\(\int \frac {1}{2} (2+e^{\frac {1}{4} (49 x^2+14 x^3+x^4) \log ^2(4)} (49 x+21 x^2+2 x^3) \log ^2(4)) \, dx\) [7566]

Optimal result

Integrand size = 49, antiderivative size = 21 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=4+e^{\frac {1}{4} x^2 (7+x)^2 \log ^2(4)}+x \]

[Out]

x+exp((x+7)^2*ln(2)^2*x^2)+4

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 1608, 6838} \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=e^{\frac {1}{4} \left (x^4+14 x^3+49 x^2\right ) \log ^2(4)}+x \]

[In]

Int[(2 + E^(((49*x^2 + 14*x^3 + x^4)*Log[4]^2)/4)*(49*x + 21*x^2 + 2*x^3)*Log[4]^2)/2,x]

[Out]

E^(((49*x^2 + 14*x^3 + x^4)*Log[4]^2)/4) + 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 1608

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 6838

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{align*} \text {integral}& = \frac {1}{2} \int \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx \\ & = x+\frac {1}{2} \log ^2(4) \int e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \, dx \\ & = x+\frac {1}{2} \log ^2(4) \int e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} x \left (49+21 x+2 x^2\right ) \, dx \\ & = e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=e^{\frac {1}{4} x^2 (7+x)^2 \log ^2(4)}+x \]

[In]

Integrate[(2 + E^(((49*x^2 + 14*x^3 + x^4)*Log[4]^2)/4)*(49*x + 21*x^2 + 2*x^3)*Log[4]^2)/2,x]

[Out]

E^((x^2*(7 + x)^2*Log[4]^2)/4) + x

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
risch	\(x +{\mathrm e}^{\left (x +7\right )^{2} \ln \left (2\right )^{2} x^{2}}\)	\(17\)
parallelrisch	\(x +{\mathrm e}^{x^{2} \left (x^{2}+14 x +49\right ) \ln \left (2\right )^{2}}\)	\(20\)
default	\(x +{\mathrm e}^{\left (x^{4}+14 x^{3}+49 x^{2}\right ) \ln \left (2\right )^{2}}\)	\(23\)
norman	\(x +{\mathrm e}^{\left (x^{4}+14 x^{3}+49 x^{2}\right ) \ln \left (2\right )^{2}}\)	\(23\)
parts	\(x +{\mathrm e}^{\left (x^{4}+14 x^{3}+49 x^{2}\right ) \ln \left (2\right )^{2}}\)	\(23\)

[In]

int(2*(2*x^3+21*x^2+49*x)*ln(2)^2*exp((x^4+14*x^3+49*x^2)*ln(2)^2)+1,x,method=_RETURNVERBOSE)

[Out]

x+exp((x+7)^2*ln(2)^2*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=x + e^{\left ({\left (x^{4} + 14 \, x^{3} + 49 \, x^{2}\right )} \log \left (2\right )^{2}\right )} \]

[In]

integrate(2*(2*x^3+21*x^2+49*x)*log(2)^2*exp((x^4+14*x^3+49*x^2)*log(2)^2)+1,x, algorithm="fricas")

[Out]

x + e^((x^4 + 14*x^3 + 49*x^2)*log(2)^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=x + e^{\left (x^{4} + 14 x^{3} + 49 x^{2}\right ) \log {\left (2 \right )}^{2}} \]

[In]

integrate(2*(2*x**3+21*x**2+49*x)*ln(2)**2*exp((x**4+14*x**3+49*x**2)*ln(2)**2)+1,x)

[Out]

x + exp((x**4 + 14*x**3 + 49*x**2)*log(2)**2)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=x + e^{\left (x^{4} \log \left (2\right )^{2} + 14 \, x^{3} \log \left (2\right )^{2} + 49 \, x^{2} \log \left (2\right )^{2}\right )} \]

[In]

integrate(2*(2*x^3+21*x^2+49*x)*log(2)^2*exp((x^4+14*x^3+49*x^2)*log(2)^2)+1,x, algorithm="maxima")

[Out]

x + e^(x^4*log(2)^2 + 14*x^3*log(2)^2 + 49*x^2*log(2)^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=x + e^{\left (x^{4} \log \left (2\right )^{2} + 14 \, x^{3} \log \left (2\right )^{2} + 49 \, x^{2} \log \left (2\right )^{2}\right )} \]

[In]

integrate(2*(2*x^3+21*x^2+49*x)*log(2)^2*exp((x^4+14*x^3+49*x^2)*log(2)^2)+1,x, algorithm="giac")

[Out]

x + e^(x^4*log(2)^2 + 14*x^3*log(2)^2 + 49*x^2*log(2)^2)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {1}{2} \left (2+e^{\frac {1}{4} \left (49 x^2+14 x^3+x^4\right ) \log ^2(4)} \left (49 x+21 x^2+2 x^3\right ) \log ^2(4)\right ) \, dx=x+{\mathrm {e}}^{x^4\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{14\,x^3\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{49\,x^2\,{\ln \left (2\right )}^2} \]

[In]

int(2*exp(log(2)^2*(49*x^2 + 14*x^3 + x^4))*log(2)^2*(49*x + 21*x^2 + 2*x^3) + 1,x)

[Out]

x + exp(x^4*log(2)^2)*exp(14*x^3*log(2)^2)*exp(49*x^2*log(2)^2)