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

Optimal. Leaf size=80 \[ -\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}-\frac{B \sqrt{b x^2+c x^4}}{x^2}+B \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right ) \]

[Out]

-((B*Sqrt[b*x^2 + c*x^4])/x^2) - (A*(b*x^2 + c*x^4)^(3/2))/(3*b*x^6) + B*Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*
x^2 + c*x^4]]

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Rubi [A]  time = 0.195615, antiderivative size = 80, normalized size of antiderivative = 1., 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 = {2034, 792, 662, 620, 206} \[ -\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}-\frac{B \sqrt{b x^2+c x^4}}{x^2}+B \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*Sqrt[b*x^2 + c*x^4])/x^2) - (A*(b*x^2 + c*x^4)^(3/2))/(3*b*x^6) + B*Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*
x^2 + c*x^4]]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), 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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 662

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}+\frac{1}{2} B \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{B \sqrt{b x^2+c x^4}}{x^2}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}+\frac{1}{2} (B c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac{B \sqrt{b x^2+c x^4}}{x^2}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}+(B c) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )\\ &=-\frac{B \sqrt{b x^2+c x^4}}{x^2}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}+B \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.122246, size = 86, normalized size = 1.08 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (-A \left (b+c x^2\right )+\frac{3 \sqrt{b} B \sqrt{c} x^3 \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{\sqrt{\frac{c x^2}{b}+1}}-3 b B x^2\right )}{3 b x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-3*b*B*x^2 - A*(b + c*x^2) + (3*Sqrt[b]*B*Sqrt[c]*x^3*ArcSinh[(Sqrt[c]*x)/Sqrt[b]])/Sq
rt[1 + (c*x^2)/b]))/(3*b*x^4)

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Maple [A]  time = 0.01, size = 109, normalized size = 1.4 \begin{align*} -{\frac{1}{3\,b{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( -3\,B{c}^{3/2}\sqrt{c{x}^{2}+b}{x}^{4}+3\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{2}-3\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){x}^{3}bc+A\sqrt{c} \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{c{x}^{2}+b}}}{\frac{1}{\sqrt{c}}}} \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^5,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.09415, size = 363, normalized size = 4.54 \begin{align*} \left [\frac{3 \, B b \sqrt{c} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \, \sqrt{c x^{4} + b x^{2}}{\left ({\left (3 \, B b + A c\right )} x^{2} + A b\right )}}{6 \, b x^{4}}, -\frac{3 \, B b \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) + \sqrt{c x^{4} + b x^{2}}{\left ({\left (3 \, B b + A c\right )} x^{2} + A b\right )}}{3 \, b x^{4}}\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^5,x, algorithm="fricas")

[Out]

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

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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^{5}}\, 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**5,x)

[Out]

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

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Giac [B]  time = 1.54875, size = 220, normalized size = 2.75 \begin{align*} -\frac{1}{2} \, B \sqrt{c} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b \sqrt{c} \mathrm{sgn}\left (x\right ) + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{2} \sqrt{c} \mathrm{sgn}\left (x\right ) + 3 \, B b^{3} \sqrt{c} \mathrm{sgn}\left (x\right ) + A b^{2} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right )\right )}}{3 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{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^5,x, algorithm="giac")

[Out]

-1/2*B*sqrt(c)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b*sqrt(c
)*sgn(x) + 3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*c^(3/2)*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^2*sqrt(c
)*sgn(x) + 3*B*b^3*sqrt(c)*sgn(x) + A*b^2*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^3