3.63 \(\int \frac {1}{x (a x^2+b x^3+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=271 \[ \frac {5 b \left (7 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{9/2}}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 x^3 \left (b^2-4 a c\right )}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 x^4 \left (b^2-4 a c\right )}-\frac {\left (256 a^2 c^2-460 a b^2 c+105 b^4\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 x^2 \left (b^2-4 a c\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a x^2 \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

5/16*b*(-12*a*c+7*b^2)*arctanh(1/2*x*(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^(1/2))/a^(9/2)+2*(b*c*x-2*a*c+b^2)/
a/(-4*a*c+b^2)/x^2/(c*x^4+b*x^3+a*x^2)^(1/2)-1/3*(-16*a*c+7*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^2/(-4*a*c+b^2)/x^
4+1/12*b*(-116*a*c+35*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^3/(-4*a*c+b^2)/x^3-1/24*(256*a^2*c^2-460*a*b^2*c+105*b^
4)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^4/(-4*a*c+b^2)/x^2

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Rubi [A]  time = 0.45, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1924, 1951, 12, 1904, 206} \[ -\frac {\left (256 a^2 c^2-460 a b^2 c+105 b^4\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 x^2 \left (b^2-4 a c\right )}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 x^3 \left (b^2-4 a c\right )}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 x^4 \left (b^2-4 a c\right )}+\frac {5 b \left (7 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{9/2}}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a x^2 \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x^2 + b*x^3 + c*x^4)^(3/2)),x]

[Out]

(2*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*x^2*Sqrt[a*x^2 + b*x^3 + c*x^4]) - ((7*b^2 - 16*a*c)*Sqrt[a*x^2 + b
*x^3 + c*x^4])/(3*a^2*(b^2 - 4*a*c)*x^4) + (b*(35*b^2 - 116*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(12*a^3*(b^2 - 4
*a*c)*x^3) - ((105*b^4 - 460*a*b^2*c + 256*a^2*c^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(24*a^4*(b^2 - 4*a*c)*x^2) +
(5*b*(7*b^2 - 12*a*c)*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(16*a^(9/2))

Rule 12

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

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])

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 1924

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(m - q + 1
)*(b^2 - 2*a*c + b*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), x]
+ Dist[1/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - q)*(b^2*(m + p*q + (n - q)*(p + 1) + 1) - 2*a*c*(m + p
*q + 2*(n - q)*(p + 1) + 1) + b*c*(m + p*q + (n - q)*(2*p + 3) + 1)*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] && LtQ[p, -1] && RationalQ[m, q] && LtQ[m + p*q + 1, n - q]

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]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {2 \int \frac {-\frac {7 b^2}{2}+8 a c-3 b c x}{x^3 \sqrt {a x^2+b x^3+c x^4}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {2 \int \frac {-\frac {1}{4} b \left (35 b^2-116 a c\right )-c \left (7 b^2-16 a c\right ) x}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx}{3 a^2 \left (b^2-4 a c\right )}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 \left (b^2-4 a c\right ) x^3}-\frac {\int \frac {\frac {1}{8} \left (-105 b^4+460 a b^2 c-256 a^2 c^2\right )-\frac {1}{4} b c \left (35 b^2-116 a c\right ) x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{3 a^3 \left (b^2-4 a c\right )}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 \left (b^2-4 a c\right ) x^3}-\frac {\left (105 b^4-460 a b^2 c+256 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 \left (b^2-4 a c\right ) x^2}+\frac {\int -\frac {15 b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )}{16 \sqrt {a x^2+b x^3+c x^4}} \, dx}{3 a^4 \left (b^2-4 a c\right )}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 \left (b^2-4 a c\right ) x^3}-\frac {\left (105 b^4-460 a b^2 c+256 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 \left (b^2-4 a c\right ) x^2}-\frac {\left (5 b \left (7 b^2-12 a c\right )\right ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{16 a^4}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 \left (b^2-4 a c\right ) x^3}-\frac {\left (105 b^4-460 a b^2 c+256 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 \left (b^2-4 a c\right ) x^2}+\frac {\left (5 b \left (7 b^2-12 a c\right )\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 )}{8 a^4}\\ &=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x^2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{3 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{12 a^3 \left (b^2-4 a c\right ) x^3}-\frac {\left (105 b^4-460 a b^2 c+256 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^4 \left (b^2-4 a c\right ) x^2}+\frac {5 b \left (7 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 225, normalized size = 0.83 \[ \frac {2 \sqrt {a} \left (-32 a^4 c+8 a^3 \left (b^2+7 b c x+16 c^2 x^2\right )+2 a^2 x \left (-7 b^3-86 b^2 c x+244 b c^2 x^2+128 c^3 x^3\right )+5 a b^2 x^2 \left (7 b^2-106 b c x-92 c^2 x^2\right )+105 b^4 x^3 (b+c x)\right )-15 b x^3 \left (48 a^2 c^2-40 a b^2 c+7 b^4\right ) \sqrt {a+x (b+c x)} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{48 a^{9/2} x^2 \left (4 a c-b^2\right ) \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a*x^2 + b*x^3 + c*x^4)^(3/2)),x]

[Out]

(2*Sqrt[a]*(-32*a^4*c + 105*b^4*x^3*(b + c*x) + 5*a*b^2*x^2*(7*b^2 - 106*b*c*x - 92*c^2*x^2) + 8*a^3*(b^2 + 7*
b*c*x + 16*c^2*x^2) + 2*a^2*x*(-7*b^3 - 86*b^2*c*x + 244*b*c^2*x^2 + 128*c^3*x^3)) - 15*b*(7*b^4 - 40*a*b^2*c
+ 48*a^2*c^2)*x^3*Sqrt[a + x*(b + c*x)]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/(48*a^(9/2)*(-
b^2 + 4*a*c)*x^2*Sqrt[x^2*(a + x*(b + c*x))])

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fricas [A]  time = 1.02, size = 716, normalized size = 2.64 \[ \left [-\frac {15 \, {\left ({\left (7 \, b^{5} c - 40 \, a b^{3} c^{2} + 48 \, a^{2} b c^{3}\right )} x^{6} + {\left (7 \, b^{6} - 40 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2}\right )} x^{5} + {\left (7 \, a b^{5} - 40 \, a^{2} b^{3} c + 48 \, a^{3} b c^{2}\right )} x^{4}\right )} \sqrt {a} \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 (8 \, a^{4} b^{2} - 32 \, a^{5} c + {\left (105 \, a b^{4} c - 460 \, a^{2} b^{2} c^{2} + 256 \, a^{3} c^{3}\right )} x^{4} + {\left (105 \, a b^{5} - 530 \, a^{2} b^{3} c + 488 \, a^{3} b c^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{4} - 172 \, a^{3} b^{2} c + 128 \, a^{4} c^{2}\right )} x^{2} - 14 \, {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, {\left ({\left (a^{5} b^{2} c - 4 \, a^{6} c^{2}\right )} x^{6} + {\left (a^{5} b^{3} - 4 \, a^{6} b c\right )} x^{5} + {\left (a^{6} b^{2} - 4 \, a^{7} c\right )} x^{4}\right )}}, -\frac {15 \, {\left ({\left (7 \, b^{5} c - 40 \, a b^{3} c^{2} + 48 \, a^{2} b c^{3}\right )} x^{6} + {\left (7 \, b^{6} - 40 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2}\right )} x^{5} + {\left (7 \, a b^{5} - 40 \, a^{2} b^{3} c + 48 \, a^{3} b c^{2}\right )} x^{4}\right )} \sqrt {-a} \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 (8 \, a^{4} b^{2} - 32 \, a^{5} c + {\left (105 \, a b^{4} c - 460 \, a^{2} b^{2} c^{2} + 256 \, a^{3} c^{3}\right )} x^{4} + {\left (105 \, a b^{5} - 530 \, a^{2} b^{3} c + 488 \, a^{3} b c^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{4} - 172 \, a^{3} b^{2} c + 128 \, a^{4} c^{2}\right )} x^{2} - 14 \, {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, {\left ({\left (a^{5} b^{2} c - 4 \, a^{6} c^{2}\right )} x^{6} + {\left (a^{5} b^{3} - 4 \, a^{6} b c\right )} x^{5} + {\left (a^{6} b^{2} - 4 \, a^{7} c\right )} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*((7*b^5*c - 40*a*b^3*c^2 + 48*a^2*b*c^3)*x^6 + (7*b^6 - 40*a*b^4*c + 48*a^2*b^2*c^2)*x^5 + (7*a*b^5
- 40*a^2*b^3*c + 48*a^3*b*c^2)*x^4)*sqrt(a)*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x - 4*sqrt(c*x^4 + b*
x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) + 4*(8*a^4*b^2 - 32*a^5*c + (105*a*b^4*c - 460*a^2*b^2*c^2 + 256*a^3*c^
3)*x^4 + (105*a*b^5 - 530*a^2*b^3*c + 488*a^3*b*c^2)*x^3 + (35*a^2*b^4 - 172*a^3*b^2*c + 128*a^4*c^2)*x^2 - 14
*(a^3*b^3 - 4*a^4*b*c)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/((a^5*b^2*c - 4*a^6*c^2)*x^6 + (a^5*b^3 - 4*a^6*b*c)*x^
5 + (a^6*b^2 - 4*a^7*c)*x^4), -1/48*(15*((7*b^5*c - 40*a*b^3*c^2 + 48*a^2*b*c^3)*x^6 + (7*b^6 - 40*a*b^4*c + 4
8*a^2*b^2*c^2)*x^5 + (7*a*b^5 - 40*a^2*b^3*c + 48*a^3*b*c^2)*x^4)*sqrt(-a)*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*(8*a^4*b^2 - 32*a^5*c + (105*a*b^4*c - 460*a^2*b^2*c
^2 + 256*a^3*c^3)*x^4 + (105*a*b^5 - 530*a^2*b^3*c + 488*a^3*b*c^2)*x^3 + (35*a^2*b^4 - 172*a^3*b^2*c + 128*a^
4*c^2)*x^2 - 14*(a^3*b^3 - 4*a^4*b*c)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/((a^5*b^2*c - 4*a^6*c^2)*x^6 + (a^5*b^3
- 4*a^6*b*c)*x^5 + (a^6*b^2 - 4*a^7*c)*x^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^3 + a*x^2)^(3/2)*x), x)

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maple [A]  time = 0.01, size = 340, normalized size = 1.25 \[ -\frac {\left (c \,x^{2}+b x +a \right ) \left (-512 a^{\frac {7}{2}} c^{3} x^{4}+920 a^{\frac {5}{2}} b^{2} c^{2} x^{4}-210 a^{\frac {3}{2}} b^{4} c \,x^{4}+720 \sqrt {c \,x^{2}+b x +a}\, a^{3} b \,c^{2} x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-600 \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{3} c \,x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+105 \sqrt {c \,x^{2}+b x +a}\, a \,b^{5} x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-976 a^{\frac {7}{2}} b \,c^{2} x^{3}+1060 a^{\frac {5}{2}} b^{3} c \,x^{3}-210 a^{\frac {3}{2}} b^{5} x^{3}-256 a^{\frac {9}{2}} c^{2} x^{2}+344 a^{\frac {7}{2}} b^{2} c \,x^{2}-70 a^{\frac {5}{2}} b^{4} x^{2}-112 a^{\frac {9}{2}} b c x +28 a^{\frac {7}{2}} b^{3} x +64 a^{\frac {11}{2}} c -16 a^{\frac {9}{2}} b^{2}\right )}{48 \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right ) a^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^4+b*x^3+a*x^2)^(3/2),x)

[Out]

-1/48*(c*x^2+b*x+a)*(-512*a^(7/2)*x^4*c^3-256*a^(9/2)*x^2*c^2-976*a^(7/2)*x^3*b*c^2+920*a^(5/2)*x^4*b^2*c^2+64
*a^(11/2)*c-112*a^(9/2)*x*b*c+344*a^(7/2)*x^2*b^2*c+1060*a^(5/2)*x^3*b^3*c-210*a^(3/2)*x^4*b^4*c-16*a^(9/2)*b^
2+28*a^(7/2)*x*b^3-70*a^(5/2)*x^2*b^4-210*a^(3/2)*x^3*b^5+720*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*(c
*x^2+b*x+a)^(1/2)*x^3*a^3*b*c^2-600*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*(c*x^2+b*x+a)^(1/2)*x^3*a^2*
b^3*c+105*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*(c*x^2+b*x+a)^(1/2)*x^3*a*b^5)/(c*x^4+b*x^3+a*x^2)^(3/
2)/a^(11/2)/(4*a*c-b^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^3 + a*x^2)^(3/2)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^2 + b*x^3 + c*x^4)^(3/2)),x)

[Out]

int(1/(x*(a*x^2 + b*x^3 + c*x^4)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(1/(x*(x**2*(a + b*x + c*x**2))**(3/2)), x)

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