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

Optimal. Leaf size=155 \[ -\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{5/2}}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4} \]

[Out]

-1/16*b*(-4*a*c+b^2)*arctanh(1/2*x*(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^(1/2))/a^(5/2)-1/3*(c*x^4+b*x^3+a*x^2
)^(1/2)/x^4-1/12*b*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^3+1/24*(-8*a*c+3*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^2/x^2

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Rubi [A]  time = 0.26, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1920, 1951, 12, 1904, 206} \[ \frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{5/2}}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}+\frac {1}{6} \int \frac {b+2 c x}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx\\ &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}-\frac {\int \frac {\frac {1}{2} \left (3 b^2-8 a c\right )+b c x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{12 a}\\ &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}+\frac {\int \frac {3 b \left (b^2-4 a c\right )}{4 \sqrt {a x^2+b x^3+c x^4}} \, dx}{12 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}+\frac {\left (b \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{16 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {\left (b \left (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 )}{8 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \left (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 )}{16 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 131, normalized size = 0.85 \[ \frac {\sqrt {x^2 (a+x (b+c x))} \left (-2 \sqrt {a} \sqrt {a+x (b+c x)} \left (8 a^2+2 a x (b+4 c x)-3 b^2 x^2\right )-3 b x^3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )\right )}{48 a^{5/2} x^4 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(a + x*(b + c*x))]*(-2*Sqrt[a]*Sqrt[a + x*(b + c*x)]*(8*a^2 - 3*b^2*x^2 + 2*a*x*(b + 4*c*x)) - 3*b*(
b^2 - 4*a*c)*x^3*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])]))/(48*a^(5/2)*x^4*Sqrt[a + x*(b + c*x)
])

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fricas [A]  time = 0.74, size = 272, normalized size = 1.75 \[ \left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {a} x^{4} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, a^{2} b x + 8 \, a^{3} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{2}\right )}}{96 \, a^{3} x^{4}}, \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, a^{2} b x + 8 \, a^{3} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{2}\right )}}{48 \, a^{3} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

[-1/96*(3*(b^3 - 4*a*b*c)*sqrt(a)*x^4*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)*(2*a^2*b*x + 8*a^3 - (3*a*b^2 - 8*a^2*c)*x^2))
/(a^3*x^4), 1/48*(3*(b^3 - 4*a*b*c)*sqrt(-a)*x^4*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)*(2*a^2*b*x + 8*a^3 - (3*a*b^2 - 8*a^2*c)*x^2))/(a^
3*x^4)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 234, normalized size = 1.51 \[ \frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (12 a^{\frac {3}{2}} b c \,x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-3 \sqrt {a}\, b^{3} x^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+6 \sqrt {c \,x^{2}+b x +a}\, b^{2} c \,x^{4}-12 \sqrt {c \,x^{2}+b x +a}\, a b c \,x^{3}+6 \sqrt {c \,x^{2}+b x +a}\, b^{3} x^{3}-6 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} x^{2}+12 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b x -16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2}\right )}{48 \sqrt {c \,x^{2}+b x +a}\, a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{3} + a x^{2}}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^5,x)

[Out]

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**3+a*x**2)**(1/2)/x**5,x)

[Out]

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

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