\(\int (k^{x/2}+x^{\sqrt {k}}) \, dx\) [742]

Optimal result

Integrand size = 15, antiderivative size = 33 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}+\frac {2 k^{x/2}}{\log (k)} \]

[Out]

2*k^(1/2*x)/ln(k)+x^(1+k^(1/2))/(1+k^(1/2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2225} \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {2 k^{x/2}}{\log (k)}+\frac {x^{\sqrt {k}+1}}{\sqrt {k}+1} \]

[In]

Int[k^(x/2) + x^Sqrt[k],x]

[Out]

x^(1 + Sqrt[k])/(1 + Sqrt[k]) + (2*k^(x/2))/Log[k]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}+\int k^{x/2} \, dx \\ & = \frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}+\frac {2 k^{x/2}}{\log (k)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}+\frac {2 k^{x/2}}{\log (k)} \]

[In]

Integrate[k^(x/2) + x^Sqrt[k],x]

[Out]

x^(1 + Sqrt[k])/(1 + Sqrt[k]) + (2*k^(x/2))/Log[k]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {2 k^{\frac {x}{2}}}{\ln \left (k \right )}+\frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}\)	\(28\)
parts	\(\frac {2 k^{\frac {x}{2}}}{\ln \left (k \right )}+\frac {x^{1+\sqrt {k}}}{1+\sqrt {k}}\)	\(28\)
norman	\(\frac {x \,{\mathrm e}^{\sqrt {k}\, \ln \left (x \right )}}{1+\sqrt {k}}+\frac {2 \,{\mathrm e}^{\frac {x \ln \left (k \right )}{2}}}{\ln \left (k \right )}\)	\(30\)
risch	\(\frac {2 k^{\frac {x}{2}}}{\ln \left (k \right )}+\frac {\left (\sqrt {k}-1\right ) x \,x^{\sqrt {k}}}{k -1}\)	\(30\)
parallelrisch	\(\frac {x \,x^{\sqrt {k}} \ln \left (k \right )+2 k^{\frac {x}{2}} \sqrt {k}+2 k^{\frac {x}{2}}}{\left (1+\sqrt {k}\right ) \ln \left (k \right )}\)	\(40\)

[In]

int(k^(1/2*x)+x^(k^(1/2)),x,method=_RETURNVERBOSE)

[Out]

2*k^(1/2*x)/ln(k)+x^(1+k^(1/2))/(1+k^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {2 \, {\left (k - 1\right )} k^{\frac {1}{2} \, x} + {\left (\sqrt {k} x \log \left (k\right ) - x \log \left (k\right )\right )} x^{\left (\sqrt {k}\right )}}{{\left (k - 1\right )} \log \left (k\right )} \]

[In]

integrate(k^(1/2*x)+x^(k^(1/2)),x, algorithm="fricas")

[Out]

(2*(k - 1)*k^(1/2*x) + (sqrt(k)*x*log(k) - x*log(k))*x^sqrt(k))/((k - 1)*log(k))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\begin {cases} \frac {2 k^{\frac {x}{2}}}{\log {\left (k \right )}} & \text {for}\: \log {\left (k \right )} \neq 0 \\x & \text {otherwise} \end {cases} + \begin {cases} \frac {x^{\sqrt {k} + 1}}{\sqrt {k} + 1} & \text {for}\: \sqrt {k} \neq -1 \\\log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(k**(1/2*x)+x**(k**(1/2)),x)

[Out]

Piecewise((2*k**(x/2)/log(k), Ne(log(k), 0)), (x, True)) + Piecewise((x**(sqrt(k) + 1)/(sqrt(k) + 1), Ne(sqrt(
k), -1)), (log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {x^{\sqrt {k} + 1}}{\sqrt {k} + 1} + \frac {2 \, k^{\frac {1}{2} \, x}}{\log \left (k\right )} \]

[In]

integrate(k^(1/2*x)+x^(k^(1/2)),x, algorithm="maxima")

[Out]

x^(sqrt(k) + 1)/(sqrt(k) + 1) + 2*k^(1/2*x)/log(k)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {x^{\sqrt {k} + 1}}{\sqrt {k} + 1} + \frac {2 \, \sqrt {k^{x}}}{\log \left (k\right )} \]

[In]

integrate(k^(1/2*x)+x^(k^(1/2)),x, algorithm="giac")

[Out]

x^(sqrt(k) + 1)/(sqrt(k) + 1) + 2*sqrt(k^x)/log(k)

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \left (k^{x/2}+x^{\sqrt {k}}\right ) \, dx=\frac {2\,k^{x/2}}{\ln \left (k\right )}+\frac {x\,x^{\sqrt {k}}}{\sqrt {k}+1} \]

[In]

int(k^(x/2) + x^(k^(1/2)),x)

[Out]

(2*k^(x/2))/log(k) + (x*x^(k^(1/2)))/(k^(1/2) + 1)