\(\int x (a+b (c x^n)^{\frac {1}{n}})^p \, dx\) [3024]

Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=-\frac {a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p}}{b^2 (1+p)}+\frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{2+p}}{b^2 (2+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {375, 45} \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac {a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

[In]

Int[x*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 375

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76 \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p} \left (-a+b (1+p) \left (c x^n\right )^{\frac {1}{n}}\right )}{b^2 (1+p) (2+p)} \]

[In]

Integrate[x*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 6.71 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.66

method	result	size
risch	\(\frac {{\left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}^{1+p} c^{-\frac {1}{n}} x^{2} \left (x^{n}\right )^{-\frac {1}{n}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}}{b \left (1+p \right )}-\frac {\left (x^{n}\right )^{-\frac {2}{n}} c^{-\frac {2}{n}} x^{2} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}} {\left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}^{2+p}}{b^{2} \left (1+p \right ) \left (2+p \right )}\)	\(304\)

[In]

int(x*(a+b*(c*x^n)^(1/n))^p,x,method=_RETURNVERBOSE)

[Out]

(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)
^(1+p)/(c^(1/n))*x^2/((x^n)^(1/n))*exp(-1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*
c*x^n))/n)/b/(1+p)-1/b^2/(1+p)/((x^n)^(1/n))^2/(c^(1/n))^2*x^2*exp(-I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*
x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)*(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^
n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)^(2+p)/(2+p)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (a b c^{\left (\frac {1}{n}\right )} p x + {\left (b^{2} p + b^{2}\right )} c^{\frac {2}{n}} x^{2} - a^{2}\right )} {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p}}{{\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )} c^{\frac {2}{n}}} \]

[In]

integrate(x*(a+b*(c*x^n)^(1/n))^p,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int x \left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{p}\, dx \]

[In]

integrate(x*(a+b*(c*x**n)**(1/n))**p,x)

[Out]

Integral(x*(a + b*(c*x**n)**(1/n))**p, x)

Maxima [F]

\[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int { {\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{p} x \,d x } \]

[In]

integrate(x*(a+b*(c*x^n)^(1/n))^p,x, algorithm="maxima")

[Out]

integrate(((c*x^n)^(1/n)*b + a)^p*x, x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.64 \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac {2}{n}} p x^{2} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b c^{\left (\frac {1}{n}\right )} p x + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac {2}{n}} x^{2} - {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2}}{b^{2} c^{\frac {2}{n}} p^{2} + 3 \, b^{2} c^{\frac {2}{n}} p + 2 \, b^{2} c^{\frac {2}{n}}} \]

[In]

integrate(x*(a+b*(c*x^n)^(1/n))^p,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int x\,{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^p \,d x \]

[In]

int(x*(a + b*(c*x^n)^(1/n))^p,x)

[Out]

int(x*(a + b*(c*x^n)^(1/n))^p, x)