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

Optimal. Leaf size=143 \[ \frac {x \left (3 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 c^2 x}+\frac {\sqrt {a x^2+b x^3+c x^4}}{2 c} \]

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Rubi [A]  time = 0.17, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1928, 1949, 12, 1914, 621, 206} \begin {gather*} \frac {x \left (3 b^2-4 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 )}{8 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 c^2 x}+\frac {\sqrt {a x^2+b x^3+c x^4}}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 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 1928

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 105, normalized size = 0.73 \begin {gather*} \frac {x \left (\left (3 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 )+2 \sqrt {c} (2 c x-3 b) (a+x (b+c x))\right )}{8 c^{5/2} \sqrt {x^2 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.37, size = 122, normalized size = 0.85 \begin {gather*} \frac {\log (x) \left (3 b^2-4 a c\right )}{8 c^{5/2}}+\frac {\left (4 a c-3 b^2\right ) \log \left (-2 c^{5/2} \sqrt {a x^2+b x^3+c x^4}+b c^2 x+2 c^3 x^2\right )}{8 c^{5/2}}+\frac {(2 c x-3 b) \sqrt {a x^2+b x^3+c x^4}}{4 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 226, normalized size = 1.58 \begin {gather*} \left [-\frac {{\left (3 \, b^{2} - 4 \, a 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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x - 3 \, b c\right )}}{16 \, c^{3} x}, -\frac {{\left (3 \, b^{2} - 4 \, a 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 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x - 3 \, b c\right )}}{8 \, c^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((3*b^2 - 4*a*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c^2*x - 3*b*c))/(c^3*x), -1/8*((3*b^2 - 4*a*c)*sq
rt(-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*sqrt(c*x
^4 + b*x^3 + a*x^2)*(2*c^2*x - 3*b*c))/(c^3*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {c x^{4} + b x^{3} + a x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 144, normalized size = 1.01 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x +a}\, \left (-4 a \,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^{2} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+4 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {5}{2}} x -6 \sqrt {c \,x^{2}+b x +a}\, b \,c^{\frac {3}{2}}\right ) x}{8 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(c*x^2+b*x+a)^(1/2)*(4*(c*x^2+b*x+a)^(1/2)*c^(5/2)*x-6*(c*x^2+b*x+a)^(1/2)*b*c^(3/2)-4*a*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^2*c*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2)))/
(c*x^4+b*x^3+a*x^2)^(1/2)/c^(7/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {c x^{4} + b x^{3} + a x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{\sqrt {c\,x^4+b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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