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

Optimal. Leaf size=78 \[ \frac{A \sqrt{b x^2+c x^4}}{x}-A \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )+\frac{B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3} \]

[Out]

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

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Rubi [A]  time = 0.145952, antiderivative size = 78, 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 = {2039, 2021, 2008, 206} \[ \frac{A \sqrt{b x^2+c x^4}}{x}-A \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )+\frac{B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2039

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

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

Mathematica [A]  time = 0.0769255, size = 84, normalized size = 1.08 \[ \frac{x \left (\left (b+c x^2\right ) \left (3 A c+b B+B c x^2\right )-3 A \sqrt{b} c \sqrt{b+c x^2} \tanh ^{-1}\left (\frac{\sqrt{b+c x^2}}{\sqrt{b}}\right )\right )}{3 c \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 85, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,cx}\sqrt{c{x}^{4}+b{x}^{2}} \left ( 3\,A\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) \sqrt{b}c-B \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}-3\,A\sqrt{c{x}^{2}+b}c \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^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} A \int \frac{\sqrt{c x^{2} + b}}{x}\,{d x} + \frac{{\left (c x^{2} + b\right )}^{\frac{3}{2}} B}{3 \, 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^2,x, algorithm="maxima")

[Out]

A*integrate(sqrt(c*x^2 + b)/x, x) + 1/3*(c*x^2 + b)^(3/2)*B/c

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Fricas [A]  time = 1.25291, size = 359, normalized size = 4.6 \begin{align*} \left [\frac{3 \, A \sqrt{b} c x \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 (B c x^{2} + B b + 3 \, A c\right )}}{6 \, c x}, \frac{3 \, A \sqrt{-b} c x \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 (B c x^{2} + B b + 3 \, A c\right )}}{3 \, c x}\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^2,x, algorithm="fricas")

[Out]

[1/6*(3*A*sqrt(b)*c*x*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)*(B*c*x
^2 + B*b + 3*A*c))/(c*x), 1/3*(3*A*sqrt(-b)*c*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^
4 + b*x^2)*(B*c*x^2 + B*b + 3*A*c))/(c*x)]

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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^{2}}\, 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**2,x)

[Out]

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

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Giac [A]  time = 1.15775, size = 157, normalized size = 2.01 \begin{align*} \frac{A b \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-b}} - \frac{{\left (3 \, A b c \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + B \sqrt{-b} b^{\frac{3}{2}} + 3 \, A \sqrt{-b} \sqrt{b} c\right )} \mathrm{sgn}\left (x\right )}{3 \, \sqrt{-b} c} + \frac{{\left (c x^{2} + b\right )}^{\frac{3}{2}} B c^{2} \mathrm{sgn}\left (x\right ) + 3 \, \sqrt{c x^{2} + b} A c^{3} \mathrm{sgn}\left (x\right )}{3 \, c^{3}} \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^2,x, algorithm="giac")

[Out]

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