3.58.58 \(\int \frac {1}{5} (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))) \, dx\)

Optimal. Leaf size=17 \[ \frac {x}{5}+\left (3+2^{2 x}\right )^2 x \]

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Rubi [B]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 3.94, number of steps used = 6, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 2176, 2194} \begin {gather*} \frac {46 x}{5}+\frac {3\ 2^{2 x} (2 x \log (2)+1)}{\log (2)}+\frac {2^{4 x-2} (4 x \log (2)+1)}{\log (2)}-\frac {3\ 2^{2 x}}{\log (2)}-\frac {2^{4 x-2}}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(46 + 2^(4*x)*(5 + 20*x*Log[2]) + 2^(2*x)*(30 + 60*x*Log[2]))/5,x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx\\ &=\frac {46 x}{5}+\frac {1}{5} \int 2^{4 x} (5+20 x \log (2)) \, dx+\frac {1}{5} \int 2^{2 x} (30+60 x \log (2)) \, dx\\ &=\frac {46 x}{5}+\frac {3\ 2^{2 x} (1+2 x \log (2))}{\log (2)}+\frac {2^{-2+4 x} (1+4 x \log (2))}{\log (2)}-6 \int 2^{2 x} \, dx-\int 2^{4 x} \, dx\\ &=\frac {46 x}{5}-\frac {3\ 2^{2 x}}{\log (2)}-\frac {2^{-2+4 x}}{\log (2)}+\frac {3\ 2^{2 x} (1+2 x \log (2))}{\log (2)}+\frac {2^{-2+4 x} (1+4 x \log (2))}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.65 \begin {gather*} \frac {1}{5} \left (46 x+5\ 16^x x+\frac {15\ 4^{1+x} x \log (2)}{\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(46 + 2^(4*x)*(5 + 20*x*Log[2]) + 2^(2*x)*(30 + 60*x*Log[2]))/5,x]

[Out]

(46*x + 5*16^x*x + (15*4^(1 + x)*x*Log[2])/Log[4])/5

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fricas [A]  time = 0.63, size = 19, normalized size = 1.12 \begin {gather*} 2^{4 \, x} x + 6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x*log(2)+5)*exp(2*x*log(2))^2+1/5*(60*x*log(2)+30)*exp(2*x*log(2))+46/5,x, algorithm="fricas
")

[Out]

2^(4*x)*x + 6*2^(2*x)*x + 46/5*x

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giac [A]  time = 0.12, size = 19, normalized size = 1.12 \begin {gather*} 2^{4 \, x} x + 6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x*log(2)+5)*exp(2*x*log(2))^2+1/5*(60*x*log(2)+30)*exp(2*x*log(2))+46/5,x, algorithm="giac")

[Out]

2^(4*x)*x + 6*2^(2*x)*x + 46/5*x

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maple [A]  time = 0.11, size = 18, normalized size = 1.06

method	result	size
risch	\(\frac {46 x}{5}+x 4^{2 x}+6 \,4^{x} x\)	\(18\)
default	\(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \relax (2)}+6 \,{\mathrm e}^{2 x \ln \relax (2)} x\)	\(24\)
norman	\(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \relax (2)}+6 \,{\mathrm e}^{2 x \ln \relax (2)} x\)	\(24\)
derivativedivides	\(\frac {92 x \ln \relax (2)+10 \ln \relax (2) {\mathrm e}^{4 x \ln \relax (2)} x +60 \ln \relax (2) {\mathrm e}^{2 x \ln \relax (2)} x}{10 \ln \relax (2)}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(20*x*ln(2)+5)*exp(2*x*ln(2))^2+1/5*(60*x*ln(2)+30)*exp(2*x*ln(2))+46/5,x,method=_RETURNVERBOSE)

[Out]

46/5*x+x*(4^x)^2+6*4^x*x

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maxima [B]  time = 0.45, size = 42, normalized size = 2.47 \begin {gather*} 6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x + \frac {{\left (4 \, x \log \relax (2) - 1\right )} 2^{4 \, x}}{4 \, \log \relax (2)} + \frac {2^{4 \, x - 2}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x*log(2)+5)*exp(2*x*log(2))^2+1/5*(60*x*log(2)+30)*exp(2*x*log(2))+46/5,x, algorithm="maxima
")

[Out]

6*2^(2*x)*x + 46/5*x + 1/4*(4*x*log(2) - 1)*2^(4*x)/log(2) + 2^(4*x - 2)/log(2)

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mupad [B]  time = 0.08, size = 19, normalized size = 1.12 \begin {gather*} \frac {x\,\left (30\,2^{2\,x}+5\,2^{4\,x}+46\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x*log(2))*(20*x*log(2) + 5))/5 + (exp(2*x*log(2))*(60*x*log(2) + 30))/5 + 46/5,x)

[Out]

(x*(30*2^(2*x) + 5*2^(4*x) + 46))/5

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sympy [A]  time = 0.13, size = 26, normalized size = 1.53 \begin {gather*} x e^{4 x \log {\relax (2 )}} + 6 x e^{2 x \log {\relax (2 )}} + \frac {46 x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x*ln(2)+5)*exp(2*x*ln(2))**2+1/5*(60*x*ln(2)+30)*exp(2*x*ln(2))+46/5,x)

[Out]

x*exp(4*x*log(2)) + 6*x*exp(2*x*log(2)) + 46*x/5

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