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

Optimal. Leaf size=262 \[ \frac {3 x \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 x \left (b^2-4 a c\right )}+\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c^2 \left (b^2-4 a c\right )}+\frac {2 x^4 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b x \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )} \]

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Rubi [A]  time = 0.51, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \begin {gather*} \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 x \left (b^2-4 a c\right )}+\frac {3 x \left (5 b^2-4 a c\right ) \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}}+\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 )} \end {gather*}

Antiderivative was successfully verified.

[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 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^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 {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 ) \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 )}{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*}

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Mathematica [A]  time = 0.23, size = 183, normalized size = 0.70 \begin {gather*} \frac {x \left (2 \sqrt {c} \left (4 a^2 c (6 c x-13 b)+a \left (15 b^3-62 b^2 c x-20 b c^2 x^2+8 c^3 x^3\right )+b^2 x \left (15 b^2+5 b c x-2 c^2 x^2\right )\right )-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\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 )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt {x^2 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[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)]*ArcTanh[(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))])

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IntegrateAlgebraic [A]  time = 2.80, size = 192, normalized size = 0.73 \begin {gather*} \frac {\left (-52 a^2 b c+24 a^2 c^2 x+15 a b^3-62 a b^2 c x-20 a b c^2 x^2+8 a c^3 x^3+15 b^4 x+5 b^3 c x^2-2 b^2 c^2 x^3\right ) \sqrt {a x^2+b x^3+c x^4}}{4 c^3 x \left (4 a c-b^2\right ) \left (a+b x+c x^2\right )}-\frac {3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a} x-\sqrt {a x^2+b x^3+c x^4}}\right )}{4 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((15*a*b^3 - 52*a^2*b*c + 15*b^4*x - 62*a*b^2*c*x + 24*a^2*c^2*x + 5*b^3*c*x^2 - 20*a*b*c^2*x^2 - 2*b^2*c^2*x^
3 + 8*a*c^3*x^3)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(4*c^3*(-b^2 + 4*a*c)*x*(a + b*x + c*x^2)) - (3*(5*b^2 - 4*a*c)*
ArcTanh[(Sqrt[c]*x^2)/(Sqrt[a]*x - Sqrt[a*x^2 + b*x^3 + c*x^4])])/(4*c^(7/2))

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fricas [A]  time = 1.55, size = 616, normalized size = 2.35 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 283, normalized size = 1.08 \begin {gather*} \frac {\left (c \,x^{2}+b x +a \right ) \left (16 a \,c^{\frac {9}{2}} x^{3}-4 b^{2} c^{\frac {7}{2}} x^{3}-40 a b \,c^{\frac {7}{2}} x^{2}+10 b^{3} c^{\frac {5}{2}} x^{2}+48 a^{2} c^{\frac {7}{2}} x -124 a \,b^{2} c^{\frac {5}{2}} x +30 b^{4} c^{\frac {3}{2}} x -48 \sqrt {c \,x^{2}+b x +a}\, 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 )+72 \sqrt {c \,x^{2}+b x +a}\, 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 )-15 \sqrt {c \,x^{2}+b x +a}\, b^{4} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-104 a^{2} b \,c^{\frac {5}{2}}+30 a \,b^{3} c^{\frac {3}{2}}\right ) x^{3}}{8 \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right ) c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^3*(c*x^2+b*x+a)/c^(9/2)*(16*c^(9/2)*x^3*a-4*c^(7/2)*x^3*b^2-40*c^(7/2)*x^2*a*b+10*c^(5/2)*x^2*b^3+48*c^(
7/2)*x*a^2-124*c^(5/2)*x*a*b^2+30*c^(3/2)*x*b^4-104*c^(5/2)*a^2*b+30*c^(3/2)*a*b^3-48*ln(1/2*(2*c*x+b+2*(c*x^2
+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*(c*x^2+b*x+a)^(1/2)*a^2*c^3+72*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/
c^(1/2))*(c*x^2+b*x+a)^(1/2)*a*b^2*c^2-15*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*(c*x^2+b*x+a
)^(1/2)*b^4*c)/(c*x^4+b*x^3+a*x^2)^(3/2)/(4*a*c-b^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^7}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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