3.1.29 \(\int x^2 \sqrt {a x^2+b x^3+c x^4} \, dx\)

Optimal. Leaf size=257 \[ -\frac {\left (256 a^2 c^2-460 a b^2 c+105 b^4\right ) \sqrt {a x^2+b x^3+c x^4}}{1920 c^4 x}+\frac {b x \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^3}-\frac {x \left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c} \]

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Rubi [A]  time = 0.59, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1919, 1949, 12, 1914, 621, 206} \begin {gather*} -\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}}{1920 c^4 x}-\frac {x \left (7 b^2-16 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^3}+\frac {b x \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

(b*(35*b^2 - 116*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(960*c^3) - ((105*b^4 - 460*a*b^2*c + 256*a^2*c^2)*Sqrt[a*x
^2 + b*x^3 + c*x^4])/(1920*c^4*x) - ((7*b^2 - 16*a*c)*x*Sqrt[a*x^2 + b*x^3 + c*x^4])/(240*c^2) + (x^2*(b + 8*c
*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(40*c) + (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(9/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])

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 621

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

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1919

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - n + q
+ 1)*(b*(n - q)*p + c*(m + p*q + (n - q)*(2*p - 1) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^p)/(c*(m +
p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1)), x] + Dist[((n - q)*p)/(c*(m + p*(2*n - q) + 1)*(m + p*q +
(n - q)*(2*p - 1) + 1)), Int[x^(m - (n - 2*q))*Simp[-(a*b*(m + p*q - n + q + 1)) + (2*a*c*(m + p*q + (n - q)*
(2*p - 1) + 1) - b^2*(m + p*q + (n - q)*(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}, 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] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) + 1, 0] && NeQ[m + p*
q + (n - q)*(2*p - 1) + 1, 0]

Rule 1949

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(B*x^(m - n + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(c*(m + p*q + (n - q)*(2*p + 1) + 1)),
x] - Dist[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m - n + q)*Simp[a*B*(m + p*q - n + q + 1) + (b*B*(m
+ p*q + (n - q)*p + 1) - 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, 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] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (
n - q)*(2*p + 1) + 1, 0]

Rubi steps

\begin {align*} \int x^2 \sqrt {a x^2+b x^3+c x^4} \, dx &=\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {\int \frac {x^3 \left (-3 a b-\frac {1}{2} \left (7 b^2-16 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{40 c}\\ &=-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}-\frac {\int \frac {x^2 \left (-a \left (7 b^2-16 a c\right )-\frac {1}{4} b \left (35 b^2-116 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{120 c^2}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^3}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {\int \frac {x \left (-\frac {1}{4} a b \left (35 b^2-116 a c\right )-\frac {1}{8} \left (105 b^4-460 a b^2 c+256 a^2 c^2\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{240 c^3}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^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}}{1920 c^4 x}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}-\frac {\int -\frac {15 b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) x}{16 \sqrt {a x^2+b x^3+c x^4}} \, dx}{240 c^4}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^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}}{1920 c^4 x}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{256 c^4}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^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}}{1920 c^4 x}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^4 \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^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}}{1920 c^4 x}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) x \sqrt {a+b x+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{128 c^4 \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {b \left (35 b^2-116 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{960 c^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}}{1920 c^4 x}-\frac {\left (7 b^2-16 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}{240 c^2}+\frac {x^2 (b+8 c x) \sqrt {a x^2+b x^3+c x^4}}{40 c}+\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2} \sqrt {a x^2+b x^3+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 180, normalized size = 0.70 \begin {gather*} \frac {15 x \left (48 a^2 b c^2-40 a b^3 c+7 b^5\right ) \sqrt {a+x (b+c x)} \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )+2 \sqrt {c} x (a+x (b+c x)) \left (128 c^2 \left (-2 a^2+a c x^2+3 c^2 x^4\right )+4 b^2 c \left (115 a-14 c x^2\right )+8 b c^2 x \left (6 c x^2-29 a\right )-105 b^4+70 b^3 c x\right )}{3840 c^{9/2} \sqrt {x^2 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

(2*Sqrt[c]*x*(a + x*(b + c*x))*(-105*b^4 + 70*b^3*c*x + 4*b^2*c*(115*a - 14*c*x^2) + 8*b*c^2*x*(-29*a + 6*c*x^
2) + 128*c^2*(-2*a^2 + a*c*x^2 + 3*c^2*x^4)) + 15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*x*Sqrt[a + x*(b + c*x)]*
Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(3840*c^(9/2)*Sqrt[x^2*(a + x*(b + c*x))])

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IntegrateAlgebraic [A]  time = 0.84, size = 211, normalized size = 0.82 \begin {gather*} \frac {\log (x) \left (48 a^2 b c^2-40 a b^3 c+7 b^5\right )}{256 c^{9/2}}+\frac {\left (-48 a^2 b c^2+40 a b^3 c-7 b^5\right ) \log \left (-2 c^{9/2} \sqrt {a x^2+b x^3+c x^4}+b c^4 x+2 c^5 x^2\right )}{256 c^{9/2}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (-256 a^2 c^2+460 a b^2 c-232 a b c^2 x+128 a c^3 x^2-105 b^4+70 b^3 c x-56 b^2 c^2 x^2+48 b c^3 x^3+384 c^4 x^4\right )}{1920 c^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

(Sqrt[a*x^2 + b*x^3 + c*x^4]*(-105*b^4 + 460*a*b^2*c - 256*a^2*c^2 + 70*b^3*c*x - 232*a*b*c^2*x - 56*b^2*c^2*x
^2 + 128*a*c^3*x^2 + 48*b*c^3*x^3 + 384*c^4*x^4))/(1920*c^4*x) + ((7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*Log[x])/
(256*c^(9/2)) + ((-7*b^5 + 40*a*b^3*c - 48*a^2*b*c^2)*Log[b*c^4*x + 2*c^5*x^2 - 2*c^(9/2)*Sqrt[a*x^2 + b*x^3 +
c*x^4]])/(256*c^(9/2))

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fricas [A]  time = 1.23, size = 390, normalized size = 1.52 \begin {gather*} \left [\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{7680 \, c^{5} x}, -\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - 2 \, {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{3840 \, c^{5} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(c)*x*log(-(8*c^2*x^3 + 8*b*c*x^2 + 4*sqrt(c*x^4 + b*x^3 +
a*x^2)*(2*c*x + b)*sqrt(c) + (b^2 + 4*a*c)*x)/x) + 4*(384*c^5*x^4 + 48*b*c^4*x^3 - 105*b^4*c + 460*a*b^2*c^2
- 256*a^2*c^3 - 8*(7*b^2*c^3 - 16*a*c^4)*x^2 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c
^5*x), -1/3840*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*
x + b)*sqrt(-c)/(c^2*x^3 + b*c*x^2 + a*c*x)) - 2*(384*c^5*x^4 + 48*b*c^4*x^3 - 105*b^4*c + 460*a*b^2*c^2 - 256
*a^2*c^3 - 8*(7*b^2*c^3 - 16*a*c^4)*x^2 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^5*x)
]

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giac [A]  time = 0.87, size = 283, normalized size = 1.10 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x - \frac {7 \, b^{2} c^{2} \mathrm {sgn}\relax (x) - 16 \, a c^{3} \mathrm {sgn}\relax (x)}{c^{4}}\right )} x + \frac {35 \, b^{3} c \mathrm {sgn}\relax (x) - 116 \, a b c^{2} \mathrm {sgn}\relax (x)}{c^{4}}\right )} x - \frac {105 \, b^{4} \mathrm {sgn}\relax (x) - 460 \, a b^{2} c \mathrm {sgn}\relax (x) + 256 \, a^{2} c^{2} \mathrm {sgn}\relax (x)}{c^{4}}\right )} - \frac {{\left (7 \, b^{5} \mathrm {sgn}\relax (x) - 40 \, a b^{3} c \mathrm {sgn}\relax (x) + 48 \, a^{2} b c^{2} \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} + \frac {{\left (105 \, b^{5} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 600 \, a b^{3} c \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 720 \, a^{2} b c^{2} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 210 \, \sqrt {a} b^{4} \sqrt {c} - 920 \, a^{\frac {3}{2}} b^{2} c^{\frac {3}{2}} + 512 \, a^{\frac {5}{2}} c^{\frac {5}{2}}\right )} \mathrm {sgn}\relax (x)}{3840 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(8*x*sgn(x) + b*sgn(x)/c)*x - (7*b^2*c^2*sgn(x) - 16*a*c^3*sgn(x))/c^4)*
x + (35*b^3*c*sgn(x) - 116*a*b*c^2*sgn(x))/c^4)*x - (105*b^4*sgn(x) - 460*a*b^2*c*sgn(x) + 256*a^2*c^2*sgn(x))
/c^4) - 1/256*(7*b^5*sgn(x) - 40*a*b^3*c*sgn(x) + 48*a^2*b*c^2*sgn(x))*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))*sqrt(c) - b))/c^(9/2) + 1/3840*(105*b^5*log(abs(-b + 2*sqrt(a)*sqrt(c))) - 600*a*b^3*c*log(abs(-b + 2*
sqrt(a)*sqrt(c))) + 720*a^2*b*c^2*log(abs(-b + 2*sqrt(a)*sqrt(c))) + 210*sqrt(a)*b^4*sqrt(c) - 920*a^(3/2)*b^2
*c^(3/2) + 512*a^(5/2)*c^(5/2))*sgn(x)/c^(9/2)

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maple [A]  time = 0.01, size = 310, normalized size = 1.21 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (720 a^{2} b \,c^{3} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-600 a \,b^{3} c^{2} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+105 b^{5} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+720 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{\frac {7}{2}} x -420 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{\frac {5}{2}} x +768 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {9}{2}} x^{2}+360 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{\frac {5}{2}}-210 \sqrt {c \,x^{2}+b x +a}\, b^{4} c^{\frac {3}{2}}-672 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,c^{\frac {7}{2}} x -512 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,c^{\frac {7}{2}}+560 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} c^{\frac {5}{2}}\right )}{3840 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {11}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3840*(c*x^4+b*x^3+a*x^2)^(1/2)*(768*x^2*(c*x^2+b*x+a)^(3/2)*c^(9/2)-672*c^(7/2)*(c*x^2+b*x+a)^(3/2)*x*b-512*
c^(7/2)*(c*x^2+b*x+a)^(3/2)*a+560*c^(5/2)*(c*x^2+b*x+a)^(3/2)*b^2+720*c^(7/2)*(c*x^2+b*x+a)^(1/2)*x*a*b-420*c^
(5/2)*(c*x^2+b*x+a)^(1/2)*x*b^3+360*c^(5/2)*(c*x^2+b*x+a)^(1/2)*a*b^2-210*c^(3/2)*(c*x^2+b*x+a)^(1/2)*b^4+720*
ln(1/2*(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)/c^(1/2))*a^2*b*c^3-600*ln(1/2*(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*
c*x+b)/c^(1/2))*a*b^3*c^2+105*ln(1/2*(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)/c^(1/2))*b^5*c)/x/(c*x^2+b*x+a)^(
1/2)/c^(11/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {c x^{4} + b x^{3} + a x^{2}} x^{2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\sqrt {c\,x^4+b\,x^3+a\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt {x^{2} \left (a + b x + c x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x**2*(a + b*x + c*x**2)), x)

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