3.57 \(\int \frac{x^5}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac{2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{2 b \sqrt{a x^2+b x^3+c x^4}}{c x \left (b^2-4 a c\right )}+\frac{x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.175293, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1923, 1949, 12, 1914, 621, 206} \[ \frac{2 x^2 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{2 b \sqrt{a x^2+b x^3+c x^4}}{c x \left (b^2-4 a c\right )}+\frac{x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

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

Rule 1923

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(m - 2*n +
q + 1)*(2*a + b*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/((n - q)*(p + 1)*(b^2 - 4*a*c)), x] + Dis
t[1/((n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - 2*n + q)*(2*a*(m + p*q - 2*(n - q) + 1) + b*(m + p*q + (n - q)
*(2*p + 1) + 1)*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] && LtQ[p, -1] && RationalQ[m, q] &
& GtQ[m + p*q + 1, 2*(n - q)]

Rule 1949

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(B*x^(m - n + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(c*(m + p*q + (n - q)*(2*p + 1) + 1)),
x] - Dist[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m - n + q)*Simp[a*B*(m + p*q - n + q + 1) + (b*B*(m
+ p*q + (n - q)*p + 1) - 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, 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] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (
n - q)*(2*p + 1) + 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 1914

Int[(x_)^(m_.)/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[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

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])

Rubi steps

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

Mathematica [A]  time = 0.122982, size = 112, normalized size = 0.73 \[ -\frac{x \left (2 \sqrt{c} \left (-a b+2 a c x+b^2 (-x)\right )+\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )}{c^{3/2} \left (4 a c-b^2\right ) \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.006, size = 166, normalized size = 1.1 \begin{align*} -{\frac{{x}^{3} \left ( c{x}^{2}+bx+a \right ) }{4\,ac-{b}^{2}} \left ( 4\,{c}^{5/2}xa-2\,{c}^{3/2}x{b}^{2}-2\,{c}^{3/2}ab-4\,\ln \left ( 1/2\,{\frac{2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b}{\sqrt{c}}} \right ) \sqrt{c{x}^{2}+bx+a}a{c}^{2}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b \right ){\frac{1}{\sqrt{c}}}} \right ) \sqrt{c{x}^{2}+bx+a}{b}^{2}c \right ){c}^{-{\frac{5}{2}}} \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\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(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^5/(c*x^4 + b*x^3 + a*x^2)^(3/2), x)

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Fricas [A]  time = 2.15883, size = 873, normalized size = 5.71 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} +{\left (b^{3} - 4 \, a b c\right )} x^{2} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt{c} \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 \, \sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (a b c +{\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{2 \,{\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} +{\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x\right )}}, -\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} +{\left (b^{3} - 4 \, a b c\right )} x^{2} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt{-c} \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 \, \sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (a b c +{\left (b^{2} c - 2 \, a c^{2}\right )} x\right )}}{{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} +{\left (a b^{2} c^{2} - 4 \, a^{2} c^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/2*(((b^2*c - 4*a*c^2)*x^3 + (b^3 - 4*a*b*c)*x^2 + (a*b^2 - 4*a^2*c)*x)*sqrt(c)*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(a*b
*c + (b^2*c - 2*a*c^2)*x))/((b^2*c^3 - 4*a*c^4)*x^3 + (b^3*c^2 - 4*a*b*c^3)*x^2 + (a*b^2*c^2 - 4*a^2*c^3)*x),
-(((b^2*c - 4*a*c^2)*x^3 + (b^3 - 4*a*b*c)*x^2 + (a*b^2 - 4*a^2*c)*x)*sqrt(-c)*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(a*b*c + (b^2*c - 2*
a*c^2)*x))/((b^2*c^3 - 4*a*c^4)*x^3 + (b^3*c^2 - 4*a*b*c^3)*x^2 + (a*b^2*c^2 - 4*a^2*c^3)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**5/(x**2*(a + b*x + c*x**2))**(3/2), x)

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Giac [A]  time = 1.19216, size = 149, normalized size = 0.97 \begin{align*} -\frac{2 \,{\left (\frac{a b c}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x} + \frac{b^{2} c - 2 \, a c^{2}}{b^{2} c^{2} - 4 \, a c^{3}}\right )}}{\sqrt{c + \frac{b}{x} + \frac{a}{x^{2}}}} - \frac{2 \, \arctan \left (\frac{\sqrt{c + \frac{b}{x} + \frac{a}{x^{2}}} - \frac{\sqrt{a}}{x}}{\sqrt{-c}}\right )}{\sqrt{-c} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

-2*(a*b*c/((b^2*c^2 - 4*a*c^3)*x) + (b^2*c - 2*a*c^2)/(b^2*c^2 - 4*a*c^3))/sqrt(c + b/x + a/x^2) - 2*arctan((s
qrt(c + b/x + a/x^2) - sqrt(a)/x)/sqrt(-c))/(sqrt(-c)*c)