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

Optimal. Leaf size=100 \[ \frac{\sqrt{b x^2+c x^4} (A c+2 b B)}{2 b x}-\frac{(A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{2 \sqrt{b}}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16086, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2038, 2021, 2008, 206} \[ \frac{\sqrt{b x^2+c x^4} (A c+2 b B)}{2 b x}-\frac{(A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{2 \sqrt{b}}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

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

Mathematica [A]  time = 0.04587, size = 94, normalized size = 0.94 \[ -\frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{b} \left (A-2 B x^2\right ) \sqrt{b+c x^2}+x^2 (A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b+c x^2}}{\sqrt{b}}\right )\right )}{2 \sqrt{b} x^3 \sqrt{b+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b]*(A - 2*B*x^2)*Sqrt[b + c*x^2] + (2*b*B + A*c)*x^2*ArcTanh[Sqrt[b + c*x^2]/Sqr
t[b]]))/(2*Sqrt[b]*x^3*Sqrt[b + c*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 135, normalized size = 1.4 \begin{align*} -{\frac{1}{2\,b{x}^{3}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( A\sqrt{b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{2}c+2\,B{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{2}-A\sqrt{c{x}^{2}+b}{x}^{2}c-2\,B\sqrt{c{x}^{2}+b}{x}^{2}b+A \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{c{x}^{2}+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2}}{\left (B x^{2} + A\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.04321, size = 375, normalized size = 3.75 \begin{align*} \left [\frac{{\left (2 \, B b + A c\right )} \sqrt{b} x^{3} \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 (2 \, B b x^{2} - A b\right )}}{4 \, b x^{3}}, \frac{{\left (2 \, B b + A c\right )} \sqrt{-b} x^{3} \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 (2 \, B b x^{2} - A b\right )}}{2 \, b x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.24104, size = 103, normalized size = 1.03 \begin{align*} \frac{2 \, \sqrt{c x^{2} + b} B c \mathrm{sgn}\left (x\right ) + \frac{{\left (2 \, B b c \mathrm{sgn}\left (x\right ) + A c^{2} \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{\sqrt{c x^{2} + b} A c \mathrm{sgn}\left (x\right )}{x^{2}}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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