Integrand size = 31, antiderivative size = 117 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {8 c (3 B c-11 A d) \left (c^2-d^2 x^2\right )^{7/2}}{693 d^2 (c+d x)^{7/2}}+\frac {2 (3 B c-11 A d) \left (c^2-d^2 x^2\right )^{7/2}}{99 d^2 (c+d x)^{5/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{11 d^2 (c+d x)^{3/2}} \] Output:
8/693*c*(-11*A*d+3*B*c)*(-d^2*x^2+c^2)^(7/2)/d^2/(d*x+c)^(7/2)+2/99*(-11*A *d+3*B*c)*(-d^2*x^2+c^2)^(7/2)/d^2/(d*x+c)^(5/2)-2/11*B*(-d^2*x^2+c^2)^(7/ 2)/d^2/(d*x+c)^(3/2)
Time = 0.74 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 (c-d x)^3 \sqrt {c^2-d^2 x^2} \left (11 A d (11 c+7 d x)+3 B \left (10 c^2+35 c d x+21 d^2 x^2\right )\right )}{693 d^2 \sqrt {c+d x}} \] Input:
Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(3/2),x]
Output:
(-2*(c - d*x)^3*Sqrt[c^2 - d^2*x^2]*(11*A*d*(11*c + 7*d*x) + 3*B*(10*c^2 + 35*c*d*x + 21*d^2*x^2)))/(693*d^2*Sqrt[c + d*x])
Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {672, 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.
\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 672 |
\(\displaystyle -\frac {(3 B c-11 A d) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}}dx}{11 d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{11 d^2 (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle -\frac {(3 B c-11 A d) \left (\frac {4}{9} c \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{5/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{7/2}}{9 d (c+d x)^{5/2}}\right )}{11 d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{11 d^2 (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle -\frac {\left (-\frac {2 \left (c^2-d^2 x^2\right )^{7/2}}{9 d (c+d x)^{5/2}}-\frac {8 c \left (c^2-d^2 x^2\right )^{7/2}}{63 d (c+d x)^{7/2}}\right ) (3 B c-11 A d)}{11 d}-\frac {2 B \left (c^2-d^2 x^2\right )^{7/2}}{11 d^2 (c+d x)^{3/2}}\) |
Input:
Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(3/2),x]
Output:
(-2*B*(c^2 - d^2*x^2)^(7/2))/(11*d^2*(c + d*x)^(3/2)) - ((3*B*c - 11*A*d)* ((-8*c*(c^2 - d^2*x^2)^(7/2))/(63*d*(c + d*x)^(7/2)) - (2*(c^2 - d^2*x^2)^ (7/2))/(9*d*(c + d*x)^(5/2))))/(11*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[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 = 0.41 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (63 B \,d^{2} x^{2}+77 A \,d^{2} x +105 B c d x +121 A c d +30 B \,c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{693 d^{2} \left (d x +c \right )^{\frac {5}{2}}}\) | \(67\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (63 B \,d^{2} x^{2}+77 A \,d^{2} x +105 B c d x +121 A c d +30 B \,c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{693 d^{2} \left (d x +c \right )^{\frac {5}{2}}}\) | \(67\) |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{3} \left (63 B \,d^{2} x^{2}+77 A \,d^{2} x +105 B c d x +121 A c d +30 B \,c^{2}\right )}{693 \sqrt {d x +c}\, d^{2}}\) | \(69\) |
risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (-63 B \,d^{5} x^{5}-77 A \,d^{5} x^{4}+84 B c \,d^{4} x^{4}+110 A c \,d^{4} x^{3}+96 B \,c^{2} d^{3} x^{3}+132 A \,c^{2} d^{3} x^{2}-162 B \,c^{3} d^{2} x^{2}-286 A \,c^{3} d^{2} x +15 B \,c^{4} d x +121 A \,c^{4} d +30 B \,c^{5}\right ) \sqrt {-d x +c}}{693 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(163\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/693*(-d*x+c)*(63*B*d^2*x^2+77*A*d^2*x+105*B*c*d*x+121*A*c*d+30*B*c^2)*( -d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(5/2)
Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (63 \, B d^{5} x^{5} - 30 \, B c^{5} - 121 \, A c^{4} d - 7 \, {\left (12 \, B c d^{4} - 11 \, A d^{5}\right )} x^{4} - 2 \, {\left (48 \, B c^{2} d^{3} + 55 \, A c d^{4}\right )} x^{3} + 6 \, {\left (27 \, B c^{3} d^{2} - 22 \, A c^{2} d^{3}\right )} x^{2} - {\left (15 \, B c^{4} d - 286 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{693 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(3/2),x, algorithm="fricas" )
Output:
2/693*(63*B*d^5*x^5 - 30*B*c^5 - 121*A*c^4*d - 7*(12*B*c*d^4 - 11*A*d^5)*x ^4 - 2*(48*B*c^2*d^3 + 55*A*c*d^4)*x^3 + 6*(27*B*c^3*d^2 - 22*A*c^2*d^3)*x ^2 - (15*B*c^4*d - 286*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d ^3*x + c*d^2)
\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(3/2),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**(3/2), x)
Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (7 \, d^{4} x^{4} - 10 \, c d^{3} x^{3} - 12 \, c^{2} d^{2} x^{2} + 26 \, c^{3} d x - 11 \, c^{4}\right )} \sqrt {-d x + c} A}{63 \, d} + \frac {2 \, {\left (21 \, d^{5} x^{5} - 28 \, c d^{4} x^{4} - 32 \, c^{2} d^{3} x^{3} + 54 \, c^{3} d^{2} x^{2} - 5 \, c^{4} d x - 10 \, c^{5}\right )} \sqrt {-d x + c} B}{231 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(3/2),x, algorithm="maxima" )
Output:
2/63*(7*d^4*x^4 - 10*c*d^3*x^3 - 12*c^2*d^2*x^2 + 26*c^3*d*x - 11*c^4)*sqr t(-d*x + c)*A/d + 2/231*(21*d^5*x^5 - 28*c*d^4*x^4 - 32*c^2*d^3*x^3 + 54*c ^3*d^2*x^2 - 5*c^4*d*x - 10*c^5)*sqrt(-d*x + c)*B/d^2
Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (99) = 198\).
Time = 0.13 (sec) , antiderivative size = 471, normalized size of antiderivative = 4.03 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{3} d - 231 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} B c^{3} + 231 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} A c^{2} d + 33 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} B c^{2} + 33 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} A c d + 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} B c - 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} A d - {\left (315 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} + 1540 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c + 2970 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{2} + 2772 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{3} - 1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{4}\right )} B\right )}}{3465 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
-2/3465*(1155*(-d*x + c)^(3/2)*A*c^3*d - 231*(3*(d*x - c)^2*sqrt(-d*x + c) - 5*(-d*x + c)^(3/2)*c)*B*c^3 + 231*(3*(d*x - c)^2*sqrt(-d*x + c) - 5*(-d *x + c)^(3/2)*c)*A*c^2*d + 33*(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)*B*c^2 + 33*(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*c*d + 11*(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*c - 11*(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)*A*d - (315* (d*x - c)^5*sqrt(-d*x + c) + 1540*(d*x - c)^4*sqrt(-d*x + c)*c + 2970*(d*x - c)^3*sqrt(-d*x + c)*c^2 + 2772*(d*x - c)^2*sqrt(-d*x + c)*c^3 - 1155*(- d*x + c)^(3/2)*c^4)*B)/d^2
Time = 9.89 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {60\,B\,c^5+242\,A\,d\,c^4}{693\,d^2}-\frac {2\,B\,d^3\,x^5}{11}+\frac {4\,c^2\,x^2\,\left (22\,A\,d-27\,B\,c\right )}{231}-\frac {x^4\,\left (154\,A\,d^5-168\,B\,c\,d^4\right )}{693\,d^2}+\frac {4\,c\,d\,x^3\,\left (55\,A\,d+48\,B\,c\right )}{693}-\frac {2\,c^3\,x\,\left (286\,A\,d-15\,B\,c\right )}{693\,d}\right )}{\sqrt {c+d\,x}} \] Input:
int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(3/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*((60*B*c^5 + 242*A*c^4*d)/(693*d^2) - (2*B*d^3*x^5 )/11 + (4*c^2*x^2*(22*A*d - 27*B*c))/231 - (x^4*(154*A*d^5 - 168*B*c*d^4)) /(693*d^2) + (4*c*d*x^3*(55*A*d + 48*B*c))/693 - (2*c^3*x*(286*A*d - 15*B* c))/(693*d)))/(c + d*x)^(1/2)
Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (63 b \,d^{5} x^{5}+77 a \,d^{5} x^{4}-84 b c \,d^{4} x^{4}-110 a c \,d^{4} x^{3}-96 b \,c^{2} d^{3} x^{3}-132 a \,c^{2} d^{3} x^{2}+162 b \,c^{3} d^{2} x^{2}+286 a \,c^{3} d^{2} x -15 b \,c^{4} d x -121 a \,c^{4} d -30 b \,c^{5}\right )}{693 d^{2}} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(3/2),x)
Output:
(2*sqrt(c - d*x)*( - 121*a*c**4*d + 286*a*c**3*d**2*x - 132*a*c**2*d**3*x* *2 - 110*a*c*d**4*x**3 + 77*a*d**5*x**4 - 30*b*c**5 - 15*b*c**4*d*x + 162* b*c**3*d**2*x**2 - 96*b*c**2*d**3*x**3 - 84*b*c*d**4*x**4 + 63*b*d**5*x**5 ))/(693*d**2)