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

Optimal. Leaf size=73 \[ \frac {x \left (c x^n\right )^{-1/n} \tan ^{-1}\left (\frac {\sqrt {b} \left (c x^n\right )^{\frac {1}{n}}}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b \left (c x^n\right )^{2/n}\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {254, 199, 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 )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b \left (c x^n\right )^{2/n}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x^n)^(2/n))^(-2),x]

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

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Mathematica [A]  time = 0.06, size = 73, 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 )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b \left (c x^n\right )^{2/n}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.26, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+b \left (c x^n\right )^{2/n}\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.30, size = 218, normalized size = 2.99 \begin {gather*} \left [\frac {2 \, a b c^{\frac {2}{n}} x - {\left (b c^{\frac {2}{n}} x^{2} + a\right )} \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 )}{4 \, {\left (a^{2} b^{2} c^{\frac {4}{n}} x^{2} + a^{3} b c^{\frac {2}{n}}\right )}}, \frac {a b c^{\frac {2}{n}} x + {\left (b c^{\frac {2}{n}} x^{2} + a\right )} \sqrt {a b c^{\frac {2}{n}}} \arctan \left (\frac {\sqrt {a b c^{\frac {2}{n}}} x}{a}\right )}{2 \, {\left (a^{2} b^{2} c^{\frac {4}{n}} x^{2} + a^{3} b c^{\frac {2}{n}}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*a*b*c^(2/n)*x - (b*c^(2/n)*x^2 + a)*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^2*b^2*c^(4/n)*x^2 + a^3*b*c^(2/n)), 1/2*(a*b*c^(2/n)*x + (b*c^(2/n)*x^2 + a)*sqrt(a
*b*c^(2/n))*arctan(sqrt(a*b*c^(2/n))*x/a))/(a^2*b^2*c^(4/n)*x^2 + a^3*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 (\left (c x^{n}\right )^{\frac {2}{n}} b + a\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.42, size = 305, normalized size = 4.18 \begin {gather*} \frac {x}{2 \left (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}}+a \right ) a}+\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 )}{2 \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}}}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*(c*x^n)^(2/n))^2,x)

[Out]

int(1/(a + b*(c*x^n)^(2/n))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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