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

Optimal. Leaf size=81 \[ \frac {3 c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}}-\frac {3 \sqrt {b x^2+c x^4}}{2 b^2 x^3}+\frac {1}{b x \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2006, 2025, 2008, 206} \begin {gather*} -\frac {3 \sqrt {b x^2+c x^4}}{2 b^2 x^3}+\frac {3 c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}}+\frac {1}{b x \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(-3/2),x]

[Out]

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

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 2006

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

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

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.49 \begin {gather*} -\frac {c x \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {c x^2}{b}+1\right )}{b^2 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(-3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.49, size = 78, normalized size = 0.96 \begin {gather*} \frac {3 c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}}+\frac {\sqrt {b x^2+c x^4} \left (-b-3 c x^2\right )}{2 b^2 x^3 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 199, normalized size = 2.46 \begin {gather*} \left [\frac {3 \, {\left (c^{2} x^{5} + b c x^{3}\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 \, \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} + b^{2}\right )}}{4 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, -\frac {3 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (3 \, b c x^{2} + b^{2}\right )}}{2 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.01, size = 77, normalized size = 0.95 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-3 \sqrt {c \,x^{2}+b}\, b c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+3 b^{\frac {3}{2}} c \,x^{2}+b^{\frac {5}{2}}\right ) x}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + b*x^2)^(-3/2), x)

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mupad [B]  time = 4.34, size = 42, normalized size = 0.52 \begin {gather*} -\frac {x\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {b}{c\,x^2}\right )}{5\,{\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/(b*x^2 + c*x^4)^(3/2),x)

[Out]

-(x*(b/(c*x^2) + 1)^(3/2)*hypergeom([3/2, 5/2], 7/2, -b/(c*x^2)))/(5*(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}{\left (b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x**2 + c*x**4)**(-3/2), x)

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