Integrand size = 17, antiderivative size = 208 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=-\frac {3 d \sqrt {c+d x}}{40 b^2 (a+b x)^4}-\frac {d^2 \sqrt {c+d x}}{80 b^2 (b c-a d) (a+b x)^3}+\frac {d^3 \sqrt {c+d x}}{64 b^2 (b c-a d)^2 (a+b x)^2}-\frac {3 d^4 \sqrt {c+d x}}{128 b^2 (b c-a d)^3 (a+b x)}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}+\frac {3 d^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{5/2} (b c-a d)^{7/2}} \]
-1/5*(d*x+c)^(3/2)/b/(b*x+a)^5+3/128*d^5*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a *d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(7/2)-3/40*d*(d*x+c)^(1/2)/b^2/(b*x+a)^4 -1/80*d^2*(d*x+c)^(1/2)/b^2/(-a*d+b*c)/(b*x+a)^3+1/64*d^3*(d*x+c)^(1/2)/b^ 2/(-a*d+b*c)^2/(b*x+a)^2-3/128*d^4*(d*x+c)^(1/2)/b^2/(-a*d+b*c)^3/(b*x+a)
Time = 1.56 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=\frac {\frac {\sqrt {b} \sqrt {c+d x} \left (-15 a^4 d^4-10 a^3 b d^3 (c+7 d x)+2 a^2 b^2 d^2 \left (124 c^2+233 c d x+64 d^2 x^2\right )-2 a b^3 d \left (168 c^3+256 c^2 d x+23 c d^2 x^2-35 d^3 x^3\right )+b^4 \left (128 c^4+176 c^3 d x+8 c^2 d^2 x^2-10 c d^3 x^3+15 d^4 x^4\right )\right )}{(-b c+a d)^3 (a+b x)^5}+\frac {15 d^5 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}}{640 b^{5/2}} \]
((Sqrt[b]*Sqrt[c + d*x]*(-15*a^4*d^4 - 10*a^3*b*d^3*(c + 7*d*x) + 2*a^2*b^ 2*d^2*(124*c^2 + 233*c*d*x + 64*d^2*x^2) - 2*a*b^3*d*(168*c^3 + 256*c^2*d* x + 23*c*d^2*x^2 - 35*d^3*x^3) + b^4*(128*c^4 + 176*c^3*d*x + 8*c^2*d^2*x^ 2 - 10*c*d^3*x^3 + 15*d^4*x^4)))/((-(b*c) + a*d)^3*(a + b*x)^5) + (15*d^5* ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2))/ (640*b^(5/2))
Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {51, 51, 52, 52, 52, 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.
| \(\displaystyle \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {3 d \int \frac {\sqrt {c+d x}}{(a+b x)^5}dx}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {3 d \left (\frac {d \int \frac {1}{(a+b x)^4 \sqrt {c+d x}}dx}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {5 d \int \frac {1}{(a+b x)^3 \sqrt {c+d x}}dx}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {5 d \left (-\frac {3 d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {5 d \left (-\frac {3 d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {5 d \left (-\frac {3 d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {5 d \left (-\frac {3 d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 b}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}\right )}{10 b}-\frac {(c+d x)^{3/2}}{5 b (a+b x)^5}\) |
-1/5*(c + d*x)^(3/2)/(b*(a + b*x)^5) + (3*d*(-1/4*Sqrt[c + d*x]/(b*(a + b* x)^4) + (d*(-1/3*Sqrt[c + d*x]/((b*c - a*d)*(a + b*x)^3) - (5*d*(-1/2*Sqrt [c + d*x]/((b*c - a*d)*(a + b*x)^2) - (3*d*(-(Sqrt[c + d*x]/((b*c - a*d)*( a + b*x))) + (d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(Sqrt[b] *(b*c - a*d)^(3/2))))/(4*(b*c - a*d))))/(6*(b*c - a*d))))/(8*b)))/(10*b)
3.14.98.3.1 Defintions of rubi rules used
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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_)^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 = 0.37 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.05
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\left (\left (-b^{4} x^{4}-\frac {14}{3} a \,b^{3} x^{3}-\frac {128}{15} a^{2} b^{2} x^{2}+\frac {14}{3} a^{3} b x +a^{4}\right ) d^{4}+\frac {2 \left (b^{3} x^{3}+\frac {23}{5} a \,b^{2} x^{2}-\frac {233}{5} a^{2} b x +a^{3}\right ) b c \,d^{3}}{3}-\frac {248 b^{2} c^{2} \left (\frac {1}{31} b^{2} x^{2}-\frac {64}{31} a b x +a^{2}\right ) d^{2}}{15}+\frac {112 b^{3} \left (-\frac {11 b x}{21}+a \right ) c^{3} d}{5}-\frac {128 b^{4} c^{4}}{15}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}-d^{5} \left (b x +a \right )^{5} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{128 \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{5} b^{2} \left (a d -b c \right )^{3}}\) | \(219\) |
| derivativedivides | \(2 d^{5} \left (\frac {\frac {3 b^{2} \left (d x +c \right )^{\frac {9}{2}}}{256 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}+\frac {7 b \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {\left (d x +c \right )^{\frac {5}{2}}}{10 a d -10 b c}-\frac {7 \left (d x +c \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a d -b c \right ) \sqrt {d x +c}}{256 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(237\) |
| default | \(2 d^{5} \left (\frac {\frac {3 b^{2} \left (d x +c \right )^{\frac {9}{2}}}{256 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}+\frac {7 b \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {\left (d x +c \right )^{\frac {5}{2}}}{10 a d -10 b c}-\frac {7 \left (d x +c \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a d -b c \right ) \sqrt {d x +c}}{256 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(237\) |
-3/128*(((-b^4*x^4-14/3*a*b^3*x^3-128/15*a^2*b^2*x^2+14/3*a^3*b*x+a^4)*d^4 +2/3*(b^3*x^3+23/5*a*b^2*x^2-233/5*a^2*b*x+a^3)*b*c*d^3-248/15*b^2*c^2*(1/ 31*b^2*x^2-64/31*a*b*x+a^2)*d^2+112/5*b^3*(-11/21*b*x+a)*c^3*d-128/15*b^4* c^4)*((a*d-b*c)*b)^(1/2)*(d*x+c)^(1/2)-d^5*(b*x+a)^5*arctan(b*(d*x+c)^(1/2 )/((a*d-b*c)*b)^(1/2)))/((a*d-b*c)*b)^(1/2)/(b*x+a)^5/b^2/(a*d-b*c)^3
Leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (176) = 352\).
Time = 0.27 (sec) , antiderivative size = 1492, normalized size of antiderivative = 7.17 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=\text {Too large to display} \]
[-1/1280*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3* b^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2* b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(128*b^6*c ^5 - 464*a*b^5*c^4*d + 584*a^2*b^4*c^3*d^2 - 258*a^3*b^3*c^2*d^3 - 5*a^4*b ^2*c*d^4 + 15*a^5*b*d^5 + 15*(b^6*c*d^4 - a*b^5*d^5)*x^4 - 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(4*b^6*c^3*d^2 - 27*a*b^5*c^2*d^ 3 + 87*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(88*b^6*c^4*d - 344*a*b^5*c ^3*d^2 + 489*a^2*b^4*c^2*d^3 - 268*a^3*b^3*c*d^4 + 35*a^4*b^2*d^5)*x)*sqrt (d*x + c))/(a^5*b^7*c^4 - 4*a^6*b^6*c^3*d + 6*a^7*b^5*c^2*d^2 - 4*a^8*b^4* c*d^3 + a^9*b^3*d^4 + (b^12*c^4 - 4*a*b^11*c^3*d + 6*a^2*b^10*c^2*d^2 - 4* a^3*b^9*c*d^3 + a^4*b^8*d^4)*x^5 + 5*(a*b^11*c^4 - 4*a^2*b^10*c^3*d + 6*a^ 3*b^9*c^2*d^2 - 4*a^4*b^8*c*d^3 + a^5*b^7*d^4)*x^4 + 10*(a^2*b^10*c^4 - 4* a^3*b^9*c^3*d + 6*a^4*b^8*c^2*d^2 - 4*a^5*b^7*c*d^3 + a^6*b^6*d^4)*x^3 + 1 0*(a^3*b^9*c^4 - 4*a^4*b^8*c^3*d + 6*a^5*b^7*c^2*d^2 - 4*a^6*b^6*c*d^3 + a ^7*b^5*d^4)*x^2 + 5*(a^4*b^8*c^4 - 4*a^5*b^7*c^3*d + 6*a^6*b^6*c^2*d^2 - 4 *a^7*b^5*c*d^3 + a^8*b^4*d^4)*x), -1/640*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^ 4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5)*sqr t(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (128*b^6*c^5 - 464*a*b^5*c^4*d + 584*a^2*b^4*c^3*d^2 - 258*a^3*b^3*c^2* d^3 - 5*a^4*b^2*c*d^4 + 15*a^5*b*d^5 + 15*(b^6*c*d^4 - a*b^5*d^5)*x^4 -...
Timed out. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (176) = 352\).
Time = 0.32 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.97 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=-\frac {3 \, d^{5} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{128 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {15 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{4} d^{5} - 70 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} c d^{5} + 128 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} c^{2} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c^{3} d^{5} - 15 \, \sqrt {d x + c} b^{4} c^{4} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {7}{2}} a b^{3} d^{6} - 256 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{3} c d^{6} - 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} c^{2} d^{6} + 60 \, \sqrt {d x + c} a b^{3} c^{3} d^{6} + 128 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} b^{2} d^{7} + 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} c d^{7} - 90 \, \sqrt {d x + c} a^{2} b^{2} c^{2} d^{7} - 70 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{3} b d^{8} + 60 \, \sqrt {d x + c} a^{3} b c d^{8} - 15 \, \sqrt {d x + c} a^{4} d^{9}}{640 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{5}} \]
-3/128*d^5*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^3 - 3*a*b^ 4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*sqrt(-b^2*c + a*b*d)) - 1/640*(15 *(d*x + c)^(9/2)*b^4*d^5 - 70*(d*x + c)^(7/2)*b^4*c*d^5 + 128*(d*x + c)^(5 /2)*b^4*c^2*d^5 + 70*(d*x + c)^(3/2)*b^4*c^3*d^5 - 15*sqrt(d*x + c)*b^4*c^ 4*d^5 + 70*(d*x + c)^(7/2)*a*b^3*d^6 - 256*(d*x + c)^(5/2)*a*b^3*c*d^6 - 2 10*(d*x + c)^(3/2)*a*b^3*c^2*d^6 + 60*sqrt(d*x + c)*a*b^3*c^3*d^6 + 128*(d *x + c)^(5/2)*a^2*b^2*d^7 + 210*(d*x + c)^(3/2)*a^2*b^2*c*d^7 - 90*sqrt(d* x + c)*a^2*b^2*c^2*d^7 - 70*(d*x + c)^(3/2)*a^3*b*d^8 + 60*sqrt(d*x + c)*a ^3*b*c*d^8 - 15*sqrt(d*x + c)*a^4*d^9)/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b ^3*c*d^2 - a^3*b^2*d^3)*((d*x + c)*b - b*c + a*d)^5)
Time = 0.51 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^6} \, dx=\frac {\frac {d^5\,{\left (c+d\,x\right )}^{5/2}}{5\,\left (a\,d-b\,c\right )}-\frac {7\,d^5\,{\left (c+d\,x\right )}^{3/2}}{64\,b}+\frac {3\,b^2\,d^5\,{\left (c+d\,x\right )}^{9/2}}{128\,{\left (a\,d-b\,c\right )}^3}-\frac {3\,d^5\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{128\,b^2}+\frac {7\,b\,d^5\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^2}}{b^5\,{\left (c+d\,x\right )}^5-{\left (c+d\,x\right )}^2\,\left (-10\,a^3\,b^2\,d^3+30\,a^2\,b^3\,c\,d^2-30\,a\,b^4\,c^2\,d+10\,b^5\,c^3\right )-\left (5\,b^5\,c-5\,a\,b^4\,d\right )\,{\left (c+d\,x\right )}^4+a^5\,d^5-b^5\,c^5+{\left (c+d\,x\right )}^3\,\left (10\,a^2\,b^3\,d^2-20\,a\,b^4\,c\,d+10\,b^5\,c^2\right )+\left (c+d\,x\right )\,\left (5\,a^4\,b\,d^4-20\,a^3\,b^2\,c\,d^3+30\,a^2\,b^3\,c^2\,d^2-20\,a\,b^4\,c^3\,d+5\,b^5\,c^4\right )-10\,a^2\,b^3\,c^3\,d^2+10\,a^3\,b^2\,c^2\,d^3+5\,a\,b^4\,c^4\,d-5\,a^4\,b\,c\,d^4}+\frac {3\,d^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{128\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{7/2}} \]
((d^5*(c + d*x)^(5/2))/(5*(a*d - b*c)) - (7*d^5*(c + d*x)^(3/2))/(64*b) + (3*b^2*d^5*(c + d*x)^(9/2))/(128*(a*d - b*c)^3) - (3*d^5*(a*d - b*c)*(c + d*x)^(1/2))/(128*b^2) + (7*b*d^5*(c + d*x)^(7/2))/(64*(a*d - b*c)^2))/(b^5 *(c + d*x)^5 - (c + d*x)^2*(10*b^5*c^3 - 10*a^3*b^2*d^3 + 30*a^2*b^3*c*d^2 - 30*a*b^4*c^2*d) - (5*b^5*c - 5*a*b^4*d)*(c + d*x)^4 + a^5*d^5 - b^5*c^5 + (c + d*x)^3*(10*b^5*c^2 + 10*a^2*b^3*d^2 - 20*a*b^4*c*d) + (c + d*x)*(5 *b^5*c^4 + 5*a^4*b*d^4 - 20*a^3*b^2*c*d^3 + 30*a^2*b^3*c^2*d^2 - 20*a*b^4* c^3*d) - 10*a^2*b^3*c^3*d^2 + 10*a^3*b^2*c^2*d^3 + 5*a*b^4*c^4*d - 5*a^4*b *c*d^4) + (3*d^5*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(128*b ^(5/2)*(a*d - b*c)^(7/2))