3.117 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{17}} \, dx\)

Optimal. Leaf size=170 \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^{14}}-\frac{\left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^{16}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(13*b*x^18) - ((13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(143*b^2*x^16) + (2*c*(13*b*
B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(429*b^3*x^14) - (8*c^2*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(3003*b^4*x^
12) + (16*c^3*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(15015*b^5*x^10)

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Rubi [A]  time = 0.321332, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^{14}}-\frac{\left (b x^2+c x^4\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^{16}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(13*b*x^18) - ((13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(143*b^2*x^16) + (2*c*(13*b*
B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(429*b^3*x^14) - (8*c^2*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(3003*b^4*x^
12) + (16*c^3*(13*b*B - 8*A*c)*(b*x^2 + c*x^4)^(5/2))/(15015*b^5*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^{17}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^9} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac{\left (-9 (-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^8} \, dx,x,x^2\right )}{13 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac{(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{(3 c (13 b B-8 A c)) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^7} \, dx,x,x^2\right )}{143 b^2}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac{(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac{2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac{\left (4 c^2 (13 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )}{429 b^3}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac{(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac{2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac{8 c^2 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}-\frac{\left (8 c^3 (13 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{3003 b^4}\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac{(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac{2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac{8 c^2 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}+\frac{16 c^3 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0704421, size = 95, normalized size = 0.56 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\left (\frac{c x^3}{b}+x\right )^2 \left (-70 b^2 c x^2+105 b^3+40 b c^2 x^4-16 c^3 x^6\right ) (8 A c-13 b B)-1155 A b^2 \left (b+c x^2\right )^2\right )}{15015 b^3 x^{14}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-1155*A*b^2*(b + c*x^2)^2 + (-13*b*B + 8*A*c)*(x + (c*x^3)/b)^2*(105*b^3 - 70*b^2*c*x^
2 + 40*b*c^2*x^4 - 16*c^3*x^6)))/(15015*b^3*x^14)

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Maple [A]  time = 0.006, size = 118, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( 128\,A{c}^{4}{x}^{8}-208\,Bb{c}^{3}{x}^{8}-320\,Ab{c}^{3}{x}^{6}+520\,B{b}^{2}{c}^{2}{x}^{6}+560\,A{b}^{2}{c}^{2}{x}^{4}-910\,B{b}^{3}c{x}^{4}-840\,A{b}^{3}c{x}^{2}+1365\,B{b}^{4}{x}^{2}+1155\,A{b}^{4} \right ) }{15015\,{x}^{16}{b}^{5}} \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^17,x)

[Out]

-1/15015*(c*x^2+b)*(128*A*c^4*x^8-208*B*b*c^3*x^8-320*A*b*c^3*x^6+520*B*b^2*c^2*x^6+560*A*b^2*c^2*x^4-910*B*b^
3*c*x^4-840*A*b^3*c*x^2+1365*B*b^4*x^2+1155*A*b^4)*(c*x^4+b*x^2)^(3/2)/x^16/b^5

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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^17,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81635, size = 356, normalized size = 2.09 \begin{align*} \frac{{\left (16 \,{\left (13 \, B b c^{5} - 8 \, A c^{6}\right )} x^{12} - 8 \,{\left (13 \, B b^{2} c^{4} - 8 \, A b c^{5}\right )} x^{10} + 6 \,{\left (13 \, B b^{3} c^{3} - 8 \, A b^{2} c^{4}\right )} x^{8} - 1155 \, A b^{6} - 5 \,{\left (13 \, B b^{4} c^{2} - 8 \, A b^{3} c^{3}\right )} x^{6} - 35 \,{\left (52 \, B b^{5} c + A b^{4} c^{2}\right )} x^{4} - 105 \,{\left (13 \, B b^{6} + 14 \, A b^{5} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15015 \, b^{5} x^{14}} \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^17,x, algorithm="fricas")

[Out]

1/15015*(16*(13*B*b*c^5 - 8*A*c^6)*x^12 - 8*(13*B*b^2*c^4 - 8*A*b*c^5)*x^10 + 6*(13*B*b^3*c^3 - 8*A*b^2*c^4)*x
^8 - 1155*A*b^6 - 5*(13*B*b^4*c^2 - 8*A*b^3*c^3)*x^6 - 35*(52*B*b^5*c + A*b^4*c^2)*x^4 - 105*(13*B*b^6 + 14*A*
b^5*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*x^14)

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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^{17}}\, 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**17,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**17, x)

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Giac [B]  time = 5.15756, size = 743, normalized size = 4.37 \begin{align*} \text{result too large to display} \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^17,x, algorithm="giac")

[Out]

32/15015*(15015*(sqrt(c)*x - sqrt(c*x^2 + b))^18*B*c^(11/2)*sgn(x) - 3003*(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*b
*c^(11/2)*sgn(x) + 48048*(sqrt(c)*x - sqrt(c*x^2 + b))^16*A*c^(13/2)*sgn(x) - 6006*(sqrt(c)*x - sqrt(c*x^2 + b
))^14*B*b^2*c^(11/2)*sgn(x) + 96096*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*b*c^(13/2)*sgn(x) - 28314*(sqrt(c)*x -
sqrt(c*x^2 + b))^12*B*b^3*c^(11/2)*sgn(x) + 109824*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b^2*c^(13/2)*sgn(x) + 13
728*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^4*c^(11/2)*sgn(x) + 37752*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b^3*c^(1
3/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^5*c^(11/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + b))^
8*A*b^4*c^(13/2)*sgn(x) + 3718*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^6*c^(11/2)*sgn(x) - 2288*(sqrt(c)*x - sqrt(
c*x^2 + b))^6*A*b^5*c^(13/2)*sgn(x) - 1014*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^7*c^(11/2)*sgn(x) + 624*(sqrt(c
)*x - sqrt(c*x^2 + b))^4*A*b^6*c^(13/2)*sgn(x) + 169*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^8*c^(11/2)*sgn(x) - 1
04*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^7*c^(13/2)*sgn(x) - 13*B*b^9*c^(11/2)*sgn(x) + 8*A*b^8*c^(13/2)*sgn(x))
/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^13