3.174 \(\int \frac{A+B x^2}{x^5 \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4} \]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2 + c*x^4])/(8*a^2*x^2) - ((3*A*b^2 - 4
*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

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Rubi [A]  time = 0.145204, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1251, 834, 806, 724, 206} \[ -\frac{\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2 + c*x^4])/(8*a^2*x^2) - ((3*A*b^2 - 4
*a*b*B - 4*a*A*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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])

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^5 \sqrt{a+b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (3 A b-4 a B)+A c x}{x^2 \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}+\frac{\left (3 A b^2-4 a b B-4 a A c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{\left (3 A b^2-4 a b B-4 a A c\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^2}{\sqrt{a+b x^2+c x^4}}\right )}{8 a^2}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2+c x^4}}{8 a^2 x^2}-\frac{\left (3 A b^2-4 a b B-4 a A c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.076057, size = 107, normalized size = 0.86 \[ \frac{\left (4 a A c+4 a b B-3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{\sqrt{a+b x^2+c x^4} \left (3 A b x^2-2 a \left (A+2 B x^2\right )\right )}{8 a^2 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(3*A*b*x^2 - 2*a*(A + 2*B*x^2)))/(8*a^2*x^4) + ((-3*A*b^2 + 4*a*b*B + 4*a*A*c)*ArcTan
h[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

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Maple [A]  time = 0.018, size = 194, normalized size = 1.6 \begin{align*} -{\frac{A}{4\,a{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,Ab}{8\,{a}^{2}{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,A{b}^{2}}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{Ac}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B}{2\,{x}^{2}a}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{bB}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

-1/4*A*(c*x^4+b*x^2+a)^(1/2)/a/x^4+3/8*A*b/a^2/x^2*(c*x^4+b*x^2+a)^(1/2)-3/16*A*b^2/a^(5/2)*ln((2*a+b*x^2+2*a^
(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)+1/4*A*c/a^(3/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-1/2*B/a/
x^2*(c*x^4+b*x^2+a)^(1/2)+1/4*B*b/a^(3/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.37909, size = 593, normalized size = 4.78 \begin{align*} \left [\frac{{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt{a} x^{4} \log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )}}{32 \, a^{3} x^{4}}, -\frac{{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )}}{16 \, a^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/32*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(a)*x^4*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 + b*x^2 +
a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*sqrt(c*x^4 + b*x^2 + a)*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x^2))/(a^3*x
^4), -1/16*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(-a)*x^4*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-
a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*sqrt(c*x^4 + b*x^2 + a)*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x^2))/(a^3*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x^{2}}{x^{5} \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**5*sqrt(a + b*x**2 + c*x**4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^5), x)