Integrand size = 25, antiderivative size = 151 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=-\frac {a f^2}{2 d^2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}-\frac {1+\frac {a f^2}{d^2}}{2 e \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}-\frac {a f^2 \log \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{d^3 e}+\frac {a f^2 \log \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{d^3 e} \]
-a*f^2*ln(e*x+f*(a+e^2*x^2/f^2)^(1/2))/d^3/e+a*f^2*ln(d+e*x+f*(a+e^2*x^2/f ^2)^(1/2))/d^3/e-1/2*a*f^2/d^2/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))+1/2*(-1-a*f ^2/d^2)/e/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))
Time = 0.70 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\frac {-d \left (d^3+d^2 e x+a f^2 \left (e x-f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+d e x \left (-e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )+a f^2 \left (-d^2+a f^2-2 d e x\right ) \log \left (-\sqrt {a} f+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+a f^2 \left (d^2-a f^2+2 d e x\right ) \log \left (-a f^2+d \left (e x-f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+\sqrt {a} f \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{d^3 e \left (d^2-a f^2+2 d e x\right )} \]
(-(d*(d^3 + d^2*e*x + a*f^2*(e*x - f*Sqrt[a + (e^2*x^2)/f^2]) + d*e*x*(-(e *x) + f*Sqrt[a + (e^2*x^2)/f^2]))) + a*f^2*(-d^2 + a*f^2 - 2*d*e*x)*Log[-( Sqrt[a]*f) + e*x + f*Sqrt[a + (e^2*x^2)/f^2]] + a*f^2*(d^2 - a*f^2 + 2*d*e *x)*Log[-(a*f^2) + d*(e*x - f*Sqrt[a + (e^2*x^2)/f^2]) + Sqrt[a]*f*(d + e* x + f*Sqrt[a + (e^2*x^2)/f^2])])/(d^3*e*(d^2 - a*f^2 + 2*d*e*x))
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2542, 1195, 2009}
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.
\(\displaystyle \int \frac {1}{\left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^2} \, dx\) |
\(\Big \downarrow \) 2542 |
\(\displaystyle \frac {\int \frac {d^2-2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{2 e}\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \frac {\int \left (\frac {2 a f^2}{d^3 \left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )}+\frac {2 a f^2}{d^3 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}+\frac {a f^2}{d^2 \left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2}+\frac {d^2+a f^2}{d^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a f^2 \log \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}{d^3}+\frac {2 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{d^3}+\frac {a f^2}{d^2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}-\frac {\frac {a f^2}{d^2}+1}{f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 e}\) |
((a*f^2)/(d^2*(-(e*x) - f*Sqrt[a + (e^2*x^2)/f^2])) - (1 + (a*f^2)/d^2)/(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - (2*a*f^2*Log[-(e*x) - f*Sqrt[a + (e^ 2*x^2)/f^2]])/d^3 + (2*a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/d^3 )/(2*e)
3.5.58.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^( n_))^(p_.), x_Symbol] :> Simp[1/(2*e) Subst[Int[(g + h*x^n)^p*((d^2 + a*f ^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; Fr eeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(3181\) vs. \(2(140)=280\).
Time = 0.08 (sec) , antiderivative size = 3182, normalized size of antiderivative = 21.07
1/2*d/(a*f^2-2*d*e*x-d^2)/e+1/2*a*f^2/(a*f^2-2*d*e*x-d^2)/e/d+2*e^2*(1/4/e ^2/d^2*x+1/4/e^3/d^3*(a*f^2-d^2)*ln(-a*f^2+2*d*e*x+d^2)-1/8*(a^2*f^4-2*a*d ^2*f^2+d^4)/e^3/d^3/(-a*f^2+2*d*e*x+d^2))+2*e*d*(1/4/e^2/d^2*ln(-a*f^2+2*d *e*x+d^2)-1/4*(a*f^2-d^2)/(-a*f^2+2*d*e*x+d^2)/e^2/d^2)-1/2/d*f/e^2*(-4/(a ^2*f^4+2*a*d^2*f^2+d^4)*d^2*f^2/(x+1/2*(-a*f^2+d^2)/e/d)*(e^2*(x+1/2*(-a*f ^2+d^2)/e/d)^2/f^2+e*(a*f^2-d^2)/d/f^2*(x+1/2*(-a*f^2+d^2)/e/d)+1/4*(a^2*f ^4+2*a*d^2*f^2+d^4)/d^2/f^2)^(3/2)+2*e*(a*f^2-d^2)*d/(a^2*f^4+2*a*d^2*f^2+ d^4)*(1/2*(4*e^2*(x+1/2*(-a*f^2+d^2)/e/d)^2/f^2+4*e*(a*f^2-d^2)/d/f^2*(x+1 /2*(-a*f^2+d^2)/e/d)+(a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2)^(1/2)+1/2*e*(a*f^2 -d^2)/d/f^2*ln((1/2*e*(a*f^2-d^2)/d/f^2+e^2/f^2*(x+1/2*(-a*f^2+d^2)/e/d))/ (e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/e/d)^2/f^2+e*(a*f^2-d^2)/d/f^2*(x +1/2*(-a*f^2+d^2)/e/d)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2)^(1/2))/(e^2/ f^2)^(1/2)-1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2/((a^2*f^4+2*a*d^2*f^2+d^4 )/d^2/f^2)^(1/2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2+e*(a*f^2-d^2)/d /f^2*(x+1/2*(-a*f^2+d^2)/e/d)+1/2*((a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2)^(1/2 )*(4*e^2*(x+1/2*(-a*f^2+d^2)/e/d)^2/f^2+4*e*(a*f^2-d^2)/d/f^2*(x+1/2*(-a*f ^2+d^2)/e/d)+(a^2*f^4+2*a*d^2*f^2+d^4)/d^2/f^2)^(1/2))/(x+1/2*(-a*f^2+d^2) /e/d)))+8*e^2/(a^2*f^4+2*a*d^2*f^2+d^4)*d^2*(1/4*(2*e^2/f^2*(x+1/2*(-a*f^2 +d^2)/e/d)+e*(a*f^2-d^2)/d/f^2)/e^2*f^2*(e^2*(x+1/2*(-a*f^2+d^2)/e/d)^2/f^ 2+e*(a*f^2-d^2)/d/f^2*(x+1/2*(-a*f^2+d^2)/e/d)+1/4*(a^2*f^4+2*a*d^2*f^2...
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (139) = 278\).
Time = 0.38 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\frac {a^{2} f^{4} - 2 \, d^{2} e^{2} x^{2} + a d^{2} f^{2} - 2 \, d^{3} e x + {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-a e f^{2} x + 2 \, d e^{2} x^{2} + a d f^{2} + {\left (a f^{3} - 2 \, d e f x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-a f^{2} + 2 \, d e x + d^{2}\right ) - {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) - 2 \, {\left (a d f^{3} - d^{2} e f x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}}{2 \, {\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}} \]
1/2*(a^2*f^4 - 2*d^2*e^2*x^2 + a*d^2*f^2 - 2*d^3*e*x + (a^2*f^4 - 2*a*d*e* f^2*x - a*d^2*f^2)*log(-a*e*f^2*x + 2*d*e^2*x^2 + a*d*f^2 + (a*f^3 - 2*d*e *f*x)*sqrt((e^2*x^2 + a*f^2)/f^2)) + (a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2) *log(-a*f^2 + 2*d*e*x + d^2) - (a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*log(- e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) - d) - 2*(a*d*f^3 - d^2*e*f*x)*sqrt((e ^2*x^2 + a*f^2)/f^2))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e)
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{2}}\, dx \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\int { \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{2}} \,d x } \]
Time = 0.46 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\frac {a f^{2} \log \left ({\left | -a f^{2} + 2 \, d e x + d^{2} \right |}\right )}{2 \, d^{3} e} + \frac {a f {\left | f \right |} \log \left ({\left | -x {\left | e \right |} + \sqrt {e^{2} x^{2} + a f^{2}} \right |}\right )}{2 \, d^{3} {\left | e \right |}} + \frac {x}{2 \, d^{2}} - \frac {\sqrt {e^{2} x^{2} + a f^{2}} {\left | f \right |}}{2 \, d^{2} e f} + \frac {a^{2} f^{4} + 2 \, a d^{2} f^{2} + d^{4}}{4 \, {\left (a f^{2} - 2 \, d e x - d^{2}\right )} d^{3} e} \]
1/2*a*f^2*log(abs(-a*f^2 + 2*d*e*x + d^2))/(d^3*e) + 1/2*a*f*abs(f)*log(ab s(-x*abs(e) + sqrt(e^2*x^2 + a*f^2)))/(d^3*abs(e)) + 1/2*x/d^2 - 1/2*sqrt( e^2*x^2 + a*f^2)*abs(f)/(d^2*e*f) + 1/4*(a^2*f^4 + 2*a*d^2*f^2 + d^4)/((a* f^2 - 2*d*e*x - d^2)*d^3*e)
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^2} \,d x \]