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

Optimal. Leaf size=205 \[ -\frac {x \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c} \]

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Rubi [A]  time = 0.37, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1919, 1949, 12, 1914, 621, 206} \begin {gather*} \frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}-\frac {x \left (b^2-4 a c\right ) \left (5 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 )}{128 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

-((5*b^2 - 12*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(96*c^2) + (b*(15*b^2 - 52*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(
192*c^3*x) + (x*(b + 6*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(24*c) - ((b^2 - 4*a*c)*(5*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])])/(128*c^(7/2)*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 \sqrt {a x^2+b x^3+c x^4} \, dx &=\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}+\frac {\int \frac {x^2 \left (-2 a b-\frac {1}{2} \left (5 b^2-12 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{24 c}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\int \frac {x \left (-\frac {1}{2} a \left (5 b^2-12 a c\right )-\frac {1}{4} b \left (15 b^2-52 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{48 c^2}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}+\frac {\int -\frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) x}{8 \sqrt {a x^2+b x^3+c x^4}} \, dx}{48 c^3}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{128 c^3}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\left (\left (b^2-4 a c\right ) \left (5 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}{128 c^3 \sqrt {a x^2+b x^3+c x^4}}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\left (\left (b^2-4 a c\right ) \left (5 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 )}{64 c^3 \sqrt {a x^2+b x^3+c x^4}}\\ &=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\left (b^2-4 a c\right ) \left (5 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 )}{128 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 150, normalized size = 0.73 \begin {gather*} \frac {2 \sqrt {c} x (a+x (b+c x)) \left (b \left (8 c^2 x^2-52 a c\right )+24 c^2 x \left (a+2 c x^2\right )+15 b^3-10 b^2 c x\right )-3 x \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \sqrt {a+x (b+c x)} \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )}{384 c^{7/2} \sqrt {x^2 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*x*(a + x*(b + c*x))*(15*b^3 - 10*b^2*c*x + 24*c^2*x*(a + 2*c*x^2) + b*(-52*a*c + 8*c^2*x^2)) - 3*(5
*b^4 - 24*a*b^2*c + 16*a^2*c^2)*x*Sqrt[a + x*(b + c*x)]*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(384
*c^(7/2)*Sqrt[x^2*(a + x*(b + c*x))])

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IntegrateAlgebraic [A]  time = 0.58, size = 173, normalized size = 0.84 \begin {gather*} \frac {\left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \log \left (-2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}+b x+2 c x^2\right )}{128 c^{7/2}}+\frac {\log (x) \left (-16 a^2 c^2+24 a b^2 c-5 b^4\right )}{128 c^{7/2}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (-52 a b c+24 a c^2 x+15 b^3-10 b^2 c x+8 b c^2 x^2+48 c^3 x^3\right )}{192 c^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((15*b^3 - 52*a*b*c - 10*b^2*c*x + 24*a*c^2*x + 8*b*c^2*x^2 + 48*c^3*x^3)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(192*c^
3*x) + ((-5*b^4 + 24*a*b^2*c - 16*a^2*c^2)*Log[x])/(128*c^(7/2)) + ((5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*Log[b*x
+ 2*c*x^2 - 2*Sqrt[c]*Sqrt[a*x^2 + b*x^3 + c*x^4]])/(128*c^(7/2))

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fricas [A]  time = 0.68, size = 326, normalized size = 1.59 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} 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 (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{768 \, c^{4} x}, \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} 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 (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{384 \, c^{4} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*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*(48*c^4*x^3 + 8*b*c^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2
*c^2 - 12*a*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^4*x), 1/384*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*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*(48*c^4*x^3 + 8
*b*c^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^4*x)]

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giac [A]  time = 0.75, size = 230, normalized size = 1.12 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, x \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x - \frac {5 \, b^{2} c \mathrm {sgn}\relax (x) - 12 \, a c^{2} \mathrm {sgn}\relax (x)}{c^{3}}\right )} x + \frac {15 \, b^{3} \mathrm {sgn}\relax (x) - 52 \, a b c \mathrm {sgn}\relax (x)}{c^{3}}\right )} + \frac {{\left (5 \, b^{4} \mathrm {sgn}\relax (x) - 24 \, a b^{2} c \mathrm {sgn}\relax (x) + 16 \, a^{2} 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 )}{128 \, c^{\frac {7}{2}}} - \frac {{\left (15 \, b^{4} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 72 \, a b^{2} c \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 48 \, a^{2} c^{2} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 30 \, \sqrt {a} b^{3} \sqrt {c} - 104 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}\right )} \mathrm {sgn}\relax (x)}{384 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*x*sgn(x) + b*sgn(x)/c)*x - (5*b^2*c*sgn(x) - 12*a*c^2*sgn(x))/c^3)*x + (1
5*b^3*sgn(x) - 52*a*b*c*sgn(x))/c^3) + 1/128*(5*b^4*sgn(x) - 24*a*b^2*c*sgn(x) + 16*a^2*c^2*sgn(x))*log(abs(-2
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2) - 1/384*(15*b^4*log(abs(-b + 2*sqrt(a)*sqrt(c))) -
72*a*b^2*c*log(abs(-b + 2*sqrt(a)*sqrt(c))) + 48*a^2*c^2*log(abs(-b + 2*sqrt(a)*sqrt(c))) + 30*sqrt(a)*b^3*sqr
t(c) - 104*a^(3/2)*b*c^(3/2))*sgn(x)/c^(7/2)

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maple [A]  time = 0.01, size = 265, normalized size = 1.29 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (-48 a^{2} c^{3} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+72 a \,b^{2} c^{2} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-15 b^{4} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-48 \sqrt {c \,x^{2}+b x +a}\, a \,c^{\frac {7}{2}} x +60 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{\frac {5}{2}} x -24 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{\frac {5}{2}}+30 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{\frac {3}{2}}+96 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}} x -80 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,c^{\frac {5}{2}}\right )}{384 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {9}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(c*x^4+b*x^3+a*x^2)^(1/2)*(96*x*(c*x^2+b*x+a)^(3/2)*c^(7/2)-80*c^(5/2)*(c*x^2+b*x+a)^(3/2)*b-48*c^(7/2)*
(c*x^2+b*x+a)^(1/2)*x*a+60*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x*b^2-24*c^(5/2)*(c*x^2+b*x+a)^(1/2)*a*b+30*c^(3/2)*(c*
x^2+b*x+a)^(1/2)*b^3-48*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*a^2*c^3+72*ln(1/2*(2*c*x+b+2*(
c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*a*b^2*c^2-15*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*b^4*
c)/x/(c*x^2+b*x+a)^(1/2)/c^(9/2)

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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\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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, x)

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

[Out]

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

[Out]

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

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