\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^3} \, dx\) [42]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}+\frac {3 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

1/8*(2*c*x+b)*(c*x^4+b*x^3+a*x^2)^(3/2)/c/x^3+3/128*(-4*a*c+b^2)^2*x*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+
a)^(1/2))*(c*x^2+b*x+a)^(1/2)/c^(5/2)/(c*x^4+b*x^3+a*x^2)^(1/2)-3/64*(-4*a*c+b^2)*(2*c*x+b)*(c*x^4+b*x^3+a*x^2
)^(1/2)/c^2/x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1932, 1928, 635, 212} \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3 x \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{64 c^2 x}+\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 1928

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 1932

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*} \text {integral}& = \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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\frac {\left (x^2 (a+x (b+c x))\right )^{3/2} \left (\frac {\sqrt {c} (b+2 c x) \left (-3 b^2+8 b c x+4 c \left (5 a+2 c x^2\right )\right )}{a+x (b+c x)}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{(a+x (b+c x))^{3/2}}\right )}{64 c^{5/2} x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(\frac {\frac {3 \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right )}{8}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, \left (b \left (\frac {b x}{10}+a \right ) c^{\frac {3}{2}}+\left (\frac {6}{5} b \,x^{2}+2 a x \right ) c^{\frac {5}{2}}-\frac {3 \sqrt {c}\, b^{3}}{20}+\frac {4 c^{\frac {7}{2}} x^{3}}{5}\right )}{16}}{c^{\frac {5}{2}}}\)	\(100\)
risch	\(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+40 a \,c^{2} x +2 b^{2} c x +20 a b c -3 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{64 c^{2} x}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{128 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(148\)
default	\(\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (32 x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}}+16 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b +48 c^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, a x -12 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{2} x +24 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a b -6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{3}+48 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{2} c^{3}-24 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,b^{2} c^{2}+3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{4} c \right )}{128 x^{3} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\)	\(265\)

[In]

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

[Out]

3/8/c^(5/2)*((a*c-1/4*b^2)^2*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)+5/6*(c*x^2+b*x+a)^(1/2)*(b*(1/10*b*x+a)
*c^(3/2)+(6/5*b*x^2+2*a*x)*c^(5/2)-3/20*c^(1/2)*b^3+4/5*c^(7/2)*x^3))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{256 \, c^{3} x}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{128 \, c^{3} x}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{3}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\int { \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{3}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{64} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c x \mathrm {sgn}\left (x\right ) + 3 \, b \mathrm {sgn}\left (x\right )\right )} x + \frac {b^{2} c^{2} \mathrm {sgn}\left (x\right ) + 20 \, a c^{3} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} x - \frac {3 \, b^{3} c \mathrm {sgn}\left (x\right ) - 20 \, a b c^{2} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} \mathrm {sgn}\left (x\right ) - 8 \, a b^{2} c \mathrm {sgn}\left (x\right ) + 16 \, a^{2} c^{2} \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} + \frac {{\left (3 \, b^{4} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 24 \, a b^{2} c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 48 \, a^{2} c^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b^{3} \sqrt {c} - 40 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{128 \, c^{\frac {5}{2}}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^3} \,d x \]

[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)