\(\int (-1+8 x+2^{2 \log ^2(x)} (3 x^2+4 x^2 \log (2) \log (x))) \, dx\) [1717]

Optimal result

Integrand size = 29, antiderivative size = 22 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=-x+x^2 \left (4+2^{2 \log ^2(x)} x\right )+\log (3) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2326} \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=x^3 2^{2 \log ^2(x)}+4 x^2-x \]

[In]

Int[-1 + 8*x + 2^(2*Log[x]^2)*(3*x^2 + 4*x^2*Log[2]*Log[x]),x]

[Out]

-x + 4*x^2 + 2^(2*Log[x]^2)*x^3

Rule 2326

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]

Rubi steps \begin{align*} \text {integral}& = -x+4 x^2+\int 2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right ) \, dx \\ & = -x+4 x^2+2^{2 \log ^2(x)} x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=-x+4 x^2+\frac {2^{-1+2 \log ^2(x)} x^3 \log (16)}{\log (4)} \]

[In]

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

[Out]

-x + 4*x^2 + (2^(-1 + 2*Log[x]^2)*x^3*Log[16])/Log[4]

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
risch	\(-x +x^{3} 2^{2 \ln \left (x \right )^{2}}+4 x^{2}\)	\(22\)
default	\(-x +x^{3} {\mathrm e}^{2 \ln \left (2\right ) \ln \left (x \right )^{2}}+4 x^{2}\)	\(24\)
norman	\(-x +x^{3} {\mathrm e}^{2 \ln \left (2\right ) \ln \left (x \right )^{2}}+4 x^{2}\)	\(24\)
parallelrisch	\(-x +x^{3} {\mathrm e}^{2 \ln \left (2\right ) \ln \left (x \right )^{2}}+4 x^{2}\)	\(24\)
parts	\(-x +x^{3} {\mathrm e}^{2 \ln \left (2\right ) \ln \left (x \right )^{2}}+4 x^{2}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=2^{2 \, \log \left (x\right )^{2}} x^{3} + 4 \, x^{2} - x \]

[In]

integrate((4*x^2*log(2)*log(x)+3*x^2)*exp(log(2)*log(x)^2)^2+8*x-1,x, algorithm="fricas")

[Out]

2^(2*log(x)^2)*x^3 + 4*x^2 - x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=x^{3} e^{2 \log {\left (2 \right )} \log {\left (x \right )}^{2}} + 4 x^{2} - x \]

[In]

integrate((4*x**2*ln(2)*ln(x)+3*x**2)*exp(ln(2)*ln(x)**2)**2+8*x-1,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=2^{2 \, \log \left (x\right )^{2}} x^{3} + 4 \, x^{2} - x \]

[In]

integrate((4*x^2*log(2)*log(x)+3*x^2)*exp(log(2)*log(x)^2)^2+8*x-1,x, algorithm="maxima")

[Out]

2^(2*log(x)^2)*x^3 + 4*x^2 - x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=2^{2 \, \log \left (x\right )^{2}} x^{3} + 4 \, x^{2} - x \]

[In]

integrate((4*x^2*log(2)*log(x)+3*x^2)*exp(log(2)*log(x)^2)^2+8*x-1,x, algorithm="giac")

[Out]

2^(2*log(x)^2)*x^3 + 4*x^2 - x

Mupad [B] (verification not implemented)

Time = 9.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (-1+8 x+2^{2 \log ^2(x)} \left (3 x^2+4 x^2 \log (2) \log (x)\right )\right ) \, dx=2^{2\,{\ln \left (x\right )}^2}\,x^3-x+4\,x^2 \]

[In]

int(8*x + exp(2*log(2)*log(x)^2)*(3*x^2 + 4*x^2*log(2)*log(x)) - 1,x)

[Out]

2^(2*log(x)^2)*x^3 - x + 4*x^2