3.116 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{15}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^{10}}+\frac{4 c \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^{12}}-\frac{\left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^{14}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(11*b*x^16) - ((11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*b^2*x^14) + (4*c*(11*b*B
- 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(693*b^3*x^12) - (8*c^2*(11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*b^4*x^1
0)

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Rubi [A]  time = 0.278839, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^{10}}+\frac{4 c \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^{12}}-\frac{\left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^{14}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(11*b*x^16) - ((11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*b^2*x^14) + (4*c*(11*b*B
- 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(693*b^3*x^12) - (8*c^2*(11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*b^4*x^1
0)

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^{15}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^8} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac{\left (-8 (-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^7} \, dx,x,x^2\right )}{11 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac{(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}-\frac{(2 c (11 b B-6 A c)) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )}{99 b^2}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac{(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}+\frac{4 c (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{693 b^3 x^{12}}+\frac{\left (4 c^2 (11 b B-6 A c)\right ) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{693 b^3}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac{(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}+\frac{4 c (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{693 b^3 x^{12}}-\frac{8 c^2 (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 b^4 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0344796, size = 89, normalized size = 0.67 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (3 A \left (-70 b^2 c x^2+105 b^3+40 b c^2 x^4-16 c^3 x^6\right )+11 b B x^2 \left (35 b^2-20 b c x^2+8 c^2 x^4\right )\right )}{3465 b^4 x^{16}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]

[Out]

-((x^2*(b + c*x^2))^(5/2)*(11*b*B*x^2*(35*b^2 - 20*b*c*x^2 + 8*c^2*x^4) + 3*A*(105*b^3 - 70*b^2*c*x^2 + 40*b*c
^2*x^4 - 16*c^3*x^6)))/(3465*b^4*x^16)

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Maple [A]  time = 0.006, size = 94, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -48\,A{c}^{3}{x}^{6}+88\,B{x}^{6}b{c}^{2}+120\,Ab{c}^{2}{x}^{4}-220\,B{x}^{4}{b}^{2}c-210\,A{b}^{2}c{x}^{2}+385\,B{x}^{2}{b}^{3}+315\,A{b}^{3} \right ) }{3465\,{x}^{14}{b}^{4}} \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^15,x)

[Out]

-1/3465*(c*x^2+b)*(-48*A*c^3*x^6+88*B*b*c^2*x^6+120*A*b*c^2*x^4-220*B*b^2*c*x^4-210*A*b^2*c*x^2+385*B*b^3*x^2+
315*A*b^3)*(c*x^4+b*x^2)^(3/2)/x^14/b^4

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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^15,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48329, size = 304, normalized size = 2.29 \begin{align*} -\frac{{\left (8 \,{\left (11 \, B b c^{4} - 6 \, A c^{5}\right )} x^{10} - 4 \,{\left (11 \, B b^{2} c^{3} - 6 \, A b c^{4}\right )} x^{8} + 3 \,{\left (11 \, B b^{3} c^{2} - 6 \, A b^{2} c^{3}\right )} x^{6} + 315 \, A b^{5} + 5 \,{\left (110 \, B b^{4} c + 3 \, A b^{3} c^{2}\right )} x^{4} + 35 \,{\left (11 \, B b^{5} + 12 \, A b^{4} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{3465 \, b^{4} x^{12}} \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^15,x, algorithm="fricas")

[Out]

-1/3465*(8*(11*B*b*c^4 - 6*A*c^5)*x^10 - 4*(11*B*b^2*c^3 - 6*A*b*c^4)*x^8 + 3*(11*B*b^3*c^2 - 6*A*b^2*c^3)*x^6
+ 315*A*b^5 + 5*(110*B*b^4*c + 3*A*b^3*c^2)*x^4 + 35*(11*B*b^5 + 12*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^4*x^
12)

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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^{15}}\, 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**15,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**15, x)

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Giac [B]  time = 4.91959, size = 662, normalized size = 4.98 \begin{align*} \frac{16 \,{\left (2310 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{16} B c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 1155 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{14} B b c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 6930 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{14} A c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 231 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} B b^{2} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 12474 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} A b c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) - 4851 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} B b^{3} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 15246 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} A b^{2} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 2475 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B b^{4} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 4950 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} A b^{3} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 495 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b^{5} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 990 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} A b^{4} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 605 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{6} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 330 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b^{5} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) - 121 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{7} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 66 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{6} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 11 \, B b^{8} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 6 \, A b^{7} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right )\right )}}{3465 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{11}} \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^15,x, algorithm="giac")

[Out]

16/3465*(2310*(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*c^(9/2)*sgn(x) - 1155*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*b*c^
(9/2)*sgn(x) + 6930*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*c^(11/2)*sgn(x) + 231*(sqrt(c)*x - sqrt(c*x^2 + b))^12*
B*b^2*c^(9/2)*sgn(x) + 12474*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b*c^(11/2)*sgn(x) - 4851*(sqrt(c)*x - sqrt(c*x
^2 + b))^10*B*b^3*c^(9/2)*sgn(x) + 15246*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b^2*c^(11/2)*sgn(x) + 2475*(sqrt(c
)*x - sqrt(c*x^2 + b))^8*B*b^4*c^(9/2)*sgn(x) + 4950*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^3*c^(11/2)*sgn(x) + 4
95*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^5*c^(9/2)*sgn(x) + 990*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^4*c^(11/2)*s
gn(x) + 605*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^6*c^(9/2)*sgn(x) - 330*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^5*c
^(11/2)*sgn(x) - 121*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^7*c^(9/2)*sgn(x) + 66*(sqrt(c)*x - sqrt(c*x^2 + b))^2
*A*b^6*c^(11/2)*sgn(x) + 11*B*b^8*c^(9/2)*sgn(x) - 6*A*b^7*c^(11/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 -
b)^11