Integrand size = 29, antiderivative size = 83 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{7/2}}{9 c d^2 (c+d x)^8}-\frac {(8 B c+A d) \left (c^2-d^2 x^2\right )^{7/2}}{63 c^2 d^2 (c+d x)^7} \] Output:
1/9*(-A*d+B*c)*(-d^2*x^2+c^2)^(7/2)/c/d^2/(d*x+c)^8-1/63*(A*d+8*B*c)*(-d^2 *x^2+c^2)^(7/2)/c^2/d^2/(d*x+c)^7
Time = 0.95 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=-\frac {(c-d x)^3 \sqrt {c^2-d^2 x^2} (A d (8 c+d x)+B c (c+8 d x))}{63 c^2 d^2 (c+d x)^5} \] Input:
Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^8,x]
Output:
-1/63*((c - d*x)^3*Sqrt[c^2 - d^2*x^2]*(A*d*(8*c + d*x) + B*c*(c + 8*d*x)) )/(c^2*d^2*(c + d*x)^5)
Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {671, 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.
\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {(A d+8 B c) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^7}dx}{9 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{9 c d^2 (c+d x)^8}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{9 c d^2 (c+d x)^8}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+8 B c)}{63 c^2 d^2 (c+d x)^7}\) |
Input:
Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^8,x]
Output:
((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(9*c*d^2*(c + d*x)^8) - ((8*B*c + A*d) *(c^2 - d^2*x^2)^(7/2))/(63*c^2*d^2*(c + d*x)^7)
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[((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 = 1.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71
method | result | size |
gosper | \(-\frac {\left (-d x +c \right ) \left (A \,d^{2} x +8 B c d x +8 A c d +B \,c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{63 \left (d x +c \right )^{7} c^{2} d^{2}}\) | \(59\) |
orering | \(-\frac {\left (-d x +c \right ) \left (A \,d^{2} x +8 B c d x +8 A c d +B \,c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{63 \left (d x +c \right )^{7} c^{2} d^{2}}\) | \(59\) |
trager | \(-\frac {\left (-A \,d^{5} x^{4}-8 B c \,d^{4} x^{4}-5 A c \,d^{4} x^{3}+23 B \,c^{2} d^{3} x^{3}+21 A \,c^{2} d^{3} x^{2}-21 B \,c^{3} d^{2} x^{2}-23 A \,c^{3} d^{2} x +5 B \,c^{4} d x +8 A \,c^{4} d +B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{63 c^{2} \left (d x +c \right )^{5} d^{2}}\) | \(126\) |
default | \(-\frac {B \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{7 d^{9} c \left (x +\frac {c}{d}\right )^{7}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{9 c d \left (x +\frac {c}{d}\right )^{8}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{63 c^{2} \left (x +\frac {c}{d}\right )^{7}}\right )}{d^{9}}\) | \(148\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^8,x,method=_RETURNVERBOSE)
Output:
-1/63*(-d*x+c)*(A*d^2*x+8*B*c*d*x+8*A*c*d+B*c^2)*(-d^2*x^2+c^2)^(5/2)/(d*x +c)^7/c^2/d^2
Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (75) = 150\).
Time = 0.15 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.48 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=-\frac {B c^{6} + 8 \, A c^{5} d + {\left (B c d^{5} + 8 \, A d^{6}\right )} x^{5} + 5 \, {\left (B c^{2} d^{4} + 8 \, A c d^{5}\right )} x^{4} + 10 \, {\left (B c^{3} d^{3} + 8 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (B c^{4} d^{2} + 8 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (B c^{5} d + 8 \, A c^{4} d^{2}\right )} x + {\left (B c^{5} + 8 \, A c^{4} d - {\left (8 \, B c d^{4} + A d^{5}\right )} x^{4} + {\left (23 \, B c^{2} d^{3} - 5 \, A c d^{4}\right )} x^{3} - 21 \, {\left (B c^{3} d^{2} - A c^{2} d^{3}\right )} x^{2} + {\left (5 \, B c^{4} d - 23 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{63 \, {\left (c^{2} d^{7} x^{5} + 5 \, c^{3} d^{6} x^{4} + 10 \, c^{4} d^{5} x^{3} + 10 \, c^{5} d^{4} x^{2} + 5 \, c^{6} d^{3} x + c^{7} d^{2}\right )}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^8,x, algorithm="fricas")
Output:
-1/63*(B*c^6 + 8*A*c^5*d + (B*c*d^5 + 8*A*d^6)*x^5 + 5*(B*c^2*d^4 + 8*A*c* d^5)*x^4 + 10*(B*c^3*d^3 + 8*A*c^2*d^4)*x^3 + 10*(B*c^4*d^2 + 8*A*c^3*d^3) *x^2 + 5*(B*c^5*d + 8*A*c^4*d^2)*x + (B*c^5 + 8*A*c^4*d - (8*B*c*d^4 + A*d ^5)*x^4 + (23*B*c^2*d^3 - 5*A*c*d^4)*x^3 - 21*(B*c^3*d^2 - A*c^2*d^3)*x^2 + (5*B*c^4*d - 23*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2))/(c^2*d^7*x^5 + 5*c^3 *d^6*x^4 + 10*c^4*d^5*x^3 + 10*c^5*d^4*x^2 + 5*c^6*d^3*x + c^7*d^2)
Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=\text {Timed out} \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**8,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1137 vs. \(2 (75) = 150\).
Time = 0.06 (sec) , antiderivative size = 1137, normalized size of antiderivative = 13.70 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^8,x, algorithm="maxima")
Output:
1/2*(-d^2*x^2 + c^2)^(5/2)*B*c/(d^9*x^7 + 7*c*d^8*x^6 + 21*c^2*d^7*x^5 + 3 5*c^3*d^6*x^4 + 35*c^4*d^5*x^3 + 21*c^5*d^4*x^2 + 7*c^6*d^3*x + c^7*d^2) - 5/6*(-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^8*x^6 + 6*c*d^7*x^5 + 15*c^2*d^6*x^4 + 20*c^3*d^5*x^3 + 15*c^4*d^4*x^2 + 6*c^5*d^3*x + c^6*d^2) + 5/9*sqrt(-d^2 *x^2 + c^2)*B*c^3/(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) - 1/2*(-d^2*x^2 + c^2)^(5/2)*A/(d^8*x^7 + 7*c*d^ 7*x^6 + 21*c^2*d^6*x^5 + 35*c^3*d^5*x^4 + 35*c^4*d^4*x^3 + 21*c^5*d^3*x^2 + 7*c^6*d^2*x + c^7*d) - (-d^2*x^2 + c^2)^(5/2)*B/(d^8*x^6 + 6*c*d^7*x^5 + 15*c^2*d^6*x^4 + 20*c^3*d^5*x^3 + 15*c^4*d^4*x^2 + 6*c^5*d^3*x + c^6*d^2) + 5/6*(-d^2*x^2 + c^2)^(3/2)*A*c/(d^7*x^6 + 6*c*d^6*x^5 + 15*c^2*d^5*x^4 + 20*c^3*d^4*x^3 + 15*c^4*d^3*x^2 + 6*c^5*d^2*x + c^6*d) + 5/2*(-d^2*x^2 + c^2)^(3/2)*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) - 5/9*sqrt(-d^2*x^2 + c^2)*A*c^2/(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) - 275/126*s qrt(-d^2*x^2 + c^2)*B*c^2/(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) + 5/126*sqrt(-d^2*x^2 + c^2)*A*c/(d^5*x^4 + 4*c*d^4*x^3 + 6*c^2*d^3*x^2 + 4*c^3*d^2*x + c^4*d) + 4/21*sqrt(-d^2*x^2 + c^2)*B*c/(d^5* x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2) - 1/63*sqrt(-d^2*x^2 + c^2)*B*c /(c*d^4*x^2 + 2*c^2*d^3*x + c^3*d^2) - 1/63*sqrt(-d^2*x^2 + c^2)*B*c/(c^2* d^3*x + c^3*d^2) + 1/42*sqrt(-d^2*x^2 + c^2)*A/(d^4*x^3 + 3*c*d^3*x^2 +...
Leaf count of result is larger than twice the leaf count of optimal. 535 vs. \(2 (75) = 150\).
Time = 0.16 (sec) , antiderivative size = 535, normalized size of antiderivative = 6.45 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=\frac {2 \, {\left (B c + 8 \, A d + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} - \frac {27 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}} + \frac {225 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}} + \frac {189 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{6} x^{3}} + \frac {189 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{5} x^{3}} - \frac {189 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} B c}{d^{8} x^{4}} + \frac {693 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{7} x^{4}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} B c}{d^{10} x^{5}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} A}{d^{9} x^{5}} - \frac {105 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{6} B c}{d^{12} x^{6}} + \frac {483 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{6} A}{d^{11} x^{6}} + \frac {63 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{7} B c}{d^{14} x^{7}} + \frac {63 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{7} A}{d^{13} x^{7}} + \frac {63 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{8} A}{d^{15} x^{8}}\right )}}{63 \, c^{2} d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{9} {\left | d \right |}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^8,x, algorithm="giac")
Output:
2/63*(B*c + 8*A*d + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x) + 9* (c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) - 27*(c*d + sqrt(-d^2*x^2 + c^ 2)*abs(d))^2*B*c/(d^4*x^2) + 225*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/( d^3*x^2) + 189*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^6*x^3) + 189*( c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^5*x^3) - 189*(c*d + sqrt(-d^2*x^ 2 + c^2)*abs(d))^4*B*c/(d^8*x^4) + 693*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d)) ^4*A/(d^7*x^4) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*B*c/(d^10*x^5) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*A/(d^9*x^5) - 105*(c*d + sqrt( -d^2*x^2 + c^2)*abs(d))^6*B*c/(d^12*x^6) + 483*(c*d + sqrt(-d^2*x^2 + c^2) *abs(d))^6*A/(d^11*x^6) + 63*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^7*B*c/(d^ 14*x^7) + 63*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^7*A/(d^13*x^7) + 63*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^8*A/(d^15*x^8))/(c^2*d*((c*d + sqrt(-d^2*x^ 2 + c^2)*abs(d))/(d^2*x) + 1)^9*abs(d))
Time = 11.04 (sec) , antiderivative size = 551, normalized size of antiderivative = 6.64 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=\frac {A\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^3\,d+2\,c^2\,d^2\,x+c\,d^3\,x^2\right )}-\frac {10\,A\,\sqrt {c^2-d^2\,x^2}}{21\,\left (c^3\,d+3\,c^2\,d^2\,x+3\,c\,d^3\,x^2+d^4\,x^3\right )}-\frac {55\,B\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^2\,d^2+2\,c\,d^3\,x+d^4\,x^2\right )}+\frac {A\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^3\,d+x\,c^2\,d^2\right )}+\frac {8\,B\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^2\,d^2+x\,c\,d^3\right )}+\frac {76\,A\,c\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^4\,d+4\,c^3\,d^2\,x+6\,c^2\,d^3\,x^2+4\,c\,d^4\,x^3+d^5\,x^4\right )}-\frac {8\,A\,c^2\,\sqrt {c^2-d^2\,x^2}}{9\,\left (c^5\,d+5\,c^4\,d^2\,x+10\,c^3\,d^3\,x^2+10\,c^2\,d^4\,x^3+5\,c\,d^5\,x^4+d^6\,x^5\right )}+\frac {8\,B\,c^3\,\sqrt {c^2-d^2\,x^2}}{9\,\left (c^5\,d^2+5\,c^4\,d^3\,x+10\,c^3\,d^4\,x^2+10\,c^2\,d^5\,x^3+5\,c\,d^6\,x^4+d^7\,x^5\right )}+\frac {46\,B\,c\,\sqrt {c^2-d^2\,x^2}}{21\,\left (c^3\,d^2+3\,c^2\,d^3\,x+3\,c\,d^4\,x^2+d^5\,x^3\right )}-\frac {148\,B\,c^2\,\sqrt {c^2-d^2\,x^2}}{63\,\left (c^4\,d^2+4\,c^3\,d^3\,x+6\,c^2\,d^4\,x^2+4\,c\,d^5\,x^3+d^6\,x^4\right )} \] Input:
int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^8,x)
Output:
(A*(c^2 - d^2*x^2)^(1/2))/(63*(c^3*d + 2*c^2*d^2*x + c*d^3*x^2)) - (10*A*( c^2 - d^2*x^2)^(1/2))/(21*(c^3*d + d^4*x^3 + 3*c^2*d^2*x + 3*c*d^3*x^2)) - (55*B*(c^2 - d^2*x^2)^(1/2))/(63*(c^2*d^2 + d^4*x^2 + 2*c*d^3*x)) + (A*(c ^2 - d^2*x^2)^(1/2))/(63*(c^3*d + c^2*d^2*x)) + (8*B*(c^2 - d^2*x^2)^(1/2) )/(63*(c^2*d^2 + c*d^3*x)) + (76*A*c*(c^2 - d^2*x^2)^(1/2))/(63*(c^4*d + d ^5*x^4 + 4*c^3*d^2*x + 4*c*d^4*x^3 + 6*c^2*d^3*x^2)) - (8*A*c^2*(c^2 - d^2 *x^2)^(1/2))/(9*(c^5*d + d^6*x^5 + 5*c^4*d^2*x + 5*c*d^5*x^4 + 10*c^3*d^3* x^2 + 10*c^2*d^4*x^3)) + (8*B*c^3*(c^2 - d^2*x^2)^(1/2))/(9*(c^5*d^2 + d^7 *x^5 + 5*c^4*d^3*x + 5*c*d^6*x^4 + 10*c^3*d^4*x^2 + 10*c^2*d^5*x^3)) + (46 *B*c*(c^2 - d^2*x^2)^(1/2))/(21*(c^3*d^2 + d^5*x^3 + 3*c^2*d^3*x + 3*c*d^4 *x^2)) - (148*B*c^2*(c^2 - d^2*x^2)^(1/2))/(63*(c^4*d^2 + d^6*x^4 + 4*c^3* d^3*x + 4*c*d^5*x^3 + 6*c^2*d^4*x^2))
Time = 0.29 (sec) , antiderivative size = 471, normalized size of antiderivative = 5.67 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^8} \, dx=\frac {14 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4}+\sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d x +57 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x^{2}+19 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{3}+5 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{4}+\sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} x -27 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d \,x^{2}+19 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{3}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{4}-14 a \,c^{5}+a \,c^{4} d x -104 a \,c^{3} d^{2} x^{2}-34 a \,c^{2} d^{3} x^{3}-34 a c \,d^{4} x^{4}-7 a \,d^{5} x^{5}+b \,c^{5} x +36 b \,c^{4} d \,x^{2}-34 b \,c^{3} d^{2} x^{3}+36 b \,c^{2} d^{3} x^{4}-7 b c \,d^{4} x^{5}}{63 c^{2} d \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +6 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}+\sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}-c^{5}-5 c^{4} d x -10 c^{3} d^{2} x^{2}-10 c^{2} d^{3} x^{3}-5 c \,d^{4} x^{4}-d^{5} x^{5}\right )} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^8,x)
Output:
(14*sqrt(c**2 - d**2*x**2)*a*c**4 + sqrt(c**2 - d**2*x**2)*a*c**3*d*x + 57 *sqrt(c**2 - d**2*x**2)*a*c**2*d**2*x**2 + 19*sqrt(c**2 - d**2*x**2)*a*c*d **3*x**3 + 5*sqrt(c**2 - d**2*x**2)*a*d**4*x**4 + sqrt(c**2 - d**2*x**2)*b *c**4*x - 27*sqrt(c**2 - d**2*x**2)*b*c**3*d*x**2 + 19*sqrt(c**2 - d**2*x* *2)*b*c**2*d**2*x**3 - 9*sqrt(c**2 - d**2*x**2)*b*c*d**3*x**4 - 14*a*c**5 + a*c**4*d*x - 104*a*c**3*d**2*x**2 - 34*a*c**2*d**3*x**3 - 34*a*c*d**4*x* *4 - 7*a*d**5*x**5 + b*c**5*x + 36*b*c**4*d*x**2 - 34*b*c**3*d**2*x**3 + 3 6*b*c**2*d**3*x**4 - 7*b*c*d**4*x**5)/(63*c**2*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))