\(\int 10^{2+5 x} \, dx\) [59]

Optimal result

Integrand size = 7, antiderivative size = 19 \[ \int 10^{2+5 x} \, dx=\frac {2^{2+5 x} 5^{1+5 x}}{\log (10)} \] Output:

2^(2+5*x)*5^(1+5*x)/ln(10)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int 10^{2+5 x} \, dx=\frac {2^{2+5 x} 5^{1+5 x}}{\log (10)} \] Input:

Integrate[10^(2 + 5*x),x]
 

Output:

(2^(2 + 5*x)*5^(1 + 5*x))/Log[10]
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2624}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int 10^{5 x+2} \, dx\)

\(\Big \downarrow \) 2624

\(\displaystyle \frac {2^{5 x+2} 5^{5 x+1}}{\log (10)}\)

Input:

Int[10^(2 + 5*x),x]
 

Output:

(2^(2 + 5*x)*5^(1 + 5*x))/Log[10]
 

Defintions of rubi rules used

rule 2624

Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; 
FreeQ[{F, n}, x] && LinearQ[v, x]
 

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {20 \,10^{5 x}}{\ln \left (10\right )}\)	\(12\)
default	\(\frac {20 \,10^{5 x}}{\ln \left (10\right )}\)	\(12\)
gosper	\(\frac {10^{2+5 x}}{5 \ln \left (10\right )}\)	\(14\)
parallelrisch	\(\frac {10^{2+5 x}}{5 \ln \left (10\right )}\)	\(14\)
norman	\(\frac {{\mathrm e}^{\left (2+5 x \right ) \ln \left (10\right )}}{5 \ln \left (10\right )}\)	\(16\)
risch	\(\frac {20 \,3125^{x} 32^{x}}{\ln \left (2\right )+\ln \left (5\right )}\)	\(16\)
meijerg	\(-\frac {20 \left (1-{\mathrm e}^{5 x \ln \left (10\right )}\right )}{\ln \left (10\right )}\)	\(17\)
orering	\(\frac {10^{2+5 x}}{5 \ln \left (2\right )+5 \ln \left (5\right )}\)	\(17\)

Input:

int(10^(2+5*x),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

20*(10^x)^5/ln(10)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int 10^{2+5 x} \, dx=\frac {10^{5 \, x + 2}}{5 \, \log \left (10\right )} \] Input:

integrate(10^(2+5*x),x, algorithm="fricas")
 

Output:

1/5*10^(5*x + 2)/log(10)
 

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int 10^{2+5 x} \, dx=\frac {10^{5 x + 2}}{5 \log {\left (10 \right )}} \] Input:

integrate(10**(2+5*x),x)
 

Output:

10**(5*x + 2)/(5*log(10))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int 10^{2+5 x} \, dx=\frac {10^{5 \, x + 2}}{5 \, \log \left (10\right )} \] Input:

integrate(10^(2+5*x),x, algorithm="maxima")
 

Output:

1/5*10^(5*x + 2)/log(10)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int 10^{2+5 x} \, dx=\frac {10^{5 \, x + 2}}{5 \, \log \left (10\right )} \] Input:

integrate(10^(2+5*x),x, algorithm="giac")
 

Output:

1/5*10^(5*x + 2)/log(10)
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int 10^{2+5 x} \, dx=\frac {20\,{10}^{5\,x}}{\ln \left (10\right )} \] Input:

int(10^(5*x + 2),x)
 

Output:

(20*10^(5*x))/log(10)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int 10^{2+5 x} \, dx=\frac {20 \,10^{5 x}}{\mathrm {log}\left (10\right )} \] Input:

int(10^(2+5*x),x)
 

Output:

(20*10**(5*x))/log(10)