Integrand size = 17, antiderivative size = 162 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {5 d^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}} \]
-5/24*d*(d*x+c)^(3/2)/b^2/(b*x+a)^3-1/4*(d*x+c)^(5/2)/b/(b*x+a)^4+5/64*d^4 *arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(3/2)- 5/32*d^2*(d*x+c)^(1/2)/b^3/(b*x+a)^2-5/64*d^3*(d*x+c)^(1/2)/b^3/(-a*d+b*c) /(b*x+a)
Time = 0.85 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=\frac {\sqrt {c+d x} \left (15 a^3 d^3+5 a^2 b d^2 (2 c+11 d x)+a b^2 d \left (8 c^2+36 c d x+73 d^2 x^2\right )-b^3 \left (48 c^3+136 c^2 d x+118 c d^2 x^2+15 d^3 x^3\right )\right )}{192 b^3 (b c-a d) (a+b x)^4}+\frac {5 d^4 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{64 b^{7/2} (-b c+a d)^{3/2}} \]
(Sqrt[c + d*x]*(15*a^3*d^3 + 5*a^2*b*d^2*(2*c + 11*d*x) + a*b^2*d*(8*c^2 + 36*c*d*x + 73*d^2*x^2) - b^3*(48*c^3 + 136*c^2*d*x + 118*c*d^2*x^2 + 15*d ^3*x^3)))/(192*b^3*(b*c - a*d)*(a + b*x)^4) + (5*d^4*ArcTan[(Sqrt[b]*Sqrt[ c + d*x])/Sqrt[-(b*c) + a*d]])/(64*b^(7/2)*(-(b*c) + a*d)^(3/2))
Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {51, 51, 51, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {5 d \int \frac {(c+d x)^{3/2}}{(a+b x)^4}dx}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {5 d \left (\frac {d \int \frac {\sqrt {c+d x}}{(a+b x)^3}dx}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {5 d \left (\frac {d \left (\frac {d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {5 d \left (\frac {d \left (\frac {d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {5 d \left (\frac {d \left (\frac {d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {5 d \left (\frac {d \left (\frac {d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}\) |
-1/4*(c + d*x)^(5/2)/(b*(a + b*x)^4) + (5*d*(-1/3*(c + d*x)^(3/2)/(b*(a + b*x)^3) + (d*(-1/2*Sqrt[c + d*x]/(b*(a + b*x)^2) + (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)))/(2*b)))/(8*b)
3.15.9.3.1 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 = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(2 d^{4} \left (\frac {\frac {5 \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {73 \left (d x +c \right )^{\frac {5}{2}}}{384 b}-\frac {55 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{384 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d x +c}}{128 b^{3}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a d -b c \right ) b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(159\) |
| default | \(2 d^{4} \left (\frac {\frac {5 \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {73 \left (d x +c \right )^{\frac {5}{2}}}{384 b}-\frac {55 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{384 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d x +c}}{128 b^{3}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a d -b c \right ) b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(159\) |
| pseudoelliptic | \(-\frac {5 \left (\sqrt {\left (a d -b c \right ) b}\, \left (\left (-b^{3} x^{3}+\frac {73}{15} a \,b^{2} x^{2}+\frac {11}{3} a^{2} b x +a^{3}\right ) d^{3}+\frac {2 b \left (-\frac {59}{5} b^{2} x^{2}+\frac {18}{5} a b x +a^{2}\right ) c \,d^{2}}{3}+\frac {8 b^{2} c^{2} \left (-17 b x +a \right ) d}{15}-\frac {16 b^{3} c^{3}}{5}\right ) \sqrt {d x +c}-d^{4} \left (b x +a \right )^{4} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{64 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right ) b^{3} \left (b x +a \right )^{4}}\) | \(170\) |
2*d^4*((5/128/(a*d-b*c)*(d*x+c)^(7/2)-73/384*(d*x+c)^(5/2)/b-55/384*(a*d-b *c)/b^2*(d*x+c)^(3/2)-5/128/b^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(1/2)) /((d*x+c)*b+a*d-b*c)^4+5/128/(a*d-b*c)/b^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d *x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (134) = 268\).
Time = 0.26 (sec) , antiderivative size = 894, normalized size of antiderivative = 5.52 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx =\text {Too large to display} \]
[-1/384*(15*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d ^4*x + a^4*d^4)*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*(48*b^5*c^4 - 56*a*b^4*c^3*d - 2* a^2*b^3*c^2*d^2 - 5*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d ^4)*x^3 + (118*b^5*c^2*d^2 - 191*a*b^4*c*d^3 + 73*a^2*b^3*d^4)*x^2 + (136* b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55*a^3*b^2*d^4)*x)*sqrt (d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c^2 - 2*a*b^ 9*c*d + a^2*b^8*d^2)*x^4 + 4*(a*b^9*c^2 - 2*a^2*b^8*c*d + a^3*b^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^2 + 4*(a^3*b^7*c^2 - 2* a^4*b^6*c*d + a^5*b^5*d^2)*x), -1/192*(15*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(-b^2*c + a*b*d)*arctan( sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (48*b^5*c^4 - 56*a*b^4 *c^3*d - 2*a^2*b^3*c^2*d^2 - 5*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^ 3 - a*b^4*d^4)*x^3 + (118*b^5*c^2*d^2 - 191*a*b^4*c*d^3 + 73*a^2*b^3*d^4)* x^2 + (136*b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55*a^3*b^2*d ^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c ^2 - 2*a*b^9*c*d + a^2*b^8*d^2)*x^4 + 4*(a*b^9*c^2 - 2*a^2*b^8*c*d + a^3*b ^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^2 + 4*(a^3*b ^7*c^2 - 2*a^4*b^6*c*d + a^5*b^5*d^2)*x)]
Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.60 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=-\frac {5 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c - a b^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} + 73 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} - 55 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} + 15 \, \sqrt {d x + c} b^{3} c^{3} d^{4} - 73 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} + 110 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} - 45 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} - 55 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} + 45 \, \sqrt {d x + c} a^{2} b c d^{6} - 15 \, \sqrt {d x + c} a^{3} d^{7}}{192 \, {\left (b^{4} c - a b^{3} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \]
-5/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c - a*b^3*d)* sqrt(-b^2*c + a*b*d)) - 1/192*(15*(d*x + c)^(7/2)*b^3*d^4 + 73*(d*x + c)^( 5/2)*b^3*c*d^4 - 55*(d*x + c)^(3/2)*b^3*c^2*d^4 + 15*sqrt(d*x + c)*b^3*c^3 *d^4 - 73*(d*x + c)^(5/2)*a*b^2*d^5 + 110*(d*x + c)^(3/2)*a*b^2*c*d^5 - 45 *sqrt(d*x + c)*a*b^2*c^2*d^5 - 55*(d*x + c)^(3/2)*a^2*b*d^6 + 45*sqrt(d*x + c)*a^2*b*c*d^6 - 15*sqrt(d*x + c)*a^3*d^7)/((b^4*c - a*b^3*d)*((d*x + c) *b - b*c + a*d)^4)
Time = 0.42 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=\frac {5\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,b^{7/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {\frac {73\,d^4\,{\left (c+d\,x\right )}^{5/2}}{192\,b}-\frac {5\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,\left (a\,d-b\,c\right )}+\frac {5\,d^4\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{64\,b^3}+\frac {55\,d^4\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{192\,b^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3} \]
(5*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(7/2)*(a*d - b*c)^(3/2)) - ((73*d^4*(c + d*x)^(5/2))/(192*b) - (5*d^4*(c + d*x)^(7/2 ))/(64*(a*d - b*c)) + (5*d^4*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c* d))/(64*b^3) + (55*d^4*(a*d - b*c)*(c + d*x)^(3/2))/(192*b^2))/(b^4*(c + d *x)^4 - (4*b^4*c - 4*a*b^3*d)*(c + d*x)^3 - (c + d*x)*(4*b^4*c^3 - 4*a^3*b *d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d) + a^4*d^4 + b^4*c^4 + (c + d*x)^ 2*(6*b^4*c^2 + 6*a^2*b^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3 *c^3*d - 4*a^3*b*c*d^3)
Time = 0.00 (sec) , antiderivative size = 661, normalized size of antiderivative = 4.08 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx=\frac {15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{4} d^{4}+60 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} b \,d^{4} x +90 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b^{2} d^{4} x^{2}+60 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{3} d^{4} x^{3}+15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{4} d^{4} x^{4}-15 \sqrt {d x +c}\, a^{4} b \,d^{4}+5 \sqrt {d x +c}\, a^{3} b^{2} c \,d^{3}-55 \sqrt {d x +c}\, a^{3} b^{2} d^{4} x +2 \sqrt {d x +c}\, a^{2} b^{3} c^{2} d^{2}+19 \sqrt {d x +c}\, a^{2} b^{3} c \,d^{3} x -73 \sqrt {d x +c}\, a^{2} b^{3} d^{4} x^{2}+56 \sqrt {d x +c}\, a \,b^{4} c^{3} d +172 \sqrt {d x +c}\, a \,b^{4} c^{2} d^{2} x +191 \sqrt {d x +c}\, a \,b^{4} c \,d^{3} x^{2}+15 \sqrt {d x +c}\, a \,b^{4} d^{4} x^{3}-48 \sqrt {d x +c}\, b^{5} c^{4}-136 \sqrt {d x +c}\, b^{5} c^{3} d x -118 \sqrt {d x +c}\, b^{5} c^{2} d^{2} x^{2}-15 \sqrt {d x +c}\, b^{5} c \,d^{3} x^{3}}{192 b^{4} \left (a^{2} b^{4} d^{2} x^{4}-2 a \,b^{5} c d \,x^{4}+b^{6} c^{2} x^{4}+4 a^{3} b^{3} d^{2} x^{3}-8 a^{2} b^{4} c d \,x^{3}+4 a \,b^{5} c^{2} x^{3}+6 a^{4} b^{2} d^{2} x^{2}-12 a^{3} b^{3} c d \,x^{2}+6 a^{2} b^{4} c^{2} x^{2}+4 a^{5} b \,d^{2} x -8 a^{4} b^{2} c d x +4 a^{3} b^{3} c^{2} x +a^{6} d^{2}-2 a^{5} b c d +a^{4} b^{2} c^{2}\right )} \]
int((sqrt(c + d*x)*(c**2 + 2*c*d*x + d**2*x**2))/(a**5 + 5*a**4*b*x + 10*a **3*b**2*x**2 + 10*a**2*b**3*x**3 + 5*a*b**4*x**4 + b**5*x**5),x)
(15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c )))*a**4*d**4 + 60*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b) *sqrt(a*d - b*c)))*a**3*b*d**4*x + 90*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b**2*d**4*x**2 + 60*sqrt(b)*sqr t(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**3*d**4 *x**3 + 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a* d - b*c)))*b**4*d**4*x**4 - 15*sqrt(c + d*x)*a**4*b*d**4 + 5*sqrt(c + d*x) *a**3*b**2*c*d**3 - 55*sqrt(c + d*x)*a**3*b**2*d**4*x + 2*sqrt(c + d*x)*a* *2*b**3*c**2*d**2 + 19*sqrt(c + d*x)*a**2*b**3*c*d**3*x - 73*sqrt(c + d*x) *a**2*b**3*d**4*x**2 + 56*sqrt(c + d*x)*a*b**4*c**3*d + 172*sqrt(c + d*x)* a*b**4*c**2*d**2*x + 191*sqrt(c + d*x)*a*b**4*c*d**3*x**2 + 15*sqrt(c + d* x)*a*b**4*d**4*x**3 - 48*sqrt(c + d*x)*b**5*c**4 - 136*sqrt(c + d*x)*b**5* c**3*d*x - 118*sqrt(c + d*x)*b**5*c**2*d**2*x**2 - 15*sqrt(c + d*x)*b**5*c *d**3*x**3)/(192*b**4*(a**6*d**2 - 2*a**5*b*c*d + 4*a**5*b*d**2*x + a**4*b **2*c**2 - 8*a**4*b**2*c*d*x + 6*a**4*b**2*d**2*x**2 + 4*a**3*b**3*c**2*x - 12*a**3*b**3*c*d*x**2 + 4*a**3*b**3*d**2*x**3 + 6*a**2*b**4*c**2*x**2 - 8*a**2*b**4*c*d*x**3 + a**2*b**4*d**2*x**4 + 4*a*b**5*c**2*x**3 - 2*a*b**5 *c*d*x**4 + b**6*c**2*x**4))