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

Optimal. Leaf size=139 \[ -\frac{b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{7/2}}-\frac{x^2 \sqrt{b x^2+c x^4} (5 b B-6 A c)}{24 c^2}+\frac{b \sqrt{b x^2+c x^4} (5 b B-6 A c)}{16 c^3}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c} \]

[Out]

(b*(5*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^3) - ((5*b*B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^2) + (B*x^4
*Sqrt[b*x^2 + c*x^4])/(6*c) - (b^2*(5*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(7/2))

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Rubi [A]  time = 0.274916, antiderivative size = 139, 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, 794, 670, 640, 620, 206} \[ -\frac{b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{7/2}}-\frac{x^2 \sqrt{b x^2+c x^4} (5 b B-6 A c)}{24 c^2}+\frac{b \sqrt{b x^2+c x^4} (5 b B-6 A c)}{16 c^3}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^3) - ((5*b*B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^2) + (B*x^4
*Sqrt[b*x^2 + c*x^4])/(6*c) - (b^2*(5*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(7/2))

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 794

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

Rule 670

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

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 )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c}+\frac{\left (2 (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac{(5 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^2}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c}+\frac{(b (5 b B-6 A c)) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=\frac{b (5 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^3}-\frac{(5 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^2}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c}-\frac{\left (b^2 (5 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^3}\\ &=\frac{b (5 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^3}-\frac{(5 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^2}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c}-\frac{\left (b^2 (5 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^3}\\ &=\frac{b (5 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^3}-\frac{(5 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^2}+\frac{B x^4 \sqrt{b x^2+c x^4}}{6 c}-\frac{b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.114082, size = 123, normalized size = 0.88 \[ \frac{x \left (\sqrt{c} x \left (b+c x^2\right ) \left (-2 b c \left (9 A+5 B x^2\right )+4 c^2 x^2 \left (3 A+2 B x^2\right )+15 b^2 B\right )-3 b^2 \sqrt{b+c x^2} (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[c]*x*(b + c*x^2)*(15*b^2*B + 4*c^2*x^2*(3*A + 2*B*x^2) - 2*b*c*(9*A + 5*B*x^2)) - 3*b^2*(5*b*B - 6*A*
c)*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/Sqrt[b + c*x^2]]))/(48*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.008, size = 169, normalized size = 1.2 \begin{align*}{\frac{x}{48}\sqrt{c{x}^{2}+b} \left ( 8\,B{c}^{7/2}\sqrt{c{x}^{2}+b}{x}^{5}+12\,A{c}^{7/2}\sqrt{c{x}^{2}+b}{x}^{3}-10\,B{c}^{5/2}\sqrt{c{x}^{2}+b}{x}^{3}b-18\,A{c}^{5/2}\sqrt{c{x}^{2}+b}xb+15\,B{c}^{3/2}\sqrt{c{x}^{2}+b}x{b}^{2}+18\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{2}{c}^{2}-15\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{3}c \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{c}^{-{\frac{9}{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)^(1/2),x)

[Out]

1/48*x*(c*x^2+b)^(1/2)*(8*B*c^(7/2)*(c*x^2+b)^(1/2)*x^5+12*A*c^(7/2)*(c*x^2+b)^(1/2)*x^3-10*B*c^(5/2)*(c*x^2+b
)^(1/2)*x^3*b-18*A*c^(5/2)*(c*x^2+b)^(1/2)*x*b+15*B*c^(3/2)*(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^2-15*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^3*c)/(c*x^4+b*x^2)^(1/2)/c^(9/2)

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.11245, size = 512, normalized size = 3.68 \begin{align*} \left [-\frac{3 \,{\left (5 \, 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} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac{3 \,{\left (5 \, 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} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{48 \, c^{4}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*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 + 1
5*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4, 1/48*(3*(5*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 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*
c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4]

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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 )}{\sqrt{x^{2} \left (b + c x^{2}\right )}}\, 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)**(1/2),x)

[Out]

Integral(x**5*(A + B*x**2)/sqrt(x**2*(b + c*x**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}}{\sqrt{c x^{4} + b x^{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)^(1/2),x, algorithm="giac")

[Out]

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