\(\int \frac {x^5}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [57]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 153 \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1937, 1963, 12, 1928, 635, 212} \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c x \left (b^2-4 a c\right )} \]

[In]

Int[x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

(2*x^2*(2*a + b*x))/((b^2 - 4*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]) - (2*b*Sqrt[a*x^2 + b*x^3 + c*x^4])/(c*(b^2 -
4*a*c)*x) + (x*Sqrt[a + b*x + c*x^2]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(c^(3/2)*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 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 1937

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

Rule 1963

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*} \text {integral}& = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 \int \frac {x (2 a+b x)}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{b^2-4 a c} \\ & = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {2 \int \frac {\left (b^2-4 a c\right ) x}{2 \sqrt {a x^2+b x^3+c x^4}} \, dx}{c \left (b^2-4 a c\right )} \\ & = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {\int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{c} \\ & = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {\left (x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {\left (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 )}{c \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right ) x}+\frac {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 )}{c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.73 \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {x \left (2 \sqrt {c} \left (-a b-b^2 x+2 a c x\right )+\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{c^{3/2} \left (-b^2+4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

Integrate[x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.55

method result size
pseudoelliptic \(-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}+\frac {b \left (b x +2 a \right )}{\sqrt {c \,x^{2}+b x +a}\, c \left (4 a c -b^{2}\right )}+\frac {\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right )}{c^{\frac {3}{2}}}\) \(84\)
default \(-\frac {x^{3} \left (c \,x^{2}+b x +a \right ) \left (4 c^{\frac {5}{2}} a x -2 c^{\frac {3}{2}} b^{2} x -2 c^{\frac {3}{2}} a b -4 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, a \,c^{2}+\ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, b^{2} c \right )}{c^{\frac {5}{2}} \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )}\) \(166\)

[In]

int(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-x/c/(c*x^2+b*x+a)^(1/2)+b*(b*x+2*a)/(c*x^2+b*x+a)^(1/2)/c/(4*a*c-b^2)+1/c^(3/2)*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1
/2)+2*c*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.71 \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {c} \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{2 \, {\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x}\right ] \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**5/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**5/(x**2*(a + b*x + c*x**2))**(3/2), x)

Maxima [F]

\[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{5}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^5/(c*x^4 + b*x^3 + a*x^2)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07 \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {{\left (b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{b^{2} c^{\frac {3}{2}} - 4 \, a c^{\frac {5}{2}}} - \frac {2 \, {\left (\frac {a b}{b^{2} c \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} \mathrm {sgn}\left (x\right )} + \frac {{\left (b^{2} - 2 \, a c\right )} x}{b^{2} c \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} \mathrm {sgn}\left (x\right )}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

(b^2*log(abs(b - 2*sqrt(a)*sqrt(c))) - 4*a*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 2*sqrt(a)*b*sqrt(c))*sgn(x)/(b^
2*c^(3/2) - 4*a*c^(5/2)) - 2*(a*b/(b^2*c*sgn(x) - 4*a*c^2*sgn(x)) + (b^2 - 2*a*c)*x/(b^2*c*sgn(x) - 4*a*c^2*sg
n(x)))/sqrt(c*x^2 + b*x + a) - log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(3/2)*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \]

[In]

int(x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2),x)

[Out]

int(x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2), x)