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

Optimal. Leaf size=364 \[ \frac {4 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}-\frac {8 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {8 c^{3/2} x^{3/2} \left (b+c x^2\right ) (A c+9 b B)}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (A c+9 b B)}{45 b x^{11/2}}-\frac {4 c \sqrt {b x^2+c x^4} (A c+9 b B)}{15 b x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}} \]

[Out]

-2/45*(A*c+9*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^(11/2)-2/9*A*(c*x^4+b*x^2)^(5/2)/b/x^(19/2)+8/15*c^(3/2)*(A*c+9*B*b)
*x^(3/2)*(c*x^2+b)/b/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)-4/15*c*(A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^(3/2)-
8/15*c^(5/4)*(A*c+9*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/
4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c
^(1/2))^2)^(1/2)/b^(3/4)/(c*x^4+b*x^2)^(1/2)+4/15*c^(5/4)*(A*c+9*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))
)^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2)
)*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b^(3/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2038, 2020, 2032, 329, 305, 220, 1196} \[ \frac {4 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}-\frac {8 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+9 b B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {8 c^{3/2} x^{3/2} \left (b+c x^2\right ) (A c+9 b B)}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (A c+9 b B)}{45 b x^{11/2}}-\frac {4 c \sqrt {b x^2+c x^4} (A c+9 b B)}{15 b x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^(3/2)*(9*b*B + A*c)*x^(3/2)*(b + c*x^2))/(15*b*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (4*c*(9*b*B +
A*c)*Sqrt[b*x^2 + c*x^4])/(15*b*x^(3/2)) - (2*(9*b*B + A*c)*(b*x^2 + c*x^4)^(3/2))/(45*b*x^(11/2)) - (2*A*(b*
x^2 + c*x^4)^(5/2))/(9*b*x^(19/2)) - (8*c^(5/4)*(9*b*B + A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b
] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[b*x^2 + c*x^4]) + (4*c
^(5/4)*(9*b*B + A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(
1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[b*x^2 + c*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx &=-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}-\frac {\left (2 \left (-\frac {9 b B}{2}-\frac {A c}{2}\right )\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx}{9 b}\\ &=-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {(2 c (9 b B+A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^{5/2}} \, dx}{15 b}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (4 c^2 (9 b B+A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{15 b}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (4 c^2 (9 b B+A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{15 b \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (8 c^2 (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}+\frac {\left (8 c^{3/2} (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {b} \sqrt {b x^2+c x^4}}-\frac {\left (8 c^{3/2} (9 b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {b} \sqrt {b x^2+c x^4}}\\ &=\frac {8 c^{3/2} (9 b B+A c) x^{3/2} \left (b+c x^2\right )}{15 b \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c (9 b B+A c) \sqrt {b x^2+c x^4}}{15 b x^{3/2}}-\frac {2 (9 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{45 b x^{11/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{19/2}}-\frac {8 c^{5/4} (9 b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}+\frac {4 c^{5/4} (9 b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 100, normalized size = 0.27 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (b x^2 (A c+9 b B) \, _2F_1\left (-\frac {3}{2},-\frac {5}{4};-\frac {1}{4};-\frac {c x^2}{b}\right )+5 A \left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1}\right )}{45 b x^{11/2} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x^2*(b + c*x^2)]*(5*A*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] + b*(9*b*B + A*c)*x^2*Hypergeometric2F1[-3/2,
-5/4, -1/4, -((c*x^2)/b)]))/(45*b*x^(11/2)*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{4} + {\left (B b + A c\right )} x^{2} + A b\right )} \sqrt {c x^{4} + b x^{2}}}{x^{\frac {13}{2}}}, x\right ) \]

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^(17/2),x, algorithm="fricas")

[Out]

integral((B*c*x^4 + (B*b + A*c)*x^2 + A*b)*sqrt(c*x^4 + b*x^2)/x^(13/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {17}{2}}}\,{d x} \]

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^(17/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 452, normalized size = 1.24 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-12 A \,c^{3} x^{6}-63 B b \,c^{2} x^{6}+12 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A b \,c^{2} x^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A b \,c^{2} x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+108 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{2} c \,x^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-54 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{2} c \,x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-23 A b \,c^{2} x^{4}-72 B \,b^{2} c \,x^{4}-16 A \,b^{2} c \,x^{2}-9 B \,b^{3} x^{2}-5 A \,b^{3}\right )}{45 \left (c \,x^{2}+b \right )^{2} b \,x^{\frac {15}{2}}} \]

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^(17/2),x)

[Out]

2/45*(c*x^4+b*x^2)^(3/2)/x^(15/2)/(c*x^2+b)^2*(12*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b
*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1
/2*2^(1/2))*x^4*b*c^2-6*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(
1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b*c^2+108*
B*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*
x)^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c-54*B*((c*x+(-b*c)^(1/2))/(-b
*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+
(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c-12*A*c^3*x^6-63*B*b*c^2*x^6-23*A*b*c^2*x^4-72*B*b^2*c
*x^4-16*A*b^2*c*x^2-9*B*b^3*x^2-5*A*b^3)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {17}{2}}}\,{d x} \]

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^(17/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{17/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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**(17/2),x)

[Out]

Timed out

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