Integrand size = 31, antiderivative size = 158 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {64 c^2 (B c+3 A d) \left (c^2-d^2 x^2\right )^{3/2}}{315 d^2 (c+d x)^{3/2}}-\frac {16 c (B c+3 A d) \left (c^2-d^2 x^2\right )^{3/2}}{105 d^2 \sqrt {c+d x}}-\frac {2 (B c+3 A d) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{21 d^2}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^2} \] Output:
-64/315*c^2*(3*A*d+B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)-16/105*c*(3 *A*d+B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(1/2)-2/21*(3*A*d+B*c)*(d*x+c)^ (1/2)*(-d^2*x^2+c^2)^(3/2)/d^2-2/9*B*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2)/d^ 2
Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.59 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (3 A d \left (71 c^2+54 c d x+15 d^2 x^2\right )+B \left (106 c^3+159 c^2 d x+120 c d^2 x^2+35 d^3 x^3\right )\right )}{315 d^2 \sqrt {c+d x}} \] Input:
Integrate[(A + B*x)*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(3*A*d*(71*c^2 + 54*c*d*x + 15*d^2*x^2) + B*(106*c^3 + 159*c^2*d*x + 120*c*d^2*x^2 + 35*d^3*x^3)))/(315*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.
\(\displaystyle \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {(3 A d+B c) \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}dx}{3 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^2}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(3 A d+B c) \left (\frac {8}{7} c \int \sqrt {c+d x} \sqrt {c^2-d^2 x^2}dx-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )}{3 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^2}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(3 A d+B c) \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )}{3 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^2}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {\left (\frac {8}{7} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}-\frac {8 c \left (c^2-d^2 x^2\right )^{3/2}}{15 d (c+d x)^{3/2}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right ) (3 A d+B c)}{3 d}-\frac {2 B (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^2}\) |
Input:
Int[(A + B*x)*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*B*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2))/(9*d^2) + ((B*c + 3*A*d)*((-2 *Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2))/(7*d) + (8*c*((-8*c*(c^2 - d^2*x^2)^ (3/2))/(15*d*(c + d*x)^(3/2)) - (2*(c^2 - d^2*x^2)^(3/2))/(5*d*Sqrt[c + d* x])))/7))/(3*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 = 0.42 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (35 B \,d^{3} x^{3}+45 A \,d^{3} x^{2}+120 B c \,d^{2} x^{2}+162 A c \,d^{2} x +159 B \,c^{2} d x +213 A \,c^{2} d +106 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{2} \sqrt {d x +c}}\) | \(91\) |
default | \(-\frac {2 \left (-d x +c \right ) \left (35 B \,d^{3} x^{3}+45 A \,d^{3} x^{2}+120 B c \,d^{2} x^{2}+162 A c \,d^{2} x +159 B \,c^{2} d x +213 A \,c^{2} d +106 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{2} \sqrt {d x +c}}\) | \(91\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (35 B \,d^{3} x^{3}+45 A \,d^{3} x^{2}+120 B c \,d^{2} x^{2}+162 A c \,d^{2} x +159 B \,c^{2} d x +213 A \,c^{2} d +106 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{2} \sqrt {d x +c}}\) | \(91\) |
risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (-35 B \,d^{4} x^{4}-45 A \,d^{4} x^{3}-85 B c \,d^{3} x^{3}-117 A c \,d^{3} x^{2}-39 x^{2} c^{2} B \,d^{2}-51 A \,c^{2} d^{2} x +53 B \,c^{3} d x +213 A \,c^{3} d +106 B \,c^{4}\right ) \sqrt {-d x +c}}{315 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(139\) |
Input:
int((B*x+A)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/315*(-d*x+c)*(35*B*d^3*x^3+45*A*d^3*x^2+120*B*c*d^2*x^2+162*A*c*d^2*x+1 59*B*c^2*d*x+213*A*c^2*d+106*B*c^3)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (35 \, B d^{4} x^{4} - 106 \, B c^{4} - 213 \, A c^{3} d + 5 \, {\left (17 \, B c d^{3} + 9 \, A d^{4}\right )} x^{3} + 39 \, {\left (B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} - {\left (53 \, B c^{3} d - 51 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{315 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" )
Output:
2/315*(35*B*d^4*x^4 - 106*B*c^4 - 213*A*c^3*d + 5*(17*B*c*d^3 + 9*A*d^4)*x ^3 + 39*(B*c^2*d^2 + 3*A*c*d^3)*x^2 - (53*B*c^3*d - 51*A*c^2*d^2)*x)*sqrt( -d^2*x^2 + c^2)*sqrt(d*x + c)/(d^3*x + c*d^2)
\[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=\int \sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right ) \left (c + d x\right )^{\frac {3}{2}}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(3/2)*(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)*(c + d*x)**(3/2), x)
Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.81 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (15 \, d^{3} x^{3} + 39 \, c d^{2} x^{2} + 17 \, c^{2} d x - 71 \, c^{3}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{105 \, {\left (d^{2} x + c d\right )}} + \frac {2 \, {\left (35 \, d^{4} x^{4} + 85 \, c d^{3} x^{3} + 39 \, c^{2} d^{2} x^{2} - 53 \, c^{3} d x - 106 \, c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{315 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" )
Output:
2/105*(15*d^3*x^3 + 39*c*d^2*x^2 + 17*c^2*d*x - 71*c^3)*(d*x + c)*sqrt(-d* x + c)*A/(d^2*x + c*d) + 2/315*(35*d^4*x^4 + 85*c*d^3*x^3 + 39*c^2*d^2*x^2 - 53*c^3*d*x - 106*c^4)*(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. 404 vs. \(2 (134) = 268\).
Time = 0.12 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.56 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 \, {\left (315 \, \sqrt {-d x + c} A c^{3} d - 105 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} B c^{3} - 105 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} A c^{2} d + 21 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-d x + c} c^{2}\right )} B c^{2} - 21 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-d x + c} c^{2}\right )} A c d - 9 \, {\left (5 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-d x + c} c^{3}\right )} B c - 9 \, {\left (5 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-d x + c} c^{3}\right )} A d - {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 180 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 378 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 420 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-d x + c} c^{4}\right )} B\right )}}{315 \, d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
-2/315*(315*sqrt(-d*x + c)*A*c^3*d - 105*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*B*c^3 - 105*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*A*c^2*d + 21*(3 *(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15*sqrt(-d*x + c)*c^ 2)*B*c^2 - 21*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15*s qrt(-d*x + c)*c^2)*A*c*d - 9*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(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 - 9*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d*x + c)*c - 35*( -d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*A*d - (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)*B)/d^2
Time = 9.74 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.91 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {2\,B\,d\,x^4\,\sqrt {c+d\,x}}{9}-\frac {\left (212\,B\,c^4+426\,A\,d\,c^3\right )\,\sqrt {c+d\,x}}{315\,d^3}+\frac {x^3\,\left (90\,A\,d^4+170\,B\,c\,d^3\right )\,\sqrt {c+d\,x}}{315\,d^3}+\frac {26\,c\,x^2\,\left (3\,A\,d+B\,c\right )\,\sqrt {c+d\,x}}{105\,d}+\frac {2\,c^2\,x\,\left (51\,A\,d-53\,B\,c\right )\,\sqrt {c+d\,x}}{315\,d^2}\right )}{x+\frac {c}{d}} \] Input:
int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x)^(3/2),x)
Output:
((c^2 - d^2*x^2)^(1/2)*((2*B*d*x^4*(c + d*x)^(1/2))/9 - ((212*B*c^4 + 426* A*c^3*d)*(c + d*x)^(1/2))/(315*d^3) + (x^3*(90*A*d^4 + 170*B*c*d^3)*(c + d *x)^(1/2))/(315*d^3) + (26*c*x^2*(3*A*d + B*c)*(c + d*x)^(1/2))/(105*d) + (2*c^2*x*(51*A*d - 53*B*c)*(c + d*x)^(1/2))/(315*d^2)))/(x + c/d)
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.59 \[ \int (A+B x) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (35 b \,d^{4} x^{4}+45 a \,d^{4} x^{3}+85 b c \,d^{3} x^{3}+117 a c \,d^{3} x^{2}+39 b \,c^{2} d^{2} x^{2}+51 a \,c^{2} d^{2} x -53 b \,c^{3} d x -213 a \,c^{3} d -106 b \,c^{4}\right )}{315 d^{2}} \] Input:
int((B*x+A)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2),x)
Output:
(2*sqrt(c - d*x)*( - 213*a*c**3*d + 51*a*c**2*d**2*x + 117*a*c*d**3*x**2 + 45*a*d**4*x**3 - 106*b*c**4 - 53*b*c**3*d*x + 39*b*c**2*d**2*x**2 + 85*b* c*d**3*x**3 + 35*b*d**4*x**4))/(315*d**2)