3.145 \(\int \frac{x^9 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{9/2}}+\frac{x^4 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{6 b c^2}-\frac{5 x^2 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{24 c^3}+\frac{5 b \sqrt{b x^2+c x^4} (7 b B-6 A c)}{16 c^4}-\frac{x^8 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

[Out]

-(((b*B - A*c)*x^8)/(b*c*Sqrt[b*x^2 + c*x^4])) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^4) - (5*(7*b*
B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^3) + ((7*b*B - 6*A*c)*x^4*Sqrt[b*x^2 + c*x^4])/(6*b*c^2) - (5*b^2*(7
*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(9/2))

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Rubi [A]  time = 0.33328, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2034, 788, 670, 640, 620, 206} \[ -\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{9/2}}+\frac{x^4 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{6 b c^2}-\frac{5 x^2 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{24 c^3}+\frac{5 b \sqrt{b x^2+c x^4} (7 b B-6 A c)}{16 c^4}-\frac{x^8 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^8)/(b*c*Sqrt[b*x^2 + c*x^4])) + (5*b*(7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^4) - (5*(7*b*
B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^3) + ((7*b*B - 6*A*c)*x^4*Sqrt[b*x^2 + c*x^4])/(6*b*c^2) - (5*b^2*(7
*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(9/2))

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 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^9 \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^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}+\frac{1}{2} \left (-\frac{6 A}{b}+\frac{7 B}{c}\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}+\frac{(7 b B-6 A c) x^4 \sqrt{b x^2+c x^4}}{6 b c^2}-\frac{(5 (7 b B-6 A c)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{12 c^2}\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}-\frac{5 (7 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^3}+\frac{(7 b B-6 A c) x^4 \sqrt{b x^2+c x^4}}{6 b c^2}+\frac{(5 b (7 b B-6 A c)) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{16 c^3}\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}+\frac{5 b (7 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^4}-\frac{5 (7 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^3}+\frac{(7 b B-6 A c) x^4 \sqrt{b x^2+c x^4}}{6 b c^2}-\frac{\left (5 b^2 (7 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^4}\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}+\frac{5 b (7 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^4}-\frac{5 (7 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^3}+\frac{(7 b B-6 A c) x^4 \sqrt{b x^2+c x^4}}{6 b c^2}-\frac{\left (5 b^2 (7 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^4}\\ &=-\frac{(b B-A c) x^8}{b c \sqrt{b x^2+c x^4}}+\frac{5 b (7 b B-6 A c) \sqrt{b x^2+c x^4}}{16 c^4}-\frac{5 (7 b B-6 A c) x^2 \sqrt{b x^2+c x^4}}{24 c^3}+\frac{(7 b B-6 A c) x^4 \sqrt{b x^2+c x^4}}{6 b c^2}-\frac{5 b^2 (7 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{16 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.179336, size = 136, normalized size = 0.74 \[ \frac{x \left (\sqrt{c} x \left (b^2 \left (35 B c x^2-90 A c\right )-2 b c^2 x^2 \left (15 A+7 B x^2\right )+4 c^3 x^4 \left (3 A+2 B x^2\right )+105 b^3 B\right )-15 b^{5/2} \sqrt{\frac{c x^2}{b}+1} (7 b B-6 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{48 c^{9/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(105*b^3*B + 4*c^3*x^4*(3*A + 2*B*x^2) - 2*b*c^2*x^2*(15*A + 7*B*x^2) + b^2*(-90*A*c + 35*B*c*x^
2)) - 15*b^(5/2)*(7*b*B - 6*A*c)*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(48*c^(9/2)*Sqrt[x^2*(b +
c*x^2)])

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Maple [A]  time = 0.015, size = 166, normalized size = 0.9 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ){x}^{3}}{48} \left ( 8\,B{c}^{9/2}{x}^{7}+12\,A{c}^{9/2}{x}^{5}-14\,B{c}^{7/2}{x}^{5}b-30\,A{c}^{7/2}{x}^{3}b+35\,B{c}^{5/2}{x}^{3}{b}^{2}-90\,A{c}^{5/2}x{b}^{2}+105\,B{c}^{3/2}x{b}^{3}+90\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) \sqrt{c{x}^{2}+b}{b}^{2}{c}^{2}-105\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) \sqrt{c{x}^{2}+b}{b}^{3}c \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/48*x^3*(c*x^2+b)*(8*B*c^(9/2)*x^7+12*A*c^(9/2)*x^5-14*B*c^(7/2)*x^5*b-30*A*c^(7/2)*x^3*b+35*B*c^(5/2)*x^3*b^
2-90*A*c^(5/2)*x*b^2+105*B*c^(3/2)*x*b^3+90*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*(c*x^2+b)^(1/2)*b^2*c^2-105*B*ln(x
*c^(1/2)+(c*x^2+b)^(1/2))*(c*x^2+b)^(1/2)*b^3*c)/(c*x^4+b*x^2)^(3/2)/c^(11/2)

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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^9*(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 = 1.53562, size = 738, normalized size = 4.01 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{4} - 6 \, A b^{3} c +{\left (7 \, B b^{3} c - 6 \, A b^{2} 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 (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \,{\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \,{\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{96 \,{\left (c^{6} x^{2} + b c^{5}\right )}}, \frac{15 \,{\left (7 \, B b^{4} - 6 \, A b^{3} c +{\left (7 \, B b^{3} c - 6 \, A b^{2} 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 (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \,{\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \,{\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{48 \,{\left (c^{6} x^{2} + b c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x^2)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x
^2)*sqrt(c)) - 2*(8*B*c^4*x^6 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*
A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^6*x^2 + b*c^5), 1/48*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2
)*x^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^4*x^6 + 105*B*b^3*c - 90*A*b^2*c^2 -
2*(7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^6*x^2 + b*c^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9} \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**9*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**9*(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^{9}}{{\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^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^9/(c*x^4 + b*x^2)^(3/2), x)