3.66.82 \(\int \frac {1}{4} (x^3+e^{4 x+4 e^2 x^3} (1+3 e^2 x^2)+e^{3 x+3 e^2 x^3} (1+3 x+9 e^2 x^3)+e^{2 x+2 e^2 x^3} (3 x+3 x^2+9 e^2 x^4)+e^{x+e^2 x^3} (3 x^2+x^3+3 e^2 x^5)) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{16} \left (e^{x+e^2 x^3}+x\right )^4 \]

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Rubi [B]  time = 0.20, antiderivative size = 147, normalized size of antiderivative = 7.74, number of steps used = 8, number of rules used = 4, integrand size = 122, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 6706, 2288, 1594} \begin {gather*} \frac {x^4}{16}+\frac {1}{16} e^{4 e^2 x^3+4 x}+\frac {e^{e^2 x^3+x} \left (3 e^2 x^3+x\right ) x^2}{4 \left (3 e^2 x^2+1\right )}+\frac {3 e^{2 e^2 x^3+2 x} \left (3 e^2 x^3+x\right ) x}{8 \left (3 e^2 x^2+1\right )}+\frac {e^{3 e^2 x^3+3 x} \left (3 e^2 x^3+x\right )}{4 \left (3 e^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3 + E^(4*x + 4*E^2*x^3)*(1 + 3*E^2*x^2) + E^(3*x + 3*E^2*x^3)*(1 + 3*x + 9*E^2*x^3) + E^(2*x + 2*E^2*x^
3)*(3*x + 3*x^2 + 9*E^2*x^4) + E^(x + E^2*x^3)*(3*x^2 + x^3 + 3*E^2*x^5))/4,x]

[Out]

E^(4*x + 4*E^2*x^3)/16 + x^4/16 + (E^(3*x + 3*E^2*x^3)*(x + 3*E^2*x^3))/(4*(1 + 3*E^2*x^2)) + (3*E^(2*x + 2*E^
2*x^3)*x*(x + 3*E^2*x^3))/(8*(1 + 3*E^2*x^2)) + (E^(x + E^2*x^3)*x^2*(x + 3*E^2*x^3))/(4*(1 + 3*E^2*x^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 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 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 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} &=\frac {1}{4} \int \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx\\ &=\frac {x^4}{16}+\frac {1}{4} \int e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right ) \, dx+\frac {1}{4} \int e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right ) \, dx+\frac {1}{4} \int e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right ) \, dx+\frac {1}{4} \int e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right ) \, dx\\ &=\frac {1}{16} e^{4 x+4 e^2 x^3}+\frac {x^4}{16}+\frac {e^{3 x+3 e^2 x^3} \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )}+\frac {1}{4} \int e^{x+e^2 x^3} x^2 \left (3+x+3 e^2 x^3\right ) \, dx+\frac {1}{4} \int e^{2 x+2 e^2 x^3} x \left (3+3 x+9 e^2 x^3\right ) \, dx\\ &=\frac {1}{16} e^{4 x+4 e^2 x^3}+\frac {x^4}{16}+\frac {e^{3 x+3 e^2 x^3} \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )}+\frac {3 e^{2 x+2 e^2 x^3} x \left (x+3 e^2 x^3\right )}{8 \left (1+3 e^2 x^2\right )}+\frac {e^{x+e^2 x^3} x^2 \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{16} \left (e^{x+e^2 x^3}+x\right )^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + E^(4*x + 4*E^2*x^3)*(1 + 3*E^2*x^2) + E^(3*x + 3*E^2*x^3)*(1 + 3*x + 9*E^2*x^3) + E^(2*x + 2*
E^2*x^3)*(3*x + 3*x^2 + 9*E^2*x^4) + E^(x + E^2*x^3)*(3*x^2 + x^3 + 3*E^2*x^5))/4,x]

[Out]

(E^(x + E^2*x^3) + x)^4/16

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fricas [B]  time = 0.54, size = 66, normalized size = 3.47 \begin {gather*} \frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^2*exp(1)^2+1)*exp(x^3*exp(1)^2+x)^4+1/4*(9*x^3*exp(1)^2+3*x+1)*exp(x^3*exp(1)^2+x)^3+1/4*(9
*x^4*exp(1)^2+3*x^2+3*x)*exp(x^3*exp(1)^2+x)^2+1/4*(3*x^5*exp(1)^2+x^3+3*x^2)*exp(x^3*exp(1)^2+x)+1/4*x^3,x, a
lgorithm="fricas")

[Out]

1/16*x^4 + 1/4*x^3*e^(x^3*e^2 + x) + 3/8*x^2*e^(2*x^3*e^2 + 2*x) + 1/4*x*e^(3*x^3*e^2 + 3*x) + 1/16*e^(4*x^3*e
^2 + 4*x)

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giac [B]  time = 0.16, size = 66, normalized size = 3.47 \begin {gather*} \frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^2*exp(1)^2+1)*exp(x^3*exp(1)^2+x)^4+1/4*(9*x^3*exp(1)^2+3*x+1)*exp(x^3*exp(1)^2+x)^3+1/4*(9
*x^4*exp(1)^2+3*x^2+3*x)*exp(x^3*exp(1)^2+x)^2+1/4*(3*x^5*exp(1)^2+x^3+3*x^2)*exp(x^3*exp(1)^2+x)+1/4*x^3,x, a
lgorithm="giac")

[Out]

1/16*x^4 + 1/4*x^3*e^(x^3*e^2 + x) + 3/8*x^2*e^(2*x^3*e^2 + 2*x) + 1/4*x*e^(3*x^3*e^2 + 3*x) + 1/16*e^(4*x^3*e
^2 + 4*x)

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maple [A]  time = 0.43, size = 69, normalized size = 3.63

method	result	size
risch	\(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x \left (x^{2} {\mathrm e}^{2}+1\right )}}{16}+\frac {x \,{\mathrm e}^{3 x \left (x^{2} {\mathrm e}^{2}+1\right )}}{4}+\frac {3 x^{2} {\mathrm e}^{2 x \left (x^{2} {\mathrm e}^{2}+1\right )}}{8}+\frac {{\mathrm e}^{x \left (x^{2} {\mathrm e}^{2}+1\right )} x^{3}}{4}\)	\(69\)
default	\(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {x \,{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x}}{4}+\frac {3 x^{2} {\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\)	\(72\)
norman	\(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {x \,{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x}}{4}+\frac {3 x^{2} {\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(3*x^2*exp(1)^2+1)*exp(x^3*exp(1)^2+x)^4+1/4*(9*x^3*exp(1)^2+3*x+1)*exp(x^3*exp(1)^2+x)^3+1/4*(9*x^4*e
xp(1)^2+3*x^2+3*x)*exp(x^3*exp(1)^2+x)^2+1/4*(3*x^5*exp(1)^2+x^3+3*x^2)*exp(x^3*exp(1)^2+x)+1/4*x^3,x,method=_
RETURNVERBOSE)

[Out]

1/16*x^4+1/16*exp(4*x*(x^2*exp(2)+1))+1/4*x*exp(3*x*(x^2*exp(2)+1))+3/8*x^2*exp(2*x*(x^2*exp(2)+1))+1/4*exp(x*
(x^2*exp(2)+1))*x^3

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maxima [B]  time = 0.45, size = 66, normalized size = 3.47 \begin {gather*} \frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^2*exp(1)^2+1)*exp(x^3*exp(1)^2+x)^4+1/4*(9*x^3*exp(1)^2+3*x+1)*exp(x^3*exp(1)^2+x)^3+1/4*(9
*x^4*exp(1)^2+3*x^2+3*x)*exp(x^3*exp(1)^2+x)^2+1/4*(3*x^5*exp(1)^2+x^3+3*x^2)*exp(x^3*exp(1)^2+x)+1/4*x^3,x, a
lgorithm="maxima")

[Out]

1/16*x^4 + 1/4*x^3*e^(x^3*e^2 + x) + 3/8*x^2*e^(2*x^3*e^2 + 2*x) + 1/4*x*e^(3*x^3*e^2 + 3*x) + 1/16*e^(4*x^3*e
^2 + 4*x)

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mupad [B]  time = 0.22, size = 15, normalized size = 0.79 \begin {gather*} \frac {{\left (x+{\mathrm {e}}^{{\mathrm {e}}^2\,x^3+x}\right )}^4}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + x^3*exp(2))*(3*x^5*exp(2) + 3*x^2 + x^3))/4 + (exp(3*x + 3*x^3*exp(2))*(3*x + 9*x^3*exp(2) + 1))/
4 + (exp(4*x + 4*x^3*exp(2))*(3*x^2*exp(2) + 1))/4 + (exp(2*x + 2*x^3*exp(2))*(3*x + 9*x^4*exp(2) + 3*x^2))/4
+ x^3/4,x)

[Out]

(x + exp(x + x^3*exp(2)))^4/16

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sympy [B]  time = 0.24, size = 71, normalized size = 3.74 \begin {gather*} \frac {x^{4}}{16} + \frac {x^{3} e^{x^{3} e^{2} + x}}{4} + \frac {3 x^{2} e^{2 x^{3} e^{2} + 2 x}}{8} + \frac {x e^{3 x^{3} e^{2} + 3 x}}{4} + \frac {e^{4 x^{3} e^{2} + 4 x}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x**2*exp(1)**2+1)*exp(x**3*exp(1)**2+x)**4+1/4*(9*x**3*exp(1)**2+3*x+1)*exp(x**3*exp(1)**2+x)
**3+1/4*(9*x**4*exp(1)**2+3*x**2+3*x)*exp(x**3*exp(1)**2+x)**2+1/4*(3*x**5*exp(1)**2+x**3+3*x**2)*exp(x**3*exp
(1)**2+x)+1/4*x**3,x)

[Out]

x**4/16 + x**3*exp(x**3*exp(2) + x)/4 + 3*x**2*exp(2*x**3*exp(2) + 2*x)/8 + x*exp(3*x**3*exp(2) + 3*x)/4 + exp
(4*x**3*exp(2) + 4*x)/16

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