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

Optimal. Leaf size=137 \[ -\frac{b^2 (b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{3/2}}+\frac{\left (b x^2+c x^4\right )^{3/2} (b B-6 A c)}{6 b}+\frac{\left (b+2 c x^2\right ) \sqrt{b x^2+c x^4} (b B-6 A c)}{16 c}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4} \]

[Out]

((b*B - 6*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(16*c) + ((b*B - 6*A*c)*(b*x^2 + c*x^4)^(3/2))/(6*b) + (A*(b
*x^2 + c*x^4)^(5/2))/(b*x^4) - (b^2*(b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(3/2))

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Rubi [A]  time = 0.295905, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2034, 792, 664, 612, 620, 206} \[ -\frac{b^2 (b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{3/2}}+\frac{\left (b x^2+c x^4\right )^{3/2} (b B-6 A c)}{6 b}+\frac{\left (b+2 c x^2\right ) \sqrt{b x^2+c x^4} (b B-6 A c)}{16 c}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - 6*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(16*c) + ((b*B - 6*A*c)*(b*x^2 + c*x^4)^(3/2))/(6*b) + (A*(b
*x^2 + c*x^4)^(5/2))/(b*x^4) - (b^2*(b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(3/2))

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*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^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac{\left (-2 (-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} \, dx,x,x^2\right )}{b}\\ &=\frac{(b B-6 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac{1}{4} (-b B+6 A c) \operatorname{Subst}\left (\int \sqrt{b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{(b B-6 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{16 c}+\frac{(b B-6 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac{\left (b^2 (b B-6 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{32 c}\\ &=\frac{(b B-6 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{16 c}+\frac{(b B-6 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac{\left (b^2 (b B-6 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c}\\ &=\frac{(b B-6 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{16 c}+\frac{(b B-6 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b}+\frac{A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac{b^2 (b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.118235, size = 130, normalized size = 0.95 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} x \sqrt{\frac{c x^2}{b}+1} \left (2 b c \left (15 A+7 B x^2\right )+4 c^2 x^2 \left (3 A+2 B x^2\right )+3 b^2 B\right )-3 b^{3/2} (b B-6 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{48 c^{3/2} x \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(3*b^2*B + 4*c^2*x^2*(3*A + 2*B*x^2) + 2*b*c*(15*A + 7*B
*x^2)) - 3*b^(3/2)*(b*B - 6*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(48*c^(3/2)*x*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.004, size = 162, normalized size = 1.2 \begin{align*}{\frac{1}{48\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 8\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}x+12\,A{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{3/2}x-2\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}xb+18\,A{c}^{3/2}\sqrt{c{x}^{2}+b}xb-3\,B\sqrt{c}\sqrt{c{x}^{2}+b}x{b}^{2}+18\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{2}c-3\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{3} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{3}{2}}}} \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^3,x)

[Out]

1/48*(c*x^4+b*x^2)^(3/2)*(8*B*c^(1/2)*(c*x^2+b)^(5/2)*x+12*A*c^(3/2)*(c*x^2+b)^(3/2)*x-2*B*c^(1/2)*(c*x^2+b)^(
3/2)*x*b+18*A*c^(3/2)*(c*x^2+b)^(1/2)*x*b-3*B*c^(1/2)*(c*x^2+b)^(1/2)*x*b^2+18*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))
*b^2*c-3*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^3)/x^3/(c*x^2+b)^(3/2)/c^(3/2)

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3026, size = 504, normalized size = 3.68 \begin{align*} \left [-\frac{3 \,{\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (8 \, B c^{3} x^{4} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \,{\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{96 \, c^{2}}, \frac{3 \,{\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) +{\left (8 \, B c^{3} x^{4} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \,{\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{48 \, c^{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^3,x, algorithm="fricas")

[Out]

[-1/96*(3*(B*b^3 - 6*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 + 3*B
*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^2, 1/48*(3*(B*b^3 - 6*A*b^2*c)*sqrt(
-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^3*x^4 + 3*B*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 +
6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^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^{3}}\, 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**3,x)

[Out]

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

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Giac [A]  time = 1.18436, size = 192, normalized size = 1.4 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, B c x^{2} \mathrm{sgn}\left (x\right ) + \frac{7 \, B b c^{4} \mathrm{sgn}\left (x\right ) + 6 \, A c^{5} \mathrm{sgn}\left (x\right )}{c^{4}}\right )} x^{2} + \frac{3 \,{\left (B b^{2} c^{3} \mathrm{sgn}\left (x\right ) + 10 \, A b c^{4} \mathrm{sgn}\left (x\right )\right )}}{c^{4}}\right )} \sqrt{c x^{2} + b} x + \frac{{\left (B b^{3} \mathrm{sgn}\left (x\right ) - 6 \, A b^{2} c \mathrm{sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right )}{16 \, c^{\frac{3}{2}}} - \frac{{\left (B b^{3} \log \left ({\left | b \right |}\right ) - 6 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm{sgn}\left (x\right )}{32 \, c^{\frac{3}{2}}} \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^3,x, algorithm="giac")

[Out]

1/48*(2*(4*B*c*x^2*sgn(x) + (7*B*b*c^4*sgn(x) + 6*A*c^5*sgn(x))/c^4)*x^2 + 3*(B*b^2*c^3*sgn(x) + 10*A*b*c^4*sg
n(x))/c^4)*sqrt(c*x^2 + b)*x + 1/16*(B*b^3*sgn(x) - 6*A*b^2*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c
^(3/2) - 1/32*(B*b^3*log(abs(b)) - 6*A*b^2*c*log(abs(b)))*sgn(x)/c^(3/2)