3.38.58 \(\int \frac {-15+10 x+e^{2 x^4} (-5-40 x^4)+\log (9)}{\log (9)} \, dx\)

Optimal. Leaf size=28 \[ x+\frac {5 x \left (-3-e^{2 x^4}+x-\frac {\log (5)}{x}\right )}{\log (9)} \]

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Rubi [C]  time = 0.07, antiderivative size = 87, normalized size of antiderivative = 3.11, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 2226, 2208, 2218} \begin {gather*} \frac {5 x \Gamma \left (\frac {1}{4},-2 x^4\right )}{4 \sqrt [4]{2} \sqrt [4]{-x^4} \log (9)}+\frac {5 x^2}{\log (9)}+\frac {5 x^5 \Gamma \left (\frac {5}{4},-2 x^4\right )}{\sqrt [4]{2} \left (-x^4\right )^{5/4} \log (9)}-\frac {x (15-\log (9))}{\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 + 10*x + E^(2*x^4)*(-5 - 40*x^4) + Log[9])/Log[9],x]

[Out]

(5*x^2)/Log[9] + (5*x*Gamma[1/4, -2*x^4])/(4*2^(1/4)*(-x^4)^(1/4)*Log[9]) + (5*x^5*Gamma[5/4, -2*x^4])/(2^(1/4
)*(-x^4)^(5/4)*Log[9]) - (x*(15 - Log[9]))/Log[9]

Rule 12

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

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-15+10 x+e^{2 x^4} \left (-5-40 x^4\right )+\log (9)\right ) \, dx}{\log (9)}\\ &=\frac {5 x^2}{\log (9)}-\frac {x (15-\log (9))}{\log (9)}+\frac {\int e^{2 x^4} \left (-5-40 x^4\right ) \, dx}{\log (9)}\\ &=\frac {5 x^2}{\log (9)}-\frac {x (15-\log (9))}{\log (9)}+\frac {\int \left (-5 e^{2 x^4}-40 e^{2 x^4} x^4\right ) \, dx}{\log (9)}\\ &=\frac {5 x^2}{\log (9)}-\frac {x (15-\log (9))}{\log (9)}-\frac {5 \int e^{2 x^4} \, dx}{\log (9)}-\frac {40 \int e^{2 x^4} x^4 \, dx}{\log (9)}\\ &=\frac {5 x^2}{\log (9)}+\frac {5 x \Gamma \left (\frac {1}{4},-2 x^4\right )}{4 \sqrt [4]{2} \sqrt [4]{-x^4} \log (9)}+\frac {5 x^5 \Gamma \left (\frac {5}{4},-2 x^4\right )}{\sqrt [4]{2} \left (-x^4\right )^{5/4} \log (9)}-\frac {x (15-\log (9))}{\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.05, size = 74, normalized size = 2.64 \begin {gather*} \frac {-15 x+5 x^2+\frac {5 x \Gamma \left (\frac {1}{4},-2 x^4\right )}{4 \sqrt [4]{2} \sqrt [4]{-x^4}}+\frac {5 x^5 \Gamma \left (\frac {5}{4},-2 x^4\right )}{\sqrt [4]{2} \left (-x^4\right )^{5/4}}+x \log (9)}{\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + 10*x + E^(2*x^4)*(-5 - 40*x^4) + Log[9])/Log[9],x]

[Out]

(-15*x + 5*x^2 + (5*x*Gamma[1/4, -2*x^4])/(4*2^(1/4)*(-x^4)^(1/4)) + (5*x^5*Gamma[5/4, -2*x^4])/(2^(1/4)*(-x^4
)^(5/4)) + x*Log[9])/Log[9]

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fricas [A]  time = 0.77, size = 29, normalized size = 1.04 \begin {gather*} \frac {5 \, x^{2} - 5 \, x e^{\left (2 \, x^{4}\right )} + 2 \, x \log \relax (3) - 15 \, x}{2 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-40*x^4-5)*exp(x^4)^2+2*log(3)+10*x-15)/log(3),x, algorithm="fricas")

[Out]

1/2*(5*x^2 - 5*x*e^(2*x^4) + 2*x*log(3) - 15*x)/log(3)

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giac [A]  time = 0.21, size = 29, normalized size = 1.04 \begin {gather*} \frac {5 \, x^{2} - 5 \, x e^{\left (2 \, x^{4}\right )} + 2 \, x \log \relax (3) - 15 \, x}{2 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-40*x^4-5)*exp(x^4)^2+2*log(3)+10*x-15)/log(3),x, algorithm="giac")

[Out]

1/2*(5*x^2 - 5*x*e^(2*x^4) + 2*x*log(3) - 15*x)/log(3)

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maple [A]  time = 0.06, size = 30, normalized size = 1.07

method	result	size
default	\(\frac {-15 x -5 x \,{\mathrm e}^{2 x^{4}}+5 x^{2}+2 x \ln \relax (3)}{2 \ln \relax (3)}\)	\(30\)
risch	\(-\frac {15 x}{2 \ln \relax (3)}-\frac {5 x \,{\mathrm e}^{2 x^{4}}}{2 \ln \relax (3)}+\frac {5 x^{2}}{2 \ln \relax (3)}+x\)	\(32\)
norman	\(\frac {5 x^{2}}{2 \ln \relax (3)}-\frac {5 x \,{\mathrm e}^{2 x^{4}}}{2 \ln \relax (3)}+\frac {\left (2 \ln \relax (3)-15\right ) x}{2 \ln \relax (3)}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-40*x^4-5)*exp(x^4)^2+2*ln(3)+10*x-15)/ln(3),x,method=_RETURNVERBOSE)

[Out]

1/2/ln(3)*(-15*x-5*x*exp(x^4)^2+5*x^2+2*x*ln(3))

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maxima [A]  time = 0.37, size = 29, normalized size = 1.04 \begin {gather*} \frac {5 \, x^{2} - 5 \, x e^{\left (2 \, x^{4}\right )} + 2 \, x \log \relax (3) - 15 \, x}{2 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-40*x^4-5)*exp(x^4)^2+2*log(3)+10*x-15)/log(3),x, algorithm="maxima")

[Out]

1/2*(5*x^2 - 5*x*e^(2*x^4) + 2*x*log(3) - 15*x)/log(3)

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mupad [B]  time = 0.06, size = 27, normalized size = 0.96 \begin {gather*} \frac {x\,\left (\ln \relax (9)-15\right )-5\,x\,{\mathrm {e}}^{2\,x^4}+5\,x^2}{2\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(3) - (exp(2*x^4)*(40*x^4 + 5))/2 - 15/2)/log(3),x)

[Out]

(x*(log(9) - 15) - 5*x*exp(2*x^4) + 5*x^2)/(2*log(3))

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sympy [A]  time = 0.13, size = 37, normalized size = 1.32 \begin {gather*} \frac {5 x^{2}}{2 \log {\relax (3 )}} - \frac {5 x e^{2 x^{4}}}{2 \log {\relax (3 )}} + \frac {x \left (-15 + 2 \log {\relax (3 )}\right )}{2 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-40*x**4-5)*exp(x**4)**2+2*ln(3)+10*x-15)/ln(3),x)

[Out]

5*x**2/(2*log(3)) - 5*x*exp(2*x**4)/(2*log(3)) + x*(-15 + 2*log(3))/(2*log(3))

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