3.62 \(\int \frac{1}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=209 \[ \frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 x^2 \left (b^2-4 a c\right )}-\frac{\left (5 b^2-12 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{2 a^2 x^3 \left (b^2-4 a c\right )}-\frac{3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{8 a^{7/2}}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a x \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.286989, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1907, 1951, 12, 1904, 206} \[ \frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 x^2 \left (b^2-4 a c\right )}-\frac{\left (5 b^2-12 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{2 a^2 x^3 \left (b^2-4 a c\right )}-\frac{3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{8 a^{7/2}}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a x \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(-3/2),x]

[Out]

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

Rule 1907

Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(-q + 1)*(b^2 - 2*a*c
+ b*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*(
n - q)*(p + 1)*(b^2 - 4*a*c)), Int[(((p*q + 1)*(b^2 - 2*a*c) + (n - q)*(p + 1)*(b^2 - 4*a*c) + b*c*(p*q + (n -
q)*(2*p + 3) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/x^q, x], x] /; FreeQ[{a, b, c, n, q}, x
] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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])

Rubi steps

\begin{align*} \int \frac{1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \sqrt{a x^2+b x^3+c x^4}}-\frac{2 \int \frac{-2 \left (b^2-2 a c\right )+\frac{1}{2} \left (-b^2+4 a c\right )-2 b c x}{x^2 \sqrt{a x^2+b x^3+c x^4}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \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 a^2 \left (b^2-4 a c\right ) x^3}+\frac{\int \frac{-\frac{1}{4} b \left (15 b^2-52 a c\right )-\frac{1}{2} c \left (5 b^2-12 a c\right ) x}{x \sqrt{a x^2+b x^3+c x^4}} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \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 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 \left (b^2-4 a c\right ) x^2}-\frac{\int -\frac{3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )}{8 \sqrt{a x^2+b x^3+c x^4}} \, dx}{a^3 \left (b^2-4 a c\right )}\\ &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \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 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 \left (b^2-4 a c\right ) x^2}+\frac{\left (3 \left (5 b^2-4 a c\right )\right ) \int \frac{1}{\sqrt{a x^2+b x^3+c x^4}} \, dx}{8 a^3}\\ &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \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 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 \left (b^2-4 a c\right ) x^2}-\frac{\left (3 \left (5 b^2-4 a c\right )\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 )}{4 a^3}\\ &=\frac{2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \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 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \left (15 b^2-52 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{4 a^3 \left (b^2-4 a c\right ) x^2}-\frac{3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{8 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.166795, size = 181, normalized size = 0.87 \[ -\frac{2 \sqrt{a} \left (2 a^2 \left (b^2+10 b c x-12 c^2 x^2\right )-8 a^3 c+a b x \left (-5 b^2+62 b c x+52 c^2 x^2\right )-15 b^3 x^2 (b+c x)\right )+3 x^2 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{8 a^{7/2} x \left (b^2-4 a c\right ) \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Sqrt[a]*(-8*a^3*c - 15*b^3*x^2*(b + c*x) + 2*a^2*(b^2 + 10*b*c*x - 12*c^2*x^2) + a*b*x*(-5*b^2 + 62*b*c*x
+ 52*c^2*x^2)) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*x^2*Sqrt[a + x*(b + c*x)]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*
Sqrt[a + x*(b + c*x)])])/(8*a^(7/2)*(b^2 - 4*a*c)*x*Sqrt[x^2*(a + x*(b + c*x))])

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Maple [A]  time = 0.007, size = 292, normalized size = 1.4 \begin{align*} -{\frac{x \left ( c{x}^{2}+bx+a \right ) }{32\,ac-8\,{b}^{2}} \left ( 48\,{a}^{7/2}{x}^{2}{c}^{2}-104\,{a}^{5/2}{x}^{3}b{c}^{2}+16\,{a}^{9/2}c-40\,{a}^{7/2}xbc-124\,{a}^{5/2}{x}^{2}{b}^{2}c+30\,{a}^{3/2}{x}^{3}{b}^{3}c-4\,{a}^{7/2}{b}^{2}+10\,{a}^{5/2}x{b}^{3}+30\,{a}^{3/2}{x}^{2}{b}^{4}-48\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}{x}^{2}{a}^{3}{c}^{2}+72\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}{x}^{2}{a}^{2}{b}^{2}c-15\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}{x}^{2}a{b}^{4} \right ) \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^4+b*x^3+a*x^2)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.20791, size = 1328, normalized size = 6.35 \begin{align*} \left [-\frac{3 \,{\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{5} +{\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{4} +{\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{3}\right )} \sqrt{a} \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 (2 \, a^{3} b^{2} - 8 \, a^{4} c -{\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{3} -{\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{2} - 5 \,{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x\right )} \sqrt{c x^{4} + b x^{3} + a x^{2}}}{16 \,{\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{5} +{\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{4} +{\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{3}\right )}}, \frac{3 \,{\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{5} +{\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{4} +{\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{3}\right )} \sqrt{-a} \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 (2 \, a^{3} b^{2} - 8 \, a^{4} c -{\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{3} -{\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{2} - 5 \,{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x\right )} \sqrt{c x^{4} + b x^{3} + a x^{2}}}{8 \,{\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{5} +{\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{4} +{\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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^5 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^4 + (5*a*b^4 - 24
*a^2*b^2*c + 16*a^3*c^2)*x^3)*sqrt(a)*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*(2*a^3*b^2 - 8*a^4*c - (15*a*b^3*c - 52*a^2*b*c^2)*x^3 - (15*a*b^4 - 62*a^
2*b^2*c + 24*a^3*c^2)*x^2 - 5*(a^2*b^3 - 4*a^3*b*c)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/((a^4*b^2*c - 4*a^5*c^2)*x
^5 + (a^4*b^3 - 4*a^5*b*c)*x^4 + (a^5*b^2 - 4*a^6*c)*x^3), 1/8*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^5 +
(5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^4 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*x^3)*sqrt(-a)*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*(2*a^3*b^2 - 8*a^4*c - (15*a*b^3
*c - 52*a^2*b*c^2)*x^3 - (15*a*b^4 - 62*a^2*b^2*c + 24*a^3*c^2)*x^2 - 5*(a^2*b^3 - 4*a^3*b*c)*x)*sqrt(c*x^4 +
b*x^3 + a*x^2))/((a^4*b^2*c - 4*a^5*c^2)*x^5 + (a^4*b^3 - 4*a^5*b*c)*x^4 + (a^5*b^2 - 4*a^6*c)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a x^{2} + b x^{3} + c x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x