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

Optimal. Leaf size=112 \[ -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}}+\frac {b^2 \sqrt {a x^2+b x^3}}{8 a^2 x^2}-\frac {b \sqrt {a x^2+b x^3}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3}}{3 x^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2020, 2025, 2008, 206} \begin {gather*} \frac {b^2 \sqrt {a x^2+b x^3}}{8 a^2 x^2}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}}-\frac {b \sqrt {a x^2+b x^3}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3}}{3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 42, normalized size = 0.38 \begin {gather*} \frac {2 b^3 \left (x^2 (a+b x)\right )^{3/2} \, _2F_1\left (\frac {3}{2},4;\frac {5}{2};\frac {b x}{a}+1\right )}{3 a^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

(2*b^3*(x^2*(a + b*x))^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x)/a])/(3*a^4*x^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.08, size = 80, normalized size = 0.71 \begin {gather*} \frac {\left (-8 a^2-2 a b x+3 b^2 x^2\right ) \sqrt {a x^2+b x^3}}{24 a^2 x^4}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 175, normalized size = 1.56 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{3} x^{4}}, \frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{3} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*sqrt(a)*b^3*x^4*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(3*a*b^2*x^2 - 2*a^2*b*x
- 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^3*x^4), 1/24*(3*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x))
+ (3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^3*x^4)]

________________________________________________________________________________________

giac [A]  time = 0.27, size = 92, normalized size = 0.82 \begin {gather*} \frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\relax (x) - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} \mathrm {sgn}\relax (x) - 3 \, \sqrt {b x + a} a^{2} b^{4} \mathrm {sgn}\relax (x)}{a^{2} b^{3} x^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*b^4*sgn(x) - 8*(b*x + a)
^(3/2)*a*b^4*sgn(x) - 3*sqrt(b*x + a)*a^2*b^4*sgn(x))/(a^2*b^3*x^3))/b

________________________________________________________________________________________

maple [A]  time = 0.06, size = 89, normalized size = 0.79 \begin {gather*} \frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (-3 a^{2} b^{3} x^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \sqrt {b x +a}\, a^{\frac {9}{2}}-8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}}+3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {5}{2}}\right )}{24 \sqrt {b x +a}\, a^{\frac {9}{2}} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x^3+a*x^2)^(1/2)*(3*(b*x+a)^(5/2)*a^(5/2)-8*(b*x+a)^(3/2)*a^(7/2)-3*arctanh((b*x+a)^(1/2)/a^(1/2))*a^2
*b^3*x^3-3*(b*x+a)^(1/2)*a^(9/2))/x^4/(b*x+a)^(1/2)/a^(9/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^5, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2)/x^5,x)

[Out]

int((a*x^2 + b*x^3)^(1/2)/x^5, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________