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

Optimal. Leaf size=94 \[ -\frac {2 b \log (x) \left (c x^n\right )^{\frac {1}{n}}}{a^3 x}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^3 x}-\frac {b \left (c x^n\right )^{\frac {1}{n}}}{a^2 x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}-\frac {1}{a^2 x} \]

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {b \left (c x^n\right )^{\frac {1}{n}}}{a^2 x \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}-\frac {2 b \log (x) \left (c x^n\right )^{\frac {1}{n}}}{a^3 x}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^3 x}-\frac {1}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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)]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 71, normalized size = 0.76 \begin {gather*} -\frac {\left (c x^n\right )^{\frac {1}{n}} \left (a \left (\frac {b}{a+b \left (c x^n\right )^{\frac {1}{n}}}+\left (c x^n\right )^{-1/n}\right )-2 b \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )+2 b \log (x)\right )}{a^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.94, size = 99, normalized size = 1.05 \begin {gather*} -\frac {2 \, b^{2} c^{\frac {2}{n}} x^{2} \log \relax (x) + a^{2} + 2 \, {\left (a b x \log \relax (x) + a b x\right )} c^{\left (\frac {1}{n}\right )} - 2 \, {\left (b^{2} c^{\frac {2}{n}} x^{2} + a b c^{\left (\frac {1}{n}\right )} x\right )} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right )}{a^{3} b c^{\left (\frac {1}{n}\right )} x^{2} + a^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{2} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.14, size = 296, normalized size = 3.15 \begin {gather*} -\frac {2 b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}} \ln \relax (x )}{a^{3} x}+\frac {2 b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}} \ln \left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}}+a \right )}{a^{3} x}+\frac {1}{\left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (-\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 n}}+a \right ) a x}-\frac {2}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{a b c^{\left (\frac {1}{n}\right )} x {\left (x^{n}\right )}^{\left (\frac {1}{n}\right )} + a^{2} x} + 2 \, \int \frac {1}{a b c^{\left (\frac {1}{n}\right )} x^{2} {\left (x^{n}\right )}^{\left (\frac {1}{n}\right )} + a^{2} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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