Integrand size = 31, antiderivative size = 115 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=-\frac {8 c (B c+7 A d) \left (c^2-d^2 x^2\right )^{3/2}}{105 d^2 (c+d x)^{3/2}}-\frac {2 (B c+7 A d) \left (c^2-d^2 x^2\right )^{3/2}}{35 d^2 \sqrt {c+d x}}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^2} \] Output:
-8/105*c*(7*A*d+B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)-2/35*(7*A*d+B* c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(1/2)-2/7*B*(d*x+c)^(1/2)*(-d^2*x^2+c^ 2)^(3/2)/d^2
Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.63 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (7 A d (7 c+3 d x)+B \left (22 c^2+33 c d x+15 d^2 x^2\right )\right )}{105 d^2 \sqrt {c+d x}} \] Input:
Integrate[(A + B*x)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(7*A*d*(7*c + 3*d*x) + B*(22*c^2 + 33*c* d*x + 15*d^2*x^2)))/(105*d^2*Sqrt[c + d*x])
Time = 0.37 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {(7 A d+B c) \int \sqrt {c+d x} \sqrt {c^2-d^2 x^2}dx}{7 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^2}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(7 A d+B c) \left (\frac {4}{5} c \int \frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}\right )}{7 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^2}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {\left (-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}-\frac {8 c \left (c^2-d^2 x^2\right )^{3/2}}{15 d (c+d x)^{3/2}}\right ) (7 A d+B c)}{7 d}-\frac {2 B \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^2}\) |
Input:
Int[(A + B*x)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*B*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2))/(7*d^2) + ((B*c + 7*A*d)*((-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*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 = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (15 B \,d^{2} x^{2}+21 A \,d^{2} x +33 B c d x +49 A c d +22 B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 d^{2} \sqrt {d x +c}}\) | \(67\) |
default | \(-\frac {2 \left (-d x +c \right ) \left (15 B \,d^{2} x^{2}+21 A \,d^{2} x +33 B c d x +49 A c d +22 B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 d^{2} \sqrt {d x +c}}\) | \(67\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (15 B \,d^{2} x^{2}+21 A \,d^{2} x +33 B c d x +49 A c d +22 B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 d^{2} \sqrt {d x +c}}\) | \(67\) |
risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (-15 B \,d^{3} x^{3}-21 A \,d^{3} x^{2}-18 B c \,d^{2} x^{2}-28 A c \,d^{2} x +11 B \,c^{2} d x +49 A \,c^{2} d +22 B \,c^{3}\right ) \sqrt {-d x +c}}{105 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(115\) |
Input:
int((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/105*(-d*x+c)*(15*B*d^2*x^2+21*A*d^2*x+33*B*c*d*x+49*A*c*d+22*B*c^2)*(-d ^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (15 \, B d^{3} x^{3} - 22 \, B c^{3} - 49 \, A c^{2} d + 3 \, {\left (6 \, B c d^{2} + 7 \, A d^{3}\right )} x^{2} - {\left (11 \, B c^{2} d - 28 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" )
Output:
2/105*(15*B*d^3*x^3 - 22*B*c^3 - 49*A*c^2*d + 3*(6*B*c*d^2 + 7*A*d^3)*x^2 - (11*B*c^2*d - 28*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^3*x + c*d^2)
\[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=\int \sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right ) \sqrt {c + d x}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(1/2)*(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)*sqrt(c + d*x), x)
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \, {\left (3 \, d^{2} x^{2} + 4 \, c d x - 7 \, c^{2}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{15 \, {\left (d^{2} x + c d\right )}} + \frac {2 \, {\left (15 \, d^{3} x^{3} + 18 \, c d^{2} x^{2} - 11 \, c^{2} d x - 22 \, c^{3}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{105 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" )
Output:
2/15*(3*d^2*x^2 + 4*c*d*x - 7*c^2)*(d*x + c)*sqrt(-d*x + c)*A/(d^2*x + c*d ) + 2/105*(15*d^3*x^3 + 18*c*d^2*x^2 - 11*c^2*d*x - 22*c^3)*(d*x + c)*sqrt (-d*x + c)*B/(d^3*x + c*d^2)
Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.43 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=-\frac {2 \, {\left (105 \, \sqrt {-d x + c} A c^{2} d - 35 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-d x + c} c\right )} B c^{2} - 7 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-d x + c} c^{2}\right )} A d - 3 \, {\left (5 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-d x + c} c^{3}\right )} B\right )}}{105 \, d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
-2/105*(105*sqrt(-d*x + c)*A*c^2*d - 35*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*B*c^2 - 7*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15 *sqrt(-d*x + c)*c^2)*A*d - 3*(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)/d ^2
Time = 9.78 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.01 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {2\,B\,x^3\,\sqrt {c+d\,x}}{7}-\frac {\left (44\,B\,c^3+98\,A\,d\,c^2\right )\,\sqrt {c+d\,x}}{105\,d^3}+\frac {x^2\,\left (42\,A\,d^3+36\,B\,c\,d^2\right )\,\sqrt {c+d\,x}}{105\,d^3}+\frac {2\,c\,x\,\left (28\,A\,d-11\,B\,c\right )\,\sqrt {c+d\,x}}{105\,d^2}\right )}{x+\frac {c}{d}} \] Input:
int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x)^(1/2),x)
Output:
((c^2 - d^2*x^2)^(1/2)*((2*B*x^3*(c + d*x)^(1/2))/7 - ((44*B*c^3 + 98*A*c^ 2*d)*(c + d*x)^(1/2))/(105*d^3) + (x^2*(42*A*d^3 + 36*B*c*d^2)*(c + d*x)^( 1/2))/(105*d^3) + (2*c*x*(28*A*d - 11*B*c)*(c + d*x)^(1/2))/(105*d^2)))/(x + c/d)
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int (A+B x) \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \, dx=\frac {2 \sqrt {-d x +c}\, \left (15 b \,d^{3} x^{3}+21 a \,d^{3} x^{2}+18 b c \,d^{2} x^{2}+28 a c \,d^{2} x -11 b \,c^{2} d x -49 a \,c^{2} d -22 b \,c^{3}\right )}{105 d^{2}} \] Input:
int((B*x+A)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2),x)
Output:
(2*sqrt(c - d*x)*( - 49*a*c**2*d + 28*a*c*d**2*x + 21*a*d**3*x**2 - 22*b*c **3 - 11*b*c**2*d*x + 18*b*c*d**2*x**2 + 15*b*d**3*x**3))/(105*d**2)