\(\int (a+b (c x^n)^{2/n}) \, dx\) [3030]

Optimal result

Integrand size = 15, antiderivative size = 21 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=a x+\frac {1}{3} b x \left (c x^n\right )^{2/n} \]

[Out]

a*x+1/3*b*x*(c*x^n)^(2/n)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {15, 30} \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=a x+\frac {1}{3} b x \left (c x^n\right )^{2/n} \]

[In]

Int[a + b*(c*x^n)^(2/n),x]

[Out]

a*x + (b*x*(c*x^n)^(2/n))/3

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = a x+b \int \left (c x^n\right )^{2/n} \, dx \\ & = a x+\frac {\left (b \left (c x^n\right )^{2/n}\right ) \int x^2 \, dx}{x^2} \\ & = a x+\frac {1}{3} b x \left (c x^n\right )^{2/n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=a x+\frac {1}{3} b x \left (c x^n\right )^{2/n} \]

[In]

Integrate[a + b*(c*x^n)^(2/n),x]

[Out]

a*x + (b*x*(c*x^n)^(2/n))/3

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(a x +\frac {b x \left (c \,x^{n}\right )^{\frac {2}{n}}}{3}\)	\(20\)
default	\(a x +\frac {b x \,{\mathrm e}^{\frac {2 \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{n}}}{3}\)	\(23\)
norman	\(a x +\frac {b x \,{\mathrm e}^{\frac {2 \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{n}}}{3}\)	\(23\)
parts	\(a x +\frac {b x \,{\mathrm e}^{\frac {2 \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{n}}}{3}\)	\(23\)

[In]

int(a+b*(c*x^n)^(2/n),x,method=_RETURNVERBOSE)

[Out]

a*x+1/3*b*x*(c*x^n)^(2/n)

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=\frac {1}{3} \, b c^{\frac {2}{n}} x^{3} + a x \]

[In]

integrate(a+b*(c*x^n)^(2/n),x, algorithm="fricas")

[Out]

1/3*b*c^(2/n)*x^3 + a*x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=a x + \frac {b x \left (c x^{n}\right )^{\frac {2}{n}}}{3} \]

[In]

integrate(a+b*(c*x**n)**(2/n),x)

[Out]

a*x + b*x*(c*x**n)**(2/n)/3

Maxima [F]

\[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=\int { \left (c x^{n}\right )^{\frac {2}{n}} b + a \,d x } \]

[In]

integrate(a+b*(c*x^n)^(2/n),x, algorithm="maxima")

[Out]

b*c^(2/n)*integrate((x^n)^(2/n), x) + a*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=\frac {1}{3} \, b c^{\frac {2}{n}} x^{3} + a x \]

[In]

integrate(a+b*(c*x^n)^(2/n),x, algorithm="giac")

[Out]

1/3*b*c^(2/n)*x^3 + a*x

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \left (a+b \left (c x^n\right )^{2/n}\right ) \, dx=a\,x+\frac {b\,x\,{\left (c\,x^n\right )}^{2/n}}{3} \]

[In]

int(a + b*(c*x^n)^(2/n),x)

[Out]

a*x + (b*x*(c*x^n)^(2/n))/3