3.1.47 \(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=197 \[ -\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{128 a^{5/2}}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7} \]

________________________________________________________________________________________

Rubi [A]  time = 0.36, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1920, 1941, 1951, 12, 1904, 206} \begin {gather*} \frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{128 a^{5/2}}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x]

[Out]

-((b^2 - 12*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(32*a*x^3) + (b*(3*b^2 - 20*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(6
4*a^2*x^2) - ((b + 6*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*x^4) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(4*x^7) - (3*(b
^2 - 4*a*c)^2*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(128*a^(5/2))

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 1904

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, (x*(2*a + b*x^(n - 2)))/Sqrt[a*x^2 + b*x^n + c*x^r]], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]

Rule 1920

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a*
x^q + b*x^n + c*x^(2*n - q))^p)/(m + p*q + 1), x] - Dist[((n - q)*p)/(m + p*q + 1), Int[x^(m + n)*(b + 2*c*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] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q + 1, -
(n - q) + 1] && NeQ[m + p*q + 1, 0]

Rule 1941

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(x^(m + 1)*(A*(m + p*q + (n - q)*(2*p + 1) + 1) + B*(m + p*q + 1)*x^(n - q))*(a*x^q + b*x^n + c*x
^(2*n - q))^p)/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)), x] + Dist[((n - q)*p)/((m + p*q + 1)*(m + p*
q + (n - q)*(2*p + 1) + 1)), Int[x^(n + m)*Simp[2*a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(2*p + 1) + 1) +
(b*B*(m + p*q + 1) - 2*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
- 1), 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] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0] && NeQ
[m + p*q + (n - q)*(2*p + 1) + 1, 0]

Rule 1951

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(A*x^(m - q + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(m + p*q + 1)), x] + Dist[1/(a*(m +
p*q + 1)), Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p + 1) + 1) - A*c*(m + p*q + 2*(
n - q)*(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] && Eq
Q[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] &&
((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*
q + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx &=-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}+\frac {3}{8} \int \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx\\ &=-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}+\frac {1}{16} \int \frac {b^2-12 a c-4 b c x}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx\\ &=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}-\frac {\int \frac {\frac {1}{2} b \left (3 b^2-20 a c\right )+c \left (b^2-12 a c\right ) x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{32 a}\\ &=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}+\frac {\int \frac {3 \left (b^2-4 a c\right )^2}{4 \sqrt {a x^2+b x^3+c x^4}} \, dx}{32 a^2}\\ &=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{128 a^2}\\ &=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}-\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x (2 a+b x)}{\sqrt {a x^2+b x^3+c x^4}}\right )}{64 a^2}\\ &=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}-\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{128 a^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 141, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {x^2 (a+x (b+c x))} \left (2 \sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)} \left (8 a^2+4 a x (2 b+5 c x)-3 b^2 x^2\right )+3 x^4 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )\right )}{128 a^{5/2} x^5 \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x]

[Out]

-1/128*(Sqrt[x^2*(a + x*(b + c*x))]*(2*Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)]*(8*a^2 - 3*b^2*x^2 + 4*a*x*(2
*b + 5*c*x)) + 3*(b^2 - 4*a*c)^2*x^4*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])]))/(a^(5/2)*x^5*Sqr
t[a + x*(b + c*x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.11, size = 175, normalized size = 0.89 \begin {gather*} \frac {3 \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \log \left (2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}-2 a x-b x^2\right )}{128 a^{5/2}}-\frac {3 \log (x) \left (16 a^2 c^2-8 a b^2 c+b^4\right )}{64 a^{5/2}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (-16 a^3-24 a^2 b x-40 a^2 c x^2-2 a b^2 x^2-20 a b c x^3+3 b^3 x^3\right )}{64 a^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x]

[Out]

((-16*a^3 - 24*a^2*b*x - 2*a*b^2*x^2 - 40*a^2*c*x^2 + 3*b^3*x^3 - 20*a*b*c*x^3)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(
64*a^2*x^5) - (3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*Log[x])/(64*a^(5/2)) + (3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*Log[-
2*a*x - b*x^2 + 2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4]])/(128*a^(5/2))

________________________________________________________________________________________

fricas [A]  time = 1.63, size = 332, normalized size = 1.69 \begin {gather*} \left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{5} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, {\left (24 \, a^{3} b x + 16 \, a^{4} - {\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{3} + 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{256 \, a^{3} x^{5}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 2 \, {\left (24 \, a^{3} b x + 16 \, a^{4} - {\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{3} + 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{128 \, a^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

[1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^5*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x - 4*sqrt(c*
x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) - 4*(24*a^3*b*x + 16*a^4 - (3*a*b^3 - 20*a^2*b*c)*x^3 + 2*(a^2*
b^2 + 20*a^3*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^3*x^5), 1/128*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-a)*
x^5*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) - 2*(24*a^3*b*x +
16*a^4 - (3*a*b^3 - 20*a^2*b*c)*x^3 + 2*(a^2*b^2 + 20*a^3*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^3*x^5)]

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.01, size = 501, normalized size = 2.54 \begin {gather*} -\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 a^{\frac {7}{2}} c^{2} x^{4} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-24 a^{\frac {5}{2}} b^{2} c \,x^{4} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+3 a^{\frac {3}{2}} b^{4} x^{4} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+24 \sqrt {c \,x^{2}+b x +a}\, a^{2} b \,c^{2} x^{5}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} c \,x^{5}-48 \sqrt {c \,x^{2}+b x +a}\, a^{3} c^{2} x^{4}+36 \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} c \,x^{4}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{4} x^{4}+24 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b \,c^{2} x^{5}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} c \,x^{5}-16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} c^{2} x^{4}+20 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{2} c \,x^{4}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{4} x^{4}-24 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a b c \,x^{3}+2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{3} x^{3}+16 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} c \,x^{2}+4 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,b^{2} x^{2}-16 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} b x +32 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{3}\right )}{128 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{4} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x)

[Out]

-1/128*(c*x^4+b*x^3+a*x^2)^(3/2)*(48*c^2*a^(7/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*x^4+24*c^2*(c*x
^2+b*x+a)^(3/2)*x^5*a*b-24*c*a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*x^4*b^2-16*c^2*(c*x^2+b*x+a
)^(3/2)*x^4*a^2+24*c^2*(c*x^2+b*x+a)^(1/2)*x^5*a^2*b-2*c*(c*x^2+b*x+a)^(3/2)*x^5*b^3-48*c^2*(c*x^2+b*x+a)^(1/2
)*x^4*a^3-24*c*(c*x^2+b*x+a)^(5/2)*x^3*a*b+20*c*(c*x^2+b*x+a)^(3/2)*x^4*a*b^2-6*c*(c*x^2+b*x+a)^(1/2)*x^5*a*b^
3+3*a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*x^4*b^4+16*c*(c*x^2+b*x+a)^(5/2)*x^2*a^2+36*c*(c*x^2
+b*x+a)^(1/2)*x^4*a^2*b^2+2*(c*x^2+b*x+a)^(5/2)*x^3*b^3-2*(c*x^2+b*x+a)^(3/2)*x^4*b^4+4*(c*x^2+b*x+a)^(5/2)*x^
2*a*b^2-6*(c*x^2+b*x+a)^(1/2)*x^4*a*b^4-16*(c*x^2+b*x+a)^(5/2)*x*a^2*b+32*(c*x^2+b*x+a)^(5/2)*a^3)/x^7/(c*x^2+
b*x+a)^(3/2)/a^4

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^8, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^8} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x)

[Out]

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**8,x)

[Out]

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**8, x)

________________________________________________________________________________________