Integrand size = 24, antiderivative size = 66 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=-\frac {(A-B) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e}}\right )}{\sqrt {d-e}}+\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d+e}}\right )}{\sqrt {d+e}} \]
-(A-B)*arctanh((e*x+d)^(1/2)/(d-e)^(1/2))/(d-e)^(1/2)+(A+B)*arctanh((e*x+d )^(1/2)/(d+e)^(1/2))/(d+e)^(1/2)
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=-\frac {(A+B) \arctan \left (\frac {\sqrt {d+e x}}{\sqrt {-d-e}}\right )}{\sqrt {-d-e}}+\frac {(A-B) \arctan \left (\frac {\sqrt {d+e x}}{\sqrt {-d+e}}\right )}{\sqrt {-d+e}} \]
-(((A + B)*ArcTan[Sqrt[d + e*x]/Sqrt[-d - e]])/Sqrt[-d - e]) + ((A - B)*Ar cTan[Sqrt[d + e*x]/Sqrt[-d + e]])/Sqrt[-d + e]
Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {654, 1480, 220}
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.
\(\displaystyle \int \frac {A+B x}{\left (1-x^2\right ) \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 654 |
\(\displaystyle 2 \int \frac {B d-A e-B (d+e x)}{d^2-2 (d+e x) d-e^2+(d+e x)^2}d\sqrt {d+e x}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle 2 \left (\frac {1}{2} (A-B) \int \frac {1}{x e+e}d\sqrt {d+e x}-\frac {1}{2} (A+B) \int \frac {1}{e x-e}d\sqrt {d+e x}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle 2 \left (\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d+e}}\right )}{2 \sqrt {d+e}}-\frac {(A-B) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e}}\right )}{2 \sqrt {d-e}}\right )\) |
2*(-1/2*((A - B)*ArcTanh[Sqrt[d + e*x]/Sqrt[d - e]])/Sqrt[d - e] + ((A + B )*ArcTanh[Sqrt[d + e*x]/Sqrt[d + e]])/(2*Sqrt[d + e]))
3.15.65.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d* x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82
method | result | size |
pseudoelliptic | \(\frac {\left (A -B \right ) \arctan \left (\frac {\sqrt {e x +d}}{\sqrt {-d +e}}\right )}{\sqrt {-d +e}}+\frac {\left (A +B \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d +e}}\right )}{\sqrt {d +e}}\) | \(54\) |
derivativedivides | \(-\frac {2 \left (-\frac {A}{2}+\frac {B}{2}\right ) \arctan \left (\frac {\sqrt {e x +d}}{\sqrt {-d +e}}\right )}{\sqrt {-d +e}}+\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d +e}}\right )}{\sqrt {d +e}}\) | \(62\) |
default | \(-\frac {2 \left (-\frac {A}{2}+\frac {B}{2}\right ) \arctan \left (\frac {\sqrt {e x +d}}{\sqrt {-d +e}}\right )}{\sqrt {-d +e}}+\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d +e}}\right )}{\sqrt {d +e}}\) | \(62\) |
(A-B)*arctan((e*x+d)^(1/2)/(-d+e)^(1/2))/(-d+e)^(1/2)+(A+B)*arctanh((e*x+d )^(1/2)/(d+e)^(1/2))/(d+e)^(1/2)
Time = 0.42 (sec) , antiderivative size = 451, normalized size of antiderivative = 6.83 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=\left [-\frac {{\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {d - e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d - e} + 2 \, d - e}{x + 1}\right ) - {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {d + e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d + e} + 2 \, d + e}{x - 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {2 \, {\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {-d + e} \arctan \left (-\frac {\sqrt {e x + d} \sqrt {-d + e}}{d - e}\right ) - {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {d + e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d + e} + 2 \, d + e}{x - 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {2 \, {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {-d - e} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d - e}}{d + e}\right ) + {\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {d - e} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d - e} + 2 \, d - e}{x + 1}\right )}{2 \, {\left (d^{2} - e^{2}\right )}}, -\frac {{\left ({\left (A - B\right )} d + {\left (A - B\right )} e\right )} \sqrt {-d + e} \arctan \left (-\frac {\sqrt {e x + d} \sqrt {-d + e}}{d - e}\right ) + {\left ({\left (A + B\right )} d - {\left (A + B\right )} e\right )} \sqrt {-d - e} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d - e}}{d + e}\right )}{d^{2} - e^{2}}\right ] \]
[-1/2*(((A - B)*d + (A - B)*e)*sqrt(d - e)*log((e*x + 2*sqrt(e*x + d)*sqrt (d - e) + 2*d - e)/(x + 1)) - ((A + B)*d - (A + B)*e)*sqrt(d + e)*log((e*x + 2*sqrt(e*x + d)*sqrt(d + e) + 2*d + e)/(x - 1)))/(d^2 - e^2), -1/2*(2*( (A - B)*d + (A - B)*e)*sqrt(-d + e)*arctan(-sqrt(e*x + d)*sqrt(-d + e)/(d - e)) - ((A + B)*d - (A + B)*e)*sqrt(d + e)*log((e*x + 2*sqrt(e*x + d)*sqr t(d + e) + 2*d + e)/(x - 1)))/(d^2 - e^2), -1/2*(2*((A + B)*d - (A + B)*e) *sqrt(-d - e)*arctan(sqrt(e*x + d)*sqrt(-d - e)/(d + e)) + ((A - B)*d + (A - B)*e)*sqrt(d - e)*log((e*x + 2*sqrt(e*x + d)*sqrt(d - e) + 2*d - e)/(x + 1)))/(d^2 - e^2), -(((A - B)*d + (A - B)*e)*sqrt(-d + e)*arctan(-sqrt(e* x + d)*sqrt(-d + e)/(d - e)) + ((A + B)*d - (A + B)*e)*sqrt(-d - e)*arctan (sqrt(e*x + d)*sqrt(-d - e)/(d + e)))/(d^2 - e^2)]
Time = 4.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {e \left (A - B\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d + e}} \right )}}{2 \sqrt {- d + e}} - \frac {e \left (A + B\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d - e}} \right )}}{2 \sqrt {- d - e}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A \left (- \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x + 1 \right )}}{2}\right ) - \frac {B \log {\left (1 - x^{2} \right )}}{2}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(e*(A - B)*atan(sqrt(d + e*x)/sqrt(-d + e))/(2*sqrt(-d + e)) - e*(A + B)*atan(sqrt(d + e*x)/sqrt(-d - e))/(2*sqrt(-d - e)))/e, Ne(e, 0) ), ((A*(-log(x - 1)/2 + log(x + 1)/2) - B*log(1 - x**2)/2)/sqrt(d), True))
Exception generated. \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d-4*e>0)', see `assume?` for m ore detail
Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=\frac {{\left (A - B\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d + e}}\right )}{\sqrt {-d + e}} - \frac {{\left (A + B\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d - e}}\right )}{\sqrt {-d - e}} \]
(A - B)*arctan(sqrt(e*x + d)/sqrt(-d + e))/sqrt(-d + e) - (A + B)*arctan(s qrt(e*x + d)/sqrt(-d - e))/sqrt(-d - e)
Time = 10.89 (sec) , antiderivative size = 773, normalized size of antiderivative = 11.71 \[ \int \frac {A+B x}{\sqrt {d+e x} \left (1-x^2\right )} \, dx=-\frac {\mathrm {atan}\left (\frac {\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{2\,\sqrt {d-e}}+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{2\,\sqrt {d-e}}}{16\,B^3\,e^2-16\,A^2\,B\,e^2+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )}{2\,\sqrt {d-e}}-\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A-B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A-B\right )\,\sqrt {d+e\,x}}{\sqrt {d-e}}\right )}{2\,\sqrt {d-e}}\right )\,\left (A-B\right )}{2\,\sqrt {d-e}}}\right )\,\left (A-B\right )\,1{}\mathrm {i}}{\sqrt {d-e}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{2\,\sqrt {d+e}}+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{2\,\sqrt {d+e}}}{16\,B^3\,e^2-16\,A^2\,B\,e^2+\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,B\,d\,e^2-32\,A\,e^3+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )}{2\,\sqrt {d+e}}-\frac {\left (\left (16\,A^2\,e^2+16\,B^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {\left (A+B\right )\,\left (32\,A\,e^3-32\,B\,d\,e^2+\frac {32\,d\,e^2\,\left (A+B\right )\,\sqrt {d+e\,x}}{\sqrt {d+e}}\right )}{2\,\sqrt {d+e}}\right )\,\left (A+B\right )}{2\,\sqrt {d+e}}}\right )\,\left (A+B\right )\,1{}\mathrm {i}}{\sqrt {d+e}} \]
- (atan(((((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*B*d*e^ 2 - 32*A*e^3 + (32*d*e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B)*1i)/(2*(d - e)^(1/2)) + (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*A*e^3 - 32*B*d*e^2 + (32*d*e^2*(A - B)*(d + e* x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B)*1i)/(2*(d - e)^(1/2)) )/(16*B^3*e^2 - 16*A^2*B*e^2 + (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B))/(2*(d - e)^(1/2)) - (((16*A^2*e^ 2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A - B)*(32*A*e^3 - 32*B*d*e^2 + (32*d* e^2*(A - B)*(d + e*x)^(1/2))/(d - e)^(1/2)))/(2*(d - e)^(1/2)))*(A - B))/( 2*(d - e)^(1/2))))*(A - B)*1i)/(d - e)^(1/2) - (atan(((((16*A^2*e^2 + 16*B ^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*B*d*e^2 - 32*A*e^3 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2*(d + e)^(1/2)))*(A + B)*1i)/(2*(d + e)^(1/2)) + (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*A *e^3 - 32*B*d*e^2 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2* (d + e)^(1/2)))*(A + B)*1i)/(2*(d + e)^(1/2)))/(16*B^3*e^2 - 16*A^2*B*e^2 + (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/2) - ((A + B)*(32*B*d*e^2 - 32* A*e^3 + (32*d*e^2*(A + B)*(d + e*x)^(1/2))/(d + e)^(1/2)))/(2*(d + e)^(1/2 )))*(A + B))/(2*(d + e)^(1/2)) - (((16*A^2*e^2 + 16*B^2*e^2)*(d + e*x)^(1/ 2) - ((A + B)*(32*A*e^3 - 32*B*d*e^2 + (32*d*e^2*(A + B)*(d + e*x)^(1/2...