3.3024 \(\int x (a+b (c x^n)^{\frac {1}{n}})^p \, dx\)

Optimal. Leaf size=83 \[ \frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac {a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {368, 43} \[ \frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac {a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

Antiderivative was successfully verified.

[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 43

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 368

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*} \int x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx &=\left (x^2 \left (c x^n\right )^{-2/n}\right ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.03, size = 63, normalized size = 0.76 \[ \frac {x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1} \left (b (p+1) \left (c x^n\right )^{\frac {1}{n}}-a\right )}{b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.92, size = 79, normalized size = 0.95 \[ \frac {{\left (a b c^{\left (\frac {1}{n}\right )} p x + {\left (b^{2} p + b^{2}\right )} c^{\frac {2}{n}} x^{2} - a^{2}\right )} {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p}}{{\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )} c^{\frac {2}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.46, size = 136, normalized size = 1.64 \[ \frac {{\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac {2}{n}} p x^{2} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b c^{\left (\frac {1}{n}\right )} p x + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{2} c^{\frac {2}{n}} x^{2} - {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2}}{b^{2} c^{\frac {2}{n}} p^{2} + 3 \, b^{2} c^{\frac {2}{n}} p + 2 \, b^{2} c^{\frac {2}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.18, size = 304, normalized size = 3.66 \[ \frac {x^{2} c^{-\frac {1}{n}} \left (x^{n}\right )^{-\frac {1}{n}} \left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}}+a \right )^{p +1} {\mathrm e}^{-\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}}}{\left (p +1\right ) b}-\frac {x^{2} c^{-\frac {2}{n}} \left (x^{n}\right )^{-\frac {2}{n}} \left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}}+a \right )^{p +2} {\mathrm e}^{-\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{n}}}{\left (p +1\right ) \left (p +2\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{p} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*(c*x^n)^(1/n))^p,x)

[Out]

int(x*(a + b*(c*x^n)^(1/n))^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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