Integrand size = 27, antiderivative size = 219 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{3/2} \sqrt {d} (3 c+2 d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-a*d*tan(f*x+e)/c/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*a^(3/2 )*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c^2/f/(a-a*sec(f*x+e) )^(1/2)/(a+a*sec(f*x+e))^(1/2)-a^(3/2)*(3*c+2*d)*arctanh(d^(1/2)*(a-a*sec( f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*d^(1/2)*tan(f*x+e)/c^2/(c+d)^(3/2)/f/(a -a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)
Time = 5.18 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\frac {(d+c \cos (e+f x))^2 \sec ^{\frac {3}{2}}(e+f x) \sqrt {a (1+\sec (e+f x))} \left (\frac {2 \left (2 (c+d)^{3/2} \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )-\sqrt {d} (3 c+2 d) \arctan \left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) \sqrt {\frac {\cos (e+f x)}{(1+\cos (e+f x))^2}} \sqrt {1+\sec (e+f x)}}{(c+d)^{3/2}}-\frac {2 c d \tan \left (\frac {1}{2} (e+f x)\right )}{(c+d) (d+c \cos (e+f x)) \sqrt {\sec (e+f x)}}\right )}{2 c^2 f (c+d \sec (e+f x))^2} \]
((d + c*Cos[e + f*x])^2*Sec[e + f*x]^(3/2)*Sqrt[a*(1 + Sec[e + f*x])]*((2* (2*(c + d)^(3/2)*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f* x])]] - Sqrt[d]*(3*c + 2*d)*ArcTan[(Sqrt[d]*Tan[(e + f*x)/2])/(Sqrt[c + d] *Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])*Sqrt[Cos[e + f*x]/(1 + Cos[e + f *x])^2]*Sqrt[1 + Sec[e + f*x]])/(c + d)^(3/2) - (2*c*d*Tan[(e + f*x)/2])/( (c + d)*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]])))/(2*c^2*f*(c + d*Sec[e + f*x])^2)
Time = 0.39 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 4428, 114, 27, 174, 73, 219, 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.
| \(\displaystyle \int \frac {\sqrt {a \sec (e+f x)+a}}{(c+d \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a \cos (e+f x) (2 (c+d)-d \sec (e+f x))}{2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{a c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {\cos (e+f x) (2 (c+d)-d \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{2 c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {2 (c+d) \int \frac {\cos (e+f x)}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{c}-\frac {d (3 c+2 d) \int \frac {1}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{c}}{2 c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {2 d (3 c+2 d) \int \frac {1}{c+d-\frac {d (a-a \sec (e+f x))}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}-\frac {4 (c+d) \int \frac {1}{1-\frac {a-a \sec (e+f x)}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}}{2 c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {2 d (3 c+2 d) \int \frac {1}{c+d-\frac {d (a-a \sec (e+f x))}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}-\frac {4 (c+d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c}}{2 c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {2 \sqrt {d} (3 c+2 d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c \sqrt {c+d}}-\frac {4 (c+d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c}}{2 c (c+d)}+\frac {d \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*(((-4*(c + d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c ) + (2*Sqrt[d]*(3*c + 2*d)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqr t[a]*Sqrt[c + d])])/(Sqrt[a]*c*Sqrt[c + d]))/(2*c*(c + d)) + (d*Sqrt[a - a *Sec[e + f*x]])/(a*c*(c + d)*(c + d*Sec[e + f*x])))*Tan[e + f*x])/(f*Sqrt[ a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.52.3.1 Defintions of rubi rules used
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0 ] && IntegerQ[m - 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(38051\) vs. \(2(189)=378\).
Time = 15.52 (sec) , antiderivative size = 38052, normalized size of antiderivative = 173.75
Time = 1.57 (sec) , antiderivative size = 1413, normalized size of antiderivative = 6.45 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\text {Too large to display} \]
[-1/2*(2*c*d*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - ((3*c^2 + 2*c*d)*cos(f*x + e)^2 + 3*c*d + 2*d^2 + (3*c^2 + 5*c*d + 2*d^2)*cos(f*x + e))*sqrt(-a*d/(c + d))*log((2*(c + d)*sqrt(-a*d/(c + d))* sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) - 2*((c^2 + c*d)*cos(f*x + e)^2 + c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*s qrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/((c^4 + c^3*d)*f*cos(f*x + e)^2 + (c^4 + 2*c^3*d + c^2*d^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f), -1/2*( 2*c*d*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 4*((c^2 + c*d)*cos(f*x + e)^2 + c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(s qrt(a)*sin(f*x + e))) - ((3*c^2 + 2*c*d)*cos(f*x + e)^2 + 3*c*d + 2*d^2 + (3*c^2 + 5*c*d + 2*d^2)*cos(f*x + e))*sqrt(-a*d/(c + d))*log((2*(c + d)*sq rt(-a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin( f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/ (c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)))/((c^4 + c^3*d)*f*cos(f*x + e)^2 + (c^4 + 2*c^3*d + c^2*d^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f), - (c*d*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) ...
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}{\left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]