3.105 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x^6} \, dx\)

Optimal. Leaf size=103 \[ -\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}-\frac{\sqrt{b x^2+c x^4} (4 b B-A c)}{8 b x^3}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7} \]

[Out]

-((4*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(8*b*x^3) - (A*(b*x^2 + c*x^4)^(3/2))/(4*b*x^7) - (c*(4*b*B - A*c)*ArcTan
h[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(3/2))

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Rubi [A]  time = 0.157621, antiderivative size = 103, 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, 2020, 2008, 206} \[ -\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}-\frac{\sqrt{b x^2+c x^4} (4 b B-A c)}{8 b x^3}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^6,x]

[Out]

-((4*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(8*b*x^3) - (A*(b*x^2 + c*x^4)^(3/2))/(4*b*x^7) - (c*(4*b*B - A*c)*ArcTan
h[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(3/2))

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 2020

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 + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*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^6} \, dx &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7}-\frac{(-4 b B+A c) \int \frac{\sqrt{b x^2+c x^4}}{x^4} \, dx}{4 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7}+\frac{(c (4 b B-A c)) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{8 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7}-\frac{(c (4 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 )}{8 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7}-\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.107429, size = 95, normalized size = 0.92 \[ -\frac{\left (b+c x^2\right ) \left (2 A b+A c x^2+4 b B x^2\right )+c x^4 \sqrt{\frac{c x^2}{b}+1} (4 b B-A c) \tanh ^{-1}\left (\sqrt{\frac{c x^2}{b}+1}\right )}{8 b x^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^6,x]

[Out]

-((b + c*x^2)*(2*A*b + 4*b*B*x^2 + A*c*x^2) + c*(4*b*B - A*c)*x^4*Sqrt[1 + (c*x^2)/b]*ArcTanh[Sqrt[1 + (c*x^2)
/b]])/(8*b*x^3*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.01, size = 174, normalized size = 1.7 \begin{align*}{\frac{1}{8\,{x}^{5}{b}^{2}}\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}^{4}{c}^{2}-4\,B{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}c-A\sqrt{c{x}^{2}+b}{x}^{4}{c}^{2}+4\,B\sqrt{c{x}^{2}+b}{x}^{4}bc+A \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{2}c-4\,B \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{2}b-2\,A \left ( c{x}^{2}+b \right ) ^{3/2}b \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^6,x)

[Out]

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

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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^{6}}\,{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^6,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^6, x)

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Fricas [A]  time = 1.07663, size = 437, normalized size = 4.24 \begin{align*} \left [-\frac{{\left (4 \, B b c - A c^{2}\right )} \sqrt{b} x^{5} \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 \, A b^{2} +{\left (4 \, B b^{2} + A b c\right )} x^{2}\right )}}{16 \, b^{2} x^{5}}, \frac{{\left (4 \, B b c - A c^{2}\right )} \sqrt{-b} x^{5} \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 \, A b^{2} +{\left (4 \, B b^{2} + A b c\right )} x^{2}\right )}}{8 \, b^{2} x^{5}}\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^6,x, algorithm="fricas")

[Out]

[-1/16*((4*B*b*c - A*c^2)*sqrt(b)*x^5*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*A*b^2 + (4*B*b^2 + A*b*c)*x^2))/(b^2*x^5), 1/8*((4*B*b*c - A*c^2)*sqrt(-b)*x^5*arctan(sqrt(c*x^4
+ b*x^2)*sqrt(-b)/(c*x^3 + b*x)) - sqrt(c*x^4 + b*x^2)*(2*A*b^2 + (4*B*b^2 + A*b*c)*x^2))/(b^2*x^5)]

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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^{6}}\, 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**6,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**6, x)

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Giac [A]  time = 1.24954, size = 178, normalized size = 1.73 \begin{align*} \frac{\frac{{\left (4 \, B b c^{2} \mathrm{sgn}\left (x\right ) - A c^{3} \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{4 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} B b c^{2} \mathrm{sgn}\left (x\right ) - 4 \, \sqrt{c x^{2} + b} B b^{2} c^{2} \mathrm{sgn}\left (x\right ) +{\left (c x^{2} + b\right )}^{\frac{3}{2}} A c^{3} \mathrm{sgn}\left (x\right ) + \sqrt{c x^{2} + b} A b c^{3} \mathrm{sgn}\left (x\right )}{b c^{2} x^{4}}}{8 \, 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^6,x, algorithm="giac")

[Out]

1/8*((4*B*b*c^2*sgn(x) - A*c^3*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) - (4*(c*x^2 + b)^(3/2)*B*
b*c^2*sgn(x) - 4*sqrt(c*x^2 + b)*B*b^2*c^2*sgn(x) + (c*x^2 + b)^(3/2)*A*c^3*sgn(x) + sqrt(c*x^2 + b)*A*b*c^3*s
gn(x))/(b*c^2*x^4))/c