Integrand size = 20, antiderivative size = 151 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}+\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}} \]
-1/2*c*(d*x+c)^(3/2)/a/x^2+2*(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x+c)^(1/2 )/(-a*d+b*c)^(1/2))/a^3/b^(1/2)-1/4*(15*a^2*d^2-20*a*b*c*d+8*b^2*c^2)*arct anh((d*x+c)^(1/2)/c^(1/2))*c^(1/2)/a^3+1/4*c*(-7*a*d+4*b*c)*(d*x+c)^(1/2)/ a^2/x
Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {\frac {a c \sqrt {c+d x} (-2 a c+4 b c x-9 a d x)}{x^2}+\frac {8 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b}}-\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3} \]
((a*c*Sqrt[c + d*x]*(-2*a*c + 4*b*c*x - 9*a*d*x))/x^2 + (8*(-(b*c) + a*d)^ (5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/Sqrt[b] - Sqrt[c ]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4 *a^3)
Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {109, 27, 166, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (4 b c-7 a d)+d (b c-4 a d) x)}{2 x^2 (a+b x)}dx}{2 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (4 b c-7 a d)+d (b c-4 a d) x)}{x^2 (a+b x)}dx}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle -\frac {\frac {\int -\frac {c \left (8 b^2 c^2-20 a b d c+15 a^2 d^2\right )+d \left (4 b^2 c^2-9 a b d c+8 a^2 d^2\right ) x}{2 x (a+b x) \sqrt {c+d x}}dx}{a}-\frac {c \sqrt {c+d x} (4 b c-7 a d)}{a x}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {c \left (8 b^2 c^2-20 a b d c+15 a^2 d^2\right )+d \left (4 b^2 c^2-9 a b d c+8 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {c+d x} (4 b c-7 a d)}{a x}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {-\frac {\frac {c \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {8 (b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{2 a}-\frac {c \sqrt {c+d x} (4 b c-7 a d)}{a x}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {-\frac {\frac {2 c \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {16 (b c-a d)^3 \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{2 a}-\frac {c \sqrt {c+d x} (4 b c-7 a d)}{a x}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {-\frac {\frac {16 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {2 \sqrt {c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}}{2 a}-\frac {c \sqrt {c+d x} (4 b c-7 a d)}{a x}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2}\) |
-1/2*(c*(c + d*x)^(3/2))/(a*x^2) - (-((c*(4*b*c - 7*a*d)*Sqrt[c + d*x])/(a *x)) - ((-2*Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a + (16*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/ Sqrt[b*c - a*d]])/(a*Sqrt[b]))/(2*a))/(4*a)
3.5.61.3.1 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[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((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 - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
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 = 124, normalized size of antiderivative = 0.82
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}+\frac {c \left (-\frac {\sqrt {d x +c}\, a \left (9 a d x -4 b c x +2 a c \right )}{x^{2}}-\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{4}}{a^{3}}\) | \(124\) |
| risch | \(-\frac {c \sqrt {d x +c}\, \left (9 a d x -4 b c x +2 a c \right )}{4 a^{2} x^{2}}+\frac {d \left (\frac {\left (8 a^{3} d^{3}-24 a^{2} b c \,d^{2}+24 a \,b^{2} c^{2} d -8 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a d \sqrt {\left (a d -b c \right ) b}}-\frac {\sqrt {c}\, \left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )}{4 a^{2}}\) | \(164\) |
| derivativedivides | \(2 d^{3} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-\frac {1}{2} a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+\frac {1}{2} b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{3} d^{3}}+\frac {\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{3} d^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(165\) |
| default | \(2 d^{3} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-\frac {1}{2} a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+\frac {1}{2} b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{3} d^{3}}+\frac {\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{3} d^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(165\) |
1/a^3*(2*(a*d-b*c)^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c) *b)^(1/2))+1/4*c*(-(d*x+c)^(1/2)*a*(9*a*d*x-4*b*c*x+2*a*c)/x^2-(15*a^2*d^2 -20*a*b*c*d+8*b^2*c^2)/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2))))
Time = 0.35 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.72 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\left [\frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {16 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}, \frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}\right ] \]
[1/8*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*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)) + (8*b^ 2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*sq rt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c)) /(a^3*x^2), 1/8*(16*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/ b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3* x^2), 1/4*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan(sqrt( d*x + c)*sqrt(-c)/c) + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*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*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^ 2), 1/4*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/b)*arctan (-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (8*b^2*c^2 - 20*a*b* c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (2*a^2*c ^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2)]
\[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{3} \left (a + b x\right )}\, dx \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=-\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{3}} + \frac {{\left (8 \, b^{2} c^{3} - 20 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, a^{3} \sqrt {-c}} + \frac {4 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d - 4 \, \sqrt {d x + c} b c^{3} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 7 \, \sqrt {d x + c} a c^{2} d^{2}}{4 \, a^{2} d^{2} x^{2}} \]
-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - 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^3) + 1/4*(8*b^2*c^3 - 20 *a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c)) + 1/4*(4*(d*x + c)^(3/2)*b*c^2*d - 4*sqrt(d*x + c)*b*c^3*d - 9*(d*x + c)^(3 /2)*a*c*d^2 + 7*sqrt(d*x + c)*a*c^2*d^2)/(a^2*d^2*x^2)
Time = 0.75 (sec) , antiderivative size = 1204, normalized size of antiderivative = 7.97 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {\frac {\left (7\,a\,c^2\,d^2-4\,b\,c^3\,d\right )\,\sqrt {c+d\,x}}{4\,a^2}-\frac {\left (9\,a\,c\,d^2-4\,b\,c^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{4\,a^2}}{{\left (c+d\,x\right )}^2-2\,c\,\left (c+d\,x\right )+c^2}+\frac {2\,\mathrm {atanh}\left (\frac {95\,b^2\,c^2\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{4\,\left (\frac {215\,b^5\,c^5\,d^6}{4}-\frac {469\,a\,b^4\,c^4\,d^7}{4}+\frac {517\,a^2\,b^3\,c^3\,d^8}{4}-\frac {287\,a^3\,b^2\,c^2\,d^9}{4}-\frac {10\,b^6\,c^6\,d^5}{a}+16\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^3\,c^3\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{-16\,a^5\,b\,c\,d^{10}+\frac {287\,a^4\,b^2\,c^2\,d^9}{4}-\frac {517\,a^3\,b^3\,c^3\,d^8}{4}+\frac {469\,a^2\,b^4\,c^4\,d^7}{4}-\frac {215\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}+\frac {16\,b\,c\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{\frac {469\,b^4\,c^4\,d^7}{4}-\frac {517\,a\,b^3\,c^3\,d^8}{4}+\frac {287\,a^2\,b^2\,c^2\,d^9}{4}-\frac {215\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-16\,a^3\,b\,c\,d^{10}}\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^5}}{a^3\,b}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {3665\,b^2\,c^{3/2}\,d^9\,\sqrt {c+d\,x}}{32\,\left (\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}\right )}+\frac {5717\,b^3\,c^{5/2}\,d^8\,\sqrt {c+d\,x}}{32\,\left (\frac {5717\,b^3\,c^3\,d^8}{32}-\frac {3665\,a\,b^2\,c^2\,d^9}{32}-\frac {1143\,b^4\,c^4\,d^7}{8\,a}+\frac {235\,b^5\,c^5\,d^6}{4\,a^2}-\frac {10\,b^6\,c^6\,d^5}{a^3}+30\,a^2\,b\,c\,d^{10}\right )}+\frac {1143\,b^4\,c^{7/2}\,d^7\,\sqrt {c+d\,x}}{8\,\left (\frac {1143\,b^4\,c^4\,d^7}{8}-\frac {5717\,a\,b^3\,c^3\,d^8}{32}+\frac {3665\,a^2\,b^2\,c^2\,d^9}{32}-\frac {235\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-30\,a^3\,b\,c\,d^{10}\right )}+\frac {235\,b^5\,c^{9/2}\,d^6\,\sqrt {c+d\,x}}{4\,\left (\frac {235\,b^5\,c^5\,d^6}{4}-\frac {1143\,a\,b^4\,c^4\,d^7}{8}+\frac {5717\,a^2\,b^3\,c^3\,d^8}{32}-\frac {3665\,a^3\,b^2\,c^2\,d^9}{32}-\frac {10\,b^6\,c^6\,d^5}{a}+30\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^6\,c^{11/2}\,d^5\,\sqrt {c+d\,x}}{-30\,a^5\,b\,c\,d^{10}+\frac {3665\,a^4\,b^2\,c^2\,d^9}{32}-\frac {5717\,a^3\,b^3\,c^3\,d^8}{32}+\frac {1143\,a^2\,b^4\,c^4\,d^7}{8}-\frac {235\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}-\frac {30\,a\,b\,\sqrt {c}\,d^{10}\,\sqrt {c+d\,x}}{\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}}\right )\,\left (15\,a^2\,d^2-20\,a\,b\,c\,d+8\,b^2\,c^2\right )}{4\,a^3} \]
(((7*a*c^2*d^2 - 4*b*c^3*d)*(c + d*x)^(1/2))/(4*a^2) - ((9*a*c*d^2 - 4*b*c ^2*d)*(c + d*x)^(3/2))/(4*a^2))/((c + d*x)^2 - 2*c*(c + d*x) + c^2) + (2*a tanh((95*b^2*c^2*d^6*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^ 4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(4*((2 15*b^5*c^5*d^6)/4 - (469*a*b^4*c^4*d^7)/4 + (517*a^2*b^3*c^3*d^8)/4 - (287 *a^3*b^2*c^2*d^9)/4 - (10*b^6*c^6*d^5)/a + 16*a^4*b*c*d^10)) + (10*b^3*c^3 *d^5*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c ^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(10*b^6*c^6*d^5 - (215 *a*b^5*c^5*d^6)/4 + (469*a^2*b^4*c^4*d^7)/4 - (517*a^3*b^3*c^3*d^8)/4 + (2 87*a^4*b^2*c^2*d^9)/4 - 16*a^5*b*c*d^10) + (16*b*c*d^7*(c + d*x)^(1/2)*(b^ 6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2* d^3 - 5*a*b^5*c^4*d)^(1/2))/((469*b^4*c^4*d^7)/4 - (517*a*b^3*c^3*d^8)/4 + (287*a^2*b^2*c^2*d^9)/4 - (215*b^5*c^5*d^6)/(4*a) + (10*b^6*c^6*d^5)/a^2 - 16*a^3*b*c*d^10))*(-b*(a*d - b*c)^5)^(1/2))/(a^3*b) - (c^(1/2)*atanh((36 65*b^2*c^(3/2)*d^9*(c + d*x)^(1/2))/(32*((3665*b^2*c^2*d^9)/32 - 30*a*b*c* d^10 - (5717*b^3*c^3*d^8)/(32*a) + (1143*b^4*c^4*d^7)/(8*a^2) - (235*b^5*c ^5*d^6)/(4*a^3) + (10*b^6*c^6*d^5)/a^4)) + (5717*b^3*c^(5/2)*d^8*(c + d*x) ^(1/2))/(32*((5717*b^3*c^3*d^8)/32 - (3665*a*b^2*c^2*d^9)/32 - (1143*b^4*c ^4*d^7)/(8*a) + (235*b^5*c^5*d^6)/(4*a^2) - (10*b^6*c^6*d^5)/a^3 + 30*a^2* b*c*d^10)) + (1143*b^4*c^(7/2)*d^7*(c + d*x)^(1/2))/(8*((1143*b^4*c^4*d...