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

Optimal. Leaf size=44 \[ \frac {x \left (c x^n\right )^{-1/n} \tan ^{-1}\left (\frac {\sqrt {b} \left (c x^n\right )^{\frac {1}{n}}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {254, 205} \begin {gather*} \frac {x \left (c x^n\right )^{-1/n} \tan ^{-1}\left (\frac {\sqrt {b} \left (c x^n\right )^{\frac {1}{n}}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x^n)^(2/n))^(-1),x]

[Out]

(x*ArcTan[(Sqrt[b]*(c*x^n)^n^(-1))/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(c*x^n)^n^(-1))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 254

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> Dist[x/(c*x^q)^(1/q), Subst[Int[(a + b*x^(n*q))
^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, n, p, q}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 44, normalized size = 1.00 \begin {gather*} \frac {x \left (c x^n\right )^{-1/n} \tan ^{-1}\left (\frac {\sqrt {b} \left (c x^n\right )^{\frac {1}{n}}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*ArcTan[(Sqrt[b]*(c*x^n)^n^(-1))/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(c*x^n)^n^(-1))

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IntegrateAlgebraic [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a+b \left (c x^n\right )^{2/n}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 127, normalized size = 2.89 \begin {gather*} \left [-\frac {\sqrt {-a b c^{\frac {2}{n}}} \log \left (\frac {b c^{\frac {2}{n}} x^{2} - 2 \, \sqrt {-a b c^{\frac {2}{n}}} x - a}{b c^{\frac {2}{n}} x^{2} + a}\right )}{2 \, a b c^{\frac {2}{n}}}, \frac {\sqrt {a b c^{\frac {2}{n}}} \arctan \left (\frac {\sqrt {a b c^{\frac {2}{n}}} x}{a}\right )}{a b c^{\frac {2}{n}}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^n)^(2/n)*b + a), x)

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maple [C]  time = 0.47, size = 222, normalized size = 5.05 \begin {gather*} \frac {\arctan \left (\frac {b \,c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{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 )}{n}}}{\sqrt {\frac {a b \,c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{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 )}{n}}}{x^{2}}}\, x}\right )}{\sqrt {\frac {a b \,c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{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 )}{n}}}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*b/x^2*(x^n)^(2/n)*c^(2/n)*exp(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)
))^(1/2)*arctan(b/x*(x^n)^(2/n)*c^(2/n)*exp(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*b/x^2*(x^n)^(2/n)*c^(2/n)*exp(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)))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^n)^(2/n)*b + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{a+b\,{\left (c\,x^n\right )}^{2/n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*(c*x^n)^(2/n)),x)

[Out]

int(1/(a + b*(c*x^n)^(2/n)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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