\(\int 3^{2^x} 4^x \, dx\) [256]

Optimal result

Integrand size = 9, antiderivative size = 33 \[ \int 3^{2^x} 4^x \, dx=-\frac {3^{2^x}}{\log (2) \log ^2(3)}+\frac {2^x 3^{2^x}}{\log (2) \log (3)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2320, 2207, 2225} \[ \int 3^{2^x} 4^x \, dx=\frac {2^x 3^{2^x}}{\log (2) \log (3)}-\frac {3^{2^x}}{\log (2) \log ^2(3)} \]

[In]

Int[3^2^x*4^x,x]

[Out]

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

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int 3^{2^x} 4^x \, dx=\frac {3^{2^x} \left (-1+2^x \log (3)\right )}{\log (2) \log ^2(3)} \]

[In]

Integrate[3^2^x*4^x,x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {\left (2^{x} \ln \left (3\right )-1\right ) 3^{2^{x}}}{\ln \left (2\right ) \ln \left (3\right )^{2}}\)	\(23\)
norman	\(\frac {{\mathrm e}^{x \ln \left (2\right )} {\mathrm e}^{{\mathrm e}^{x \ln \left (2\right )} \ln \left (3\right )}}{\ln \left (2\right ) \ln \left (3\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x \ln \left (2\right )} \ln \left (3\right )}}{\ln \left (2\right ) \ln \left (3\right )^{2}}\)	\(44\)

[In]

int(4^x*3^(2^x),x,method=_RETURNVERBOSE)

[Out]

(2^x*ln(3)-1)/ln(2)/ln(3)^2*3^(2^x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int 3^{2^x} 4^x \, dx=\frac {{\left (2^{x} \log \left (3\right ) - 1\right )} 3^{\left (2^{x}\right )}}{\log \left (3\right )^{2} \log \left (2\right )} \]

[In]

integrate(4^x*3^(2^x),x, algorithm="fricas")

[Out]

(2^x*log(3) - 1)*3^(2^x)/(log(3)^2*log(2))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int 3^{2^x} 4^x \, dx=\frac {\left (e^{\frac {x \log {\left (4 \right )}}{2}} \log {\left (3 \right )} - 1\right ) e^{e^{\frac {x \log {\left (4 \right )}}{2}} \log {\left (3 \right )}}}{\log {\left (2 \right )} \log {\left (3 \right )}^{2}} \]

[In]

integrate(4**x*3**(2**x),x)

[Out]

(exp(x*log(4)/2)*log(3) - 1)*exp(exp(x*log(4)/2)*log(3))/(log(2)*log(3)**2)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int 3^{2^x} 4^x \, dx=-\frac {4^{x} \Gamma \left (2, -4^{\frac {1}{2} \, x} \log \left (3\right )\right )}{4^{x} \log \left (3\right )^{2} \log \left (2\right )} \]

[In]

integrate(4^x*3^(2^x),x, algorithm="maxima")

[Out]

-4^x*gamma(2, -4^(1/2*x)*log(3))/(4^x*log(3)^2*log(2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int 3^{2^x} 4^x \, dx=\frac {2^{x} e^{\left (2^{x} \log \left (3\right ) + 2 \, x \log \left (2\right )\right )} \log \left (3\right ) - e^{\left (2^{x} \log \left (3\right ) + 2 \, x \log \left (2\right )\right )}}{2^{2 \, x} \log \left (3\right )^{2} \log \left (2\right )} \]

[In]

integrate(4^x*3^(2^x),x, algorithm="giac")

[Out]

(2^x*e^(2^x*log(3) + 2*x*log(2))*log(3) - e^(2^x*log(3) + 2*x*log(2)))/(2^(2*x)*log(3)^2*log(2))

Mupad [B] (verification not implemented)

Time = 15.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int 3^{2^x} 4^x \, dx=\frac {3^{2^x}\,\left (2^x\,\ln \left (3\right )-1\right )}{\ln \left (2\right )\,{\ln \left (3\right )}^2} \]

[In]

int(3^(2^x)*4^x,x)

[Out]

(3^(2^x)*(2^x*log(3) - 1))/(log(2)*log(3)^2)