3.6.93 \(\int \frac {1}{x (a+b x^n+c x^{2 n})^{3/2}} \, dx\) [593]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {1371, 754, 12, 738, 212} \begin {gather*} \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} \end {gather*}

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 12

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

Rule 212

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

Rule 738

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 754

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 1371

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]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx &=\frac {\text {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 \text {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 {\text {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 \text {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*}

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Mathematica [A]
time = 0.63, size = 98, normalized size = 1.00 \begin {gather*} \frac {2 \left (-\frac {\sqrt {a} \left (-b^2+2 a c-b c x^n\right )}{\left (b^2-4 a c\right ) \sqrt {a+x^n \left (b+c x^n\right )}}+\tanh ^{-1}\left (\frac {\sqrt {c} x^n-\sqrt {a+x^n \left (b+c x^n\right )}}{\sqrt {a}}\right )\right )}{a^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (88) = 176\).
time = 0.46, size = 449, normalized size = 4.58 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \end {gather*}

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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