Integrand size = 29, antiderivative size = 99 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=-\frac {(B c-2 A d) \sqrt {c^2-d^2 x^2}}{2 d^2}-\frac {B \left (c^2-d^2 x^2\right )^{3/2}}{2 d^2 (c+d x)}-\frac {c (B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:
-1/2*(-2*A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2-1/2*B*(-d^2*x^2+c^2)^(3/2)/d^2/ (d*x+c)-1/2*c*(-2*A*d+B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
Time = 0.43 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=\frac {(-2 B c+2 A d+B d x) \sqrt {c^2-d^2 x^2}}{2 d^2}+\frac {c (B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Input:
Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x),x]
Output:
((-2*B*c + 2*A*d + B*d*x)*Sqrt[c^2 - d^2*x^2])/(2*d^2) + (c*(B*c - 2*A*d)* ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^2
Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {667, 676, 224, 216}
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.
\(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx\) |
\(\Big \downarrow \) 667 |
\(\displaystyle \int \frac {(A+B x) (c-d x)}{\sqrt {c^2-d^2 x^2}}dx\) |
\(\Big \downarrow \) 676 |
\(\displaystyle -\frac {c (B c-2 A d) \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx}{2 d}-\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{d^2}+\frac {B x \sqrt {c^2-d^2 x^2}}{2 d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {c (B c-2 A d) \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}}{2 d}-\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{d^2}+\frac {B x \sqrt {c^2-d^2 x^2}}{2 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {c (B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2}-\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{d^2}+\frac {B x \sqrt {c^2-d^2 x^2}}{2 d}\) |
Input:
Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x),x]
Output:
-(((B*c - A*d)*Sqrt[c^2 - d^2*x^2])/d^2) + (B*x*Sqrt[c^2 - d^2*x^2])/(2*d) - (c*(B*c - 2*A*d)*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d^2)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*( x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + c*x^2)^(p - 1), x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {\left (B d x +2 A d -2 B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c \left (2 A d -B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d \sqrt {d^{2}}}\) | \(76\) |
default | \(\frac {B \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{d}+\frac {\left (A d -B c \right ) \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{d^{2}}\) | \(142\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/2*(B*d*x+2*A*d-2*B*c)/d^2*(-d^2*x^2+c^2)^(1/2)+1/2*c/d*(2*A*d-B*c)/(d^2) ^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=\frac {2 \, {\left (B c^{2} - 2 \, A c d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + \sqrt {-d^{2} x^{2} + c^{2}} {\left (B d x - 2 \, B c + 2 \, A d\right )}}{2 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c),x, algorithm="fricas")
Output:
1/2*(2*(B*c^2 - 2*A*c*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + sqrt( -d^2*x^2 + c^2)*(B*d*x - 2*B*c + 2*A*d))/d^2
\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=\int \frac {\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right )}{c + d x}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c),x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x), x)
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=-\frac {B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {A c \arcsin \left (\frac {d x}{c}\right )}{d} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B x}{2 \, d} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{d^{2}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{d} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c),x, algorithm="maxima")
Output:
-1/2*B*c^2*arcsin(d*x/c)/d^2 + A*c*arcsin(d*x/c)/d + 1/2*sqrt(-d^2*x^2 + c ^2)*B*x/d - sqrt(-d^2*x^2 + c^2)*B*c/d^2 + sqrt(-d^2*x^2 + c^2)*A/d
Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=-\frac {{\left (B c^{2} - 2 \, A c d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d {\left | d \right |}} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {B x}{d} - \frac {2 \, {\left (B c d - A d^{2}\right )}}{d^{3}}\right )} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c),x, algorithm="giac")
Output:
-1/2*(B*c^2 - 2*A*c*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/2*sqrt(- d^2*x^2 + c^2)*(B*x/d - 2*(B*c*d - A*d^2)/d^3)
Timed out. \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )}{c+d\,x} \,d x \] Input:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x),x)
Output:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x), x)
Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{c+d x} \, dx=\frac {2 \mathit {asin} \left (\frac {d x}{c}\right ) a c d -\mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, a d -2 \sqrt {-d^{2} x^{2}+c^{2}}\, b c +\sqrt {-d^{2} x^{2}+c^{2}}\, b d x -2 a c d +2 b \,c^{2}}{2 d^{2}} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c),x)
Output:
(2*asin((d*x)/c)*a*c*d - asin((d*x)/c)*b*c**2 + 2*sqrt(c**2 - d**2*x**2)*a *d - 2*sqrt(c**2 - d**2*x**2)*b*c + sqrt(c**2 - d**2*x**2)*b*d*x - 2*a*c*d + 2*b*c**2)/(2*d**2)