Integrand size = 17, antiderivative size = 218 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}-\frac {d \sqrt {c+d x}}{40 b (b c-a d) (a+b x)^4}+\frac {7 d^2 \sqrt {c+d x}}{240 b (b c-a d)^2 (a+b x)^3}-\frac {7 d^3 \sqrt {c+d x}}{192 b (b c-a d)^3 (a+b x)^2}+\frac {7 d^4 \sqrt {c+d x}}{128 b (b c-a d)^4 (a+b x)}-\frac {7 d^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{3/2} (b c-a d)^{9/2}} \]
-7/128*d^5*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b *c)^(9/2)-1/5*(d*x+c)^(1/2)/b/(b*x+a)^5-1/40*d*(d*x+c)^(1/2)/b/(-a*d+b*c)/ (b*x+a)^4+7/240*d^2*(d*x+c)^(1/2)/b/(-a*d+b*c)^2/(b*x+a)^3-7/192*d^3*(d*x+ c)^(1/2)/b/(-a*d+b*c)^3/(b*x+a)^2+7/128*d^4*(d*x+c)^(1/2)/b/(-a*d+b*c)^4/( b*x+a)
Time = 1.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\frac {\sqrt {c+d x} \left (-105 a^4 d^4+10 a^3 b d^3 (121 c+79 d x)+2 a^2 b^2 d^2 \left (-1052 c^2-289 c d x+448 d^2 x^2\right )+2 a b^3 d \left (744 c^3+128 c^2 d x-161 c d^2 x^2+245 d^3 x^3\right )+b^4 \left (-384 c^4-48 c^3 d x+56 c^2 d^2 x^2-70 c d^3 x^3+105 d^4 x^4\right )\right )}{1920 b (b c-a d)^4 (a+b x)^5}+\frac {7 d^5 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{128 b^{3/2} (-b c+a d)^{9/2}} \]
(Sqrt[c + d*x]*(-105*a^4*d^4 + 10*a^3*b*d^3*(121*c + 79*d*x) + 2*a^2*b^2*d ^2*(-1052*c^2 - 289*c*d*x + 448*d^2*x^2) + 2*a*b^3*d*(744*c^3 + 128*c^2*d* x - 161*c*d^2*x^2 + 245*d^3*x^3) + b^4*(-384*c^4 - 48*c^3*d*x + 56*c^2*d^2 *x^2 - 70*c*d^3*x^3 + 105*d^4*x^4)))/(1920*b*(b*c - a*d)^4*(a + b*x)^5) + (7*d^5*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(128*b^(3/2)*(- (b*c) + a*d)^(9/2))
Time = 0.28 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {51, 52, 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.
| \(\displaystyle \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {d \int \frac {1}{(a+b x)^5 \sqrt {c+d x}}dx}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {d \left (-\frac {7 d \int \frac {1}{(a+b x)^4 \sqrt {c+d x}}dx}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {d \left (-\frac {7 d \left (-\frac {5 d \int \frac {1}{(a+b x)^3 \sqrt {c+d x}}dx}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {d \left (-\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {d \left (-\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {d \left (-\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {d \left (-\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 (b c-a d)}-\frac {\sqrt {c+d x}}{3 (a+b x)^3 (b c-a d)}\right )}{8 (b c-a d)}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)}\right )}{10 b}-\frac {\sqrt {c+d x}}{5 b (a+b x)^5}\) |
-1/5*Sqrt[c + d*x]/(b*(a + b*x)^5) + (d*(-1/4*Sqrt[c + d*x]/((b*c - a*d)*( a + b*x)^4) - (7*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*c - a*d))))/(10*b)
3.14.86.3.1 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 = 0.38 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.01
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {7 \sqrt {\left (a d -b c \right ) b}\, \left (\left (-b^{4} x^{4}-\frac {14}{3} a \,b^{3} x^{3}-\frac {128}{15} a^{2} b^{2} x^{2}-\frac {158}{21} a^{3} b x +a^{4}\right ) d^{4}-\frac {242 \left (-\frac {7}{121} b^{3} x^{3}-\frac {161}{605} a \,b^{2} x^{2}-\frac {289}{605} a^{2} b x +a^{3}\right ) b c \,d^{3}}{21}+\frac {2104 b^{2} c^{2} \left (-\frac {7}{263} b^{2} x^{2}-\frac {32}{263} a b x +a^{2}\right ) d^{2}}{105}-\frac {496 \left (-\frac {b x}{31}+a \right ) b^{3} c^{3} d}{35}+\frac {128 b^{4} c^{4}}{35}\right ) \sqrt {d x +c}}{128}+\frac {7 d^{5} \left (b x +a \right )^{5} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128}}{\sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{5} b \left (a d -b c \right )^{4}}\) | \(220\) |
| derivativedivides | \(2 d^{5} \left (\frac {\frac {7 b^{3} \left (d x +c \right )^{\frac {9}{2}}}{256 \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}+\frac {49 b^{2} \left (d x +c \right )^{\frac {7}{2}}}{384 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}+\frac {7 b \left (d x +c \right )^{\frac {5}{2}}}{30 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {79 \left (d x +c \right )^{\frac {3}{2}}}{384 \left (a d -b c \right )}-\frac {7 \sqrt {d x +c}}{256 b}}{\left (\left (d x +c \right ) b +a d -b c \right )^{5}}+\frac {7 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\) | \(293\) |
| default | \(2 d^{5} \left (\frac {\frac {7 b^{3} \left (d x +c \right )^{\frac {9}{2}}}{256 \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}+\frac {49 b^{2} \left (d x +c \right )^{\frac {7}{2}}}{384 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}+\frac {7 b \left (d x +c \right )^{\frac {5}{2}}}{30 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {79 \left (d x +c \right )^{\frac {3}{2}}}{384 \left (a d -b c \right )}-\frac {7 \sqrt {d x +c}}{256 b}}{\left (\left (d x +c \right ) b +a d -b c \right )^{5}}+\frac {7 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\) | \(293\) |
7/128*(-((a*d-b*c)*b)^(1/2)*((-b^4*x^4-14/3*a*b^3*x^3-128/15*a^2*b^2*x^2-1 58/21*a^3*b*x+a^4)*d^4-242/21*(-7/121*b^3*x^3-161/605*a*b^2*x^2-289/605*a^ 2*b*x+a^3)*b*c*d^3+2104/105*b^2*c^2*(-7/263*b^2*x^2-32/263*a*b*x+a^2)*d^2- 496/35*(-1/31*b*x+a)*b^3*c^3*d+128/35*b^4*c^4)*(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/(a*d-b*c)^4
Leaf count of result is larger than twice the leaf count of optimal. 830 vs. \(2 (186) = 372\).
Time = 0.27 (sec) , antiderivative size = 1673, normalized size of antiderivative = 7.67 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\text {Too large to display} \]
[1/3840*(105*(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*(384*b^6*c ^5 - 1872*a*b^5*c^4*d + 3592*a^2*b^4*c^3*d^2 - 3314*a^3*b^3*c^2*d^3 + 1315 *a^4*b^2*c*d^4 - 105*a^5*b*d^5 - 105*(b^6*c*d^4 - a*b^5*d^5)*x^4 + 70*(b^6 *c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 - 14*(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*(24*b^6*c^4*d - 15 2*a*b^5*c^3*d^2 + 417*a^2*b^4*c^2*d^3 - 684*a^3*b^3*c*d^4 + 395*a^4*b^2*d^ 5)*x)*sqrt(d*x + c))/(a^5*b^7*c^5 - 5*a^6*b^6*c^4*d + 10*a^7*b^5*c^3*d^2 - 10*a^8*b^4*c^2*d^3 + 5*a^9*b^3*c*d^4 - a^10*b^2*d^5 + (b^12*c^5 - 5*a*b^1 1*c^4*d + 10*a^2*b^10*c^3*d^2 - 10*a^3*b^9*c^2*d^3 + 5*a^4*b^8*c*d^4 - a^5 *b^7*d^5)*x^5 + 5*(a*b^11*c^5 - 5*a^2*b^10*c^4*d + 10*a^3*b^9*c^3*d^2 - 10 *a^4*b^8*c^2*d^3 + 5*a^5*b^7*c*d^4 - a^6*b^6*d^5)*x^4 + 10*(a^2*b^10*c^5 - 5*a^3*b^9*c^4*d + 10*a^4*b^8*c^3*d^2 - 10*a^5*b^7*c^2*d^3 + 5*a^6*b^6*c*d ^4 - a^7*b^5*d^5)*x^3 + 10*(a^3*b^9*c^5 - 5*a^4*b^8*c^4*d + 10*a^5*b^7*c^3 *d^2 - 10*a^6*b^6*c^2*d^3 + 5*a^7*b^5*c*d^4 - a^8*b^4*d^5)*x^2 + 5*(a^4*b^ 8*c^5 - 5*a^5*b^7*c^4*d + 10*a^6*b^6*c^3*d^2 - 10*a^7*b^5*c^2*d^3 + 5*a^8* b^4*c*d^4 - a^9*b^3*d^5)*x), 1/1920*(105*(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)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) ...
Timed out. \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\text {Exception raised: ValueError} \]
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. 432 vs. \(2 (186) = 372\).
Time = 0.32 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.98 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\frac {7 \, d^{5} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{128 \, {\left (b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}\right )} \sqrt {-b^{2} c + a b d}} + \frac {105 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{4} d^{5} - 490 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} c d^{5} + 896 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} c^{2} d^{5} - 790 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c^{3} d^{5} - 105 \, \sqrt {d x + c} b^{4} c^{4} d^{5} + 490 \, {\left (d x + c\right )}^{\frac {7}{2}} a b^{3} d^{6} - 1792 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{3} c d^{6} + 2370 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} c^{2} d^{6} + 420 \, \sqrt {d x + c} a b^{3} c^{3} d^{6} + 896 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} b^{2} d^{7} - 2370 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} c d^{7} - 630 \, \sqrt {d x + c} a^{2} b^{2} c^{2} d^{7} + 790 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{3} b d^{8} + 420 \, \sqrt {d x + c} a^{3} b c d^{8} - 105 \, \sqrt {d x + c} a^{4} d^{9}}{1920 \, {\left (b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{5}} \]
7/128*d^5*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^4 - 4*a*b^4 *c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)*sqrt(-b^2*c + a* b*d)) + 1/1920*(105*(d*x + c)^(9/2)*b^4*d^5 - 490*(d*x + c)^(7/2)*b^4*c*d^ 5 + 896*(d*x + c)^(5/2)*b^4*c^2*d^5 - 790*(d*x + c)^(3/2)*b^4*c^3*d^5 - 10 5*sqrt(d*x + c)*b^4*c^4*d^5 + 490*(d*x + c)^(7/2)*a*b^3*d^6 - 1792*(d*x + c)^(5/2)*a*b^3*c*d^6 + 2370*(d*x + c)^(3/2)*a*b^3*c^2*d^6 + 420*sqrt(d*x + c)*a*b^3*c^3*d^6 + 896*(d*x + c)^(5/2)*a^2*b^2*d^7 - 2370*(d*x + c)^(3/2) *a^2*b^2*c*d^7 - 630*sqrt(d*x + c)*a^2*b^2*c^2*d^7 + 790*(d*x + c)^(3/2)*a ^3*b*d^8 + 420*sqrt(d*x + c)*a^3*b*c*d^8 - 105*sqrt(d*x + c)*a^4*d^9)/((b^ 5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)*( (d*x + c)*b - b*c + a*d)^5)
Time = 0.55 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx=\frac {\frac {79\,d^5\,{\left (c+d\,x\right )}^{3/2}}{192\,\left (a\,d-b\,c\right )}-\frac {7\,d^5\,\sqrt {c+d\,x}}{128\,b}+\frac {49\,b^2\,d^5\,{\left (c+d\,x\right )}^{7/2}}{192\,{\left (a\,d-b\,c\right )}^3}+\frac {7\,b^3\,d^5\,{\left (c+d\,x\right )}^{9/2}}{128\,{\left (a\,d-b\,c\right )}^4}+\frac {7\,b\,d^5\,{\left (c+d\,x\right )}^{5/2}}{15\,{\left (a\,d-b\,c\right )}^2}}{b^5\,{\left (c+d\,x\right )}^5-{\left (c+d\,x\right )}^2\,\left (-10\,a^3\,b^2\,d^3+30\,a^2\,b^3\,c\,d^2-30\,a\,b^4\,c^2\,d+10\,b^5\,c^3\right )-\left (5\,b^5\,c-5\,a\,b^4\,d\right )\,{\left (c+d\,x\right )}^4+a^5\,d^5-b^5\,c^5+{\left (c+d\,x\right )}^3\,\left (10\,a^2\,b^3\,d^2-20\,a\,b^4\,c\,d+10\,b^5\,c^2\right )+\left (c+d\,x\right )\,\left (5\,a^4\,b\,d^4-20\,a^3\,b^2\,c\,d^3+30\,a^2\,b^3\,c^2\,d^2-20\,a\,b^4\,c^3\,d+5\,b^5\,c^4\right )-10\,a^2\,b^3\,c^3\,d^2+10\,a^3\,b^2\,c^2\,d^3+5\,a\,b^4\,c^4\,d-5\,a^4\,b\,c\,d^4}+\frac {7\,d^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{128\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{9/2}} \]
((79*d^5*(c + d*x)^(3/2))/(192*(a*d - b*c)) - (7*d^5*(c + d*x)^(1/2))/(128 *b) + (49*b^2*d^5*(c + d*x)^(7/2))/(192*(a*d - b*c)^3) + (7*b^3*d^5*(c + d *x)^(9/2))/(128*(a*d - b*c)^4) + (7*b*d^5*(c + d*x)^(5/2))/(15*(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) + (7*d^5*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2) ))/(128*b^(3/2)*(a*d - b*c)^(9/2))
Time = 0.01 (sec) , antiderivative size = 1167, normalized size of antiderivative = 5.35 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^6} \, dx =\text {Too large to display} \]
int(sqrt(c + d*x)/(a**6 + 6*a**5*b*x + 15*a**4*b**2*x**2 + 20*a**3*b**3*x* *3 + 15*a**2*b**4*x**4 + 6*a*b**5*x**5 + b**6*x**6),x)
(105*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b* c)))*a**5*d**5 + 525*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt( b)*sqrt(a*d - b*c)))*a**4*b*d**5*x + 1050*sqrt(b)*sqrt(a*d - b*c)*atan((sq rt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**3*b**2*d**5*x**2 + 1050*sqrt( b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2* b**3*d**5*x**3 + 525*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt( b)*sqrt(a*d - b*c)))*a*b**4*d**5*x**4 + 105*sqrt(b)*sqrt(a*d - b*c)*atan(( sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**5*d**5*x**5 - 105*sqrt(c + d*x)*a**5*b*d**5 + 1315*sqrt(c + d*x)*a**4*b**2*c*d**4 + 790*sqrt(c + d*x) *a**4*b**2*d**5*x - 3314*sqrt(c + d*x)*a**3*b**3*c**2*d**3 - 1368*sqrt(c + d*x)*a**3*b**3*c*d**4*x + 896*sqrt(c + d*x)*a**3*b**3*d**5*x**2 + 3592*sq rt(c + d*x)*a**2*b**4*c**3*d**2 + 834*sqrt(c + d*x)*a**2*b**4*c**2*d**3*x - 1218*sqrt(c + d*x)*a**2*b**4*c*d**4*x**2 + 490*sqrt(c + d*x)*a**2*b**4*d **5*x**3 - 1872*sqrt(c + d*x)*a*b**5*c**4*d - 304*sqrt(c + d*x)*a*b**5*c** 3*d**2*x + 378*sqrt(c + d*x)*a*b**5*c**2*d**3*x**2 - 560*sqrt(c + d*x)*a*b **5*c*d**4*x**3 + 105*sqrt(c + d*x)*a*b**5*d**5*x**4 + 384*sqrt(c + d*x)*b **6*c**5 + 48*sqrt(c + d*x)*b**6*c**4*d*x - 56*sqrt(c + d*x)*b**6*c**3*d** 2*x**2 + 70*sqrt(c + d*x)*b**6*c**2*d**3*x**3 - 105*sqrt(c + d*x)*b**6*c*d **4*x**4)/(1920*b**2*(a**10*d**5 - 5*a**9*b*c*d**4 + 5*a**9*b*d**5*x + 10* a**8*b**2*c**2*d**3 - 25*a**8*b**2*c*d**4*x + 10*a**8*b**2*d**5*x**2 - ...