3.3075 \(\int \frac{(c x^n)^{\frac{1}{n}}}{(a+b (c x^n)^{\frac{1}{n}})^2} \, dx\)

Optimal. Leaf size=63 \[ \frac{a x \left (c x^n\right )^{-1/n}}{b^2 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}+\frac{x \left (c x^n\right )^{-1/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0316528, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {15, 368, 43} \[ \frac{a x \left (c x^n\right )^{-1/n}}{b^2 \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}+\frac{x \left (c x^n\right )^{-1/n} \log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.0310488, size = 48, normalized size = 0.76 \[ \frac{x \left (c x^n\right )^{-1/n} \left (\frac{a}{a+b \left (c x^n\right )^{\frac{1}{n}}}+\log \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.291, size = 322, normalized size = 5.1 \begin{align*} -{\frac{x}{b} \left ( a+b{{\rm e}^{-{\frac{i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{n} \right ) +i\pi \,{\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{n} \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{n} \right ) }{2\,n}}}} \right ) ^{-1}}+{\frac{1}{\sqrt [n]{c}{b}^{2}}\ln \left ( b{{\rm e}^{-{\frac{i\pi \,{\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{n} \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{n} \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+2\,n\ln \left ( x \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{n} \right ) }{2\,n}}}}x+a \right ){{\rm e}^{{\frac{i\pi \,{\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{n} \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{n} \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +i\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+2\,n\ln \left ( x \right ) -2\,\ln \left ({x}^{n} \right ) }{2\,n}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/b/(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))+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)/(c^(1/n))/b^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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.66002, size = 105, normalized size = 1.67 \begin{align*} \frac{{\left (b c^{\left (\frac{1}{n}\right )} x + a\right )} \log \left (b c^{\left (\frac{1}{n}\right )} x + a\right ) + a}{b^{3} c^{\frac{2}{n}} x + a b^{2} c^{\left (\frac{1}{n}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral((c*x**n)**(1/n)/(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{\left (c x^{n}\right )^{\left (\frac{1}{n}\right )}}{{\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((c*x^n)^(1/n)/(a+b*(c*x^n)^(1/n))^2,x, algorithm="giac")

[Out]

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