Integrand size = 20, antiderivative size = 142 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=-\frac {d (b c-3 a d) \sqrt {c+d x}}{a b^2}+\frac {(b c-a d) (c+d x)^{3/2}}{a b (a+b x)}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {(b c-a d)^{3/2} (2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{5/2}} \]
(-a*d+b*c)*(d*x+c)^(3/2)/a/b/(b*x+a)-2*c^(5/2)*arctanh((d*x+c)^(1/2)/c^(1/ 2))/a^2+(-a*d+b*c)^(3/2)*(3*a*d+2*b*c)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d +b*c)^(1/2))/a^2/b^(5/2)-d*(-3*a*d+b*c)*(d*x+c)^(1/2)/a/b^2
Time = 0.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\frac {\frac {a \sqrt {c+d x} \left (b^2 c^2+3 a^2 d^2+2 a b d (-c+d x)\right )}{b^2 (a+b x)}+\frac {\sqrt {-b c+a d} \left (2 b^2 c^2+a b c d-3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2}}-2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \]
((a*Sqrt[c + d*x]*(b^2*c^2 + 3*a^2*d^2 + 2*a*b*d*(-c + d*x)))/(b^2*(a + b* x)) + (Sqrt[-(b*c) + a*d]*(2*b^2*c^2 + a*b*c*d - 3*a^2*d^2)*ArcTan[(Sqrt[b ]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(5/2) - 2*c^(5/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^2
Time = 0.31 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {109, 27, 171, 27, 174, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (2 b c^2-d (b c-3 a d) x\right )}{2 x (a+b x)}dx}{a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (2 b c^2-d (b c-3 a d) x\right )}{x (a+b x)}dx}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {\frac {2 \int \frac {2 b^2 c^3+d \left (b^2 c^2+4 a b d c-3 a^2 d^2\right ) x}{2 x (a+b x) \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {c+d x} (b c-3 a d)}{b}}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {2 b^2 c^3+d \left (b^2 c^2+4 a b d c-3 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {c+d x} (b c-3 a d)}{b}}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {\frac {\frac {2 b^2 c^3 \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {(b c-a d)^2 (3 a d+2 b c) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{b}-\frac {2 d \sqrt {c+d x} (b c-3 a d)}{b}}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {4 b^2 c^3 \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {2 (b c-a d)^2 (3 a d+2 b c) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{b}-\frac {2 d \sqrt {c+d x} (b c-3 a d)}{b}}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 (b c-a d)^{3/2} (3 a d+2 b c) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {4 b^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}}{b}-\frac {2 d \sqrt {c+d x} (b c-3 a d)}{b}}{2 a b}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)}\) |
((b*c - a*d)*(c + d*x)^(3/2))/(a*b*(a + b*x)) + ((-2*d*(b*c - 3*a*d)*Sqrt[ c + d*x])/b + ((-4*b^2*c^(5/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a + (2*(b*c - a*d)^(3/2)*(2*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a *d]])/(a*Sqrt[b]))/b)/(2*a*b)
3.5.72.3.1 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[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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_) )^(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] && IntegersQ[2*m, 2*n, 2*p]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
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]
Time = 0.62 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06
| method | result | size |
| pseudoelliptic | \(\frac {-3 \left (a d -b c \right )^{2} \left (b x +a \right ) \left (a d +\frac {2 b c}{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+3 \sqrt {\left (a d -b c \right ) b}\, \left (-\frac {2 b^{2} c^{\frac {5}{2}} \left (b x +a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{3}+\left (\frac {b^{2} c^{2}}{3}-\frac {2 a d \left (-d x +c \right ) b}{3}+a^{2} d^{2}\right ) a \sqrt {d x +c}\right )}{b^{2} \left (b x +a \right ) a^{2} \sqrt {\left (a d -b c \right ) b}}\) | \(151\) |
| derivativedivides | \(2 d^{2} \left (\frac {\sqrt {d x +c}}{b^{2}}-\frac {c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2} d^{2}}-\frac {\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (3 a^{3} d^{3}-4 a^{2} b c \,d^{2}-a \,b^{2} c^{2} d +2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2} a^{2} d^{2}}\right )\) | \(179\) |
| default | \(2 d^{2} \left (\frac {\sqrt {d x +c}}{b^{2}}-\frac {c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2} d^{2}}-\frac {\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (3 a^{3} d^{3}-4 a^{2} b c \,d^{2}-a \,b^{2} c^{2} d +2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2} a^{2} d^{2}}\right )\) | \(179\) |
3/((a*d-b*c)*b)^(1/2)*(-(a*d-b*c)^2*(b*x+a)*(a*d+2/3*b*c)*arctan(b*(d*x+c) ^(1/2)/((a*d-b*c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*(-2/3*b^2*c^(5/2)*(b*x+a)* arctanh((d*x+c)^(1/2)/c^(1/2))+(1/3*b^2*c^2-2/3*a*d*(-d*x+c)*b+a^2*d^2)*a* (d*x+c)^(1/2)))/b^2/(b*x+a)/a^2
Time = 0.33 (sec) , antiderivative size = 865, normalized size of antiderivative = 6.09 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\left [-\frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - 2 \, {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{a^{2} b^{3} x + a^{3} b^{2}}, \frac {4 \, {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + 2 \, {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{a^{2} b^{3} x + a^{3} b^{2}}\right ] \]
[-1/2*((2*a*b^2*c^2 + a^2*b*c*d - 3*a^3*d^2 + (2*b^3*c^2 + a*b^2*c*d - 3*a ^2*b*d^2)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c )*b*sqrt((b*c - a*d)/b))/(b*x + a)) - 2*(b^3*c^2*x + a*b^2*c^2)*sqrt(c)*lo g((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^2*b*d^2*x + a*b^2*c^2 - 2*a^2*b*c*d + 3*a^3*d^2)*sqrt(d*x + c))/(a^2*b^3*x + a^3*b^2), ((2*a*b^2 *c^2 + a^2*b*c*d - 3*a^3*d^2 + (2*b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2)*x)*sq rt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d )) + (b^3*c^2*x + a*b^2*c^2)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + (2*a^2*b*d^2*x + a*b^2*c^2 - 2*a^2*b*c*d + 3*a^3*d^2)*sqrt(d*x + c))/(a^2*b^3*x + a^3*b^2), 1/2*(4*(b^3*c^2*x + a*b^2*c^2)*sqrt(-c)*arctan (sqrt(d*x + c)*sqrt(-c)/c) - (2*a*b^2*c^2 + a^2*b*c*d - 3*a^3*d^2 + (2*b^3 *c^2 + a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(2*a^2*b*d^2 *x + a*b^2*c^2 - 2*a^2*b*c*d + 3*a^3*d^2)*sqrt(d*x + c))/(a^2*b^3*x + a^3* b^2), ((2*a*b^2*c^2 + a^2*b*c*d - 3*a^3*d^2 + (2*b^3*c^2 + a*b^2*c*d - 3*a ^2*b*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d )/b)/(b*c - a*d)) + 2*(b^3*c^2*x + a*b^2*c^2)*sqrt(-c)*arctan(sqrt(d*x + c )*sqrt(-c)/c) + (2*a^2*b*d^2*x + a*b^2*c^2 - 2*a^2*b*c*d + 3*a^3*d^2)*sqrt (d*x + c))/(a^2*b^3*x + a^3*b^2)]
\[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x \left (a + b x\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.36 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\frac {2 \, c^{3} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, \sqrt {d x + c} d^{2}}{b^{2}} - \frac {{\left (2 \, b^{3} c^{3} - a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b^{2}} + \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b^{2}} \]
2*c^3*arctan(sqrt(d*x + c)/sqrt(-c))/(a^2*sqrt(-c)) + 2*sqrt(d*x + c)*d^2/ b^2 - (2*b^3*c^3 - a*b^2*c^2*d - 4*a^2*b*c*d^2 + 3*a^3*d^3)*arctan(sqrt(d* x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^2*b^2) + (sqrt(d*x + c)*b^2*c^2*d - 2*sqrt(d*x + c)*a*b*c*d^2 + sqrt(d*x + c)*a^2*d^3)/(((d*x + c)*b - b*c + a*d)*a*b^2)
Time = 0.87 (sec) , antiderivative size = 1295, normalized size of antiderivative = 9.12 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^2} \, dx=\frac {2\,d^2\,\sqrt {c+d\,x}}{b^2}+\frac {\mathrm {atan}\left (\frac {a^2\,d^8\,\sqrt {c^5}\,\sqrt {c+d\,x}\,36{}\mathrm {i}}{36\,a^2\,c^3\,d^8+40\,b^2\,c^5\,d^6+\frac {80\,b^3\,c^6\,d^5}{a}-\frac {60\,b^4\,c^7\,d^4}{a^2}-96\,a\,b\,c^4\,d^7}+\frac {c^2\,d^6\,\sqrt {c^5}\,\sqrt {c+d\,x}\,40{}\mathrm {i}}{40\,c^5\,d^6-\frac {96\,a\,c^4\,d^7}{b}+\frac {80\,b\,c^6\,d^5}{a}-\frac {60\,b^2\,c^7\,d^4}{a^2}+\frac {36\,a^2\,c^3\,d^8}{b^2}}+\frac {c^3\,d^5\,\sqrt {c^5}\,\sqrt {c+d\,x}\,80{}\mathrm {i}}{80\,c^6\,d^5+\frac {40\,a\,c^5\,d^6}{b}-\frac {60\,b\,c^7\,d^4}{a}-\frac {96\,a^2\,c^4\,d^7}{b^2}+\frac {36\,a^3\,c^3\,d^8}{b^3}}-\frac {b\,c^4\,d^4\,\sqrt {c^5}\,\sqrt {c+d\,x}\,60{}\mathrm {i}}{80\,a\,c^6\,d^5-60\,b\,c^7\,d^4+\frac {40\,a^2\,c^5\,d^6}{b}-\frac {96\,a^3\,c^4\,d^7}{b^2}+\frac {36\,a^4\,c^3\,d^8}{b^3}}-\frac {a\,c\,d^7\,\sqrt {c^5}\,\sqrt {c+d\,x}\,96{}\mathrm {i}}{40\,b\,c^5\,d^6-96\,a\,c^4\,d^7+\frac {80\,b^2\,c^6\,d^5}{a}+\frac {36\,a^2\,c^3\,d^8}{b}-\frac {60\,b^3\,c^7\,d^4}{a^2}}\right )\,\sqrt {c^5}\,2{}\mathrm {i}}{a^2}+\frac {\sqrt {c+d\,x}\,\left (a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{a\,\left (b^3\,\left (c+d\,x\right )-b^3\,c+a\,b^2\,d\right )}-\frac {\mathrm {atan}\left (\frac {c^4\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,70{}\mathrm {i}}{72\,a^3\,b\,c^3\,d^8-50\,b^4\,c^6\,d^5-170\,a\,b^3\,c^5\,d^6-162\,a^4\,c^2\,d^9+\frac {54\,a^5\,c\,d^{10}}{b}+196\,a^2\,b^2\,c^4\,d^7+\frac {60\,b^5\,c^7\,d^4}{a}}-\frac {c^3\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,90{}\mathrm {i}}{54\,a^4\,c\,d^{10}-170\,b^4\,c^5\,d^6+196\,a\,b^3\,c^4\,d^7-162\,a^3\,b\,c^2\,d^9+72\,a^2\,b^2\,c^3\,d^8-\frac {50\,b^5\,c^6\,d^5}{a}+\frac {60\,b^6\,c^7\,d^4}{a^2}}+\frac {c^5\,d^4\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,60{}\mathrm {i}}{72\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-50\,a\,b^3\,c^6\,d^5+196\,a^3\,b\,c^4\,d^7+\frac {54\,a^6\,c\,d^{10}}{b^2}-170\,a^2\,b^2\,c^5\,d^6-\frac {162\,a^5\,c^2\,d^9}{b}}-\frac {a\,c^2\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,54{}\mathrm {i}}{196\,a\,b^4\,c^4\,d^7-170\,b^5\,c^5\,d^6+72\,a^2\,b^3\,c^3\,d^8-162\,a^3\,b^2\,c^2\,d^9-\frac {50\,b^6\,c^6\,d^5}{a}+\frac {60\,b^7\,c^7\,d^4}{a^2}+54\,a^4\,b\,c\,d^{10}}+\frac {a^2\,c\,d^8\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,54{}\mathrm {i}}{196\,a\,b^5\,c^4\,d^7-170\,b^6\,c^5\,d^6+54\,a^4\,b^2\,c\,d^{10}+72\,a^2\,b^4\,c^3\,d^8-162\,a^3\,b^3\,c^2\,d^9-\frac {50\,b^7\,c^6\,d^5}{a}+\frac {60\,b^8\,c^7\,d^4}{a^2}}\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^3}\,\left (3\,a\,d+2\,b\,c\right )\,1{}\mathrm {i}}{a^2\,b^5} \]
(2*d^2*(c + d*x)^(1/2))/b^2 + (atan((a^2*d^8*(c^5)^(1/2)*(c + d*x)^(1/2)*3 6i)/(36*a^2*c^3*d^8 + 40*b^2*c^5*d^6 + (80*b^3*c^6*d^5)/a - (60*b^4*c^7*d^ 4)/a^2 - 96*a*b*c^4*d^7) + (c^2*d^6*(c^5)^(1/2)*(c + d*x)^(1/2)*40i)/(40*c ^5*d^6 - (96*a*c^4*d^7)/b + (80*b*c^6*d^5)/a - (60*b^2*c^7*d^4)/a^2 + (36* a^2*c^3*d^8)/b^2) + (c^3*d^5*(c^5)^(1/2)*(c + d*x)^(1/2)*80i)/(80*c^6*d^5 + (40*a*c^5*d^6)/b - (60*b*c^7*d^4)/a - (96*a^2*c^4*d^7)/b^2 + (36*a^3*c^3 *d^8)/b^3) - (b*c^4*d^4*(c^5)^(1/2)*(c + d*x)^(1/2)*60i)/(80*a*c^6*d^5 - 6 0*b*c^7*d^4 + (40*a^2*c^5*d^6)/b - (96*a^3*c^4*d^7)/b^2 + (36*a^4*c^3*d^8) /b^3) - (a*c*d^7*(c^5)^(1/2)*(c + d*x)^(1/2)*96i)/(40*b*c^5*d^6 - 96*a*c^4 *d^7 + (80*b^2*c^6*d^5)/a + (36*a^2*c^3*d^8)/b - (60*b^3*c^7*d^4)/a^2))*(c ^5)^(1/2)*2i)/a^2 + ((c + d*x)^(1/2)*(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))/ (a*(b^3*(c + d*x) - b^3*c + a*b^2*d)) - (atan((c^4*d^5*(c + d*x)^(1/2)*(b^ 8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*70i)/(72*a^3* b*c^3*d^8 - 50*b^4*c^6*d^5 - 170*a*b^3*c^5*d^6 - 162*a^4*c^2*d^9 + (54*a^5 *c*d^10)/b + 196*a^2*b^2*c^4*d^7 + (60*b^5*c^7*d^4)/a) - (c^3*d^6*(c + d*x )^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*90 i)/(54*a^4*c*d^10 - 170*b^4*c^5*d^6 + 196*a*b^3*c^4*d^7 - 162*a^3*b*c^2*d^ 9 + 72*a^2*b^2*c^3*d^8 - (50*b^5*c^6*d^5)/a + (60*b^6*c^7*d^4)/a^2) + (c^5 *d^4*(c + d*x)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^ 2*d)^(1/2)*60i)/(72*a^4*c^3*d^8 + 60*b^4*c^7*d^4 - 50*a*b^3*c^6*d^5 + 1...