Integrand size = 29, antiderivative size = 107 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=-\frac {\sqrt {1-d^2 x^2}}{f}-\frac {(d e-2 f) \arcsin (d x)}{f^2}+\frac {(d e-f)^2 \arctan \left (\frac {f+d^2 e x}{\sqrt {d^2 e^2-f^2} \sqrt {1-d^2 x^2}}\right )}{f^2 \sqrt {d^2 e^2-f^2}} \] Output:
-(-d^2*x^2+1)^(1/2)/f-(d*e-2*f)*arcsin(d*x)/f^2+(d*e-f)^2*arctan((d^2*e*x+ f)/(d^2*e^2-f^2)^(1/2)/(-d^2*x^2+1)^(1/2))/f^2/(d^2*e^2-f^2)^(1/2)
Time = 0.66 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=\frac {-f \sqrt {1-d^2 x^2}+(-2 d e+4 f) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )-\frac {2 (d e-f) \sqrt {d^2 e^2-f^2} \arctan \left (\frac {\sqrt {d^2 e^2-f^2} x}{e+f x-e \sqrt {1-d^2 x^2}}\right )}{d e+f}}{f^2} \] Input:
Integrate[(1 + d*x)^2/((e + f*x)*Sqrt[1 - d^2*x^2]),x]
Output:
(-(f*Sqrt[1 - d^2*x^2]) + (-2*d*e + 4*f)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x ^2])] - (2*(d*e - f)*Sqrt[d^2*e^2 - f^2]*ArcTan[(Sqrt[d^2*e^2 - f^2]*x)/(e + f*x - e*Sqrt[1 - d^2*x^2])])/(d*e + f))/f^2
Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {716, 25, 27, 719, 223, 488, 217}
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.
\(\displaystyle \int \frac {(d x+1)^2}{\sqrt {1-d^2 x^2} (e+f x)} \, dx\) |
\(\Big \downarrow \) 716 |
\(\displaystyle -\frac {\int -\frac {d^2 f (f-d (d e-2 f) x)}{(e+f x) \sqrt {1-d^2 x^2}}dx}{d^2 f^2}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {d^2 f (f-d (d e-2 f) x)}{(e+f x) \sqrt {1-d^2 x^2}}dx}{d^2 f^2}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {f-d (d e-2 f) x}{(e+f x) \sqrt {1-d^2 x^2}}dx}{f}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {\frac {(d e-f)^2 \int \frac {1}{(e+f x) \sqrt {1-d^2 x^2}}dx}{f}-\frac {d (d e-2 f) \int \frac {1}{\sqrt {1-d^2 x^2}}dx}{f}}{f}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {(d e-f)^2 \int \frac {1}{(e+f x) \sqrt {1-d^2 x^2}}dx}{f}-\frac {\arcsin (d x) (d e-2 f)}{f}}{f}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {-\frac {(d e-f)^2 \int \frac {1}{-d^2 e^2+f^2-\frac {\left (e x d^2+f\right )^2}{1-d^2 x^2}}d\frac {e x d^2+f}{\sqrt {1-d^2 x^2}}}{f}-\frac {\arcsin (d x) (d e-2 f)}{f}}{f}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {(d e-f)^2 \arctan \left (\frac {d^2 e x+f}{\sqrt {1-d^2 x^2} \sqrt {d^2 e^2-f^2}}\right )}{f \sqrt {d^2 e^2-f^2}}-\frac {\arcsin (d x) (d e-2 f)}{f}}{f}-\frac {\sqrt {1-d^2 x^2}}{f}\) |
Input:
Int[(1 + d*x)^2/((e + f*x)*Sqrt[1 - d^2*x^2]),x]
Output:
-(Sqrt[1 - d^2*x^2]/f) + (-(((d*e - 2*f)*ArcSin[d*x])/f) + ((d*e - f)^2*Ar cTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(f*Sqrt[d^2*e ^2 - f^2]))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_) ^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + c*x^2)^(p + 1)/ (c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) I nt[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x)^ n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d + e*x)^(n - 2)*(a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) - 2*c*d*e*(m + n + p)*x), x], x], x] /; F reeQ[{a, c, d, e, f, g, m, p}, x] && IGtQ[n, 1] && IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(228\) vs. \(2(99)=198\).
Time = 0.88 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.14
method | result | size |
risch | \(\frac {d^{2} x^{2}-1}{f \sqrt {-d^{2} x^{2}+1}}-\frac {\frac {d \left (d e -2 f \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{f \sqrt {d^{2}}}+\frac {\left (d^{2} e^{2}-2 d e f +f^{2}\right ) \ln \left (\frac {-\frac {2 \left (d^{2} e^{2}-f^{2}\right )}{f^{2}}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}+2 \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, \sqrt {-d^{2} \left (x +\frac {e}{f}\right )^{2}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}{x +\frac {e}{f}}\right )}{f^{2} \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}}{f}\) | \(229\) |
default | \(-\frac {\left (d^{2} e^{2}-2 d e f +f^{2}\right ) \ln \left (\frac {-\frac {2 \left (d^{2} e^{2}-f^{2}\right )}{f^{2}}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}+2 \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, \sqrt {-d^{2} \left (x +\frac {e}{f}\right )^{2}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}{x +\frac {e}{f}}\right )}{f^{3} \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}-\frac {d \left (\frac {d e \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{\sqrt {d^{2}}}-\frac {2 f \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{\sqrt {d^{2}}}+\frac {f \sqrt {-d^{2} x^{2}+1}}{d}\right )}{f^{2}}\) | \(242\) |
Input:
int((d*x+1)^2/(f*x+e)/(-d^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*(d^2*x^2-1)/(-d^2*x^2+1)^(1/2)-1/f*(d*(d*e-2*f)/f/(d^2)^(1/2)*arctan(( d^2)^(1/2)*x/(-d^2*x^2+1)^(1/2))+(d^2*e^2-2*d*e*f+f^2)/f^2/(-(d^2*e^2-f^2) /f^2)^(1/2)*ln((-2*(d^2*e^2-f^2)/f^2+2*d^2*e/f*(x+e/f)+2*(-(d^2*e^2-f^2)/f ^2)^(1/2)*(-d^2*(x+e/f)^2+2*d^2*e/f*(x+e/f)-(d^2*e^2-f^2)/f^2)^(1/2))/(x+e /f)))
Time = 0.16 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.97 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=\left [-\frac {{\left (d e - f\right )} \sqrt {-\frac {d e - f}{d e + f}} \log \left (\frac {d^{2} e f x + f^{2} - {\left (d^{2} e^{2} - f^{2}\right )} \sqrt {-d^{2} x^{2} + 1} - {\left (d e f + f^{2} + {\left (d^{3} e^{2} + d^{2} e f\right )} x + \sqrt {-d^{2} x^{2} + 1} {\left (d e f + f^{2}\right )}\right )} \sqrt {-\frac {d e - f}{d e + f}}}{f x + e}\right ) - 2 \, {\left (d e - 2 \, f\right )} \arctan \left (\frac {\sqrt {-d^{2} x^{2} + 1} - 1}{d x}\right ) + \sqrt {-d^{2} x^{2} + 1} f}{f^{2}}, \frac {2 \, {\left (d e - f\right )} \sqrt {\frac {d e - f}{d e + f}} \arctan \left (\frac {{\left (f x - \sqrt {-d^{2} x^{2} + 1} e + e\right )} \sqrt {\frac {d e - f}{d e + f}}}{{\left (d e - f\right )} x}\right ) + 2 \, {\left (d e - 2 \, f\right )} \arctan \left (\frac {\sqrt {-d^{2} x^{2} + 1} - 1}{d x}\right ) - \sqrt {-d^{2} x^{2} + 1} f}{f^{2}}\right ] \] Input:
integrate((d*x+1)^2/(f*x+e)/(-d^2*x^2+1)^(1/2),x, algorithm="fricas")
Output:
[-((d*e - f)*sqrt(-(d*e - f)/(d*e + f))*log((d^2*e*f*x + f^2 - (d^2*e^2 - f^2)*sqrt(-d^2*x^2 + 1) - (d*e*f + f^2 + (d^3*e^2 + d^2*e*f)*x + sqrt(-d^2 *x^2 + 1)*(d*e*f + f^2))*sqrt(-(d*e - f)/(d*e + f)))/(f*x + e)) - 2*(d*e - 2*f)*arctan((sqrt(-d^2*x^2 + 1) - 1)/(d*x)) + sqrt(-d^2*x^2 + 1)*f)/f^2, (2*(d*e - f)*sqrt((d*e - f)/(d*e + f))*arctan((f*x - sqrt(-d^2*x^2 + 1)*e + e)*sqrt((d*e - f)/(d*e + f))/((d*e - f)*x)) + 2*(d*e - 2*f)*arctan((sqrt (-d^2*x^2 + 1) - 1)/(d*x)) - sqrt(-d^2*x^2 + 1)*f)/f^2]
\[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=\int \frac {\left (d x + 1\right )^{2}}{\sqrt {- \left (d x - 1\right ) \left (d x + 1\right )} \left (e + f x\right )}\, dx \] Input:
integrate((d*x+1)**2/(f*x+e)/(-d**2*x**2+1)**(1/2),x)
Output:
Integral((d*x + 1)**2/(sqrt(-(d*x - 1)*(d*x + 1))*(e + f*x)), x)
Exception generated. \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+1)^2/(f*x+e)/(-d^2*x^2+1)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=-\frac {{\left (d^{2} e - 2 \, d f\right )} \arcsin \left (d x\right ) \mathrm {sgn}\left (d\right )}{f^{2} {\left | d \right |}} - \frac {\sqrt {-d^{2} x^{2} + 1}}{f} - \frac {2 \, {\left (d^{3} e^{2} - 2 \, d^{2} e f + d f^{2}\right )} \arctan \left (\frac {f + \frac {{\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )} e}{d x}}{\sqrt {d^{2} e^{2} - f^{2}}}\right )}{\sqrt {d^{2} e^{2} - f^{2}} f^{2} {\left | d \right |}} \] Input:
integrate((d*x+1)^2/(f*x+e)/(-d^2*x^2+1)^(1/2),x, algorithm="giac")
Output:
-(d^2*e - 2*d*f)*arcsin(d*x)*sgn(d)/(f^2*abs(d)) - sqrt(-d^2*x^2 + 1)/f - 2*(d^3*e^2 - 2*d^2*e*f + d*f^2)*arctan((f + (sqrt(-d^2*x^2 + 1)*abs(d) + d )*e/(d*x))/sqrt(d^2*e^2 - f^2))/(sqrt(d^2*e^2 - f^2)*f^2*abs(d))
Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.38 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=-\frac {\sqrt {1-d^2\,x^2}}{f}-\frac {\mathrm {asinh}\left (x\,\sqrt {-d^2}\right )\,\left (2\,d\,\sqrt {-d^2}-\frac {d^2\,e\,\sqrt {-d^2}}{f}\right )}{d^2\,f}-\frac {\left (\ln \left (\sqrt {1-\frac {d^2\,e^2}{f^2}}\,\sqrt {1-d^2\,x^2}+\frac {d^2\,e\,x}{f}+1\right )-\ln \left (e+f\,x\right )\right )\,\left (d^2\,e^2-2\,d\,e\,f+f^2\right )}{f^3\,\sqrt {1-\frac {d^2\,e^2}{f^2}}} \] Input:
int((d*x + 1)^2/((e + f*x)*(1 - d^2*x^2)^(1/2)),x)
Output:
- (1 - d^2*x^2)^(1/2)/f - (asinh(x*(-d^2)^(1/2))*(2*d*(-d^2)^(1/2) - (d^2* e*(-d^2)^(1/2))/f))/(d^2*f) - ((log((1 - (d^2*e^2)/f^2)^(1/2)*(1 - d^2*x^2 )^(1/2) + (d^2*e*x)/f + 1) - log(e + f*x))*(f^2 + d^2*e^2 - 2*d*e*f))/(f^3 *(1 - (d^2*e^2)/f^2)^(1/2))
Time = 0.19 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.64 \[ \int \frac {(1+d x)^2}{(e+f x) \sqrt {1-d^2 x^2}} \, dx=\frac {-\mathit {asin} \left (d x \right ) d^{2} e^{2}+\mathit {asin} \left (d x \right ) d e f +2 \mathit {asin} \left (d x \right ) f^{2}+2 \sqrt {d^{2} e^{2}-f^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (d x \right )}{2}\right ) d e +f}{\sqrt {d^{2} e^{2}-f^{2}}}\right ) d e -2 \sqrt {d^{2} e^{2}-f^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (d x \right )}{2}\right ) d e +f}{\sqrt {d^{2} e^{2}-f^{2}}}\right ) f -\sqrt {-d^{2} x^{2}+1}\, d e f -\sqrt {-d^{2} x^{2}+1}\, f^{2}+d e f +f^{2}}{f^{2} \left (d e +f \right )} \] Input:
int((d*x+1)^2/(f*x+e)/(-d^2*x^2+1)^(1/2),x)
Output:
( - asin(d*x)*d**2*e**2 + asin(d*x)*d*e*f + 2*asin(d*x)*f**2 + 2*sqrt(d**2 *e**2 - f**2)*atan((tan(asin(d*x)/2)*d*e + f)/sqrt(d**2*e**2 - f**2))*d*e - 2*sqrt(d**2*e**2 - f**2)*atan((tan(asin(d*x)/2)*d*e + f)/sqrt(d**2*e**2 - f**2))*f - sqrt( - d**2*x**2 + 1)*d*e*f - sqrt( - d**2*x**2 + 1)*f**2 + d*e*f + f**2)/(f**2*(d*e + f))