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

Optimal. Leaf size=114 \[ \frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{3/2}}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{4 a x^2}-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{3/2}}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{4 a x^2}-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 112, normalized size = 0.98 \[ \frac {\sqrt {x^2 (a+x (b+c x))} \left (x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)}\right )}{8 a^{3/2} x^3 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 226, normalized size = 1.98 \[ \left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {a} x^{3} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{16 \, a^{2} x^{3}}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{8 \, a^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.01, size = 207, normalized size = 1.82 \[ -\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (4 a^{\frac {3}{2}} c \,x^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-\sqrt {a}\, b^{2} x^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+2 \sqrt {c \,x^{2}+b x +a}\, b c \,x^{3}-4 \sqrt {c \,x^{2}+b x +a}\, a c \,x^{2}+2 \sqrt {c \,x^{2}+b x +a}\, b^{2} x^{2}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b x +4 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \right )}{8 \sqrt {c \,x^{2}+b x +a}\, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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