3.65.58 \(\int \frac {e^{3 x} (3+15 x+12 x^2)+e^{2 x} (-4-x+14 x^2+8 x^3+e^3 (2+10 x+8 x^2))+e^{2 x} (-2-10 x-8 x^2) \log (1+5 x+4 x^2)}{1+5 x+4 x^2} \, dx\)

Optimal. Leaf size=27 \[ e^{2 x} \left (e^3+e^x+x-\log \left (1+5 x+4 x^2\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 1.92, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 36, number of rules used = 9, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {6728, 2194, 6688, 6742, 2268, 2178, 2270, 2176, 2554} \begin {gather*} -e^{2 x} \log \left (4 x^2+5 x+1\right )+e^{2 x} x+e^{3 x}+e^{2 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3*x)*(3 + 15*x + 12*x^2) + E^(2*x)*(-4 - x + 14*x^2 + 8*x^3 + E^3*(2 + 10*x + 8*x^2)) + E^(2*x)*(-2 -
10*x - 8*x^2)*Log[1 + 5*x + 4*x^2])/(1 + 5*x + 4*x^2),x]

[Out]

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2268

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[F^(g*(d + e*x)^n), 1/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x]

Rule 2270

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_.) + (b_.)*(x_) + (c_)*(x_)^2), x_Symbol] :> Int[
ExpandIntegrand[F^(g*(d + e*x)^n), u^m/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x] && Poly
nomialQ[u, x] && IntegerQ[m]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

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 (3 e^{3 x}+\frac {e^{2 x} \left (-4 \left (1-\frac {e^3}{2}\right )-\left (1-10 e^3\right ) x+14 \left (1+\frac {4 e^3}{7}\right ) x^2+8 x^3-2 \log \left (1+5 x+4 x^2\right )-10 x \log \left (1+5 x+4 x^2\right )-8 x^2 \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)}\right ) \, dx\\ &=3 \int e^{3 x} \, dx+\int \frac {e^{2 x} \left (-4 \left (1-\frac {e^3}{2}\right )-\left (1-10 e^3\right ) x+14 \left (1+\frac {4 e^3}{7}\right ) x^2+8 x^3-2 \log \left (1+5 x+4 x^2\right )-10 x \log \left (1+5 x+4 x^2\right )-8 x^2 \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)} \, dx\\ &=e^{3 x}+\int \frac {e^{2 x} \left (-4-x+14 x^2+8 x^3+2 e^3 \left (1+5 x+4 x^2\right )-2 \left (1+5 x+4 x^2\right ) \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)} \, dx\\ &=e^{3 x}+\int \left (2 e^{3+2 x}-\frac {4 e^{2 x}}{1+5 x+4 x^2}-\frac {e^{2 x} x}{1+5 x+4 x^2}+\frac {14 e^{2 x} x^2}{1+5 x+4 x^2}+\frac {8 e^{2 x} x^3}{1+5 x+4 x^2}-2 e^{2 x} \log \left (1+5 x+4 x^2\right )\right ) \, dx\\ &=e^{3 x}+2 \int e^{3+2 x} \, dx-2 \int e^{2 x} \log \left (1+5 x+4 x^2\right ) \, dx-4 \int \frac {e^{2 x}}{1+5 x+4 x^2} \, dx+8 \int \frac {e^{2 x} x^3}{1+5 x+4 x^2} \, dx+14 \int \frac {e^{2 x} x^2}{1+5 x+4 x^2} \, dx-\int \frac {e^{2 x} x}{1+5 x+4 x^2} \, dx\\ &=e^{3 x}+e^{3+2 x}-e^{2 x} \log \left (1+5 x+4 x^2\right )+2 \int \frac {e^{2 x} (5+8 x)}{2+10 x+8 x^2} \, dx-4 \int \left (-\frac {8 e^{2 x}}{3 (-2-8 x)}-\frac {8 e^{2 x}}{3 (8+8 x)}\right ) \, dx+8 \int \left (-\frac {5 e^{2 x}}{16}+\frac {1}{4} e^{2 x} x+\frac {e^{2 x} (5+21 x)}{16 \left (1+5 x+4 x^2\right )}\right ) \, dx+14 \int \left (\frac {e^{2 x}}{4}-\frac {e^{2 x} (1+5 x)}{4 \left (1+5 x+4 x^2\right )}\right ) \, dx-\int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {8 e^{2 x}}{3 (8+8 x)}\right ) \, dx\\ &=e^{3 x}+e^{3+2 x}-e^{2 x} \log \left (1+5 x+4 x^2\right )+\frac {1}{2} \int \frac {e^{2 x} (5+21 x)}{1+5 x+4 x^2} \, dx+\frac {2}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+2 \int e^{2 x} x \, dx+2 \int \left (\frac {8 e^{2 x}}{4+16 x}+\frac {8 e^{2 x}}{16+16 x}\right ) \, dx-\frac {5}{2} \int e^{2 x} \, dx-\frac {8}{3} \int \frac {e^{2 x}}{8+8 x} \, dx+\frac {7}{2} \int e^{2 x} \, dx-\frac {7}{2} \int \frac {e^{2 x} (1+5 x)}{1+5 x+4 x^2} \, dx+\frac {32}{3} \int \frac {e^{2 x}}{-2-8 x} \, dx+\frac {32}{3} \int \frac {e^{2 x}}{8+8 x} \, dx\\ &=\frac {e^{2 x}}{2}+e^{3 x}+e^{3+2 x}+e^{2 x} x+\frac {\text {Ei}(2 (1+x))}{e^2}-\frac {5 \text {Ei}\left (\frac {1}{2} (1+4 x)\right )}{4 \sqrt {e}}-e^{2 x} \log \left (1+5 x+4 x^2\right )+\frac {1}{2} \int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {128 e^{2 x}}{3 (8+8 x)}\right ) \, dx-\frac {7}{2} \int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {32 e^{2 x}}{3 (8+8 x)}\right ) \, dx+16 \int \frac {e^{2 x}}{4+16 x} \, dx+16 \int \frac {e^{2 x}}{16+16 x} \, dx-\int e^{2 x} \, dx\\ &=e^{3 x}+e^{3+2 x}+e^{2 x} x+\frac {2 \text {Ei}(2 (1+x))}{e^2}-\frac {\text {Ei}\left (\frac {1}{2} (1+4 x)\right )}{4 \sqrt {e}}-e^{2 x} \log \left (1+5 x+4 x^2\right )-\frac {1}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+\frac {7}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+\frac {64}{3} \int \frac {e^{2 x}}{8+8 x} \, dx-\frac {112}{3} \int \frac {e^{2 x}}{8+8 x} \, dx\\ &=e^{3 x}+e^{3+2 x}+e^{2 x} x-e^{2 x} \log \left (1+5 x+4 x^2\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 27, normalized size = 1.00 \begin {gather*} e^{2 x} \left (e^3+e^x+x-\log \left (1+5 x+4 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*x)*(3 + 15*x + 12*x^2) + E^(2*x)*(-4 - x + 14*x^2 + 8*x^3 + E^3*(2 + 10*x + 8*x^2)) + E^(2*x)*
(-2 - 10*x - 8*x^2)*Log[1 + 5*x + 4*x^2])/(1 + 5*x + 4*x^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 31, normalized size = 1.15 \begin {gather*} {\left (x + e^{3}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x^{2} + 5 \, x + 1\right ) + e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-10*x-2)*exp(x)^2*log(4*x^2+5*x+1)+(12*x^2+15*x+3)*exp(x)^3+((8*x^2+10*x+2)*exp(3)+8*x^3+14*
x^2-x-4)*exp(x)^2)/(4*x^2+5*x+1),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.15, size = 34, normalized size = 1.26 \begin {gather*} x e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x^{2} + 5 \, x + 1\right ) + e^{\left (3 \, x\right )} + e^{\left (2 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-10*x-2)*exp(x)^2*log(4*x^2+5*x+1)+(12*x^2+15*x+3)*exp(x)^3+((8*x^2+10*x+2)*exp(3)+8*x^3+14*
x^2-x-4)*exp(x)^2)/(4*x^2+5*x+1),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.51, size = 35, normalized size = 1.30

method	result	size
default	\({\mathrm e}^{3} {\mathrm e}^{2 x}+x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (\left (x +1\right ) \left (4 x +1\right )\right )+{\mathrm e}^{3 x}\)	\(35\)
risch	\({\mathrm e}^{2 x +3}+x \,{\mathrm e}^{2 x}+{\mathrm e}^{3 x}-{\mathrm e}^{2 x} \ln \left (4 x^{2}+5 x +1\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^2-10*x-2)*exp(x)^2*ln(4*x^2+5*x+1)+(12*x^2+15*x+3)*exp(x)^3+((8*x^2+10*x+2)*exp(3)+8*x^3+14*x^2-x-4
)*exp(x)^2)/(4*x^2+5*x+1),x,method=_RETURNVERBOSE)

[Out]

exp(3)*exp(2*x)+x*exp(2*x)-exp(2*x)*ln((x+1)*(4*x+1))+exp(3*x)

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 36, normalized size = 1.33 \begin {gather*} {\left (x + e^{3}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x + 1\right ) - e^{\left (2 \, x\right )} \log \left (x + 1\right ) + e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-10*x-2)*exp(x)^2*log(4*x^2+5*x+1)+(12*x^2+15*x+3)*exp(x)^3+((8*x^2+10*x+2)*exp(3)+8*x^3+14*
x^2-x-4)*exp(x)^2)/(4*x^2+5*x+1),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.38, size = 24, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^3-\ln \left (4\,x^2+5\,x+1\right )+{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x)*(15*x + 12*x^2 + 3) + exp(2*x)*(exp(3)*(10*x + 8*x^2 + 2) - x + 14*x^2 + 8*x^3 - 4) - exp(2*x)*l
og(5*x + 4*x^2 + 1)*(10*x + 8*x^2 + 2))/(5*x + 4*x^2 + 1),x)

[Out]

exp(2*x)*(x + exp(3) - log(5*x + 4*x^2 + 1) + exp(x))

________________________________________________________________________________________

sympy [A]  time = 0.57, size = 26, normalized size = 0.96 \begin {gather*} \left (x - \log {\left (4 x^{2} + 5 x + 1 \right )} + e^{3}\right ) e^{2 x} + e^{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**2-10*x-2)*exp(x)**2*ln(4*x**2+5*x+1)+(12*x**2+15*x+3)*exp(x)**3+((8*x**2+10*x+2)*exp(3)+8*x*
*3+14*x**2-x-4)*exp(x)**2)/(4*x**2+5*x+1),x)

[Out]

(x - log(4*x**2 + 5*x + 1) + exp(3))*exp(2*x) + exp(3*x)

________________________________________________________________________________________