Integrand size = 31, antiderivative size = 75 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 (5 B c-9 A d) \left (c^2-d^2 x^2\right )^{7/2}}{63 d^2 (c+d x)^{7/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{9 d^2 (c+d x)^{5/2}} \] Output:
2/63*(-9*A*d+5*B*c)*(-d^2*x^2+c^2)^(7/2)/d^2/(d*x+c)^(7/2)-2/9*B*(-d^2*x^2 +c^2)^(7/2)/d^2/(d*x+c)^(5/2)
Time = 0.77 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 (-5 B c+9 A d+7 B (c+d x)) \left (2 c (c+d x)-(c+d x)^2\right )^{7/2}}{63 d^2 (c+d x)^{7/2}} \] Input:
Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(5/2),x]
Output:
(-2*(-5*B*c + 9*A*d + 7*B*(c + d*x))*(2*c*(c + d*x) - (c + d*x)^2)^(7/2))/ (63*d^2*(c + d*x)^(7/2))
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {672, 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.
\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 672 |
\(\displaystyle -\frac {(5 B c-9 A d) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}}dx}{9 d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{9 d^2 (c+d x)^{5/2}}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {2 \left (c^2-d^2 x^2\right )^{7/2} (5 B c-9 A d)}{63 d^2 (c+d x)^{7/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{9 d^2 (c+d x)^{5/2}}\) |
Input:
Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(5/2),x]
Output:
(2*(5*B*c - 9*A*d)*(c^2 - d^2*x^2)^(7/2))/(63*d^2*(c + d*x)^(7/2)) - (2*B* (c^2 - d^2*x^2)^(7/2))/(9*d^2*(c + d*x)^(5/2))
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[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)) Int[(d + e*x )^m*(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] && NeQ[m + 2*p + 2, 0]
Time = 0.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (7 B d x +9 A d +2 B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{63 d^{2} \left (d x +c \right )^{\frac {5}{2}}}\) | \(47\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (7 B d x +9 A d +2 B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{63 d^{2} \left (d x +c \right )^{\frac {5}{2}}}\) | \(47\) |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{3} \left (7 B d x +9 A d +2 B c \right )}{63 \sqrt {d x +c}\, d^{2}}\) | \(49\) |
risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (-7 B \,d^{4} x^{4}-9 A \,d^{4} x^{3}+19 B c \,d^{3} x^{3}+27 A c \,d^{3} x^{2}-15 x^{2} c^{2} B \,d^{2}-27 A \,c^{2} d^{2} x +B \,c^{3} d x +9 A \,c^{3} d +2 B \,c^{4}\right ) \sqrt {-d x +c}}{63 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(138\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/63*(-d*x+c)*(7*B*d*x+9*A*d+2*B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(5/2 )
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.59 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (7 \, B d^{4} x^{4} - 2 \, B c^{4} - 9 \, A c^{3} d - {\left (19 \, B c d^{3} - 9 \, A d^{4}\right )} x^{3} + 3 \, {\left (5 \, B c^{2} d^{2} - 9 \, A c d^{3}\right )} x^{2} - {\left (B c^{3} d - 27 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{63 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(5/2),x, algorithm="fricas" )
Output:
2/63*(7*B*d^4*x^4 - 2*B*c^4 - 9*A*c^3*d - (19*B*c*d^3 - 9*A*d^4)*x^3 + 3*( 5*B*c^2*d^2 - 9*A*c*d^3)*x^2 - (B*c^3*d - 27*A*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^3*x + c*d^2)
\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(5/2),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**(5/2), x)
Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (d^{3} x^{3} - 3 \, c d^{2} x^{2} + 3 \, c^{2} d x - c^{3}\right )} \sqrt {-d x + c} A}{7 \, d} + \frac {2 \, {\left (7 \, d^{4} x^{4} - 19 \, c d^{3} x^{3} + 15 \, c^{2} d^{2} x^{2} - c^{3} d x - 2 \, c^{4}\right )} \sqrt {-d x + c} B}{63 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(5/2),x, algorithm="maxima" )
Output:
2/7*(d^3*x^3 - 3*c*d^2*x^2 + 3*c^2*d*x - c^3)*sqrt(-d*x + c)*A/d + 2/63*(7 *d^4*x^4 - 19*c*d^3*x^3 + 15*c^2*d^2*x^2 - c^3*d*x - 2*c^4)*sqrt(-d*x + c) *B/d^2
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (63) = 126\).
Time = 0.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.81 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (105 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{2} d - 21 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} B c^{2} + 42 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} A c d + 6 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} B c - 3 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} A d - {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} B\right )}}{315 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
-2/315*(105*(-d*x + c)^(3/2)*A*c^2*d - 21*(3*(d*x - c)^2*sqrt(-d*x + c) - 5*(-d*x + c)^(3/2)*c)*B*c^2 + 42*(3*(d*x - c)^2*sqrt(-d*x + c) - 5*(-d*x + c)^(3/2)*c)*A*c*d + 6*(15*(d*x - c)^3*sqrt(-d*x + c) + 42*(d*x - c)^2*sqr t(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2)*B*c - 3*(15*(d*x - c)^3*sqrt(-d*x + c) + 42*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2)*A*d - ( 35*(d*x - c)^4*sqrt(-d*x + c) + 135*(d*x - c)^3*sqrt(-d*x + c)*c + 189*(d* x - c)^2*sqrt(-d*x + c)*c^2 - 105*(-d*x + c)^(3/2)*c^3)*B)/d^2
Time = 9.78 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.41 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {2\,B\,d^2\,x^4}{9}-\frac {2\,c\,x^2\,\left (9\,A\,d-5\,B\,c\right )}{21}-\frac {4\,B\,c^4+18\,A\,d\,c^3}{63\,d^2}+\frac {x^3\,\left (18\,A\,d^4-38\,B\,c\,d^3\right )}{63\,d^2}+\frac {2\,c^2\,x\,\left (27\,A\,d-B\,c\right )}{63\,d}\right )}{\sqrt {c+d\,x}} \] Input:
int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(5/2),x)
Output:
((c^2 - d^2*x^2)^(1/2)*((2*B*d^2*x^4)/9 - (2*c*x^2*(9*A*d - 5*B*c))/21 - ( 4*B*c^4 + 18*A*c^3*d)/(63*d^2) + (x^3*(18*A*d^4 - 38*B*c*d^3))/(63*d^2) + (2*c^2*x*(27*A*d - B*c))/(63*d)))/(c + d*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (7 b \,d^{4} x^{4}+9 a \,d^{4} x^{3}-19 b c \,d^{3} x^{3}-27 a c \,d^{3} x^{2}+15 b \,c^{2} d^{2} x^{2}+27 a \,c^{2} d^{2} x -b \,c^{3} d x -9 a \,c^{3} d -2 b \,c^{4}\right )}{63 d^{2}} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(5/2),x)
Output:
(2*sqrt(c - d*x)*( - 9*a*c**3*d + 27*a*c**2*d**2*x - 27*a*c*d**3*x**2 + 9* a*d**4*x**3 - 2*b*c**4 - b*c**3*d*x + 15*b*c**2*d**2*x**2 - 19*b*c*d**3*x* *3 + 7*b*d**4*x**4))/(63*d**2)