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

Optimal. Leaf size=257 \[ \frac {b x \left (b^2-12 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{3/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 a x^2}+\frac {c^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5} \]

[Out]

-1/3*(c*x^4+b*x^3+a*x^2)^(3/2)/x^6-1/4*b*(c*x^4+b*x^3+a*x^2)^(3/2)/a/x^5+1/16*b*(-12*a*c+b^2)*x*arctanh(1/2*(b
*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))*(c*x^2+b*x+a)^(1/2)/a^(3/2)/(c*x^4+b*x^3+a*x^2)^(1/2)+c^(3/2)*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*x^4+b*x^3+a*x^2)^(1/2)+1/8*(2*b*c*x-8*a*c+b^2
)*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^2

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Rubi [A]  time = 0.35, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1920, 1951, 1941, 1933, 843, 621, 206, 724} \[ \frac {b x \left (b^2-12 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{3/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 a x^2}+\frac {c^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 8*a*c + 2*b*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*a*x^2) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(3*x^6) - (b*(
a*x^2 + b*x^3 + c*x^4)^(3/2))/(4*a*x^5) + (b*(b^2 - 12*a*c)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(2*a + b*x)/(2*Sqr
t[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(3/2)*Sqrt[a*x^2 + b*x^3 + c*x^4]) + (c^(3/2)*x*Sqrt[a + b*x + c*x^2]*ArcT
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[a*x^2 + b*x^3 + c*x^4]

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 1933

Int[((A_) + (B_.)*(x_)^(j_.))/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[(A + B*x^(n - q))/(
x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; FreeQ[{a, b, c, A, B, n, q}, x] && EqQ[j, n - q] &
& EqQ[r, 2*n - q] && PosQ[n - q] && EqQ[n, 3] && EqQ[q, 2]

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

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Mathematica [A]  time = 0.29, size = 175, normalized size = 0.68 \[ \frac {\sqrt {x^2 (a+x (b+c x))} \left (3 b x^3 \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {a} \left (\sqrt {a+x (b+c x)} \left (8 a^2+2 a x (7 b+16 c x)+3 b^2 x^2\right )-24 a c^{3/2} x^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )\right )}{48 a^{3/2} x^4 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(a + x*(b + c*x))]*(3*b*(b^2 - 12*a*c)*x^3*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])] -
2*Sqrt[a]*(Sqrt[a + x*(b + c*x)]*(8*a^2 + 3*b^2*x^2 + 2*a*x*(7*b + 16*c*x)) - 24*a*c^(3/2)*x^3*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])))/(48*a^(3/2)*x^4*Sqrt[a + x*(b + c*x)])

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fricas [A]  time = 0.81, size = 815, normalized size = 3.17 \[ \left [\frac {48 \, a^{2} c^{\frac {3}{2}} x^{4} \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 ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {a} x^{4} \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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (14 \, a^{2} b x + 8 \, a^{3} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{2}\right )}}{96 \, a^{2} x^{4}}, -\frac {96 \, a^{2} \sqrt {-c} c x^{4} \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 ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {a} x^{4} \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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (14 \, a^{2} b x + 8 \, a^{3} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{2}\right )}}{96 \, a^{2} x^{4}}, \frac {24 \, a^{2} c^{\frac {3}{2}} x^{4} \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 ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-a} x^{4} \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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (14 \, a^{2} b x + 8 \, a^{3} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{2}\right )}}{48 \, a^{2} x^{4}}, -\frac {48 \, a^{2} \sqrt {-c} c x^{4} \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 ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-a} x^{4} \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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (14 \, a^{2} b x + 8 \, a^{3} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{2}\right )}}{48 \, a^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*a^2*c^(3/2)*x^4*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) - 3*(b^3 - 12*a*b*c)*sqrt(a)*x^4*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*a^2*b*x + 8*a^3 + (3*a*b^2 +
32*a^2*c)*x^2))/(a^2*x^4), -1/96*(96*a^2*sqrt(-c)*c*x^4*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqr
t(-c)/(c^2*x^3 + b*c*x^2 + a*c*x)) + 3*(b^3 - 12*a*b*c)*sqrt(a)*x^4*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*a^2*b*x + 8*
a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4), 1/48*(24*a^2*c^(3/2)*x^4*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) - 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^4*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*
a^2*b*x + 8*a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4), -1/48*(48*a^2*sqrt(-c)*c*x^4*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)) + 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^4*arctan(1/2*s
qrt(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*sqrt(c*x^4 + b*x^3 + a*x^2)*(
14*a^2*b*x + 8*a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-2,[1,0,0,2]%%%}+%%%{-2,[0,1,0,1]%%%}+%%%{-4,[0,0,1,0]%%%},0,%%%{1,[2,
0,0,4]%%%}+%%%{2,[1,1,0,3]%%%}+%%%{1,[0,2,0,2]%%%}] at parameters values [-97,-82,63.4443001123,-27]Warning, c
hoosing root of [1,0,%%%{-2,[1,0,0,2]%%%}+%%%{-2,[0,1,0,1]%%%}+%%%{-4,[0,0,1,0]%%%},0,%%%{1,[2,0,0,4]%%%}+%%%{
2,[1,1,0,3]%%%}+%%%{1,[0,2,0,2]%%%}] at parameters values [63,-49,35.2935628123,-64]Warning, choosing root of
[1,0,%%%{-2,[2,1,0,0]%%%}+%%%{-2,[1,0,1,0]%%%}+%%%{-4,[0,0,0,1]%%%},0,%%%{1,[4,2,0,0]%%%}+%%%{2,[3,1,1,0]%%%}+
%%%{1,[2,0,2,0]%%%}] at parameters values [22,42,56,43.9628838282]Sign error (%%%{b-2*sqrt(a)*sqrt(c),0%%%}+%%
%{-(-2*a*c+b*sqrt(a)*sqrt(c))/a,1%%%}+%%%{-(4*a*c*sqrt(a)*sqrt(c)-b^2*sqrt(a)*sqrt(c))/(4*a^2),2%%%}+%%%{undef
,3%%%})Evaluation time: 0.45Limit: Max order reached or unable to make series expansion Error: Bad Argument Va
lue

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maple [A]  time = 0.01, size = 435, normalized size = 1.69 \[ \frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 a^{3} c^{3} x^{3} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-36 a^{\frac {5}{2}} b \,c^{\frac {5}{2}} x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+3 a^{\frac {3}{2}} b^{3} c^{\frac {3}{2}} x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+48 \sqrt {c \,x^{2}+b x +a}\, a^{2} c^{\frac {7}{2}} x^{4}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c^{\frac {5}{2}} x^{4}+60 \sqrt {c \,x^{2}+b x +a}\, a^{2} b \,c^{\frac {5}{2}} x^{3}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} c^{\frac {3}{2}} x^{3}+32 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,c^{\frac {7}{2}} x^{4}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} c^{\frac {5}{2}} x^{4}+28 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b \,c^{\frac {5}{2}} x^{3}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} c^{\frac {3}{2}} x^{3}-32 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,c^{\frac {5}{2}} x^{2}+2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{2} c^{\frac {3}{2}} x^{2}+4 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a b \,c^{\frac {3}{2}} x -16 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} c^{\frac {3}{2}}\right )}{48 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{3} c^{\frac {3}{2}} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(c*x^4+b*x^3+a*x^2)^(3/2)*(32*c^(7/2)*(c*x^2+b*x+a)^(3/2)*x^4*a-36*c^(5/2)*a^(5/2)*ln((b*x+2*a+2*(c*x^2+b
*x+a)^(1/2)*a^(1/2))/x)*x^3*b+48*c^(7/2)*(c*x^2+b*x+a)^(1/2)*x^4*a^2-2*c^(5/2)*(c*x^2+b*x+a)^(3/2)*x^4*b^2-32*
c^(5/2)*(c*x^2+b*x+a)^(5/2)*x^2*a+28*c^(5/2)*(c*x^2+b*x+a)^(3/2)*x^3*a*b-6*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x^4*a*b
^2+3*c^(3/2)*a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*x^3*b^3+60*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x^3*
a^2*b+2*c^(3/2)*(c*x^2+b*x+a)^(5/2)*x^2*b^2-2*c^(3/2)*(c*x^2+b*x+a)^(3/2)*x^3*b^3+4*c^(3/2)*(c*x^2+b*x+a)^(5/2
)*x*a*b-6*c^(3/2)*(c*x^2+b*x+a)^(1/2)*x^3*a*b^3-16*(c*x^2+b*x+a)^(5/2)*a^2*c^(3/2)+48*ln(1/2*(2*c*x+b+2*(c*x^2
+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*x^3*a^3*c^3)/x^6/(c*x^2+b*x+a)^(3/2)/a^3/c^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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