Integrand size = 29, antiderivative size = 165 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{9 c d^2 (c+d x)^6}-\frac {(2 B c+A d) \left (c^2-d^2 x^2\right )^{3/2}}{21 c^2 d^2 (c+d x)^5}-\frac {2 (2 B c+A d) \left (c^2-d^2 x^2\right )^{3/2}}{105 c^3 d^2 (c+d x)^4}-\frac {2 (2 B c+A d) \left (c^2-d^2 x^2\right )^{3/2}}{315 c^4 d^2 (c+d x)^3} \] Output:
1/9*(-A*d+B*c)*(-d^2*x^2+c^2)^(3/2)/c/d^2/(d*x+c)^6-1/21*(A*d+2*B*c)*(-d^2 *x^2+c^2)^(3/2)/c^2/d^2/(d*x+c)^5-2/105*(A*d+2*B*c)*(-d^2*x^2+c^2)^(3/2)/c ^3/d^2/(d*x+c)^4-2/315*(A*d+2*B*c)*(-d^2*x^2+c^2)^(3/2)/c^4/d^2/(d*x+c)^3
Time = 0.85 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=-\frac {(c-d x) \sqrt {c^2-d^2 x^2} \left (A d \left (58 c^3+33 c^2 d x+12 c d^2 x^2+2 d^3 x^3\right )+B c \left (11 c^3+66 c^2 d x+24 c d^2 x^2+4 d^3 x^3\right )\right )}{315 c^4 d^2 (c+d x)^5} \] Input:
Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^6,x]
Output:
-1/315*((c - d*x)*Sqrt[c^2 - d^2*x^2]*(A*d*(58*c^3 + 33*c^2*d*x + 12*c*d^2 *x^2 + 2*d^3*x^3) + B*c*(11*c^3 + 66*c^2*d*x + 24*c*d^2*x^2 + 4*d^3*x^3))) /(c^4*d^2*(c + d*x)^5)
Time = 0.44 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 461, 461, 460}
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.
| \(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx\) |
| \(\Big \downarrow \) 671 |
| \(\displaystyle \frac {(A d+2 B c) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^5}dx}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{9 c d^2 (c+d x)^6}\) |
| \(\Big \downarrow \) 461 |
| \(\displaystyle \frac {(A d+2 B c) \left (\frac {2 \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^4}dx}{7 c}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{7 c d (c+d x)^5}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{9 c d^2 (c+d x)^6}\) |
| \(\Big \downarrow \) 461 |
| \(\displaystyle \frac {(A d+2 B c) \left (\frac {2 \left (\frac {\int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^3}dx}{5 c}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{5 c d (c+d x)^4}\right )}{7 c}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{7 c d (c+d x)^5}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{9 c d^2 (c+d x)^6}\) |
| \(\Big \downarrow \) 460 |
| \(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{9 c d^2 (c+d x)^6}+\frac {\left (\frac {2 \left (-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{15 c^2 d (c+d x)^3}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{5 c d (c+d x)^4}\right )}{7 c}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{7 c d (c+d x)^5}\right ) (A d+2 B c)}{3 c d}\) |
Input:
Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^6,x]
Output:
((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(9*c*d^2*(c + d*x)^6) + ((2*B*c + A*d) *(-1/7*(c^2 - d^2*x^2)^(3/2)/(c*d*(c + d*x)^5) + (2*(-1/5*(c^2 - d^2*x^2)^ (3/2)/(c*d*(c + d*x)^4) - (c^2 - d^2*x^2)^(3/2)/(15*c^2*d*(c + d*x)^3)))/( 7*c)))/(3*c*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Time = 0.52 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66
| method | result | size |
| gosper | \(-\frac {\left (-d x +c \right ) \left (2 A \,d^{4} x^{3}+4 B c \,d^{3} x^{3}+12 A c \,d^{3} x^{2}+24 x^{2} c^{2} B \,d^{2}+33 A \,c^{2} d^{2} x +66 B \,c^{3} d x +58 A \,c^{3} d +11 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 \left (d x +c \right )^{5} c^{4} d^{2}}\) | \(109\) |
| orering | \(-\frac {\left (-d x +c \right ) \left (2 A \,d^{4} x^{3}+4 B c \,d^{3} x^{3}+12 A c \,d^{3} x^{2}+24 x^{2} c^{2} B \,d^{2}+33 A \,c^{2} d^{2} x +66 B \,c^{3} d x +58 A \,c^{3} d +11 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 \left (d x +c \right )^{5} c^{4} d^{2}}\) | \(109\) |
| trager | \(-\frac {\left (-2 A \,d^{5} x^{4}-4 B c \,d^{4} x^{4}-10 A c \,d^{4} x^{3}-20 B \,c^{2} d^{3} x^{3}-21 A \,c^{2} d^{3} x^{2}-42 B \,c^{3} d^{2} x^{2}-25 A \,c^{3} d^{2} x +55 B \,c^{4} d x +58 A \,c^{4} d +11 B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 c^{4} \left (d x +c \right )^{5} d^{2}}\) | \(127\) |
| default | \(\frac {B \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{7 c d \left (x +\frac {c}{d}\right )^{5}}+\frac {2 d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{7 c}\right )}{d^{6}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{9 c d \left (x +\frac {c}{d}\right )^{6}}+\frac {d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{7 c d \left (x +\frac {c}{d}\right )^{5}}+\frac {2 d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{7 c}\right )}{3 c}\right )}{d^{7}}\) | \(351\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^6,x,method=_RETURNVERBOSE)
Output:
-1/315*(-d*x+c)*(2*A*d^4*x^3+4*B*c*d^3*x^3+12*A*c*d^3*x^2+24*B*c^2*d^2*x^2 +33*A*c^2*d^2*x+66*B*c^3*d*x+58*A*c^3*d+11*B*c^4)*(-d^2*x^2+c^2)^(1/2)/(d* x+c)^5/c^4/d^2
Time = 0.12 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=-\frac {11 \, B c^{6} + 58 \, A c^{5} d + {\left (11 \, B c d^{5} + 58 \, A d^{6}\right )} x^{5} + 5 \, {\left (11 \, B c^{2} d^{4} + 58 \, A c d^{5}\right )} x^{4} + 10 \, {\left (11 \, B c^{3} d^{3} + 58 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (11 \, B c^{4} d^{2} + 58 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (11 \, B c^{5} d + 58 \, A c^{4} d^{2}\right )} x + {\left (11 \, B c^{5} + 58 \, A c^{4} d - 2 \, {\left (2 \, B c d^{4} + A d^{5}\right )} x^{4} - 10 \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} x^{3} - 21 \, {\left (2 \, B c^{3} d^{2} + A c^{2} d^{3}\right )} x^{2} + 5 \, {\left (11 \, B c^{4} d - 5 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{315 \, {\left (c^{4} d^{7} x^{5} + 5 \, c^{5} d^{6} x^{4} + 10 \, c^{6} d^{5} x^{3} + 10 \, c^{7} d^{4} x^{2} + 5 \, c^{8} d^{3} x + c^{9} d^{2}\right )}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^6,x, algorithm="fricas")
Output:
-1/315*(11*B*c^6 + 58*A*c^5*d + (11*B*c*d^5 + 58*A*d^6)*x^5 + 5*(11*B*c^2* d^4 + 58*A*c*d^5)*x^4 + 10*(11*B*c^3*d^3 + 58*A*c^2*d^4)*x^3 + 10*(11*B*c^ 4*d^2 + 58*A*c^3*d^3)*x^2 + 5*(11*B*c^5*d + 58*A*c^4*d^2)*x + (11*B*c^5 + 58*A*c^4*d - 2*(2*B*c*d^4 + A*d^5)*x^4 - 10*(2*B*c^2*d^3 + A*c*d^4)*x^3 - 21*(2*B*c^3*d^2 + A*c^2*d^3)*x^2 + 5*(11*B*c^4*d - 5*A*c^3*d^2)*x)*sqrt(-d ^2*x^2 + c^2))/(c^4*d^7*x^5 + 5*c^5*d^6*x^4 + 10*c^6*d^5*x^3 + 10*c^7*d^4* x^2 + 5*c^8*d^3*x + c^9*d^2)
\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=\int \frac {\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right )}{\left (c + d x\right )^{6}}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**6,x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x)**6, x)
Leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (149) = 298\).
Time = 0.05 (sec) , antiderivative size = 751, normalized size of antiderivative = 4.55 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=\frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{9 \, {\left (d^{7} x^{5} + 5 \, c d^{6} x^{4} + 10 \, c^{2} d^{5} x^{3} + 10 \, c^{3} d^{4} x^{2} + 5 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{63 \, {\left (c d^{6} x^{4} + 4 \, c^{2} d^{5} x^{3} + 6 \, c^{3} d^{4} x^{2} + 4 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{105 \, {\left (c^{2} d^{5} x^{3} + 3 \, c^{3} d^{4} x^{2} + 3 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{315 \, {\left (c^{3} d^{4} x^{2} + 2 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{315 \, {\left (c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{9 \, {\left (d^{6} x^{5} + 5 \, c d^{5} x^{4} + 10 \, c^{2} d^{4} x^{3} + 10 \, c^{3} d^{3} x^{2} + 5 \, c^{4} d^{2} x + c^{5} d\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{63 \, {\left (c d^{5} x^{4} + 4 \, c^{2} d^{4} x^{3} + 6 \, c^{3} d^{3} x^{2} + 4 \, c^{4} d^{2} x + c^{5} d\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{105 \, {\left (c^{2} d^{4} x^{3} + 3 \, c^{3} d^{3} x^{2} + 3 \, c^{4} d^{2} x + c^{5} d\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{315 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{4} d^{2} x + c^{5} d\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{315 \, {\left (c^{4} d^{2} x + c^{5} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{7 \, {\left (d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{35 \, {\left (c d^{5} x^{3} + 3 \, c^{2} d^{4} x^{2} + 3 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{105 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{105 \, {\left (c^{3} d^{3} x + c^{4} d^{2}\right )}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^6,x, algorithm="maxima")
Output:
2/9*sqrt(-d^2*x^2 + c^2)*B*c/(d^7*x^5 + 5*c*d^6*x^4 + 10*c^2*d^5*x^3 + 10* c^3*d^4*x^2 + 5*c^4*d^3*x + c^5*d^2) - 1/63*sqrt(-d^2*x^2 + c^2)*B*c/(c*d^ 6*x^4 + 4*c^2*d^5*x^3 + 6*c^3*d^4*x^2 + 4*c^4*d^3*x + c^5*d^2) - 1/105*sqr t(-d^2*x^2 + c^2)*B*c/(c^2*d^5*x^3 + 3*c^3*d^4*x^2 + 3*c^4*d^3*x + c^5*d^2 ) - 2/315*sqrt(-d^2*x^2 + c^2)*B*c/(c^3*d^4*x^2 + 2*c^4*d^3*x + c^5*d^2) - 2/315*sqrt(-d^2*x^2 + c^2)*B*c/(c^4*d^3*x + c^5*d^2) - 2/9*sqrt(-d^2*x^2 + c^2)*A/(d^6*x^5 + 5*c*d^5*x^4 + 10*c^2*d^4*x^3 + 10*c^3*d^3*x^2 + 5*c^4* d^2*x + c^5*d) + 1/63*sqrt(-d^2*x^2 + c^2)*A/(c*d^5*x^4 + 4*c^2*d^4*x^3 + 6*c^3*d^3*x^2 + 4*c^4*d^2*x + c^5*d) + 1/105*sqrt(-d^2*x^2 + c^2)*A/(c^2*d ^4*x^3 + 3*c^3*d^3*x^2 + 3*c^4*d^2*x + c^5*d) + 2/315*sqrt(-d^2*x^2 + c^2) *A/(c^3*d^3*x^2 + 2*c^4*d^2*x + c^5*d) + 2/315*sqrt(-d^2*x^2 + c^2)*A/(c^4 *d^2*x + c^5*d) - 2/7*sqrt(-d^2*x^2 + c^2)*B/(d^6*x^4 + 4*c*d^5*x^3 + 6*c^ 2*d^4*x^2 + 4*c^3*d^3*x + c^4*d^2) + 1/35*sqrt(-d^2*x^2 + c^2)*B/(c*d^5*x^ 3 + 3*c^2*d^4*x^2 + 3*c^3*d^3*x + c^4*d^2) + 2/105*sqrt(-d^2*x^2 + c^2)*B/ (c^2*d^4*x^2 + 2*c^3*d^3*x + c^4*d^2) + 2/105*sqrt(-d^2*x^2 + c^2)*B/(c^3* d^3*x + c^4*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (149) = 298\).
Time = 0.13 (sec) , antiderivative size = 536, normalized size of antiderivative = 3.25 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=\frac {2 \, {\left (11 \, B c + 58 \, A d + \frac {99 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {207 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} + \frac {81 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}} + \frac {1143 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}} + \frac {609 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{6} x^{3}} + \frac {2247 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{5} x^{3}} + \frac {441 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} B c}{d^{8} x^{4}} + \frac {3843 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{7} x^{4}} + \frac {945 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} B c}{d^{10} x^{5}} + \frac {3465 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} A}{d^{9} x^{5}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{6} B c}{d^{12} x^{6}} + \frac {2625 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{6} A}{d^{11} x^{6}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{7} B c}{d^{14} x^{7}} + \frac {945 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{7} A}{d^{13} x^{7}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{8} A}{d^{15} x^{8}}\right )}}{315 \, c^{4} d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{9} {\left | d \right |}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^6,x, algorithm="giac")
Output:
2/315*(11*B*c + 58*A*d + 99*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x ) + 207*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) + 81*(c*d + sqrt(-d^2* x^2 + c^2)*abs(d))^2*B*c/(d^4*x^2) + 1143*(c*d + sqrt(-d^2*x^2 + c^2)*abs( d))^2*A/(d^3*x^2) + 609*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^6*x^3 ) + 2247*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^5*x^3) + 441*(c*d + sq rt(-d^2*x^2 + c^2)*abs(d))^4*B*c/(d^8*x^4) + 3843*(c*d + sqrt(-d^2*x^2 + c ^2)*abs(d))^4*A/(d^7*x^4) + 945*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*B*c/ (d^10*x^5) + 3465*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*A/(d^9*x^5) + 315* (c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*B*c/(d^12*x^6) + 2625*(c*d + sqrt(-d ^2*x^2 + c^2)*abs(d))^6*A/(d^11*x^6) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs (d))^7*B*c/(d^14*x^7) + 945*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^7*A/(d^13* x^7) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^8*A/(d^15*x^8))/(c^4*d*((c* d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^9*abs(d))
Time = 10.68 (sec) , antiderivative size = 847, normalized size of antiderivative = 5.13 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx =\text {Too large to display} \] Input:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^6,x)
Output:
((c^2 - d^2*x^2)^(1/2)*(B/(189*c^2*d^2) - (3*A*d - 39*B*c)/(945*c^3*d^2))) /(c + d*x)^2 - ((c^2 - d^2*x^2)^(1/2)*(B/(45*c*d^2) - (4*A*d - 12*B*c)/(31 5*c^2*d^2)))/(c + d*x)^3 - ((c^2 - d^2*x^2)^(1/2)*(B/(315*c^3*d^2) - (2*A* d - 42*B*c)/(945*c^4*d^2)))/(c + d*x) - ((c^2 - d^2*x^2)^(1/2)*(A/(9*d) - (c*(B/(9*d) - (A*d - B*c)/(9*c*d)))/d))/(c + d*x)^5 - ((c^2 - d^2*x^2)^(1/ 2)*((10*B*c^2 - A*c*d)/(63*c^2*d^2) - (c*((A*d^2 - 11*B*c*d)/(63*c^2*d^2) - B/(63*c*d)))/d))/(c + d*x)^4 + ((c^2 - d^2*x^2)^(1/2)*((18*B*c^2 - A*c*d )/(315*c^3*d^2) - (c*((A*d^2 - 19*B*c*d)/(315*c^3*d^2) - B/(315*c^2*d)))/d ))/(c + d*x)^3 - ((c^2 - d^2*x^2)^(1/2)*((24*B*c^2 - A*c*d)/(945*c^4*d^2) - (c*((A*d^2 - 25*B*c*d)/(945*c^4*d^2) - B/(945*c^3*d)))/d))/(c + d*x)^2 + ((c^2 - d^2*x^2)^(1/2)*((28*B*c^2 - A*c*d)/(945*c^5*d^2) - (c*((A*d^2 - 2 9*B*c*d)/(945*c^5*d^2) - B/(945*c^4*d)))/d))/(c + d*x) + ((c^2 - d^2*x^2)^ (1/2)*(8*A*d - 3*B*c))/(945*c^3*d^2*(c + d*x)^2) + ((c^2 - d^2*x^2)^(1/2)* (8*A*d - 3*B*c))/(945*c^4*d^2*(c + d*x)) - ((c^2 - d^2*x^2)^(1/2)*(3*A*d - 29*B*c))/(945*c^4*d^2*(c + d*x)) + (c*(c^2 - d^2*x^2)^(1/2)*(B/(63*c*d) - (A*d - 2*B*c)/(63*c^2*d)))/(d*(c + d*x)^4) - (c*(c^2 - d^2*x^2)^(1/2)*(B/ (315*c^2*d) - (A*d - 12*B*c)/(315*c^3*d)))/(d*(c + d*x)^3) + (c*(c^2 - d^2 *x^2)^(1/2)*(B/(945*c^3*d) - (A*d - 20*B*c)/(945*c^4*d)))/(d*(c + d*x)^2) - (c*(c^2 - d^2*x^2)^(1/2)*(B/(945*c^4*d) - (A*d - 26*B*c)/(945*c^5*d)))/( d*(c + d*x))
Time = 0.23 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.88 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^6} \, dx=\frac {70 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4}+23 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d x +51 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x^{2}+38 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{3}+10 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{4}+11 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} x -108 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d \,x^{2}-64 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{3}-15 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{4}-70 a \,c^{5}+23 a \,c^{4} d x -124 a \,c^{3} d^{2} x^{2}-131 a \,c^{2} d^{3} x^{3}-68 a c \,d^{4} x^{4}-14 a \,d^{5} x^{5}+11 b \,c^{5} x +207 b \,c^{4} d \,x^{2}+88 b \,c^{3} d^{2} x^{3}+39 b \,c^{2} d^{3} x^{4}+7 b c \,d^{4} x^{5}}{315 c^{4} d \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +6 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}+\sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}-c^{5}-5 c^{4} d x -10 c^{3} d^{2} x^{2}-10 c^{2} d^{3} x^{3}-5 c \,d^{4} x^{4}-d^{5} x^{5}\right )} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^6,x)
Output:
(70*sqrt(c**2 - d**2*x**2)*a*c**4 + 23*sqrt(c**2 - d**2*x**2)*a*c**3*d*x + 51*sqrt(c**2 - d**2*x**2)*a*c**2*d**2*x**2 + 38*sqrt(c**2 - d**2*x**2)*a* c*d**3*x**3 + 10*sqrt(c**2 - d**2*x**2)*a*d**4*x**4 + 11*sqrt(c**2 - d**2* x**2)*b*c**4*x - 108*sqrt(c**2 - d**2*x**2)*b*c**3*d*x**2 - 64*sqrt(c**2 - d**2*x**2)*b*c**2*d**2*x**3 - 15*sqrt(c**2 - d**2*x**2)*b*c*d**3*x**4 - 7 0*a*c**5 + 23*a*c**4*d*x - 124*a*c**3*d**2*x**2 - 131*a*c**2*d**3*x**3 - 6 8*a*c*d**4*x**4 - 14*a*d**5*x**5 + 11*b*c**5*x + 207*b*c**4*d*x**2 + 88*b* c**3*d**2*x**3 + 39*b*c**2*d**3*x**4 + 7*b*c*d**4*x**5)/(315*c**4*d*(sqrt( c**2 - d**2*x**2)*c**4 + 4*sqrt(c**2 - d**2*x**2)*c**3*d*x + 6*sqrt(c**2 - d**2*x**2)*c**2*d**2*x**2 + 4*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 + sqrt(c **2 - d**2*x**2)*d**4*x**4 - c**5 - 5*c**4*d*x - 10*c**3*d**2*x**2 - 10*c* *2*d**3*x**3 - 5*c*d**4*x**4 - d**5*x**5))