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

Optimal. Leaf size=137 \[ \frac {3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}-\frac {3 \sqrt {b x^2+c x^4} (4 b B-5 A c)}{8 b^3 x^3}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}} \]

[Out]

3/8*c*(-5*A*c+4*B*b)*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(7/2)-1/4*A/b/x^3/(c*x^4+b*x^2)^(1/2)+1/4*(-5*A*
c+4*B*b)/b^2/x/(c*x^4+b*x^2)^(1/2)-3/8*(-5*A*c+4*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^3

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Rubi [A]  time = 0.19, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2038, 2006, 2025, 2008, 206} \[ -\frac {3 \sqrt {b x^2+c x^4} (4 b B-5 A c)}{8 b^3 x^3}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}+\frac {3 c (4 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(4*b*x^3*Sqrt[b*x^2 + c*x^4]) + (4*b*B - 5*A*c)/(4*b^2*x*Sqrt[b*x^2 + c*x^4]) - (3*(4*b*B - 5*A*c)*Sqrt[b*x
^2 + c*x^4])/(8*b^3*x^3) + (3*c*(4*b*B - 5*A*c)*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 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]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}}-\frac {(-4 b B+5 A c) \int \frac {1}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{4 b}\\ &=-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}+\frac {(3 (4 b B-5 A c)) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{4 b^2}\\ &=-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}-\frac {3 (4 b B-5 A c) \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {(3 c (4 b B-5 A c)) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{8 b^3}\\ &=-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}-\frac {3 (4 b B-5 A c) \sqrt {b x^2+c x^4}}{8 b^3 x^3}+\frac {(3 c (4 b B-5 A c)) \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 {A}{4 b x^3 \sqrt {b x^2+c x^4}}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}-\frac {3 (4 b B-5 A c) \sqrt {b x^2+c x^4}}{8 b^3 x^3}+\frac {3 c (4 b B-5 A c) \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 = 0.05, size = 64, normalized size = 0.47 \[ \frac {c x^4 (5 A c-4 b B) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {c x^2}{b}+1\right )-A b^2}{4 b^3 x^3 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b^2) + c*(-4*b*B + 5*A*c)*x^4*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (c*x^2)/b])/(4*b^3*x^3*Sqrt[x^2*(b + c*
x^2)])

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fricas [A]  time = 0.92, size = 315, normalized size = 2.30 \[ \left [-\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^2*c - 5*A*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*(3*(4*B*b^2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^2*c)*x^2)*sqrt(c*x^4 +
b*x^2))/(b^4*c*x^7 + b^5*x^5), -1/8*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^2*c - 5*A*b*c^2)*x^5)*sqrt(-b)*arc
tan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (3*(4*B*b^2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^
2*c)*x^2)*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 \[ \int \frac {B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 157, normalized size = 1.15 \[ -\frac {\left (c \,x^{2}+b \right ) \left (15 \sqrt {c \,x^{2}+b}\, A b \,c^{2} x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-12 \sqrt {c \,x^{2}+b}\, B \,b^{2} c \,x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 A \,b^{\frac {3}{2}} c^{2} x^{4}+12 B \,b^{\frac {5}{2}} c \,x^{4}-5 A \,b^{\frac {5}{2}} c \,x^{2}+4 B \,b^{\frac {7}{2}} x^{2}+2 A \,b^{\frac {7}{2}}\right )}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {9}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/x*(c*x^2+b)*(15*A*(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*A*b^(3/2)*x^4*c^2-12*B
*(c*x^2+b)^(1/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*b^2*c+12*B*b^(5/2)*x^4*c-5*A*b^(5/2)*x^2*c+4*B*b^(7/2
)*x^2+2*A*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 \[ \int \frac {B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^2/(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^2), x)

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mupad [B]  time = 1.26, size = 89, normalized size = 0.65 \[ -\frac {A\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {b}{c\,x^2}\right )}{7\,x\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}-\frac {B\,x\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {b}{c\,x^2}\right )}{5\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (A*(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)) - (B*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 \[ \int \frac {A + B x^{2}}{x^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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