3.3017 \(\int x \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^p \, dx\)

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)^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))

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Rubi [A]  time = 0.0694753, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118 \[ \frac{x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+2}}{b^2 (p+2)}-\frac{a x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^{p+1}}{b^2 (p+1)} \]

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))

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Rubi in Sympy [A]  time = 11.3773, size = 68, normalized size = 0.82 \[ - \frac{a x^{2} \left (c x^{n}\right )^{- \frac{2}{n}} \left (a + b \left (c x^{n}\right )^{\frac{1}{n}}\right )^{p + 1}}{b^{2} \left (p + 1\right )} + \frac{x^{2} \left (c x^{n}\right )^{- \frac{2}{n}} \left (a + b \left (c x^{n}\right )^{\frac{1}{n}}\right )^{p + 2}}{b^{2} \left (p + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x*(a+b*(c*x**n)**(1/n))**p,x)

[Out]

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

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Mathematica [A]  time = 0.211161, size = 156, normalized size = 1.88 \[ \frac{x^2 \left (c x^n\right )^{-2/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^p \left (\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a}+1\right )^{-p} \left (-a^2 \left (\left (\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a}+1\right )^p-1\right )+b^2 (p+1) \left (c x^n\right )^{2/n} \left (\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a}+1\right )^p+a b p \left (c x^n\right )^{\frac{1}{n}} \left (\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a}+1\right )^p\right )}{b^2 (p+1) (p+2)} \]

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

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Maple [C]  time = 0.244, size = 1007, normalized size = 12.1 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x*(a+b*(c*x^n)^(1/n))^p,x)

[Out]

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

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Maxima [A]  time = 1.82564, size = 89, normalized size = 1.07 \[ \frac{{\left (b^{2} c^{\frac{2}{n}}{\left (p + 1\right )} x^{2} + a b c^{\left (\frac{1}{n}\right )} p x - a^{2}\right )}{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p} c^{-\frac{2}{n}}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [A]  time = 0.241639, size = 107, normalized size = 1.29 \[ \frac{{\left (a b c^{\left (\frac{1}{n}\right )} p x +{\left (b^{2} p + b^{2}\right )} c^{\frac{2}{n}} x^{2} - a^{2}\right )}{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )}^{p}}{{\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )} c^{\frac{2}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(((c*x^n)^(1/n)*b + a)^p*x,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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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int x \left (a + b \left (c x^{n}\right )^{\frac{1}{n}}\right )^{p}\, dx \]

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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GIAC/XCAS [A]  time = 23.3805, size = 215, normalized size = 2.59 \[ \frac{b^{2} p x^{2} e^{\left (p{\rm ln}\left (b x e^{\left (\frac{{\rm ln}\left (c\right )}{n}\right )} + a\right ) + \frac{2 \,{\rm ln}\left (c\right )}{n}\right )} + b^{2} x^{2} e^{\left (p{\rm ln}\left (b x e^{\left (\frac{{\rm ln}\left (c\right )}{n}\right )} + a\right ) + \frac{2 \,{\rm ln}\left (c\right )}{n}\right )} + a b p x e^{\left (p{\rm ln}\left (b x e^{\left (\frac{{\rm ln}\left (c\right )}{n}\right )} + a\right ) + \frac{{\rm ln}\left (c\right )}{n}\right )} - a^{2} e^{\left (p{\rm ln}\left (b x e^{\left (\frac{{\rm ln}\left (c\right )}{n}\right )} + a\right )\right )}}{b^{2} p^{2} e^{\left (\frac{2 \,{\rm ln}\left (c\right )}{n}\right )} + 3 \, b^{2} p e^{\left (\frac{2 \,{\rm ln}\left (c\right )}{n}\right )} + 2 \, b^{2} e^{\left (\frac{2 \,{\rm ln}\left (c\right )}{n}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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