2)*exp(x^3+x)^2+((-6*x^4-2*x^2-4*x)*ln(5)+12*x^4+4*x^2+8*x)*exp(x^3+x)+4*x^3*ln(5)^2-16*x^3*ln(5)+1
6*x^3,x,method=_RETURNVERBOSE)

[Out]

exp(x^3+x)^2+(ln(5)^2-4*ln(5)+4)*x^4+exp(x^3+x)*(-2*ln(5)+4)*x^2

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maxima [B]  time = 0.48, size = 46, normalized size = 2.19 \begin {gather*} x^{4} \log \relax (5)^{2} - 4 \, x^{4} \log \relax (5) + 4 \, x^{4} - 2 \, x^{2} {\left (\log \relax (5) - 2\right )} e^{\left (x^{3} + x\right )} + e^{\left (2 \, x^{3} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2+2)*exp(x^3+x)^2+((-6*x^4-2*x^2-4*x)*log(5)+12*x^4+4*x^2+8*x)*exp(x^3+x)+4*x^3*log(5)^2-16*x^3
*log(5)+16*x^3,x, algorithm="maxima")

[Out]

x^4*log(5)^2 - 4*x^4*log(5) + 4*x^4 - 2*x^2*(log(5) - 2)*e^(x^3 + x) + e^(2*x^3 + 2*x)

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mupad [B]  time = 0.10, size = 40, normalized size = 1.90 \begin {gather*} {\mathrm {e}}^{2\,x^3+2\,x}+x^4\,\left ({\ln \relax (5)}^2-4\,\ln \relax (5)+4\right )-x^2\,{\mathrm {e}}^{x^3+x}\,\left (\ln \left (25\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*x^3*log(5)^2 + exp(x + x^3)*(8*x - log(5)*(4*x + 2*x^2 + 6*x^4) + 4*x^2 + 12*x^4) - 16*x^3*log(5) + exp(
2*x + 2*x^3)*(6*x^2 + 2) + 16*x^3,x)

[Out]

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

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sympy [B]  time = 0.17, size = 44, normalized size = 2.10 \begin {gather*} x^{4} \left (- 4 \log {\relax (5 )} + \log {\relax (5 )}^{2} + 4\right ) + \left (- 2 x^{2} \log {\relax (5 )} + 4 x^{2}\right ) e^{x^{3} + x} + e^{2 x^{3} + 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**2+2)*exp(x**3+x)**2+((-6*x**4-2*x**2-4*x)*ln(5)+12*x**4+4*x**2+8*x)*exp(x**3+x)+4*x**3*ln(5)**
2-16*x**3*ln(5)+16*x**3,x)

[Out]

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

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