\(\int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx\) [1722]

Optimal result

Integrand size = 15, antiderivative size = 120 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=-\frac {35 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^4}+\frac {35 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^3}-\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}+\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{9/2}} \]

[Out]

35/64*b^4*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(9/2)-35/64*b^3*x*(a+b/x)^(1/2)/a^4+35/96*b^2*x^2*(a+b/x)^(1/2)/a^3
-7/24*b*x^3*(a+b/x)^(1/2)/a^2+1/4*x^4*(a+b/x)^(1/2)/a

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 44, 65, 214} \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{9/2}}-\frac {35 b^3 x \sqrt {a+\frac {b}{x}}}{64 a^4}+\frac {35 b^2 x^2 \sqrt {a+\frac {b}{x}}}{96 a^3}-\frac {7 b x^3 \sqrt {a+\frac {b}{x}}}{24 a^2}+\frac {x^4 \sqrt {a+\frac {b}{x}}}{4 a} \]

[In]

Int[x^3/Sqrt[a + b/x],x]

[Out]

(-35*b^3*Sqrt[a + b/x]*x)/(64*a^4) + (35*b^2*Sqrt[a + b/x]*x^2)/(96*a^3) - (7*b*Sqrt[a + b/x]*x^3)/(24*a^2) +
(Sqrt[a + b/x]*x^4)/(4*a) + (35*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(64*a^(9/2))

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 {1}{x^5 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}+\frac {(7 b) \text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{8 a} \\ & = -\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}-\frac {\left (35 b^2\right ) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{48 a^2} \\ & = \frac {35 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^3}-\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}+\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{64 a^3} \\ & = -\frac {35 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^4}+\frac {35 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^3}-\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}-\frac {\left (35 b^4\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{128 a^4} \\ & = -\frac {35 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^4}+\frac {35 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^3}-\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}-\frac {\left (35 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^4} \\ & = -\frac {35 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^4}+\frac {35 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^3}-\frac {7 b \sqrt {a+\frac {b}{x}} x^3}{24 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^4}{4 a}+\frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (-105 b^3+70 a b^2 x-56 a^2 b x^2+48 a^3 x^3\right )+105 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{192 a^{9/2}} \]

[In]

Integrate[x^3/Sqrt[a + b/x],x]

[Out]

(Sqrt[a]*Sqrt[a + b/x]*x*(-105*b^3 + 70*a*b^2*x - 56*a^2*b*x^2 + 48*a^3*x^3) + 105*b^4*ArcTanh[Sqrt[a + b/x]/S
qrt[a]])/(192*a^(9/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\left (48 a^{3} x^{3}-56 a^{2} b \,x^{2}+70 a \,b^{2} x -105 b^{3}\right ) \left (a x +b \right )}{192 a^{4} \sqrt {\frac {a x +b}{x}}}+\frac {35 b^{4} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {x \left (a x +b \right )}}{128 a^{\frac {9}{2}} x \sqrt {\frac {a x +b}{x}}}\)	\(108\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (96 x \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}}-208 a^{\frac {5}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b +348 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x +174 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b^{3}-384 a^{\frac {3}{2}} \sqrt {x \left (a x +b \right )}\, b^{3}+192 a \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{4}-87 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4}\right )}{384 \sqrt {x \left (a x +b \right )}\, a^{\frac {11}{2}}}\)	\(184\)

[In]

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

[Out]

1/192*(48*a^3*x^3-56*a^2*b*x^2+70*a*b^2*x-105*b^3)*(a*x+b)/a^4/((a*x+b)/x)^(1/2)+35/128*b^4/a^(9/2)*ln((1/2*b+
a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))/x/((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.45 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\left [\frac {105 \, \sqrt {a} b^{4} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (48 \, a^{4} x^{4} - 56 \, a^{3} b x^{3} + 70 \, a^{2} b^{2} x^{2} - 105 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{384 \, a^{5}}, -\frac {105 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (48 \, a^{4} x^{4} - 56 \, a^{3} b x^{3} + 70 \, a^{2} b^{2} x^{2} - 105 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{192 \, a^{5}}\right ] \]

[In]

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

[Out]

[1/384*(105*sqrt(a)*b^4*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(48*a^4*x^4 - 56*a^3*b*x^3 + 70*a^2
*b^2*x^2 - 105*a*b^3*x)*sqrt((a*x + b)/x))/a^5, -1/192*(105*sqrt(-a)*b^4*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a)
- (48*a^4*x^4 - 56*a^3*b*x^3 + 70*a^2*b^2*x^2 - 105*a*b^3*x)*sqrt((a*x + b)/x))/a^5]

Sympy [A] (verification not implemented)

Time = 34.22 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.29 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\frac {x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {\sqrt {b} x^{\frac {7}{2}}}{24 a \sqrt {\frac {a x}{b} + 1}} + \frac {7 b^{\frac {3}{2}} x^{\frac {5}{2}}}{96 a^{2} \sqrt {\frac {a x}{b} + 1}} - \frac {35 b^{\frac {5}{2}} x^{\frac {3}{2}}}{192 a^{3} \sqrt {\frac {a x}{b} + 1}} - \frac {35 b^{\frac {7}{2}} \sqrt {x}}{64 a^{4} \sqrt {\frac {a x}{b} + 1}} + \frac {35 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {9}{2}}} \]

[In]

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

[Out]

x**(9/2)/(4*sqrt(b)*sqrt(a*x/b + 1)) - sqrt(b)*x**(7/2)/(24*a*sqrt(a*x/b + 1)) + 7*b**(3/2)*x**(5/2)/(96*a**2*
sqrt(a*x/b + 1)) - 35*b**(5/2)*x**(3/2)/(192*a**3*sqrt(a*x/b + 1)) - 35*b**(7/2)*sqrt(x)/(64*a**4*sqrt(a*x/b +
1)) + 35*b**4*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(64*a**(9/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.38 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=-\frac {35 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {9}{2}}} - \frac {105 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} - 385 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} + 511 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} - 279 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{192 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a^{4} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{5} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{6} - 4 \, {\left (a + \frac {b}{x}\right )} a^{7} + a^{8}\right )}} \]

[In]

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

[Out]

-35/128*b^4*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(9/2) - 1/192*(105*(a + b/x)^(7/2)*b^4
- 385*(a + b/x)^(5/2)*a*b^4 + 511*(a + b/x)^(3/2)*a^2*b^4 - 279*sqrt(a + b/x)*a^3*b^4)/((a + b/x)^4*a^4 - 4*(a
+ b/x)^3*a^5 + 6*(a + b/x)^2*a^6 - 4*(a + b/x)*a^7 + a^8)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\frac {1}{192} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, x {\left (\frac {6 \, x}{a \mathrm {sgn}\left (x\right )} - \frac {7 \, b}{a^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {35 \, b^{2}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} x - \frac {105 \, b^{3}}{a^{4} \mathrm {sgn}\left (x\right )}\right )} + \frac {35 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {9}{2}}} - \frac {35 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right )}{128 \, a^{\frac {9}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/192*sqrt(a*x^2 + b*x)*(2*(4*x*(6*x/(a*sgn(x)) - 7*b/(a^2*sgn(x))) + 35*b^2/(a^3*sgn(x)))*x - 105*b^3/(a^4*sg
n(x))) + 35/128*b^4*log(abs(b))*sgn(x)/a^(9/2) - 35/128*b^4*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)
+ b))/(a^(9/2)*sgn(x))

Mupad [B] (verification not implemented)

Time = 6.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x}}} \, dx=\frac {93\,x^4\,\sqrt {a+\frac {b}{x}}}{64\,a}-\frac {511\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{192\,a^2}+\frac {385\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{192\,a^3}-\frac {35\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a^4}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,35{}\mathrm {i}}{64\,a^{9/2}} \]

[In]

int(x^3/(a + b/x)^(1/2),x)

[Out]

(93*x^4*(a + b/x)^(1/2))/(64*a) - (b^4*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*35i)/(64*a^(9/2)) - (511*x^4*(a + b/
x)^(3/2))/(192*a^2) + (385*x^4*(a + b/x)^(5/2))/(192*a^3) - (35*x^4*(a + b/x)^(7/2))/(64*a^4)