3.45 \(\int \frac{(a x^2+b x^3+c x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=219 \[ -\frac{3 x \left (4 a c+b^2\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{2 x^5}-\frac{3 (b-2 c x) \sqrt{a x^2+b x^3+c x^4}}{4 x^2}+\frac{3 b \sqrt{c} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.237785, antiderivative size = 219, normalized size of antiderivative = 1., 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 = {1920, 1941, 1933, 843, 621, 206, 724} \[ -\frac{3 x \left (4 a c+b^2\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{2 x^5}-\frac{3 (b-2 c x) \sqrt{a x^2+b x^3+c x^4}}{4 x^2}+\frac{3 b \sqrt{c} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6,x]

[Out]

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

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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.199347, size = 162, normalized size = 0.74 \[ -\frac{\sqrt{x^2 (a+x (b+c x))} \left (3 x^2 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )+2 \sqrt{a} \left ((2 a+x (5 b-4 c x)) \sqrt{a+x (b+c x)}-6 b \sqrt{c} x^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )\right )}{8 \sqrt{a} x^3 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 338, normalized size = 1.5 \begin{align*} -{\frac{1}{8\,{a}^{2}{x}^{5}} \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 12\,{c}^{5/2}{a}^{5/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{2}-2\,{c}^{5/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{3}b-4\,{c}^{5/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{2}a-6\,{c}^{5/2}\sqrt{c{x}^{2}+bx+a}{x}^{3}ab+3\,{c}^{3/2}{a}^{3/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{2}{b}^{2}-12\,{c}^{5/2}\sqrt{c{x}^{2}+bx+a}{x}^{2}{a}^{2}+2\,{c}^{3/2} \left ( c{x}^{2}+bx+a \right ) ^{5/2}xb-2\,{c}^{3/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{2}{b}^{2}+4\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}a{c}^{3/2}-6\,{c}^{3/2}\sqrt{c{x}^{2}+bx+a}{x}^{2}a{b}^{2}-12\,{c}^{2}\ln \left ( 1/2\,{\frac{2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b}{\sqrt{c}}} \right ){x}^{2}{a}^{2}b \right ) \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^6, x)

________________________________________________________________________________________

Fricas [A]  time = 2.24241, size = 1737, normalized size = 7.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[1/16*(12*a*b*sqrt(c)*x^3*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^2 + 4*a*c)*sqrt(a)*x^3*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)*(4*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^
3), -1/16*(24*a*b*sqrt(-c)*x^3*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^2 + 4*a*c)*sqrt(a)*x^3*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)*(4*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^3), 1/8*
(6*a*b*sqrt(c)*x^3*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^2 + 4*a*c)*sqrt(-a)*x^3*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)*(4*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^3), -1/8*(12*a*b*sqrt
(-c)*x^3*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^2 + 4
*a*c)*sqrt(-a)*x^3*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)*(4*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac{3}{2}}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**6,x)

[Out]

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**6, x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

Timed out