3.2.57 \(\int \frac {A+B x^2}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac {(2 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}}+\frac {\sqrt {b x^2+c x^4} (2 b B-3 A c)}{2 b^2 c x^3}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}-\frac {B}{3 c x \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.09, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1145, 2006, 2025, 2008, 206} \begin {gather*} \frac {\sqrt {b x^2+c x^4} (2 b B-3 A c)}{2 b^2 c x^3}-\frac {(2 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{5/2}}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}-\frac {B}{3 c x \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-B/(3*c*x*Sqrt[b*x^2 + c*x^4]) - (2*b*B - 3*A*c)/(3*b*c*x*Sqrt[b*x^2 + c*x^4]) + ((2*b*B - 3*A*c)*Sqrt[b*x^2 +
c*x^4])/(2*b^2*c*x^3) - ((2*b*B - 3*A*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 1145

Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*(b*x^2 + c*x^4)^(p + 1))/(c
*(4*p + 3)*x), x] - Dist[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)), Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ
[{b, c, d, e, p}, x] &&  !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 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 {A+B x^2}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {B}{3 c x \sqrt {b x^2+c x^4}}+\frac {(-2 b B+3 A c) \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{3 c}\\ &=-\frac {B}{3 c x \sqrt {b x^2+c x^4}}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}+\frac {(-2 b B+3 A c) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{b c}\\ &=-\frac {B}{3 c x \sqrt {b x^2+c x^4}}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}+\frac {(2 b B-3 A c) \sqrt {b x^2+c x^4}}{2 b^2 c x^3}+\frac {(2 b B-3 A c) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{2 b^2}\\ &=-\frac {B}{3 c x \sqrt {b x^2+c x^4}}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}+\frac {(2 b B-3 A c) \sqrt {b x^2+c x^4}}{2 b^2 c x^3}-\frac {(2 b B-3 A 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 {B}{3 c x \sqrt {b x^2+c x^4}}-\frac {2 b B-3 A c}{3 b c x \sqrt {b x^2+c x^4}}+\frac {(2 b B-3 A c) \sqrt {b x^2+c x^4}}{2 b^2 c x^3}-\frac {(2 b B-3 A 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 = 0.03, size = 61, normalized size = 0.43 \begin {gather*} \frac {x^2 (2 b B-3 A c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c x^2}{b}+1\right )-A b}{2 b^2 x \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.88, size = 95, normalized size = 0.67 \begin {gather*} \frac {(3 A c-2 b B) \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 (-A b-3 A c x^2+2 b B x^2\right )}{2 b^2 x^3 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 260, normalized size = 1.83 \begin {gather*} \left [-\frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{5} + {\left (2 \, B b^{2} - 3 \, A b c\right )} 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 (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{4 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, \frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{5} + {\left (2 \, B b^{2} - 3 \, A b c\right )} 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 (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{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((B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(((2*B*b*c - 3*A*c^2)*x^5 + (2*B*b^2 - 3*A*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)*(A*b^2 - (2*B*b^2 - 3*A*b*c)*x^2))/(b^3*c*x^5 + b^4*x^3), 1/2*(((2*B*b*
c - 3*A*c^2)*x^5 + (2*B*b^2 - 3*A*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)*(A*b^2 - (2*B*b^2 - 3*A*b*c)*x^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((B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.06, size = 129, normalized size = 0.91 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-3 \sqrt {c \,x^{2}+b}\, A b c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+2 \sqrt {c \,x^{2}+b}\, B \,b^{2} x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+3 A \,b^{\frac {3}{2}} c \,x^{2}-2 B \,b^{\frac {5}{2}} x^{2}+A \,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((B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/2*x*(c*x^2+b)*(3*A*b^(3/2)*x^2*c-3*A*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*(c*x^2+b)^(1/2)*x^2*b*c-2*B*b^(5/2
)*x^2+2*B*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*(c*x^2+b)^(1/2)*x^2*b^2+A*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 {B x^{2} + A}{{\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((B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {B\,x^2+A}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B 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((B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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