3.3012 \(\int \frac{1}{x^3 (a+b (c x^n)^{\frac{1}{n}})} \, dx\)

Optimal. Leaf size=87 \[ \frac{b^2 \log (x) \left (c x^n\right )^{2/n}}{a^3 x^2}-\frac{b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{a^3 x^2}+\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a^2 x^2}-\frac{1}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0321322, antiderivative size = 87, 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, 44} \[ \frac{b^2 \log (x) \left (c x^n\right )^{2/n}}{a^3 x^2}-\frac{b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{a^3 x^2}+\frac{b \left (c x^n\right )^{\frac{1}{n}}}{a^2 x^2}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*(c*x^n)^n^(-1))),x]

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0423497, size = 75, normalized size = 0.86 \[ -\frac{a^2+2 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )-2 a b \left (c x^n\right )^{\frac{1}{n}}-2 b^2 \log (x) \left (c x^n\right )^{2/n}}{2 a^3 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.107, size = 448, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.56592, size = 146, normalized size = 1.68 \begin{align*} -\frac{2 \, b^{2} c^{\frac{2}{n}} x^{2} \log \left (b c^{\left (\frac{1}{n}\right )} x + a\right ) - 2 \, b^{2} c^{\frac{2}{n}} x^{2} \log \left (x\right ) - 2 \, a b c^{\left (\frac{1}{n}\right )} x + a^{2}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(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{1}{{\left (\left (c x^{n}\right )^{\left (\frac{1}{n}\right )} b + a\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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