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

Optimal. Leaf size=165 \[ \frac {3 x \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3} \]

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Rubi [A]  time = 0.13, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1918, 1914, 621, 206} \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {3 x \left (b^2-4 a c\right )^2 \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^{5/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(64*c^2*x) + ((b + 2*c*x)*(a*x^2 + b*x^3 + c*x^4)^(
3/2))/(8*c*x^3) + (3*(b^2 - 4*a*c)^2*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^(5/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])

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 1918

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - n + q
+ 1)*(b + 2*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^p)/(2*c*(n - q)*(2*p + 1)), x] - Dist[(p*(b^2 - 4*a*c
))/(2*c*(2*p + 1)), Int[x^(m + q)*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && Eq
Q[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] && EqQ[m + p*q + 1, n - q]

Rubi steps

\begin {align*} \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx &=\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx}{16 c}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{128 c^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}+\frac {\left (3 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2 \sqrt {a x^2+b x^3+c x^4}}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}+\frac {\left (3 \left (b^2-4 a c\right )^2 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^2 \sqrt {a x^2+b x^3+c x^4}}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}+\frac {3 \left (b^2-4 a c\right )^2 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^{5/2} \sqrt {a x^2+b x^3+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 132, normalized size = 0.80 \begin {gather*} \frac {x \sqrt {a+x (b+c x)} \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{128 c^{5/2} \sqrt {x^2 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + x*(b + c*x)]*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2))
+ 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(128*c^(5/2)*Sqrt[x^2*(a + x*(b
+ c*x))])

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IntegrateAlgebraic [A]  time = 1.58, size = 143, normalized size = 0.87 \begin {gather*} \frac {\left (20 a b c+40 a c^2 x-3 b^3+2 b^2 c x+24 b c^2 x^2+16 c^3 x^3\right ) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}-\frac {3 \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a} x-\sqrt {a x^2+b x^3+c x^4}}\right )}{64 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*b^3 + 20*a*b*c + 2*b^2*c*x + 40*a*c^2*x + 24*b*c^2*x^2 + 16*c^3*x^3)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(64*c^2
*x) - (3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*ArcTanh[(Sqrt[c]*x^2)/(Sqrt[a]*x - Sqrt[a*x^2 + b*x^3 + c*x^4])])/(64*
c^(5/2))

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fricas [A]  time = 1.31, size = 320, normalized size = 1.94 \begin {gather*} \left [\frac {3 \, {\left (b^{4} - 8 \, 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 (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{256 \, c^{3} x}, -\frac {3 \, {\left (b^{4} - 8 \, 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 (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{128 \, c^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/256*(3*(b^4 - 8*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*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2
+ 20*a*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^3*x), -1/128*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*x*arcta
n(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*(16*c^4*x^3 + 24*b*c^3
*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^3*x)]

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giac [A]  time = 1.01, size = 232, normalized size = 1.41 \begin {gather*} \frac {1}{64} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c x \mathrm {sgn}\relax (x) + 3 \, b \mathrm {sgn}\relax (x)\right )} x + \frac {b^{2} c^{2} \mathrm {sgn}\relax (x) + 20 \, a c^{3} \mathrm {sgn}\relax (x)}{c^{3}}\right )} x - \frac {3 \, b^{3} c \mathrm {sgn}\relax (x) - 20 \, a b c^{2} \mathrm {sgn}\relax (x)}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} \mathrm {sgn}\relax (x) - 8 \, 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 {5}{2}}} + \frac {{\left (3 \, b^{4} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 24 \, 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 ) + 6 \, \sqrt {a} b^{3} \sqrt {c} - 40 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}\right )} \mathrm {sgn}\relax (x)}{128 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c*x*sgn(x) + 3*b*sgn(x))*x + (b^2*c^2*sgn(x) + 20*a*c^3*sgn(x))/c^3)*x - (
3*b^3*c*sgn(x) - 20*a*b*c^2*sgn(x))/c^3) - 3/128*(b^4*sgn(x) - 8*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^(5/2) + 1/128*(3*b^4*log(abs(-b + 2*sqrt(a)*sqrt(c))) -
24*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))) + 6*sqrt(a)*b^3*sqrt
(c) - 40*a^(3/2)*b*c^(3/2))*sgn(x)/c^(5/2)

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maple [A]  time = 0.00, size = 265, normalized size = 1.61 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{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 )-24 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 )+3 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 -12 \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}}-6 \sqrt {c \,x^{2}+b x +a}\, b^{3} c^{\frac {3}{2}}+32 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}} x +16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,c^{\frac {5}{2}}\right )}{128 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(c*x^4+b*x^3+a*x^2)^(3/2)*(32*(c*x^2+b*x+a)^(3/2)*c^(7/2)*x+16*(c*x^2+b*x+a)^(3/2)*b*c^(5/2)+48*(c*x^2+b
*x+a)^(1/2)*a*c^(7/2)*x-12*(c*x^2+b*x+a)^(1/2)*b^2*c^(5/2)*x+24*(c*x^2+b*x+a)^(1/2)*a*b*c^(5/2)-6*(c*x^2+b*x+a
)^(1/2)*b^3*c^(3/2)+48*a^2*c^3*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))-24*a*b^2*c^2*ln(1/2*(2*
c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))+3*b^4*c*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2)))
/x^3/(c*x^2+b*x+a)^(3/2)/c^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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