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

Optimal. Leaf size=84 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{3/2}}-\frac {b \sqrt {a x^2+b x^3}}{4 a x^2}-\frac {\sqrt {a x^2+b x^3}}{2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2020, 2025, 2008, 206} \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{3/2}}-\frac {b \sqrt {a x^2+b x^3}}{4 a x^2}-\frac {\sqrt {a x^2+b x^3}}{2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^4,x]

[Out]

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

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 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.50 \[ -\frac {2 b^2 \left (x^2 (a+b x)\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x}{a}+1\right )}{3 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^4,x]

[Out]

(-2*b^2*(x^2*(a + b*x))^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*x)/a])/(3*a^3*x^3)

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fricas [A]  time = 0.41, size = 149, normalized size = 1.77 \[ \left [\frac {\sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{8 \, a^{2} x^{3}}, -\frac {\sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{4 \, a^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a)*b^2*x^3*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*sqrt(b*x^3 + a*x^2)*(a*b*x
+ 2*a^2))/(a^2*x^3), -1/4*(sqrt(-a)*b^2*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + sqrt(b*x^3 + a*x^2)*(
a*b*x + 2*a^2))/(a^2*x^3)]

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giac [A]  time = 0.21, size = 72, normalized size = 0.86 \[ -\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a} a} + \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\relax (x) + \sqrt {b x + a} a b^{3} \mathrm {sgn}\relax (x)}{a b^{2} x^{2}}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a) + ((b*x + a)^(3/2)*b^3*sgn(x) + sqrt(b*x + a)*a*b
^3*sgn(x))/(a*b^2*x^2))/b

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maple [A]  time = 0.06, size = 73, normalized size = 0.87 \[ -\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (-a \,b^{2} x^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\sqrt {b x +a}\, a^{\frac {5}{2}}+\left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}}\right )}{4 \sqrt {b x +a}\, a^{\frac {5}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2)/x^4,x)

[Out]

int((a*x^2 + b*x^3)^(1/2)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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