3.115 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{13}} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 c \left (b x^2+c x^4\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^{10}}-\frac{\left (b x^2+c x^4\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^{12}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}} \]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(9*b*x^14) - ((9*b*B - 4*A*c)*(b*x^2 + c*x^4)^(5/2))/(63*b^2*x^12) + (2*c*(9*b*B -
4*A*c)*(b*x^2 + c*x^4)^(5/2))/(315*b^3*x^10)

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Rubi [A]  time = 0.233431, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \[ \frac{2 c \left (b x^2+c x^4\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^{10}}-\frac{\left (b x^2+c x^4\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^{12}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^13,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(9*b*x^14) - ((9*b*B - 4*A*c)*(b*x^2 + c*x^4)^(5/2))/(63*b^2*x^12) + (2*c*(9*b*B -
4*A*c)*(b*x^2 + c*x^4)^(5/2))/(315*b^3*x^10)

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 658

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

Rule 650

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

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}+\frac{\left (-7 (-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^6} \, dx,x,x^2\right )}{9 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}-\frac{(9 b B-4 A c) \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}-\frac{(c (9 b B-4 A c)) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{63 b^2}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}-\frac{(9 b B-4 A c) \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}+\frac{2 c (9 b B-4 A c) \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0306484, size = 66, normalized size = 0.69 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (A \left (-35 b^2+20 b c x^2-8 c^2 x^4\right )+9 b B x^2 \left (2 c x^2-5 b\right )\right )}{315 b^3 x^{14}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^13,x]

[Out]

((x^2*(b + c*x^2))^(5/2)*(9*b*B*x^2*(-5*b + 2*c*x^2) + A*(-35*b^2 + 20*b*c*x^2 - 8*c^2*x^4)))/(315*b^3*x^14)

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Maple [A]  time = 0.005, size = 70, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( 8\,A{c}^{2}{x}^{4}-18\,B{x}^{4}bc-20\,Abc{x}^{2}+45\,B{x}^{2}{b}^{2}+35\,A{b}^{2} \right ) }{315\,{x}^{12}{b}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\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^13,x)

[Out]

-1/315*(c*x^2+b)*(8*A*c^2*x^4-18*B*b*c*x^4-20*A*b*c*x^2+45*B*b^2*x^2+35*A*b^2)*(c*x^4+b*x^2)^(3/2)/x^12/b^3

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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^13,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33211, size = 236, normalized size = 2.46 \begin{align*} \frac{{\left (2 \,{\left (9 \, B b c^{3} - 4 \, A c^{4}\right )} x^{8} -{\left (9 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{6} - 35 \, A b^{4} - 3 \,{\left (24 \, B b^{3} c + A b^{2} c^{2}\right )} x^{4} - 5 \,{\left (9 \, B b^{4} + 10 \, A b^{3} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, b^{3} x^{10}} \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^13,x, algorithm="fricas")

[Out]

1/315*(2*(9*B*b*c^3 - 4*A*c^4)*x^8 - (9*B*b^2*c^2 - 4*A*b*c^3)*x^6 - 35*A*b^4 - 3*(24*B*b^3*c + A*b^2*c^2)*x^4
- 5*(9*B*b^4 + 10*A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^3*x^10)

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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^{13}}\, 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**13,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**13, x)

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Giac [B]  time = 4.78231, size = 581, normalized size = 6.05 \begin{align*} \frac{4 \,{\left (315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{14} B c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} B b c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 840 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} A c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} B b^{2} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 1260 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} A b c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 819 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B b^{3} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 1764 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} A b^{2} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 441 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b^{4} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 504 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} A b^{3} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{5} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 144 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b^{4} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 81 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{6} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{5} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 9 \, B b^{7} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 4 \, A b^{6} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right )\right )}}{315 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{9}} \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^13,x, algorithm="giac")

[Out]

4/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*c^(7/2)*sgn(x) - 315*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*b*c^(7/2
)*sgn(x) + 840*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*c^(9/2)*sgn(x) + 315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^2*
c^(7/2)*sgn(x) + 1260*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b*c^(9/2)*sgn(x) - 819*(sqrt(c)*x - sqrt(c*x^2 + b))^
8*B*b^3*c^(7/2)*sgn(x) + 1764*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^2*c^(9/2)*sgn(x) + 441*(sqrt(c)*x - sqrt(c*x
^2 + b))^6*B*b^4*c^(7/2)*sgn(x) + 504*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^3*c^(9/2)*sgn(x) - 9*(sqrt(c)*x - sq
rt(c*x^2 + b))^4*B*b^5*c^(7/2)*sgn(x) + 144*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^4*c^(9/2)*sgn(x) + 81*(sqrt(c)
*x - sqrt(c*x^2 + b))^2*B*b^6*c^(7/2)*sgn(x) - 36*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^5*c^(9/2)*sgn(x) - 9*B*b
^7*c^(7/2)*sgn(x) + 4*A*b^6*c^(9/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9