\(\int \frac {x^7}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [55]

Optimal result

Integrand size = 24, antiderivative size = 262 \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {3 \left (5 b^2-4 a c\right ) x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

2*x^4*(b*x+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^3+a*x^2)^(1/2)+3/8*(-4*a*c+5*b^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^(7/2)/(c*x^4+b*x^3+a*x^2)^(1/2)+1/2*(-12*a*c+5*b^2)*(c*x^4+b*x^3+a*x^2)
^(1/2)/c^2/(-4*a*c+b^2)-1/4*b*(-52*a*c+15*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/c^3/(-4*a*c+b^2)/x-2*b*x*(c*x^4+b*x^3
+a*x^2)^(1/2)/c/(-4*a*c+b^2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1937, 1963, 12, 1928, 635, 212} \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {3 x \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 x \left (b^2-4 a c\right )}+\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c^2 \left (b^2-4 a c\right )}+\frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )} \]

[In]

Int[x^7/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

(2*x^4*(2*a + b*x))/((b^2 - 4*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]) + ((5*b^2 - 12*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4
])/(2*c^2*(b^2 - 4*a*c)) - (b*(15*b^2 - 52*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(4*c^3*(b^2 - 4*a*c)*x) - (2*b*x*
Sqrt[a*x^2 + b*x^3 + c*x^4])/(c*(b^2 - 4*a*c)) + (3*(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])])/(8*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 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^4 (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^3 (6 a+3 b x)}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{b^2-4 a c} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {2 \int \frac {x^2 \left (6 a b+\frac {3}{2} \left (5 b^2-12 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{3 c \left (b^2-4 a c\right )} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\int \frac {x \left (\frac {3}{2} a \left (5 b^2-12 a c\right )+\frac {3}{4} b \left (15 b^2-52 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{3 c^2 \left (b^2-4 a c\right )} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\int \frac {9 \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}{3 c^3 \left (b^2-4 a c\right )} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 b^2-4 a c\right )\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{8 c^3} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 \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}{8 c^3 \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 b^2-4 a c\right ) 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 )}{4 c^3 \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \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}}{2 c^2 \left (b^2-4 a c\right )}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 \left (b^2-4 a c\right ) x}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {3 \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 )}{8 c^{7/2} \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.71 \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {x \left (2 \sqrt {c} \left (4 a^2 c (-13 b+6 c x)+b^2 x \left (15 b^2+5 b c x-2 c^2 x^2\right )+a \left (15 b^3-62 b^2 c x-20 b c^2 x^2+8 c^3 x^3\right )\right )+3 \left (5 b^4-24 a b^2 c+16 a^2 c^2\right ) \sqrt {a+x (b+c x)} \log \left (c^3 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{8 c^{7/2} \left (-b^2+4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

Integrate[x^7/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

(x*(2*Sqrt[c]*(4*a^2*c*(-13*b + 6*c*x) + b^2*x*(15*b^2 + 5*b*c*x - 2*c^2*x^2) + a*(15*b^3 - 62*b^2*c*x - 20*b*
c^2*x^2 + 8*c^3*x^3)) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*Sqrt[a + x*(b + c*x)]*Log[c^3*(b + 2*c*x - 2*Sqrt[
c]*Sqrt[a + x*(b + c*x)])]))/(8*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[x^2*(a + x*(b + c*x))])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(-\frac {48 \left (\left (-\frac {5}{24} b^{3} x^{2}+\frac {31}{12} b^{2} a x +\frac {13}{6} a^{2} b \right ) c^{\frac {3}{2}}+\left (\frac {1}{12} b^{2} x^{3}+\frac {5}{6} a b \,x^{2}-a^{2} x \right ) c^{\frac {5}{2}}-\frac {a \,c^{\frac {7}{2}} x^{3}}{3}-\frac {5 b^{3} \sqrt {c}\, \left (b x +a \right )}{8}+\frac {\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, \left (16 a^{2} c^{2}-24 a \,b^{2} c +5 b^{4}\right )}{16}\right )}{\sqrt {c \,x^{2}+b x +a}\, c^{\frac {7}{2}} \left (32 a c -8 b^{2}\right )}\)	\(164\)
default	\(\frac {x^{3} \left (c \,x^{2}+b x +a \right ) \left (16 c^{\frac {9}{2}} a \,x^{3}-4 c^{\frac {7}{2}} b^{2} x^{3}-40 c^{\frac {7}{2}} a b \,x^{2}+48 c^{\frac {7}{2}} a^{2} x +10 c^{\frac {5}{2}} b^{3} x^{2}-124 c^{\frac {5}{2}} a \,b^{2} x -104 c^{\frac {5}{2}} a^{2} b +30 c^{\frac {3}{2}} b^{4} x +30 c^{\frac {3}{2}} a \,b^{3}-48 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{3}+72 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{2}-15 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, b^{4} c \right )}{8 c^{\frac {9}{2}} \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )}\)	\(283\)
risch	\(-\frac {\left (-2 c x +7 b \right ) \left (c \,x^{2}+b x +a \right ) x}{4 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}-\frac {\left (\frac {8 c \,a^{2} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {14 b^{2} a \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\left (-4 a b c -7 b^{3}\right ) \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )+\left (12 a \,c^{2}-15 b^{2} c \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )\right ) x \sqrt {c \,x^{2}+b x +a}}{8 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(337\)

[In]

int(x^7/(c*x^4+b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-48/(c*x^2+b*x+a)^(1/2)/c^(7/2)*((-5/24*b^3*x^2+31/12*b^2*a*x+13/6*a^2*b)*c^(3/2)+(1/12*b^2*x^3+5/6*a*b*x^2-a^
2*x)*c^(5/2)-1/3*a*c^(7/2)*x^3-5/8*b^3*c^(1/2)*(b*x+a)+1/16*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*(c*x^2+b
*x+a)^(1/2)*(16*a^2*c^2-24*a*b^2*c+5*b^4))/(32*a*c-8*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 616, normalized size of antiderivative = 2.35 \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{3} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2} - 2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{16 \, {\left ({\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{3} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2} + {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5}\right )} x\right )}}, -\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{3} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, {\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2} - 2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{8 \, {\left ({\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{3} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2} + {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5}\right )} x\right )}}\right ] \]

[In]

integrate(x^7/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^3 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^2 + (5*a*b^4 - 24
*a^2*b^2*c + 16*a^3*c^2)*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)*sq
rt(c) + (b^2 + 4*a*c)*x)/x) + 4*(15*a*b^3*c - 52*a^2*b*c^2 - 2*(b^2*c^3 - 4*a*c^4)*x^3 + 5*(b^3*c^2 - 4*a*b*c^
3)*x^2 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/((b^2*c^5 - 4*a*c^6)*x^3 + (b^
3*c^4 - 4*a*b*c^5)*x^2 + (a*b^2*c^4 - 4*a^2*c^5)*x), -1/8*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^3 + (5*b
^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^2 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*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*(15*a*b^3*c - 52*a^2*b*c^2 - 2*(b^2*c^3
- 4*a*c^4)*x^3 + 5*(b^3*c^2 - 4*a*b*c^3)*x^2 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x)*sqrt(c*x^4 + b*x^3 +
a*x^2))/((b^2*c^5 - 4*a*c^6)*x^3 + (b^3*c^4 - 4*a*b*c^5)*x^2 + (a*b^2*c^4 - 4*a^2*c^5)*x)]

Sympy [F]

\[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**7/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**7/(x**2*(a + b*x + c*x**2))**(3/2), x)

Maxima [F]

\[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{7}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^7/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^7/(c*x^4 + b*x^3 + a*x^2)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.21 \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {{\left (15 \, b^{4} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 72 \, a b^{2} c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 48 \, a^{2} c^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 30 \, \sqrt {a} b^{3} \sqrt {c} - 104 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{8 \, {\left (b^{2} c^{\frac {7}{2}} - 4 \, a c^{\frac {9}{2}}\right )}} + \frac {{\left ({\left (\frac {2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x}{b^{2} c^{3} \mathrm {sgn}\left (x\right ) - 4 \, a c^{4} \mathrm {sgn}\left (x\right )} - \frac {5 \, {\left (b^{3} c - 4 \, a b c^{2}\right )}}{b^{2} c^{3} \mathrm {sgn}\left (x\right ) - 4 \, a c^{4} \mathrm {sgn}\left (x\right )}\right )} x - \frac {15 \, b^{4} - 62 \, a b^{2} c + 24 \, a^{2} c^{2}}{b^{2} c^{3} \mathrm {sgn}\left (x\right ) - 4 \, a c^{4} \mathrm {sgn}\left (x\right )}\right )} x - \frac {15 \, a b^{3} - 52 \, a^{2} b c}{b^{2} c^{3} \mathrm {sgn}\left (x\right ) - 4 \, a c^{4} \mathrm {sgn}\left (x\right )}}{4 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (5 \, b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^7/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

1/8*(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*sqrt(c) - 104*a^(3/2)*b*c^(3/2))*sgn(x)/(b^2*c^(7/2) - 4*a*c^(9/2)) +
1/4*(((2*(b^2*c^2 - 4*a*c^3)*x/(b^2*c^3*sgn(x) - 4*a*c^4*sgn(x)) - 5*(b^3*c - 4*a*b*c^2)/(b^2*c^3*sgn(x) - 4*
a*c^4*sgn(x)))*x - (15*b^4 - 62*a*b^2*c + 24*a^2*c^2)/(b^2*c^3*sgn(x) - 4*a*c^4*sgn(x)))*x - (15*a*b^3 - 52*a^
2*b*c)/(b^2*c^3*sgn(x) - 4*a*c^4*sgn(x)))/sqrt(c*x^2 + b*x + a) - 3/8*(5*b^2 - 4*a*c)*log(abs(2*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(7/2)*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \]

[In]

int(x^7/(a*x^2 + b*x^3 + c*x^4)^(3/2),x)

[Out]

int(x^7/(a*x^2 + b*x^3 + c*x^4)^(3/2), x)