3.3.84 \(\int \frac {1}{(b x^2+c x^4)^{3/2}} \, dx\) [284]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {2031, 2050, 2033, 212} \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 {3 \sqrt {b x^2+c x^4}}{2 b^2 x^3}+\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 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 2031

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 2033

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 2050

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) \text {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 [A]
time = 0.07, size = 78, normalized size = 0.96 \begin {gather*} \frac {-\sqrt {b} \left (b+3 c x^2\right )+3 c x^2 \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{2 b^{5/2} x \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]

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

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Maple [A]
time = 0.13, size = 77, normalized size = 0.95

method	result	size
default	\(-\frac {x \left (c \,x^{2}+b \right ) \left (3 b^{\frac {3}{2}} c \,x^{2}-3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {c \,x^{2}+b}\, b c \,x^{2}+b^{\frac {5}{2}}\right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {7}{2}}}\)	\(77\)
risch	\(-\frac {c \,x^{2}+b}{2 b^{2} x \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (-\frac {c}{b^{2} \sqrt {c \,x^{2}+b}}+\frac {3 c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{2 b^{\frac {5}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x*(c*x^2+b)*(3*b^(3/2)*c*x^2-3*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*(c*x^2+b)^(1/2)*b*c*x^2+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*} \text {Failed to integrate} \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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Fricas [A]
time = 0.36, 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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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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Giac [A]
time = 5.28, size = 80, normalized size = 0.99 \begin {gather*} -\frac {3 \, c \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{2 \, \sqrt {-b} b^{2} \mathrm {sgn}\left (x\right )} - \frac {3 \, {\left (c x^{2} + b\right )} c - 2 \, b c}{2 \, {\left ({\left (c x^{2} + b\right )}^{\frac {3}{2}} - \sqrt {c x^{2} + b} b\right )} b^{2} \mathrm {sgn}\left (x\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="giac")

[Out]

-3/2*c*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2*sgn(x)) - 1/2*(3*(c*x^2 + b)*c - 2*b*c)/(((c*x^2 + b)^(3
/2) - sqrt(c*x^2 + b)*b)*b^2*sgn(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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