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

Optimal. Leaf size=227 \[ -\frac {a^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b x \left (b^2-12 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3} \]

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Rubi [A]  time = 0.26, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1921, 1945, 1933, 843, 621, 206, 724} \begin {gather*} -\frac {a^{3/2} 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 )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b x \left (b^2-12 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 )}{16 c^{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 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 + 8*a*c + 2*b*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*c*x) + (a*x^2 + b*x^3 + c*x^4)^(3/2)/(3*x^3) - (a^(3/
2)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/Sqrt[a*x^2 + b*x^3 + c*x^4]
- (b*(b^2 - 12*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])])/(16*c^(3/
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 1921

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*(2*n - q) + 1), x] + Dist[((n - q)*p)/(m + p*(2*n - q) + 1), Int[x^(m +
q)*(2*a + b*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] && Gt
Q[m + p*q + 1, -(n - q)] && NeQ[m + p*(2*n - 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 1945

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 166, normalized size = 0.73 \begin {gather*} \frac {x \sqrt {a+x (b+c x)} \left (-48 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )+2 \sqrt {c} \sqrt {a+x (b+c x)} \left (8 c \left (4 a+c x^2\right )+3 b^2+14 b c x\right )-3 b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{48 c^{3/2} \sqrt {x^2 (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^4,x]

[Out]

(x*Sqrt[a + x*(b + c*x)]*(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(3*b^2 + 14*b*c*x + 8*c*(4*a + c*x^2)) - 48*a^(3/2)*
c^(3/2)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])] - 3*b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqr
t[c]*Sqrt[a + x*(b + c*x)])]))/(48*c^(3/2)*Sqrt[x^2*(a + x*(b + c*x))])

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IntegrateAlgebraic [A]  time = 2.19, size = 168, normalized size = 0.74 \begin {gather*} a^{3/2} \log \left (2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}-2 a x-b x^2\right )-2 a^{3/2} \log (x)+\frac {\left (b^3-12 a b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a} x-\sqrt {a x^2+b x^3+c x^4}}\right )}{8 c^{3/2}}+\frac {\left (32 a c+3 b^2+14 b c x+8 c^2 x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{24 c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2 + 32*a*c + 14*b*c*x + 8*c^2*x^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(24*c*x) + ((b^3 - 12*a*b*c)*ArcTanh[(Sq
rt[c]*x^2)/(Sqrt[a]*x - Sqrt[a*x^2 + b*x^3 + c*x^4])])/(8*c^(3/2)) - 2*a^(3/2)*Log[x] + a^(3/2)*Log[-2*a*x - b
*x^2 + 2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4]]

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fricas [A]  time = 1.42, size = 791, normalized size = 3.48 \begin {gather*} \left [\frac {48 \, a^{\frac {3}{2}} c^{2} x \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 ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} x \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 (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{2} x}, \frac {24 \, a^{\frac {3}{2}} c^{2} x \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 ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} x \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 (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{2} x}, \frac {96 \, \sqrt {-a} a c^{2} x \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 ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} x \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 (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{2} x}, \frac {48 \, \sqrt {-a} a c^{2} x \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 ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} x \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 (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{2} x}\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^4,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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^4,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, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by `
u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable sho
uld perhaps be purged.index.cc index_m operator + Error: Bad Argument Value

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maple [A]  time = 0.01, size = 222, normalized size = 0.98 \begin {gather*} -\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 a^{\frac {3}{2}} c^{\frac {5}{2}} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-36 a b \,c^{2} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+3 b^{3} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-12 \sqrt {c \,x^{2}+b x +a}\, b \,c^{\frac {5}{2}} x -48 \sqrt {c \,x^{2}+b x +a}\, a \,c^{\frac {5}{2}}-6 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{\frac {3}{2}}-16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}\right )}{48 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{3}} \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^4,x)

[Out]

-1/48*(c*x^4+b*x^3+a*x^2)^(3/2)*(48*c^(5/2)*a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-16*(c*x^2+b*
x+a)^(3/2)*c^(5/2)-12*(c*x^2+b*x+a)^(1/2)*b*c^(5/2)*x-48*c^(5/2)*(c*x^2+b*x+a)^(1/2)*a-6*(c*x^2+b*x+a)^(1/2)*b
^2*c^(3/2)-36*a*b*c^2*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))+3*b^3*c*ln(1/2*(2*c*x+b+2*(c*x^2
+b*x+a)^(1/2)*c^(1/2))/c^(1/2)))/x^3/(c*x^2+b*x+a)^(3/2)/c^(5/2)

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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^{4}}\,{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^4,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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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^{4}}\, 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**4,x)

[Out]

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

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