3.593 \(\int \frac{1}{x (a+b x^n+c x^{2 n})^{3/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 \left (-2 a c+b^2+b c x^n\right )}{a n \left (b^2-4 a c\right ) \sqrt{a+b x^n+c x^{2 n}}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{a^{3/2} n} \]

[Out]

(2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*Sqrt[a + b*x^n + c*x^(2*n)]) - ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]
*Sqrt[a + b*x^n + c*x^(2*n)])]/(a^(3/2)*n)

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Rubi [A]  time = 0.0729641, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1357, 740, 12, 724, 206} \[ \frac{2 \left (-2 a c+b^2+b c x^n\right )}{a n \left (b^2-4 a c\right ) \sqrt{a+b x^n+c x^{2 n}}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{a^{3/2} n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*Sqrt[a + b*x^n + c*x^(2*n)]) - ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]
*Sqrt[a + b*x^n + c*x^(2*n)])]/(a^(3/2)*n)

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^n\right )}{n}\\ &=\frac{2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \sqrt{a+b x^n+c x^{2 n}}}-\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{b^2}{2}+2 a c}{x \sqrt{a+b x+c x^2}} \, dx,x,x^n\right )}{a \left (b^2-4 a c\right ) n}\\ &=\frac{2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \sqrt{a+b x^n+c x^{2 n}}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^n\right )}{a n}\\ &=\frac{2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \sqrt{a+b x^n+c x^{2 n}}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^n}{\sqrt{a+b x^n+c x^{2 n}}}\right )}{a n}\\ &=\frac{2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \sqrt{a+b x^n+c x^{2 n}}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{a^{3/2} n}\\ \end{align*}

Mathematica [A]  time = 0.314725, size = 94, normalized size = 0.96 \[ \frac{\frac{2 \left (-2 a c+b^2+b c x^n\right )}{a \left (b^2-4 a c\right ) \sqrt{a+x^n \left (b+c x^n\right )}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+x^n \left (b+c x^n\right )}}\right )}{a^{3/2}}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*Sqrt[a + x^n*(b + c*x^n)]) - ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]*Sq
rt[a + x^n*(b + c*x^n)])]/a^(3/2))/n

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Maple [F]  time = 0.013, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.56634, size = 984, normalized size = 10.04 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{a} x^{2 \, n} +{\left (b^{3} - 4 \, a b c\right )} \sqrt{a} x^{n} +{\left (a b^{2} - 4 \, a^{2} c\right )} \sqrt{a}\right )} \log \left (-\frac{8 \, a b x^{n} + 8 \, a^{2} +{\left (b^{2} + 4 \, a c\right )} x^{2 \, n} - 4 \,{\left (\sqrt{a} b x^{n} + 2 \, a^{\frac{3}{2}}\right )} \sqrt{c x^{2 \, n} + b x^{n} + a}}{x^{2 \, n}}\right ) + 4 \,{\left (a b c x^{n} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{c x^{2 \, n} + b x^{n} + a}}{2 \,{\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} n x^{2 \, n} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} n x^{n} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n\right )}}, \frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{-a} x^{2 \, n} +{\left (b^{3} - 4 \, a b c\right )} \sqrt{-a} x^{n} +{\left (a b^{2} - 4 \, a^{2} c\right )} \sqrt{-a}\right )} \arctan \left (\frac{{\left (\sqrt{-a} b x^{n} + 2 \, \sqrt{-a} a\right )} \sqrt{c x^{2 \, n} + b x^{n} + a}}{2 \,{\left (a c x^{2 \, n} + a b x^{n} + a^{2}\right )}}\right ) + 2 \,{\left (a b c x^{n} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{c x^{2 \, n} + b x^{n} + a}}{{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} n x^{2 \, n} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} n x^{n} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="fricas")

[Out]

[1/2*(((b^2*c - 4*a*c^2)*sqrt(a)*x^(2*n) + (b^3 - 4*a*b*c)*sqrt(a)*x^n + (a*b^2 - 4*a^2*c)*sqrt(a))*log(-(8*a*
b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) - 4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x^(2*n) + b*x^n + a))/x^(2*n)) +
4*(a*b*c*x^n + a*b^2 - 2*a^2*c)*sqrt(c*x^(2*n) + b*x^n + a))/((a^2*b^2*c - 4*a^3*c^2)*n*x^(2*n) + (a^2*b^3 - 4
*a^3*b*c)*n*x^n + (a^3*b^2 - 4*a^4*c)*n), (((b^2*c - 4*a*c^2)*sqrt(-a)*x^(2*n) + (b^3 - 4*a*b*c)*sqrt(-a)*x^n
+ (a*b^2 - 4*a^2*c)*sqrt(-a))*arctan(1/2*(sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt(c*x^(2*n) + b*x^n + a)/(a*c*x^(2
*n) + a*b*x^n + a^2)) + 2*(a*b*c*x^n + a*b^2 - 2*a^2*c)*sqrt(c*x^(2*n) + b*x^n + a))/((a^2*b^2*c - 4*a^3*c^2)*
n*x^(2*n) + (a^2*b^3 - 4*a^3*b*c)*n*x^n + (a^3*b^2 - 4*a^4*c)*n)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (a + b x^{n} + c x^{2 n}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*x**n+c*x**(2*n))**(3/2),x)

[Out]

Integral(1/(x*(a + b*x**n + c*x**(2*n))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="giac")

[Out]

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