3.3023 \(\int x^2 (a+b (c x^n)^{\frac {1}{n}})^p \, dx\)

Optimal. Leaf size=126 \[ \frac {a^2 x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^3 (p+1)}-\frac {2 a x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^3 (p+2)}+\frac {x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+3}}{b^3 (p+3)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {368, 43} \[ \frac {a^2 x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^3 (p+1)}-\frac {2 a x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^3 (p+2)}+\frac {x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+3}}{b^3 (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

(a^2*x^3*(a + b*(c*x^n)^n^(-1))^(1 + p))/(b^3*(1 + p)*(c*x^n)^(3/n)) - (2*a*x^3*(a + b*(c*x^n)^n^(-1))^(2 + p)
)/(b^3*(2 + p)*(c*x^n)^(3/n)) + (x^3*(a + b*(c*x^n)^n^(-1))^(3 + p))/(b^3*(3 + p)*(c*x^n)^(3/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^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx &=\left (x^3 \left (c x^n\right )^{-3/n}\right ) \operatorname {Subst}\left (\int x^2 (a+b x)^p \, dx,x,\left (c x^n\right )^{\frac {1}{n}}\right )\\ &=\left (x^3 \left (c x^n\right )^{-3/n}\right ) \operatorname {Subst}\left (\int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{1+p}}{b^2}+\frac {(a+b x)^{2+p}}{b^2}\right ) \, dx,x,\left (c x^n\right )^{\frac {1}{n}}\right )\\ &=\frac {a^2 x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p}}{b^3 (1+p)}-\frac {2 a x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{2+p}}{b^3 (2+p)}+\frac {x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{3+p}}{b^3 (3+p)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 95, normalized size = 0.75 \[ \frac {x^3 \left (c x^n\right )^{-3/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1} \left (2 a^2-2 a b (p+1) \left (c x^n\right )^{\frac {1}{n}}+b^2 \left (p^2+3 p+2\right ) \left (c x^n\right )^{2/n}\right )}{b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*(c*x^n)^n^(-1))^p,x]

[Out]

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

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fricas [A]  time = 0.90, size = 129, normalized size = 1.02 \[ -\frac {{\left (2 \, a^{2} b c^{\left (\frac {1}{n}\right )} p x - {\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} c^{\frac {3}{n}} x^{3} - {\left (a b^{2} p^{2} + a b^{2} p\right )} c^{\frac {2}{n}} x^{2} - 2 \, a^{3}\right )} {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p}}{{\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )} c^{\frac {3}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="fricas")

[Out]

-(2*a^2*b*c^(1/n)*p*x - (b^3*p^2 + 3*b^3*p + 2*b^3)*c^(3/n)*x^3 - (a*b^2*p^2 + a*b^2*p)*c^(2/n)*x^2 - 2*a^3)*(
b*c^(1/n)*x + a)^p/((b^3*p^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)*c^(3/n))

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giac [A]  time = 0.46, size = 243, normalized size = 1.93 \[ \frac {{\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac {3}{n}} p^{2} x^{3} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{2} c^{\frac {2}{n}} p^{2} x^{2} + 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac {3}{n}} p x^{3} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{2} c^{\frac {2}{n}} p x^{2} + 2 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{3} c^{\frac {3}{n}} x^{3} - 2 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2} b c^{\left (\frac {1}{n}\right )} p x + 2 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{3}}{b^{3} c^{\frac {3}{n}} p^{3} + 6 \, b^{3} c^{\frac {3}{n}} p^{2} + 11 \, b^{3} c^{\frac {3}{n}} p + 6 \, b^{3} c^{\frac {3}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="giac")

[Out]

((b*c^(1/n)*x + a)^p*b^3*c^(3/n)*p^2*x^3 + (b*c^(1/n)*x + a)^p*a*b^2*c^(2/n)*p^2*x^2 + 3*(b*c^(1/n)*x + a)^p*b
^3*c^(3/n)*p*x^3 + (b*c^(1/n)*x + a)^p*a*b^2*c^(2/n)*p*x^2 + 2*(b*c^(1/n)*x + a)^p*b^3*c^(3/n)*x^3 - 2*(b*c^(1
/n)*x + a)^p*a^2*b*c^(1/n)*p*x + 2*(b*c^(1/n)*x + a)^p*a^3)/(b^3*c^(3/n)*p^3 + 6*b^3*c^(3/n)*p^2 + 11*b^3*c^(3
/n)*p + 6*b^3*c^(3/n))

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maple [C]  time = 0.31, size = 787, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(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^3*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)-2/b/(p+1)*(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)/(p+3)*x^3/((x^n)^(1/n))/(c^(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))-2/b^2/(p+1)*(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)/(p+3)/(p+2)*a/((x^n)^(1/n))^2*x^3/(c^(1/n))^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))*p-2/b^2/(p+1)*(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)/(p+3)/
(p+2)*a/((x^n)^(1/n))^2*x^3/(c^(1/n))^2*exp(-I*Pi*(csgn(I*c)-csgn(I*c*x^n))*(-csgn(I*x^n)+csgn(I*c*x^n))/n*csg
n(I*c*x^n))+2/b^3/(p+1)*(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)/(p+3)*a^2/((x^n)^(1/n))^3*x^3/(c^(1/n))^3/(p+2)*exp(-3/2*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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{p} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*(c*x^n)^(1/n))^p,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*(c*x^n)^(1/n))^p,x)

[Out]

int(x^2*(a + b*(c*x^n)^(1/n))^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*(c*x**n)**(1/n))**p,x)

[Out]

Integral(x**2*(a + b*(c*x**n)**(1/n))**p, x)

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