Integrand size = 29, antiderivative size = 91 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {B \sqrt {c^2-d^2 x^2}}{d^2}+\frac {2 (B c-A d) \sqrt {c^2-d^2 x^2}}{d^2 (c+d x)}+\frac {(2 B c-A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:
B*(-d^2*x^2+c^2)^(1/2)/d^2+2*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)+( -A*d+2*B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {(3 B c-2 A d+B d x) \sqrt {c^2-d^2 x^2}}{d^2 (c+d x)}-\frac {\sqrt {-d^2} (-2 B c+A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{d^3} \] Input:
Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^2,x]
Output:
((3*B*c - 2*A*d + B*d*x)*Sqrt[c^2 - d^2*x^2])/(d^2*(c + d*x)) - (Sqrt[-d^2 ]*(-2*B*c + A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d^2*x^2]])/d^3
Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 466, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {(2 B c-A d) \int \frac {\sqrt {c^2-d^2 x^2}}{c+d x}dx}{c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{c d^2 (c+d x)^2}\) |
\(\Big \downarrow \) 466 |
\(\displaystyle \frac {(2 B c-A d) \left (c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{c d^2 (c+d x)^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(2 B c-A d) \left (c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{c d^2 (c+d x)^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {(2 B c-A d) \left (\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{c d^2 (c+d x)^2}\) |
Input:
Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^2,x]
Output:
((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(c*d^2*(c + d*x)^2) + ((2*B*c - A*d)*( Sqrt[c^2 - d^2*x^2]/d + (c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/(c*d)
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 2*(n + 2*p + 1))) Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 ] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(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] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Time = 0.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {B \sqrt {-d^{2} x^{2}+c^{2}}}{d^{2}}-\frac {\left (A d -2 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d \sqrt {d^{2}}}-\frac {2 \left (A d -B c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{3} \left (x +\frac {c}{d}\right )}\) | \(111\) |
default | \(\frac {B \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{d^{2}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}\right )}{d^{3}}\) | \(217\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
B*(-d^2*x^2+c^2)^(1/2)/d^2-(A*d-2*B*c)/d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/ (-d^2*x^2+c^2)^(1/2))-2*(A*d-B*c)/d^3/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d ))^(1/2)
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {3 \, B c^{2} - 2 \, A c d + {\left (3 \, B c d - 2 \, A d^{2}\right )} x - 2 \, {\left (2 \, B c^{2} - A c d + {\left (2 \, B c d - A d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + \sqrt {-d^{2} x^{2} + c^{2}} {\left (B d x + 3 \, B c - 2 \, A d\right )}}{d^{3} x + c d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^2,x, algorithm="fricas")
Output:
(3*B*c^2 - 2*A*c*d + (3*B*c*d - 2*A*d^2)*x - 2*(2*B*c^2 - A*c*d + (2*B*c*d - A*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + sqrt(-d^2*x^2 + c ^2)*(B*d*x + 3*B*c - 2*A*d))/(d^3*x + c*d^2)
\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\int \frac {\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right )}{\left (c + d x\right )^{2}}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**2,x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x)**2, x)
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{d^{3} x + c d^{2}} + \frac {2 \, B c \arcsin \left (\frac {d x}{c}\right )}{d^{2}} - \frac {A \arcsin \left (\frac {d x}{c}\right )}{d} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{d^{2} x + c d} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^2,x, algorithm="maxima")
Output:
2*sqrt(-d^2*x^2 + c^2)*B*c/(d^3*x + c*d^2) + 2*B*c*arcsin(d*x/c)/d^2 - A*a rcsin(d*x/c)/d - 2*sqrt(-d^2*x^2 + c^2)*A/(d^2*x + c*d) + sqrt(-d^2*x^2 + c^2)*B/d^2
Time = 0.14 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {{\left ({\left (d x + c\right )} B c d^{2} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 2 \, B c^{2} d^{2} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 2 \, A c d^{3} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 2 \, {\left (2 \, B c^{2} d^{2} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - A c d^{3} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right )\right )} {\left | d \right |}}{c d^{5}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^2,x, algorithm="giac")
Output:
((d*x + c)*B*c*d^2*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 2*B*c ^2*d^2*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 2*A*c*d^3*sqrt(2* c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 2*(2*B*c^2*d^2*sgn(1/(d*x + c)) *sgn(d) - A*c*d^3*sgn(1/(d*x + c))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 1)) )*abs(d)/(c*d^5)
Timed out. \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^2} \,d x \] Input:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^2,x)
Output:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^2, x)
Time = 0.22 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^2} \, dx=\frac {-\sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a d +2 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c +\mathit {asin} \left (\frac {d x}{c}\right ) a c d +\mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{2} x -2 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2}-2 \mathit {asin} \left (\frac {d x}{c}\right ) b c d x +4 \sqrt {-d^{2} x^{2}+c^{2}}\, a d -4 \sqrt {-d^{2} x^{2}+c^{2}}\, b c -\sqrt {-d^{2} x^{2}+c^{2}}\, b d x -4 a c d +4 b \,c^{2}-b c d x -b \,d^{2} x^{2}}{d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^2,x)
Output:
( - sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*d + 2*sqrt(c**2 - d**2*x**2)*as in((d*x)/c)*b*c + asin((d*x)/c)*a*c*d + asin((d*x)/c)*a*d**2*x - 2*asin((d *x)/c)*b*c**2 - 2*asin((d*x)/c)*b*c*d*x + 4*sqrt(c**2 - d**2*x**2)*a*d - 4 *sqrt(c**2 - d**2*x**2)*b*c - sqrt(c**2 - d**2*x**2)*b*d*x - 4*a*c*d + 4*b *c**2 - b*c*d*x - b*d**2*x**2)/(d**2*(sqrt(c**2 - d**2*x**2) - c - d*x))