3.48 \(\int \frac{(a x^2+b x^3+c x^4)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=249 \[ -\frac{\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}}-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5} \]

[Out]

-((b^2 - 8*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(80*a*x^4) + (b*(5*b^2 - 28*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(32
0*a^2*x^3) - ((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(640*a^3*x^2) - (3*(b + 4*c*x)
*Sqrt[a*x^2 + b*x^3 + c*x^4])/(40*x^5) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(5*x^8) + (3*b*(b^2 - 4*a*c)^2*ArcTanh[
(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(256*a^(7/2))

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Rubi [A]  time = 0.503853, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1920, 1941, 1951, 12, 1904, 206} \[ -\frac{\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}}-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 8*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(80*a*x^4) + (b*(5*b^2 - 28*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(32
0*a^2*x^3) - ((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(640*a^3*x^2) - (3*(b + 4*c*x)
*Sqrt[a*x^2 + b*x^3 + c*x^4])/(40*x^5) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(5*x^8) + (3*b*(b^2 - 4*a*c)^2*ArcTanh[
(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(256*a^(7/2))

Rule 1920

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a*
x^q + b*x^n + c*x^(2*n - q))^p)/(m + p*q + 1), x] - Dist[((n - q)*p)/(m + p*q + 1), Int[x^(m + n)*(b + 2*c*x^(
n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n -
q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q + 1, -
(n - q) + 1] && NeQ[m + p*q + 1, 0]

Rule 1941

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(x^(m + 1)*(A*(m + p*q + (n - q)*(2*p + 1) + 1) + B*(m + p*q + 1)*x^(n - q))*(a*x^q + b*x^n + c*x
^(2*n - q))^p)/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)), x] + Dist[((n - q)*p)/((m + p*q + 1)*(m + p*
q + (n - q)*(2*p + 1) + 1)), Int[x^(n + m)*Simp[2*a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(2*p + 1) + 1) +
(b*B*(m + p*q + 1) - 2*A*c*(m + p*q + (n - q)*(2*p + 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p
- 1), x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4
*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0] && NeQ
[m + p*q + (n - q)*(2*p + 1) + 1, 0]

Rule 1951

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(A*x^(m - q + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(m + p*q + 1)), x] + Dist[1/(a*(m +
p*q + 1)), Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p + 1) + 1) - A*c*(m + p*q + 2*(
n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && Eq
Q[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] &&
((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*
q + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1904

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, (x*(2*a + b*x^(n - 2)))/Sqrt[a*x^2 + b*x^n + c*x^r]], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]

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{\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx &=-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac{3}{10} \int \frac{(b+2 c x) \sqrt{a x^2+b x^3+c x^4}}{x^6} \, dx\\ &=-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac{3}{160} \int \frac{2 \left (b^2-8 a c\right )-4 b c x}{x^3 \sqrt{a x^2+b x^3+c x^4}} \, dx\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac{\int \frac{b \left (5 b^2-28 a c\right )+4 c \left (b^2-8 a c\right ) x}{x^2 \sqrt{a x^2+b x^3+c x^4}} \, dx}{160 a}\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac{\int \frac{\frac{1}{2} \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )+b c \left (5 b^2-28 a c\right ) x}{x \sqrt{a x^2+b x^3+c x^4}} \, dx}{320 a^2}\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac{\int \frac{15 b \left (b^2-4 a c\right )^2}{4 \sqrt{a x^2+b x^3+c x^4}} \, dx}{320 a^3}\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac{\left (3 b \left (b^2-4 a c\right )^2\right ) \int \frac{1}{\sqrt{a x^2+b x^3+c x^4}} \, dx}{256 a^3}\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac{\left (3 b \left (b^2-4 a c\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{x (2 a+b x)}{\sqrt{a x^2+b x^3+c x^4}}\right )}{128 a^3}\\ &=-\frac{\left (b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{80 a x^4}+\frac{b \left (5 b^2-28 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac{3 (b+4 c x) \sqrt{a x^2+b x^3+c x^4}}{40 x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.176466, size = 177, normalized size = 0.71 \[ \frac{\sqrt{x^2 (a+x (b+c x))} \left (15 b x^5 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{a} \sqrt{a+x (b+c x)} \left (8 a^2 x^2 \left (b^2+7 b c x+16 c^2 x^2\right )+16 a^3 x (11 b+16 c x)+128 a^4-10 a b^2 x^3 (b+10 c x)+15 b^4 x^4\right )\right )}{1280 a^{7/2} x^6 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(a + x*(b + c*x))]*(-2*Sqrt[a]*Sqrt[a + x*(b + c*x)]*(128*a^4 + 15*b^4*x^4 - 10*a*b^2*x^3*(b + 10*c*
x) + 16*a^3*x*(11*b + 16*c*x) + 8*a^2*x^2*(b^2 + 7*b*c*x + 16*c^2*x^2)) + 15*b*(b^2 - 4*a*c)^2*x^5*ArcTanh[(2*
a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])]))/(1280*a^(7/2)*x^6*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.008, size = 534, normalized size = 2.1 \begin{align*} -{\frac{1}{1280\,{x}^{8}{a}^{5}} \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -240\,{c}^{2}{a}^{7/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{5}b-120\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}{c}^{2}{x}^{6}a{b}^{2}+120\,c{a}^{5/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{5}{b}^{3}+80\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}{c}^{2}{x}^{5}{a}^{2}b+10\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}c{x}^{6}{b}^{4}-120\,\sqrt{c{x}^{2}+bx+a}{c}^{2}{x}^{6}{a}^{2}{b}^{2}+120\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}c{x}^{4}a{b}^{2}-100\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}c{x}^{5}a{b}^{3}+240\,\sqrt{c{x}^{2}+bx+a}{c}^{2}{x}^{5}{a}^{3}b+30\,\sqrt{c{x}^{2}+bx+a}c{x}^{6}a{b}^{4}-15\,{a}^{3/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{5}{b}^{5}-80\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}c{x}^{3}{a}^{2}b-10\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}{x}^{4}{b}^{4}+10\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{5}{b}^{5}-180\,\sqrt{c{x}^{2}+bx+a}c{x}^{5}{a}^{2}{b}^{3}-20\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}{x}^{3}a{b}^{3}+30\,\sqrt{c{x}^{2}+bx+a}{x}^{5}a{b}^{5}+80\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}{x}^{2}{a}^{2}{b}^{2}-160\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}x{a}^{3}b+256\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}{a}^{4} \right ) \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(c*x^4+b*x^3+a*x^2)^(3/2)*(-240*c^2*a^(7/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*x^5*b-120*(c
*x^2+b*x+a)^(3/2)*c^2*x^6*a*b^2+120*c*a^(5/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*x^5*b^3+80*(c*x^2+
b*x+a)^(3/2)*c^2*x^5*a^2*b+10*(c*x^2+b*x+a)^(3/2)*c*x^6*b^4-120*(c*x^2+b*x+a)^(1/2)*c^2*x^6*a^2*b^2+120*(c*x^2
+b*x+a)^(5/2)*c*x^4*a*b^2-100*(c*x^2+b*x+a)^(3/2)*c*x^5*a*b^3+240*(c*x^2+b*x+a)^(1/2)*c^2*x^5*a^3*b+30*(c*x^2+
b*x+a)^(1/2)*c*x^6*a*b^4-15*a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*x^5*b^5-80*(c*x^2+b*x+a)^(5/
2)*c*x^3*a^2*b-10*(c*x^2+b*x+a)^(5/2)*x^4*b^4+10*(c*x^2+b*x+a)^(3/2)*x^5*b^5-180*(c*x^2+b*x+a)^(1/2)*c*x^5*a^2
*b^3-20*(c*x^2+b*x+a)^(5/2)*x^3*a*b^3+30*(c*x^2+b*x+a)^(1/2)*x^5*a*b^5+80*(c*x^2+b*x+a)^(5/2)*x^2*a^2*b^2-160*
(c*x^2+b*x+a)^(5/2)*x*a^3*b+256*(c*x^2+b*x+a)^(5/2)*a^4)/x^8/(c*x^2+b*x+a)^(3/2)/a^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}{x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.71046, size = 898, normalized size = 3.61 \begin{align*} \left [\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{a} x^{6} \log \left (-\frac{8 \, a b x^{2} +{\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (b x + 2 \, a\right )} \sqrt{a}}{x^{3}}\right ) - 4 \,{\left (176 \, a^{4} b x + 128 \, a^{5} +{\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{4} - 2 \,{\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{3} + 8 \,{\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{3} + a x^{2}}}{2560 \, a^{4} x^{6}}, -\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{-a} x^{6} \arctan \left (\frac{\sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (b x + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \,{\left (176 \, a^{4} b x + 128 \, a^{5} +{\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{4} - 2 \,{\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{3} + 8 \,{\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{3} + a x^{2}}}{1280 \, a^{4} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^6*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x + 4*sqr
t(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) - 4*(176*a^4*b*x + 128*a^5 + (15*a*b^4 - 100*a^2*b^2*c + 12
8*a^3*c^2)*x^4 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^
4*x^6), -1/1280*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-a)*x^6*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x
+ 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) + 2*(176*a^4*b*x + 128*a^5 + (15*a*b^4 - 100*a^2*b^2*c + 128*a^3*
c^2)*x^4 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^4*x^6)
]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**9,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

Timed out