3.111 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=128 \[ \frac{\left (b x^2+c x^4\right )^{3/2} (4 A c+b B)}{4 b x^2}+\frac{3}{8} \sqrt{b x^2+c x^4} (4 A c+b B)+\frac{3 b (4 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{c}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6} \]

[Out]

(3*(b*B + 4*A*c)*Sqrt[b*x^2 + c*x^4])/8 + ((b*B + 4*A*c)*(b*x^2 + c*x^4)^(3/2))/(4*b*x^2) - (A*(b*x^2 + c*x^4)
^(5/2))/(b*x^6) + (3*b*(b*B + 4*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*Sqrt[c])

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Rubi [A]  time = 0.277322, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2034, 792, 664, 620, 206} \[ \frac{\left (b x^2+c x^4\right )^{3/2} (4 A c+b B)}{4 b x^2}+\frac{3}{8} \sqrt{b x^2+c x^4} (4 A c+b B)+\frac{3 b (4 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{c}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^5,x]

[Out]

(3*(b*B + 4*A*c)*Sqrt[b*x^2 + c*x^4])/8 + ((b*B + 4*A*c)*(b*x^2 + c*x^4)^(3/2))/(4*b*x^2) - (A*(b*x^2 + c*x^4)
^(5/2))/(b*x^6) + (3*b*(b*B + 4*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*Sqrt[c])

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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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 ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac{\left (-3 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )}{b}\\ &=\frac{(b B+4 A c) \left (b x^2+c x^4\right )^{3/2}}{4 b x^2}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac{1}{8} (3 (b B+4 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x} \, dx,x,x^2\right )\\ &=\frac{3}{8} (b B+4 A c) \sqrt{b x^2+c x^4}+\frac{(b B+4 A c) \left (b x^2+c x^4\right )^{3/2}}{4 b x^2}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac{1}{16} (3 b (b B+4 A c)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{8} (b B+4 A c) \sqrt{b x^2+c x^4}+\frac{(b B+4 A c) \left (b x^2+c x^4\right )^{3/2}}{4 b x^2}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac{1}{8} (3 b (b B+4 A 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{3}{8} (b B+4 A c) \sqrt{b x^2+c x^4}+\frac{(b B+4 A c) \left (b x^2+c x^4\right )^{3/2}}{4 b x^2}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac{3 b (b B+4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.174219, size = 96, normalized size = 0.75 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\frac{3 \sqrt{b} x (4 A c+b B) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{\sqrt{c} \sqrt{\frac{c x^2}{b}+1}}-8 A b+4 A c x^2+5 b B x^2+2 B c x^4\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^5,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-8*A*b + 5*b*B*x^2 + 4*A*c*x^2 + 2*B*c*x^4 + (3*Sqrt[b]*(b*B + 4*A*c)*x*ArcSinh[(Sqrt[
c]*x)/Sqrt[b]])/(Sqrt[c]*Sqrt[1 + (c*x^2)/b])))/(8*x^2)

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Maple [A]  time = 0.009, size = 174, normalized size = 1.4 \begin{align*}{\frac{1}{8\,b{x}^{4}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 8\,A{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{2}+12\,A{c}^{3/2}\sqrt{c{x}^{2}+b}{x}^{2}b+2\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{2}b-8\,A\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}+3\,B\sqrt{c}\sqrt{c{x}^{2}+b}{x}^{2}{b}^{2}+12\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) x{b}^{2}c+3\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) x{b}^{3} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{\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)^(3/2)/x^5,x)

[Out]

1/8*(c*x^4+b*x^2)^(3/2)*(8*A*c^(3/2)*(c*x^2+b)^(3/2)*x^2+12*A*c^(3/2)*(c*x^2+b)^(1/2)*x^2*b+2*B*c^(1/2)*(c*x^2
+b)^(3/2)*x^2*b-8*A*c^(1/2)*(c*x^2+b)^(5/2)+3*B*c^(1/2)*(c*x^2+b)^(1/2)*x^2*b^2+12*A*ln(x*c^(1/2)+(c*x^2+b)^(1
/2))*x*b^2*c+3*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*x*b^3)/x^4/(c*x^2+b)^(3/2)/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)^(3/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.11389, size = 467, normalized size = 3.65 \begin{align*} \left [\frac{3 \,{\left (B b^{2} + 4 \, A b c\right )} \sqrt{c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) + 2 \,{\left (2 \, B c^{2} x^{4} - 8 \, A b c +{\left (5 \, B b c + 4 \, A c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{16 \, c x^{2}}, -\frac{3 \,{\left (B b^{2} + 4 \, A b c\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (2 \, B c^{2} x^{4} - 8 \, A b c +{\left (5 \, B b c + 4 \, A c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{8 \, c x^{2}}\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)^(3/2)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \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)**(3/2)/x**5,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**5, x)

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Giac [A]  time = 1.23806, size = 170, normalized size = 1.33 \begin{align*} \frac{2 \, A b^{2} \sqrt{c} \mathrm{sgn}\left (x\right )}{{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b} + \frac{1}{8} \,{\left (2 \, B c x^{2} \mathrm{sgn}\left (x\right ) + \frac{5 \, B b c^{2} \mathrm{sgn}\left (x\right ) + 4 \, A c^{3} \mathrm{sgn}\left (x\right )}{c^{2}}\right )} \sqrt{c x^{2} + b} x - \frac{3 \,{\left (B b^{2} \sqrt{c} \mathrm{sgn}\left (x\right ) + 4 \, A b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right )\right )} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2}\right )}{16 \, 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)^(3/2)/x^5,x, algorithm="giac")

[Out]

2*A*b^2*sqrt(c)*sgn(x)/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b) + 1/8*(2*B*c*x^2*sgn(x) + (5*B*b*c^2*sgn(x) + 4*A
*c^3*sgn(x))/c^2)*sqrt(c*x^2 + b)*x - 3/16*(B*b^2*sqrt(c)*sgn(x) + 4*A*b*c^(3/2)*sgn(x))*log((sqrt(c)*x - sqrt
(c*x^2 + b))^2)/c