Integrand size = 31, antiderivative size = 91 \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}+\frac {\sqrt {2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c} d^2} \] Output:
-2*B*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)+2^(1/2)*(-A*d+B*c)*arctanh(2^( 1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))/c^(1/2)/d^2
Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-\frac {2 B \sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {\sqrt {2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{d^2} \] Input:
Integrate[(A + B*x)/(Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]),x]
Output:
((-2*B*Sqrt[c^2 - d^2*x^2])/Sqrt[c + d*x] + (Sqrt[2]*(B*c - A*d)*ArcTanh[( Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/Sqrt[c])/d^2
Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {600, 458, 471, 221}
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.
\(\displaystyle \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx\) |
\(\Big \downarrow \) 600 |
\(\displaystyle \frac {B \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx}{d}-\frac {(B c-A d) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{d}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle -\frac {(B c-A d) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{d}-\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle -2 (B c-A d) \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}-\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\sqrt {2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {c} d^2}-\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}\) |
Input:
Int[(A + B*x)/(Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]),x]
Output:
(-2*B*Sqrt[c^2 - d^2*x^2])/(d^2*Sqrt[c + d*x]) + (Sqrt[2]*(B*c - A*d)*ArcT anh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/(Sqrt[c]*d^2)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[((A_.) + (B_.)*(x_))/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2] ), x_Symbol] :> Simp[B/d Int[Sqrt[c + d*x]/Sqrt[a + b*x^2], x], x] - Simp [(B*c - A*d)/d Int[1/(Sqrt[c + d*x]*Sqrt[a + b*x^2]), x], x] /; FreeQ[{a, b, c, d, A, B}, x] && NegQ[b/a]
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {\left (-A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, d +B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c -2 B \sqrt {-d x +c}\, \sqrt {c}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{\sqrt {d x +c}\, \sqrt {-d x +c}\, d^{2} \sqrt {c}}\) | \(99\) |
risch | \(-\frac {2 B \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {\left (A d -B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {c}\, \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(138\) |
Input:
int((B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
(-A*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*d+B*arctanh(1/2*(- d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c-2*B*(-d*x+c)^(1/2)*c^(1/2))/(d*x+c )^(1/2)*(-d^2*x^2+c^2)^(1/2)/(-d*x+c)^(1/2)/d^2/c^(1/2)
Time = 0.09 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.89 \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\left [-\frac {4 \, \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} B + \frac {\sqrt {2} {\left (B c^{2} - A c d + {\left (B c d - A d^{2}\right )} x\right )} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{\sqrt {c}}}{2 \, {\left (d^{3} x + c d^{2}\right )}}, \frac {\sqrt {2} {\left (B c^{2} - A c d + {\left (B c d - A d^{2}\right )} x\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} c \sqrt {-\frac {1}{c}}}{d^{2} x^{2} - c^{2}}\right ) - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} B}{d^{3} x + c d^{2}}\right ] \] Input:
integrate((B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" )
Output:
[-1/2*(4*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*B + sqrt(2)*(B*c^2 - A*c*d + ( B*c*d - A*d^2)*x)*log(-(d^2*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2*x^2 + c^2) *sqrt(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2))/sqrt(c))/(d^3*x + c*d^2), (sqrt(2)*(B*c^2 - A*c*d + (B*c*d - A*d^2)*x)*sqrt(-1/c)*arctan( sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*c*sqrt(-1/c)/(d^2*x^2 - c^2)) - 2*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*B)/(d^3*x + c*d^2)]
\[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\int \frac {A + B x}{\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \sqrt {c + d x}}\, dx \] Input:
integrate((B*x+A)/(d*x+c)**(1/2)/(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral((A + B*x)/(sqrt(-(-c + d*x)*(c + d*x))*sqrt(c + d*x)), x)
\[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\int { \frac {B x + A}{\sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}} \,d x } \] Input:
integrate((B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" )
Output:
integrate((B*x + A)/(sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)), x)
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\frac {\sqrt {2} {\left (B c - A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + 2 \, \sqrt {-d x + c} B}{d^{2}} \] Input:
integrate((B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
-(sqrt(2)*(B*c - A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sqrt(-c) + 2*sqrt(-d*x + c)*B)/d^2
Timed out. \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {c^2-d^2\,x^2}\,\sqrt {c+d\,x}} \,d x \] Input:
int((A + B*x)/((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(1/2)),x)
Output:
int((A + B*x)/((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(1/2)), x)
Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-4 \sqrt {-d x +c}\, b c +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a d -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a d +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c}{2 c \,d^{2}} \] Input:
int((B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x)
Output:
( - 4*sqrt(c - d*x)*b*c + sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt (2))*a*d - sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c - sqrt (c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d + sqrt(c)*sqrt(2)*log (sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c)/(2*c*d**2)