Integrand size = 21, antiderivative size = 143 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (57 b^2 c^2+15 a b c d-2 a^2 d^2\right )+\frac {b d (33 b c+2 a d)}{x}\right )}{15 b^2}+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x+\frac {c^2 (b c+6 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]
c^2*(6*a*d+b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)-7/5*d*(c+d/x)^2*(a+ b/x)^(1/2)-1/15*d*(-4*a^2*d^2+30*a*b*c*d+114*b^2*c^2+b*d*(2*a*d+33*b*c)/x) *(a+b/x)^(1/2)/b^2+(c+d/x)^3*x*(a+b/x)^(1/2)
Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.83 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=\frac {\sqrt {a+\frac {b}{x}} \left (4 a^2 d^3 x^2-2 a b d^2 x (d+15 c x)-3 b^2 \left (2 d^3+10 c d^2 x+30 c^2 d x^2-5 c^3 x^3\right )\right )}{15 b^2 x^2}+\frac {c^2 (b c+6 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]
(Sqrt[a + b/x]*(4*a^2*d^3*x^2 - 2*a*b*d^2*x*(d + 15*c*x) - 3*b^2*(2*d^3 + 10*c*d^2*x + 30*c^2*d*x^2 - 5*c^3*x^3)))/(15*b^2*x^2) + (c^2*(b*c + 6*a*d) *ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/Sqrt[a]
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {899, 108, 27, 170, 27, 164, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx\) |
| \(\Big \downarrow \) 899 |
| \(\displaystyle -\int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x^2d\frac {1}{x}\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3-\int \frac {\left (c+\frac {d}{x}\right )^2 \left (b c+6 a d+\frac {7 b d}{x}\right ) x}{2 \sqrt {a+\frac {b}{x}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3-\frac {1}{2} \int \frac {\left (c+\frac {d}{x}\right )^2 \left (b c+6 a d+\frac {7 b d}{x}\right ) x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{2} \left (-\frac {2 \int \frac {b \left (c+\frac {d}{x}\right ) \left (\frac {d (33 b c+2 a d)}{x}+5 c (b c+6 a d)\right ) x}{2 \sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{5 b}-\frac {14}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2\right )+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{5} \int \frac {\left (c+\frac {d}{x}\right ) \left (\frac {d (33 b c+2 a d)}{x}+5 c (b c+6 a d)\right ) x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}-\frac {14}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2\right )+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (-5 c^2 (6 a d+b c) \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}-\frac {2 d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+15 a b c d+57 b^2 c^2\right )+\frac {b d (2 a d+33 b c)}{x}\right )}{3 b^2}\right )-\frac {14}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2\right )+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (-\frac {10 c^2 (6 a d+b c) \int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{b}-\frac {2 d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+15 a b c d+57 b^2 c^2\right )+\frac {b d (2 a d+33 b c)}{x}\right )}{3 b^2}\right )-\frac {14}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2\right )+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {10 c^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) (6 a d+b c)}{\sqrt {a}}-\frac {2 d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+15 a b c d+57 b^2 c^2\right )+\frac {b d (2 a d+33 b c)}{x}\right )}{3 b^2}\right )-\frac {14}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2\right )+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3\) |
Sqrt[a + b/x]*(c + d/x)^3*x + ((-14*d*Sqrt[a + b/x]*(c + d/x)^2)/5 + ((-2* d*Sqrt[a + b/x]*(2*(57*b^2*c^2 + 15*a*b*c*d - 2*a^2*d^2) + (b*d*(33*b*c + 2*a*d))/x))/(3*b^2) + (10*c^2*(b*c + 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]] )/Sqrt[a])/5)/2
3.3.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(\frac {\left (15 b^{2} c^{3} x^{3}+4 a^{2} d^{3} x^{2}-30 a b c \,d^{2} x^{2}-90 b^{2} c^{2} d \,x^{2}-2 a \,d^{3} x b -30 c \,d^{2} x \,b^{2}-6 b^{2} d^{3}\right ) \sqrt {\frac {a x +b}{x}}}{15 x^{2} b^{2}}+\frac {\left (6 a d +b c \right ) c^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{2 \sqrt {a}\, \left (a x +b \right )}\) | \(160\) |
| default | \(\frac {\sqrt {\frac {a x +b}{x}}\, \left (180 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b \,c^{2} d \,x^{4}+30 \sqrt {a}\, \sqrt {a \,x^{2}+b x}\, b^{2} c^{3} x^{4}+90 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{2} c^{2} d \,x^{4}+15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{3} c^{3} x^{4}-180 \sqrt {a}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b \,c^{2} d \,x^{2}+8 a^{\frac {3}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} d^{3} x -60 c \,d^{2} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b x -12 \sqrt {a}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b \,d^{3}\right )}{30 x^{3} \sqrt {x \left (a x +b \right )}\, \sqrt {a}\, b^{2}}\) | \(248\) |
1/15*(15*b^2*c^3*x^3+4*a^2*d^3*x^2-30*a*b*c*d^2*x^2-90*b^2*c^2*d*x^2-2*a*b *d^3*x-30*b^2*c*d^2*x-6*b^2*d^3)/x^2/b^2*((a*x+b)/x)^(1/2)+1/2*(6*a*d+b*c) *c^2*ln((1/2*b+a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))/a^(1/2)*((a*x+b)/x)^(1/2)*( x*(a*x+b))^(1/2)/(a*x+b)
Time = 0.32 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.14 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=\left [\frac {15 \, {\left (b^{3} c^{3} + 6 \, a b^{2} c^{2} d\right )} \sqrt {a} x^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (15 \, a b^{2} c^{3} x^{3} - 6 \, a b^{2} d^{3} - 2 \, {\left (45 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x^{2} - 2 \, {\left (15 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{30 \, a b^{2} x^{2}}, -\frac {15 \, {\left (b^{3} c^{3} + 6 \, a b^{2} c^{2} d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (15 \, a b^{2} c^{3} x^{3} - 6 \, a b^{2} d^{3} - 2 \, {\left (45 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x^{2} - 2 \, {\left (15 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{15 \, a b^{2} x^{2}}\right ] \]
[1/30*(15*(b^3*c^3 + 6*a*b^2*c^2*d)*sqrt(a)*x^2*log(2*a*x + 2*sqrt(a)*x*sq rt((a*x + b)/x) + b) + 2*(15*a*b^2*c^3*x^3 - 6*a*b^2*d^3 - 2*(45*a*b^2*c^2 *d + 15*a^2*b*c*d^2 - 2*a^3*d^3)*x^2 - 2*(15*a*b^2*c*d^2 + a^2*b*d^3)*x)*s qrt((a*x + b)/x))/(a*b^2*x^2), -1/15*(15*(b^3*c^3 + 6*a*b^2*c^2*d)*sqrt(-a )*x^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (15*a*b^2*c^3*x^3 - 6*a*b^2*d ^3 - 2*(45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 2*a^3*d^3)*x^2 - 2*(15*a*b^2*c*d ^2 + a^2*b*d^3)*x)*sqrt((a*x + b)/x))/(a*b^2*x^2)]
Time = 14.65 (sec) , antiderivative size = 461, normalized size of antiderivative = 3.22 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=\frac {4 a^{\frac {11}{2}} b^{\frac {3}{2}} d^{3} x^{3} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + \frac {2 a^{\frac {9}{2}} b^{\frac {5}{2}} d^{3} x^{2} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {8 a^{\frac {7}{2}} b^{\frac {7}{2}} d^{3} x \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {6 a^{\frac {5}{2}} b^{\frac {9}{2}} d^{3} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {4 a^{6} b d^{3} x^{\frac {7}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {4 a^{5} b^{2} d^{3} x^{\frac {5}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + \sqrt {b} c^{3} \sqrt {x} \sqrt {\frac {a x}{b} + 1} - 3 c^{2} d \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + \frac {b}{x}} & \text {for}\: b \neq 0 \\- \sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 3 c d^{2} \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{\sqrt {a}} \]
4*a**(11/2)*b**(3/2)*d**3*x**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + 2*a**(9/2)*b**(5/2)*d**3*x**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**(7/2)*b* *(7/2)*d**3*x*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b** 4*x**(5/2)) - 6*a**(5/2)*b**(9/2)*d**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x **(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**6*b*d**3*x**(7/2)/(15*a**(7/2) *b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**5*b**2*d**3*x**(5/2)/(1 5*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + sqrt(b)*c**3*sqrt( x)*sqrt(a*x/b + 1) - 3*c**2*d*Piecewise((2*a*atan(sqrt(a + b/x)/sqrt(-a))/ sqrt(-a) + 2*sqrt(a + b/x), Ne(b, 0)), (-sqrt(a)*log(x), True)) + 3*c*d**2 *Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) + b* c**3*asinh(sqrt(a)*sqrt(x)/sqrt(b))/sqrt(a)
Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=\frac {1}{2} \, {\left (2 \, \sqrt {a + \frac {b}{x}} x - \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{\sqrt {a}}\right )} c^{3} - 3 \, {\left (\sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) + 2 \, \sqrt {a + \frac {b}{x}}\right )} c^{2} d - \frac {2}{15} \, d^{3} {\left (\frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a}{b^{2}}\right )} - \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} c d^{2}}{b} \]
1/2*(2*sqrt(a + b/x)*x - b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a))*c^3 - 3*(sqrt(a)*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) + 2*sqrt(a + b/x))*c^2*d - 2/15*d^3*(3*(a + b/x)^(5/2)/ b^2 - 5*(a + b/x)^(3/2)*a/b^2) - 2*(a + b/x)^(3/2)*c*d^2/b
Exception generated. \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 6.79 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.21 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx={\left (a+\frac {b}{x}\right )}^{3/2}\,\left (\frac {6\,a\,d^3-6\,b\,c\,d^2}{3\,b^2}-\frac {4\,a\,d^3}{3\,b^2}\right )+\sqrt {a+\frac {b}{x}}\,\left (2\,a\,\left (\frac {6\,a\,d^3-6\,b\,c\,d^2}{b^2}-\frac {4\,a\,d^3}{b^2}\right )-\frac {6\,d\,{\left (a\,d-b\,c\right )}^2}{b^2}+\frac {2\,a^2\,d^3}{b^2}\right )+c^3\,x\,\sqrt {a+\frac {b}{x}}-\frac {2\,d^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5\,b^2}-\frac {c^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\left (6\,a\,d+b\,c\right )\,1{}\mathrm {i}}{\sqrt {a}} \]
(a + b/x)^(3/2)*((6*a*d^3 - 6*b*c*d^2)/(3*b^2) - (4*a*d^3)/(3*b^2)) + (a + b/x)^(1/2)*(2*a*((6*a*d^3 - 6*b*c*d^2)/b^2 - (4*a*d^3)/b^2) - (6*d*(a*d - b*c)^2)/b^2 + (2*a^2*d^3)/b^2) + c^3*x*(a + b/x)^(1/2) - (2*d^3*(a + b/x) ^(5/2))/(5*b^2) - (c^2*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*(6*a*d + b*c)*1i )/a^(1/2)