Integrand size = 31, antiderivative size = 158 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {64 c^2 (B c+11 A d) \left (c^2-d^2 x^2\right )^{5/2}}{3465 d^2 (c+d x)^{5/2}}-\frac {16 c (B c+11 A d) \left (c^2-d^2 x^2\right )^{5/2}}{693 d^2 (c+d x)^{3/2}}-\frac {2 (B c+11 A d) \left (c^2-d^2 x^2\right )^{5/2}}{99 d^2 \sqrt {c+d x}}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^2} \] Output:
-64/3465*c^2*(11*A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(5/2)-16/693*c* (11*A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(3/2)-2/99*(11*A*d+B*c)*(-d^ 2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(1/2)-2/11*B*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(5/ 2)/d^2
Time = 0.68 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.61 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 (c-d x)^2 \sqrt {c^2-d^2 x^2} \left (11 A d \left (107 c^2+110 c d x+35 d^2 x^2\right )+B \left (422 c^3+1055 c^2 d x+980 c d^2 x^2+315 d^3 x^3\right )\right )}{3465 d^2 \sqrt {c+d x}} \] Input:
Integrate[(A + B*x)*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2),x]
Output:
(-2*(c - d*x)^2*Sqrt[c^2 - d^2*x^2]*(11*A*d*(107*c^2 + 110*c*d*x + 35*d^2* x^2) + B*(422*c^3 + 1055*c^2*d*x + 980*c*d^2*x^2 + 315*d^3*x^3)))/(3465*d^ 2*Sqrt[c + d*x])
Time = 0.43 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {672, 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.
| \(\displaystyle \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 672 |
| \(\displaystyle \frac {(11 A d+B c) \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}dx}{11 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^2}\) |
| \(\Big \downarrow \) 459 |
| \(\displaystyle \frac {(11 A d+B c) \left (\frac {8}{9} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )}{11 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^2}\) |
| \(\Big \downarrow \) 459 |
| \(\displaystyle \frac {(11 A d+B c) \left (\frac {8}{9} c \left (\frac {4}{7} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{3/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right )}{11 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^2}\) |
| \(\Big \downarrow \) 458 |
| \(\displaystyle \frac {\left (\frac {8}{9} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}-\frac {8 c \left (c^2-d^2 x^2\right )^{5/2}}{35 d (c+d x)^{5/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{9 d \sqrt {c+d x}}\right ) (11 A d+B c)}{11 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^2}\) |
Input:
Int[(A + B*x)*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2),x]
Output:
(-2*B*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(5/2))/(11*d^2) + ((B*c + 11*A*d)*((-2 *(c^2 - d^2*x^2)^(5/2))/(9*d*Sqrt[c + d*x]) + (8*c*((-8*c*(c^2 - d^2*x^2)^ (5/2))/(35*d*(c + d*x)^(5/2)) - (2*(c^2 - d^2*x^2)^(5/2))/(7*d*(c + d*x)^( 3/2))))/9))/(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 = 0.43 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58
| method | result | size |
| gosper | \(-\frac {2 \left (-d x +c \right ) \left (315 B \,d^{3} x^{3}+385 A \,d^{3} x^{2}+980 B c \,d^{2} x^{2}+1210 A c \,d^{2} x +1055 B \,c^{2} d x +1177 A \,c^{2} d +422 B \,c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3465 d^{2} \left (d x +c \right )^{\frac {3}{2}}}\) | \(91\) |
| orering | \(-\frac {2 \left (-d x +c \right ) \left (315 B \,d^{3} x^{3}+385 A \,d^{3} x^{2}+980 B c \,d^{2} x^{2}+1210 A c \,d^{2} x +1055 B \,c^{2} d x +1177 A \,c^{2} d +422 B \,c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3465 d^{2} \left (d x +c \right )^{\frac {3}{2}}}\) | \(91\) |
| default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{2} \left (315 B \,d^{3} x^{3}+385 A \,d^{3} x^{2}+980 B c \,d^{2} x^{2}+1210 A c \,d^{2} x +1055 B \,c^{2} d x +1177 A \,c^{2} d +422 B \,c^{3}\right )}{3465 \sqrt {d x +c}\, d^{2}}\) | \(93\) |
| risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (315 B \,d^{5} x^{5}+385 A \,d^{5} x^{4}+350 B c \,d^{4} x^{4}+440 A c \,d^{4} x^{3}-590 B \,c^{2} d^{3} x^{3}-858 A \,c^{2} d^{3} x^{2}-708 B \,c^{3} d^{2} x^{2}-1144 A \,c^{3} d^{2} x +211 B \,c^{4} d x +1177 A \,c^{4} d +422 B \,c^{5}\right ) \sqrt {-d x +c}}{3465 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(163\) |
Input:
int((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3465*(-d*x+c)*(315*B*d^3*x^3+385*A*d^3*x^2+980*B*c*d^2*x^2+1210*A*c*d^2 *x+1055*B*c^2*d*x+1177*A*c^2*d+422*B*c^3)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c) ^(3/2)
Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.91 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (315 \, B d^{5} x^{5} + 422 \, B c^{5} + 1177 \, A c^{4} d + 35 \, {\left (10 \, B c d^{4} + 11 \, A d^{5}\right )} x^{4} - 10 \, {\left (59 \, B c^{2} d^{3} - 44 \, A c d^{4}\right )} x^{3} - 6 \, {\left (118 \, B c^{3} d^{2} + 143 \, A c^{2} d^{3}\right )} x^{2} + {\left (211 \, B c^{4} d - 1144 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3465 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2),x, algorithm="fricas" )
Output:
-2/3465*(315*B*d^5*x^5 + 422*B*c^5 + 1177*A*c^4*d + 35*(10*B*c*d^4 + 11*A* d^5)*x^4 - 10*(59*B*c^2*d^3 - 44*A*c*d^4)*x^3 - 6*(118*B*c^3*d^2 + 143*A*c ^2*d^3)*x^2 + (211*B*c^4*d - 1144*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt( d*x + c)/(d^3*x + c*d^2)
\[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=\int \left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (A + B x\right ) \sqrt {c + d x}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(1/2)*(-d**2*x**2+c**2)**(3/2),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x)*sqrt(c + d*x), x)
Time = 0.06 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.95 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (35 \, d^{4} x^{4} + 40 \, c d^{3} x^{3} - 78 \, c^{2} d^{2} x^{2} - 104 \, c^{3} d x + 107 \, c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{315 \, {\left (d^{2} x + c d\right )}} - \frac {2 \, {\left (315 \, d^{5} x^{5} + 350 \, c d^{4} x^{4} - 590 \, c^{2} d^{3} x^{3} - 708 \, c^{3} d^{2} x^{2} + 211 \, c^{4} d x + 422 \, c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{3465 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima" )
Output:
-2/315*(35*d^4*x^4 + 40*c*d^3*x^3 - 78*c^2*d^2*x^2 - 104*c^3*d*x + 107*c^4 )*(d*x + c)*sqrt(-d*x + c)*A/(d^2*x + c*d) - 2/3465*(315*d^5*x^5 + 350*c*d ^4*x^4 - 590*c^2*d^3*x^3 - 708*c^3*d^2*x^2 + 211*c^4*d*x + 422*c^5)*(d*x + c)*sqrt(-d*x + c)*B/(d^3*x + c*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (134) = 268\).
Time = 0.11 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.37 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (3465 \, \sqrt {-d x + c} A c^{4} d - 1155 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} B c^{4} - 462 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-d x + c} c^{2}\right )} A c^{2} d - 198 \, {\left (5 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-d x + c} c^{3}\right )} B c^{2} + 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 180 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 378 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 420 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-d x + c} c^{4}\right )} A d + 5 \, {\left (63 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} + 385 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c + 990 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{2} + 1386 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{3} - 1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{4} + 693 \, \sqrt {-d x + c} c^{5}\right )} B\right )}}{3465 \, d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2),x, algorithm="giac")
Output:
-2/3465*(3465*sqrt(-d*x + c)*A*c^4*d - 1155*((-d*x + c)^(3/2) - 3*sqrt(-d* x + c)*c)*B*c^4 - 462*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)* c + 15*sqrt(-d*x + c)*c^2)*A*c^2*d - 198*(5*(d*x - c)^3*sqrt(-d*x + c) + 2 1*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*B*c^2 + 11*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*(d*x - c)^3*sqrt(- d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + c)*c^4)*A*d + 5*(63*(d*x - c)^5*sqrt(-d*x + c) + 385*(d *x - c)^4*sqrt(-d*x + c)*c + 990*(d*x - c)^3*sqrt(-d*x + c)*c^2 + 1386*(d* x - c)^2*sqrt(-d*x + c)*c^3 - 1155*(-d*x + c)^(3/2)*c^4 + 693*sqrt(-d*x + c)*c^5)*B)/d^2
Time = 9.46 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {\left (844\,B\,c^5+2354\,A\,d\,c^4\right )\,\sqrt {c+d\,x}}{3465\,d^3}+\frac {4\,c\,x^3\,\left (44\,A\,d-59\,B\,c\right )\,\sqrt {c+d\,x}}{693}+\frac {2\,B\,d^2\,x^5\,\sqrt {c+d\,x}}{11}+\frac {x^4\,\left (770\,A\,d^5+700\,B\,c\,d^4\right )\,\sqrt {c+d\,x}}{3465\,d^3}-\frac {2\,c^3\,x\,\left (1144\,A\,d-211\,B\,c\right )\,\sqrt {c+d\,x}}{3465\,d^2}-\frac {4\,c^2\,x^2\,\left (143\,A\,d+118\,B\,c\right )\,\sqrt {c+d\,x}}{1155\,d}\right )}{x+\frac {c}{d}} \] Input:
int((c^2 - d^2*x^2)^(3/2)*(A + B*x)*(c + d*x)^(1/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(((844*B*c^5 + 2354*A*c^4*d)*(c + d*x)^(1/2))/(346 5*d^3) + (4*c*x^3*(44*A*d - 59*B*c)*(c + d*x)^(1/2))/693 + (2*B*d^2*x^5*(c + d*x)^(1/2))/11 + (x^4*(770*A*d^5 + 700*B*c*d^4)*(c + d*x)^(1/2))/(3465* d^3) - (2*c^3*x*(1144*A*d - 211*B*c)*(c + d*x)^(1/2))/(3465*d^2) - (4*c^2* x^2*(143*A*d + 118*B*c)*(c + d*x)^(1/2))/(1155*d)))/(x + c/d)
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.75 \[ \int (A+B x) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-315 b \,d^{5} x^{5}-385 a \,d^{5} x^{4}-350 b c \,d^{4} x^{4}-440 a c \,d^{4} x^{3}+590 b \,c^{2} d^{3} x^{3}+858 a \,c^{2} d^{3} x^{2}+708 b \,c^{3} d^{2} x^{2}+1144 a \,c^{3} d^{2} x -211 b \,c^{4} d x -1177 a \,c^{4} d -422 b \,c^{5}\right )}{3465 d^{2}} \] Input:
int((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2),x)
Output:
(2*sqrt(c - d*x)*( - 1177*a*c**4*d + 1144*a*c**3*d**2*x + 858*a*c**2*d**3* x**2 - 440*a*c*d**4*x**3 - 385*a*d**5*x**4 - 422*b*c**5 - 211*b*c**4*d*x + 708*b*c**3*d**2*x**2 + 590*b*c**2*d**3*x**3 - 350*b*c*d**4*x**4 - 315*b*d **5*x**5))/(3465*d**2)