\(\int \sqrt {a+\frac {b}{x}} x^3 \, dx\) [1691]

Optimal result

Integrand size = 15, antiderivative size = 117 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^3}-\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{7/2}} \]

[Out]

-5/64*b^4*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)+5/64*b^3*x*(a+b/x)^(1/2)/a^3-5/96*b^2*x^2*(a+b/x)^(1/2)/a^2+1
/24*b*x^3*(a+b/x)^(1/2)/a+1/4*x^4*(a+b/x)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 214} \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{7/2}}+\frac {5 b^3 x \sqrt {a+\frac {b}{x}}}{64 a^3}-\frac {5 b^2 x^2 \sqrt {a+\frac {b}{x}}}{96 a^2}+\frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}+\frac {b x^3 \sqrt {a+\frac {b}{x}}}{24 a} \]

[In]

Int[Sqrt[a + b/x]*x^3,x]

[Out]

(5*b^3*Sqrt[a + b/x]*x)/(64*a^3) - (5*b^2*Sqrt[a + b/x]*x^2)/(96*a^2) + (b*Sqrt[a + b/x]*x^3)/(24*a) + (Sqrt[a
+ b/x]*x^4)/4 - (5*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(64*a^(7/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^5} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{4} \sqrt {a+\frac {b}{x}} x^4-\frac {1}{8} b \text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{48 a} \\ & = -\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{64 a^2} \\ & = \frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^3}-\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4+\frac {\left (5 b^4\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{128 a^3} \\ & = \frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^3}-\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{64 a^3} \\ & = \frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^3}-\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (15 b^3-10 a b^2 x+8 a^2 b x^2+48 a^3 x^3\right )-15 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{192 a^{7/2}} \]

[In]

Integrate[Sqrt[a + b/x]*x^3,x]

[Out]

(Sqrt[a]*Sqrt[a + b/x]*x*(15*b^3 - 10*a*b^2*x + 8*a^2*b*x^2 + 48*a^3*x^3) - 15*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[
a]])/(192*a^(7/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\left (48 a^{3} x^{3}+8 a^{2} b \,x^{2}-10 a \,b^{2} x +15 b^{3}\right ) x \sqrt {\frac {a x +b}{x}}}{192 a^{3}}-\frac {5 b^{4} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{128 a^{\frac {7}{2}} \left (a x +b \right )}\)	\(108\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (96 x \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}}-80 a^{\frac {5}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b +60 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x +30 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b^{3}-15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4}\right )}{384 \sqrt {x \left (a x +b \right )}\, a^{\frac {9}{2}}}\)	\(135\)

[In]

int(x^3*(a+b/x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/192*(48*a^3*x^3+8*a^2*b*x^2-10*a*b^2*x+15*b^3)*x/a^3*((a*x+b)/x)^(1/2)-5/128*b^4/a^(7/2)*ln((1/2*b+a*x)/a^(1
/2)+(a*x^2+b*x)^(1/2))*((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)/(a*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.48 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\left [\frac {15 \, \sqrt {a} b^{4} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (48 \, a^{4} x^{4} + 8 \, a^{3} b x^{3} - 10 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{384 \, a^{4}}, \frac {15 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (48 \, a^{4} x^{4} + 8 \, a^{3} b x^{3} - 10 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{192 \, a^{4}}\right ] \]

[In]

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

[Out]

[1/384*(15*sqrt(a)*b^4*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(48*a^4*x^4 + 8*a^3*b*x^3 - 10*a^2*b
^2*x^2 + 15*a*b^3*x)*sqrt((a*x + b)/x))/a^4, 1/192*(15*sqrt(-a)*b^4*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (48
*a^4*x^4 + 8*a^3*b*x^3 - 10*a^2*b^2*x^2 + 15*a*b^3*x)*sqrt((a*x + b)/x))/a^4]

Sympy [A] (verification not implemented)

Time = 18.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {a x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {7 \sqrt {b} x^{\frac {7}{2}}}{24 \sqrt {\frac {a x}{b} + 1}} - \frac {b^{\frac {3}{2}} x^{\frac {5}{2}}}{96 a \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {5}{2}} x^{\frac {3}{2}}}{192 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {7}{2}} \sqrt {x}}{64 a^{3} \sqrt {\frac {a x}{b} + 1}} - \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {7}{2}}} \]

[In]

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

[Out]

a*x**(9/2)/(4*sqrt(b)*sqrt(a*x/b + 1)) + 7*sqrt(b)*x**(7/2)/(24*sqrt(a*x/b + 1)) - b**(3/2)*x**(5/2)/(96*a*sqr
t(a*x/b + 1)) + 5*b**(5/2)*x**(3/2)/(192*a**2*sqrt(a*x/b + 1)) + 5*b**(7/2)*sqrt(x)/(64*a**3*sqrt(a*x/b + 1))
- 5*b**4*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(64*a**(7/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.42 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {7}{2}}} + \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{192 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a^{3} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{4} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{5} - 4 \, {\left (a + \frac {b}{x}\right )} a^{6} + a^{7}\right )}} \]

[In]

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

[Out]

5/128*b^4*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2) + 1/192*(15*(a + b/x)^(7/2)*b^4 - 5
5*(a + b/x)^(5/2)*a*b^4 + 73*(a + b/x)^(3/2)*a^2*b^4 + 15*sqrt(a + b/x)*a^3*b^4)/((a + b/x)^4*a^3 - 4*(a + b/x
)^3*a^4 + 6*(a + b/x)^2*a^5 - 4*(a + b/x)*a^6 + a^7)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {7}{2}}} - \frac {5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {7}{2}}} + \frac {1}{192} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, x \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{a}\right )} x - \frac {5 \, b^{2} \mathrm {sgn}\left (x\right )}{a^{2}}\right )} x + \frac {15 \, b^{3} \mathrm {sgn}\left (x\right )}{a^{3}}\right )} \]

[In]

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

[Out]

5/128*b^4*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/a^(7/2) - 5/128*b^4*log(abs(b))*sgn(x
)/a^(7/2) + 1/192*sqrt(a*x^2 + b*x)*(2*(4*(6*x*sgn(x) + b*sgn(x)/a)*x - 5*b^2*sgn(x)/a^2)*x + 15*b^3*sgn(x)/a^
3)

Mupad [B] (verification not implemented)

Time = 6.50 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5\,x^4\,\sqrt {a+\frac {b}{x}}}{64}+\frac {73\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{192\,a}-\frac {55\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{192\,a^2}+\frac {5\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a^3}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{7/2}} \]

[In]

int(x^3*(a + b/x)^(1/2),x)

[Out]

(5*x^4*(a + b/x)^(1/2))/64 + (b^4*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*5i)/(64*a^(7/2)) + (73*x^4*(a + b/x)^(3/2
))/(192*a) - (55*x^4*(a + b/x)^(5/2))/(192*a^2) + (5*x^4*(a + b/x)^(7/2))/(64*a^3)