Integrand size = 31, antiderivative size = 156 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 (B c+A d) (c+d x)^{7/2}}{3 d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {8 c (5 B c+2 A d) \sqrt {c+d x}}{3 d^2 \sqrt {c^2-d^2 x^2}}-\frac {4 (7 B c+2 A d) \sqrt {c^2-d^2 x^2}}{3 d^2 \sqrt {c+d x}}-\frac {2 B \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^2} \] Output:
2/3*(A*d+B*c)*(d*x+c)^(7/2)/d^2/(-d^2*x^2+c^2)^(3/2)-8/3*c*(2*A*d+5*B*c)*( d*x+c)^(1/2)/d^2/(-d^2*x^2+c^2)^(1/2)-4/3*(2*A*d+7*B*c)*(-d^2*x^2+c^2)^(1/ 2)/d^2/(d*x+c)^(1/2)-2/3*B*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^2
Time = 1.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.61 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c+d x} \left (A d \left (11 c^2-18 c d x+3 d^2 x^2\right )+B \left (34 c^3-51 c^2 d x+12 c d^2 x^2+d^3 x^3\right )\right )}{3 d^2 (-c+d x) \sqrt {c^2-d^2 x^2}} \] Input:
Integrate[((A + B*x)*(c + d*x)^(9/2))/(c^2 - d^2*x^2)^(5/2),x]
Output:
(2*Sqrt[c + d*x]*(A*d*(11*c^2 - 18*c*d*x + 3*d^2*x^2) + B*(34*c^3 - 51*c^2 *d*x + 12*c*d^2*x^2 + d^3*x^3)))/(3*d^2*(-c + d*x)*Sqrt[c^2 - d^2*x^2])
Time = 0.45 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {669, 459, 459, 458}
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.
\(\displaystyle \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 669 |
\(\displaystyle \frac {(c+d x)^{9/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(A d+3 B c) \int \frac {(c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{2 c d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(c+d x)^{9/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(A d+3 B c) \left (\frac {8}{3} c \int \frac {(c+d x)^{5/2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx-\frac {2 (c+d x)^{5/2}}{3 d \sqrt {c^2-d^2 x^2}}\right )}{2 c d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(c+d x)^{9/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(A d+3 B c) \left (\frac {8}{3} c \left (4 c \int \frac {(c+d x)^{3/2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx-\frac {2 (c+d x)^{3/2}}{d \sqrt {c^2-d^2 x^2}}\right )-\frac {2 (c+d x)^{5/2}}{3 d \sqrt {c^2-d^2 x^2}}\right )}{2 c d}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {(c+d x)^{9/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (\frac {8}{3} c \left (\frac {8 c \sqrt {c+d x}}{d \sqrt {c^2-d^2 x^2}}-\frac {2 (c+d x)^{3/2}}{d \sqrt {c^2-d^2 x^2}}\right )-\frac {2 (c+d x)^{5/2}}{3 d \sqrt {c^2-d^2 x^2}}\right ) (A d+3 B c)}{2 c d}\) |
Input:
Int[((A + B*x)*(c + d*x)^(9/2))/(c^2 - d^2*x^2)^(5/2),x]
Output:
((B*c + A*d)*(c + d*x)^(9/2))/(3*c*d^2*(c^2 - d^2*x^2)^(3/2)) - ((3*B*c + A*d)*((-2*(c + d*x)^(5/2))/(3*d*Sqrt[c^2 - d^2*x^2]) + (8*c*((8*c*Sqrt[c + d*x])/(d*Sqrt[c^2 - d^2*x^2]) - (2*(c + d*x)^(3/2))/(d*Sqrt[c^2 - d^2*x^2 ])))/3))/(2*c*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*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] && IGtQ[Simplif y[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* (p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (B \,d^{3} x^{3}+3 A \,d^{3} x^{2}+12 B c \,d^{2} x^{2}-18 A c \,d^{2} x -51 B \,c^{2} d x +11 A \,c^{2} d +34 B \,c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(90\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (B \,d^{3} x^{3}+3 A \,d^{3} x^{2}+12 B c \,d^{2} x^{2}-18 A c \,d^{2} x -51 B \,c^{2} d x +11 A \,c^{2} d +34 B \,c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(90\) |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (B \,d^{3} x^{3}+3 A \,d^{3} x^{2}+12 B c \,d^{2} x^{2}-18 A c \,d^{2} x -51 B \,c^{2} d x +11 A \,c^{2} d +34 B \,c^{3}\right )}{3 \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{2}}\) | \(92\) |
risch | \(-\frac {2 \left (B d x +3 A d +14 B c \right ) \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{3 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {8 c \left (-3 A \,d^{2} x -6 B c d x +2 A c d +5 B \,c^{2}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{3 d^{2} \left (-d x +c \right )^{\frac {3}{2}} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(153\) |
Input:
int((B*x+A)*(d*x+c)^(9/2)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(-d*x+c)*(B*d^3*x^3+3*A*d^3*x^2+12*B*c*d^2*x^2-18*A*c*d^2*x-51*B*c^2* d*x+11*A*c^2*d+34*B*c^3)*(d*x+c)^(5/2)/d^2/(-d^2*x^2+c^2)^(5/2)
Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (B d^{3} x^{3} + 34 \, B c^{3} + 11 \, A c^{2} d + 3 \, {\left (4 \, B c d^{2} + A d^{3}\right )} x^{2} - 3 \, {\left (17 \, B c^{2} d + 6 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3 \, {\left (d^{5} x^{3} - c d^{4} x^{2} - c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(9/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas" )
Output:
-2/3*(B*d^3*x^3 + 34*B*c^3 + 11*A*c^2*d + 3*(4*B*c*d^2 + A*d^3)*x^2 - 3*(1 7*B*c^2*d + 6*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x^3 - c* d^4*x^2 - c^2*d^3*x + c^3*d^2)
\[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{\frac {9}{2}}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(9/2)/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral((A + B*x)*(c + d*x)**(9/2)/(-(-c + d*x)*(c + d*x))**(5/2), x)
Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, d^{2} x^{2} - 18 \, c d x + 11 \, c^{2}\right )} A}{3 \, {\left (d^{2} x - c d\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (d^{3} x^{3} + 12 \, c d^{2} x^{2} - 51 \, c^{2} d x + 34 \, c^{3}\right )} B}{3 \, {\left (d^{3} x - c d^{2}\right )} \sqrt {-d x + c}} \] Input:
integrate((B*x+A)*(d*x+c)^(9/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima" )
Output:
2/3*(3*d^2*x^2 - 18*c*d*x + 11*c^2)*A/((d^2*x - c*d)*sqrt(-d*x + c)) + 2/3 *(d^3*x^3 + 12*c*d^2*x^2 - 51*c^2*d*x + 34*c^3)*B/((d^3*x - c*d^2)*sqrt(-d *x + c))
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.72 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {4 \, {\left (6 \, {\left (d x - c\right )} B c^{2} + B c^{3} + 3 \, {\left (d x - c\right )} A c d + A c^{2} d\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} d} - \frac {{\left (-d x + c\right )}^{\frac {3}{2}} B d^{2} - 15 \, \sqrt {-d x + c} B c d^{2} - 3 \, \sqrt {-d x + c} A d^{3}}{d^{3}}\right )}}{3 \, d} \] Input:
integrate((B*x+A)*(d*x+c)^(9/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
Output:
-2/3*(4*(6*(d*x - c)*B*c^2 + B*c^3 + 3*(d*x - c)*A*c*d + A*c^2*d)/((d*x - c)*sqrt(-d*x + c)*d) - ((-d*x + c)^(3/2)*B*d^2 - 15*sqrt(-d*x + c)*B*c*d^2 - 3*sqrt(-d*x + c)*A*d^3)/d^3)/d
Time = 9.64 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {\left (68\,B\,c^3+22\,A\,d\,c^2\right )\,\sqrt {c+d\,x}}{3\,d^5}+\frac {2\,B\,x^3\,\sqrt {c+d\,x}}{3\,d^2}+\frac {x^2\,\left (6\,A\,d^3+24\,B\,c\,d^2\right )\,\sqrt {c+d\,x}}{3\,d^5}-\frac {2\,c\,x\,\left (6\,A\,d+17\,B\,c\right )\,\sqrt {c+d\,x}}{d^4}\right )}{x^3+\frac {c^3}{d^3}-\frac {c\,x^2}{d}-\frac {c^2\,x}{d^2}} \] Input:
int(((A + B*x)*(c + d*x)^(9/2))/(c^2 - d^2*x^2)^(5/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(((68*B*c^3 + 22*A*c^2*d)*(c + d*x)^(1/2))/(3*d^5) + (2*B*x^3*(c + d*x)^(1/2))/(3*d^2) + (x^2*(6*A*d^3 + 24*B*c*d^2)*(c + d* x)^(1/2))/(3*d^5) - (2*c*x*(6*A*d + 17*B*c)*(c + d*x)^(1/2))/d^4))/(x^3 + c^3/d^3 - (c*x^2)/d - (c^2*x)/d^2)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.51 \[ \int \frac {(A+B x) (c+d x)^{9/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2}{3} b \,d^{3} x^{3}-2 a \,d^{3} x^{2}-8 b c \,d^{2} x^{2}+12 a c \,d^{2} x +34 b \,c^{2} d x -\frac {22}{3} a \,c^{2} d -\frac {68}{3} b \,c^{3}}{\sqrt {-d x +c}\, d^{2} \left (-d x +c \right )} \] Input:
int((B*x+A)*(d*x+c)^(9/2)/(-d^2*x^2+c^2)^(5/2),x)
Output:
(2*( - 11*a*c**2*d + 18*a*c*d**2*x - 3*a*d**3*x**2 - 34*b*c**3 + 51*b*c**2 *d*x - 12*b*c*d**2*x**2 - b*d**3*x**3))/(3*sqrt(c - d*x)*d**2*(c - d*x))