3.2.67 \(\int \frac {1}{x^2 (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=109 \[ -\frac {15 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {1}{b x^3 \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.15, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2023, 2025, 2008, 206} \begin {gather*} -\frac {15 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {1}{b x^3 \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^3*Sqrt[b*x^2 + c*x^4]) - (5*Sqrt[b*x^2 + c*x^4])/(4*b^2*x^5) + (15*c*Sqrt[b*x^2 + c*x^4])/(8*b^3*x^3) -
(15*c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(7/2))

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])

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2023

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}+\frac {5 \int \frac {1}{x^4 \sqrt {b x^2+c x^4}} \, dx}{b}\\ &=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}-\frac {(15 c) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{4 b^2}\\ &=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}+\frac {\left (15 c^2\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{8 b^3}\\ &=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {\left (15 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{8 b^3}\\ &=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {15 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.38 \begin {gather*} \frac {c^2 x \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {c x^2}{b}+1\right )}{b^3 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(c^2*x*Hypergeometric2F1[-1/2, 3, 1/2, 1 + (c*x^2)/b])/(b^3*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.59, size = 91, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-2 b^2+5 b c x^2+15 c^2 x^4\right )}{8 b^3 x^5 \left (b+c x^2\right )}-\frac {15 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-2*b^2 + 5*b*c*x^2 + 15*c^2*x^4))/(8*b^3*x^5*(b + c*x^2)) - (15*c^2*ArcTanh[(Sqrt[b]*x)/
Sqrt[b*x^2 + c*x^4]])/(8*b^(7/2))

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fricas [A]  time = 0.61, size = 229, normalized size = 2.10 \begin {gather*} \left [\frac {15 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, \frac {15 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*(c^3*x^7 + b*c^2*x^5)*sqrt(b)*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(15*b*c^
2*x^4 + 5*b^2*c*x^2 - 2*b^3)*sqrt(c*x^4 + b*x^2))/(b^4*c*x^7 + b^5*x^5), 1/8*(15*(c^3*x^7 + b*c^2*x^5)*sqrt(-b
)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (15*b*c^2*x^4 + 5*b^2*c*x^2 - 2*b^3)*sqrt(c*x^4 + b*x^2
))/(b^4*c*x^7 + b^5*x^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^2), x)

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maple [A]  time = 0.01, size = 94, normalized size = 0.86 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (15 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 b^{\frac {3}{2}} c^{2} x^{4}-5 b^{\frac {5}{2}} c \,x^{2}+2 b^{\frac {7}{2}}\right )}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {9}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/8/x*(c*x^2+b)*(15*(c*x^2+b)^(1/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*b*c^2-15*b^(3/2)*x^4*c^2-5*b^(5/2
)*x^2*c+2*b^(7/2))/(c*x^4+b*x^2)^(3/2)/b^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^2), x)

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mupad [B]  time = 4.64, size = 44, normalized size = 0.40 \begin {gather*} -\frac {{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {b}{c\,x^2}\right )}{7\,x\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

-((b/(c*x^2) + 1)^(3/2)*hypergeom([3/2, 7/2], 9/2, -b/(c*x^2)))/(7*x*(b*x^2 + c*x^4)^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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