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

Optimal. Leaf size=112 \[ \frac{\sqrt{b x^2+c x^4} (3 b B-2 A c)}{2 b c^2}-\frac{(3 b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2 c^{5/2}}-\frac{x^4 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.241677, antiderivative size = 112, 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, 788, 640, 620, 206} \[ \frac{\sqrt{b x^2+c x^4} (3 b B-2 A c)}{2 b c^2}-\frac{(3 b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2 c^{5/2}}-\frac{x^4 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 788

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

Rule 640

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

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

Mathematica [A]  time = 0.112973, size = 91, normalized size = 0.81 \[ \frac{x \left (\sqrt{c} x \left (-2 A c+3 b B+B c x^2\right )-\sqrt{b} \sqrt{\frac{c x^2}{b}+1} (3 b B-2 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{2 c^{5/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 115, normalized size = 1. \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ){x}^{3}}{2} \left ( -B{c}^{{\frac{5}{2}}}{x}^{3}+2\,A{c}^{5/2}x-3\,B{c}^{3/2}xb-2\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) \sqrt{c{x}^{2}+b}{c}^{2}+3\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) \sqrt{c{x}^{2}+b}bc \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^3*(c*x^2+b)*(-B*c^(5/2)*x^3+2*A*c^(5/2)*x-3*B*c^(3/2)*x*b-2*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*(c*x^2+b)^(
1/2)*c^2+3*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*(c*x^2+b)^(1/2)*b*c)/(c*x^4+b*x^2)^(3/2)/c^(7/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.42984, size = 504, normalized size = 4.5 \begin{align*} \left [-\frac{{\left (3 \, B b^{2} - 2 \, A b c +{\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{4 \,{\left (c^{4} x^{2} + b c^{3}\right )}}, \frac{{\left (3 \, B b^{2} - 2 \, A b c +{\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) +{\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{2 \,{\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{5}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

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