3.3006 \(\int \frac{x^3}{a+b (c x^n)^{\frac{1}{n}}} \, dx\)

Optimal. Leaf size=101 \[ \frac{a^2 x^4 \left (c x^n\right )^{-3/n}}{b^3}-\frac{a^3 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^4}-\frac{a x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac{x^4 \left (c x^n\right )^{-1/n}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0370195, antiderivative size = 101, 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} \[ \frac{a^2 x^4 \left (c x^n\right )^{-3/n}}{b^3}-\frac{a^3 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^4}-\frac{a x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac{x^4 \left (c x^n\right )^{-1/n}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*(c*x^n)^n^(-1)),x]

[Out]

(a^2*x^4)/(b^3*(c*x^n)^(3/n)) - (a*x^4)/(2*b^2*(c*x^n)^(2/n)) + x^4/(3*b*(c*x^n)^n^(-1)) - (a^3*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}{a+b \left (c x^n\right )^{\frac{1}{n}}} \, dx &=\left (x^4 \left (c x^n\right )^{-4/n}\right ) \operatorname{Subst}\left (\int \frac{x^3}{a+b x} \, 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{a^2}{b^3}-\frac{a x}{b^2}+\frac{x^2}{b}-\frac{a^3}{b^3 (a+b x)}\right ) \, dx,x,\left (c x^n\right )^{\frac{1}{n}}\right )\\ &=\frac{a^2 x^4 \left (c x^n\right )^{-3/n}}{b^3}-\frac{a x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac{x^4 \left (c x^n\right )^{-1/n}}{3 b}-\frac{a^3 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 = 0.0504328, size = 87, normalized size = 0.86 \[ \frac{x^4 \left (c x^n\right )^{-4/n} \left (b \left (c x^n\right )^{\frac{1}{n}} \left (6 a^2-3 a b \left (c x^n\right )^{\frac{1}{n}}+2 b^2 \left (c x^n\right )^{2/n}\right )-6 a^3 \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )\right )}{6 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.164, size = 552, normalized size = 5.5 \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)),x)

[Out]

x*a^2/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)+1/3*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)-1/2*x^2/b^2/(c^(1/n))^2*a*exp((I*Pi*csgn(I*c*x^n)*csgn(I*c)*csgn(I*x^n)-I*Pi*csg
n(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)-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)*a^3/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*} \int \frac{x^{3}}{\left (c x^{n}\right )^{\left (\frac{1}{n}\right )} b + a}\,{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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.32212, size = 149, normalized size = 1.48 \begin{align*} \frac{2 \, b^{3} c^{\frac{3}{n}} x^{3} - 3 \, a b^{2} c^{\frac{2}{n}} x^{2} + 6 \, a^{2} b c^{\left (\frac{1}{n}\right )} x - 6 \, a^{3} \log \left (b c^{\left (\frac{1}{n}\right )} x + a\right )}{6 \, b^{4} c^{\frac{4}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*c^(3/n)*x^3 - 3*a*b^2*c^(2/n)*x^2 + 6*a^2*b*c^(1/n)*x - 6*a^3*log(b*c^(1/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}}{a + b \left (c x^{n}\right )^{\frac{1}{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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