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

Optimal. Leaf size=153 \[ \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c x \left (b^2-4 a c\right )}+\frac {x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[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]  time = 0.18, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1923, 1949, 12, 1914, 621, 206} \[ \frac {2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c x \left (b^2-4 a c\right )}+\frac {x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[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 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 1923

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] + Dis
t[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 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 \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 \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 ) \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 )}{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]  time = 0.12, size = 112, normalized size = 0.73 \[ -\frac {x \left (2 \sqrt {c} \left (-a b+2 a c x+b^2 (-x)\right )+\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{c^{3/2} \left (4 a c-b^2\right ) \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[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))]))

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 414, normalized size = 2.71 \[ \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {c} \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{2 \, {\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c + {\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

________________________________________________________________________________________

giac [A]  time = 0.98, size = 110, normalized size = 0.72 \[ -\frac {2 \, {\left (\frac {a b c}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x} + \frac {b^{2} c - 2 \, a c^{2}}{b^{2} c^{2} - 4 \, a c^{3}}\right )}}{\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}}} - \frac {2 \, \arctan \left (\frac {\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}} - \frac {\sqrt {a}}{x}}{\sqrt {-c}}\right )}{\sqrt {-c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*b*c/((b^2*c^2 - 4*a*c^3)*x) + (b^2*c - 2*a*c^2)/(b^2*c^2 - 4*a*c^3))/sqrt(c + b/x + a/x^2) - 2*arctan((s
qrt(c + b/x + a/x^2) - sqrt(a)/x)/sqrt(-c))/(sqrt(-c)*c)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 166, normalized size = 1.08 \[ \frac {\left (c \,x^{2}+b x +a \right ) \left (-4 a \,c^{\frac {5}{2}} x +2 b^{2} c^{\frac {3}{2}} x +4 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-\sqrt {c \,x^{2}+b x +a}\, b^{2} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+2 a b \,c^{\frac {3}{2}}\right ) x^{3}}{\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right ) c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*(c*x^2+b*x+a)/c^(5/2)*(-4*c^(5/2)*x*a+2*c^(3/2)*x*b^2+4*(c*x^2+b*x+a)^(1/2)*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a
)^(1/2)*c^(1/2))/c^(1/2))*a*c^2-(c*x^2+b*x+a)^(1/2)*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*b^
2*c+2*c^(3/2)*a*b)/(c*x^4+b*x^3+a*x^2)^(3/2)/(4*a*c-b^2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________