3.3013 \(\int \frac{x^3}{(a+b (c x^n)^{\frac{1}{n}})^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^4}-\frac{2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac{x^4 \left (c x^n\right )^{-2/n}}{2 b^2} \]

[Out]

(-2*a*x^4)/(b^3*(c*x^n)^(3/n)) + x^4/(2*b^2*(c*x^n)^(2/n)) + (a^3*x^4)/(b^4*(c*x^n)^(4/n)*(a + b*(c*x^n)^n^(-1
))) + (3*a^2*x^4*Log[a + b*(c*x^n)^n^(-1)])/(b^4*(c*x^n)^(4/n))

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Rubi [A]  time = 0.0437005, antiderivative size = 114, normalized size of antiderivative = 1., 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} \[ \frac{a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^4}-\frac{2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac{x^4 \left (c x^n\right )^{-2/n}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*(c*x^n)^n^(-1))^2,x]

[Out]

(-2*a*x^4)/(b^3*(c*x^n)^(3/n)) + x^4/(2*b^2*(c*x^n)^(2/n)) + (a^3*x^4)/(b^4*(c*x^n)^(4/n)*(a + b*(c*x^n)^n^(-1
))) + (3*a^2*x^4*Log[a + b*(c*x^n)^n^(-1)])/(b^4*(c*x^n)^(4/n))

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)]

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])

Rubi steps

\begin{align*} \int \frac{x^3}{\left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^2} \, dx &=\left (x^4 \left (c x^n\right )^{-4/n}\right ) \operatorname{Subst}\left (\int \frac{x^3}{(a+b x)^2} \, dx,x,\left (c x^n\right )^{\frac{1}{n}}\right )\\ &=\left (x^4 \left (c x^n\right )^{-4/n}\right ) \operatorname{Subst}\left (\int \left (-\frac{2 a}{b^3}+\frac{x}{b^2}-\frac{a^3}{b^3 (a+b x)^2}+\frac{3 a^2}{b^3 (a+b x)}\right ) \, dx,x,\left (c x^n\right )^{\frac{1}{n}}\right )\\ &=-\frac{2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac{x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac{a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}+\frac{3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.0721593, size = 89, normalized size = 0.78 \[ \frac{x^4 \left (c x^n\right )^{-4/n} \left (\frac{2 a^3}{a+b \left (c x^n\right )^{\frac{1}{n}}}+6 a^2 \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )-4 a b \left (c x^n\right )^{\frac{1}{n}}+b^2 \left (c x^n\right )^{2/n}\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*(c*x^n)^n^(-1))^2,x]

[Out]

(x^4*(-4*a*b*(c*x^n)^n^(-1) + b^2*(c*x^n)^(2/n) + (2*a^3)/(a + b*(c*x^n)^n^(-1)) + 6*a^2*Log[a + b*(c*x^n)^n^(
-1)]))/(2*b^4*(c*x^n)^(4/n))

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Maple [C]  time = 0.121, size = 661, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4/a/(a+b*exp(-1/2*(I*Pi*csgn(I*c*x^n)^3-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)+I*Pi
*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-2*ln(c)-2*ln(x^n))/n))-3*a*x/b^3/(c^(1/n))^3*exp(3/2*(I*Pi*csgn(I*c*x^n)*
csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n
*ln(x)-2*ln(x^n))/n)-1/a*x^3/b/(c^(1/n))*exp(1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^
2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)+3/2*x^2/b^2/(c^(1/n)
)^2*exp((I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I
*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)+3*a^2*ln(b*exp(-1/2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)
-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(c)-2*ln(x
^n))/n)*x+a)/b^4/(c^(1/n))^4*exp(2*(I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csgn(I*c*x^n)^2*csgn(I*x^n)-
I*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*csgn(I*c*x^n)^3+2*n*ln(x)-2*ln(x^n))/n)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{4}}{a b c^{\left (\frac{1}{n}\right )}{\left (x^{n}\right )}^{\left (\frac{1}{n}\right )} + a^{2}} - 3 \, \int \frac{x^{3}}{a b c^{\left (\frac{1}{n}\right )}{\left (x^{n}\right )}^{\left (\frac{1}{n}\right )} + a^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*(c*x^n)^(1/n))^2,x, algorithm="maxima")

[Out]

x^4/(a*b*c^(1/n)*(x^n)^(1/n) + a^2) - 3*integrate(x^3/(a*b*c^(1/n)*(x^n)^(1/n) + a^2), x)

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Fricas [A]  time = 1.53457, size = 208, normalized size = 1.82 \begin{align*} \frac{b^{3} c^{\frac{3}{n}} x^{3} - 3 \, a b^{2} c^{\frac{2}{n}} x^{2} - 4 \, a^{2} b c^{\left (\frac{1}{n}\right )} x + 2 \, a^{3} + 6 \,{\left (a^{2} b c^{\left (\frac{1}{n}\right )} x + a^{3}\right )} \log \left (b c^{\left (\frac{1}{n}\right )} x + a\right )}{2 \,{\left (b^{5} c^{\frac{5}{n}} x + a b^{4} c^{\frac{4}{n}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*(c*x^n)^(1/n))^2,x, algorithm="fricas")

[Out]

1/2*(b^3*c^(3/n)*x^3 - 3*a*b^2*c^(2/n)*x^2 - 4*a^2*b*c^(1/n)*x + 2*a^3 + 6*(a^2*b*c^(1/n)*x + a^3)*log(b*c^(1/
n)*x + a))/(b^5*c^(5/n)*x + a*b^4*c^(4/n))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b \left (c x^{n}\right )^{\frac{1}{n}}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (\left (c x^{n}\right )^{\left (\frac{1}{n}\right )} b + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*(c*x^n)^(1/n))^2,x, algorithm="giac")

[Out]

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