Integrand size = 27, antiderivative size = 329 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\frac {2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{7/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c d^{3/2} (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} (c-d) \sqrt {c+d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 d^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^3 (c-d)^2 \tan (e+f x)}{c d (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-a^3*(c-d)^2*tan(f*x+e)/c/d/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2 )+2*a^(7/2)*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^(7/2)*(c-d)^2*arctanh(d^(1/2)*( a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c/d^(3/2)/(c+d)^(3/2 )/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*(c-d)*arctanh( d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*(c+d)^(1/2)*tan(f*x+e) /c^2/d^(3/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)
Time = 3.26 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.85 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\frac {\sqrt {\cos (e+f x)} (d+c \cos (e+f x))^2 \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (1+\sec (e+f x)))^{5/2} \left (2 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {4 \sqrt {2} (c-d) \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\cos (e+f x)}}\right )}{\sqrt {d} \sqrt {c+d}}-\frac {(c-d)^2 \left (2 c \cos (e+f x)-\frac {2 (c+2 d) \text {arctanh}\left (\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right ) (d+c \cos (e+f x))}{(c+d) \sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}}\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{d (c+d) \sqrt {\cos (e+f x)} (d+c \cos (e+f x))}\right )}{8 c^2 f (c+d \sec (e+f x))^2} \]
(Sqrt[Cos[e + f*x]]*(d + c*Cos[e + f*x])^2*Sec[(e + f*x)/2]^5*(a*(1 + Sec[ e + f*x]))^(5/2)*(2*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]] + (4*Sqrt[2]* (c - d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]])])/(Sqrt[d]*Sqrt[c + d]) - ((c - d)^2*(2*c*Cos[e + f*x] - (2*(c + 2*d)*ArcTanh[Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]]*(d + c*Cos[e + f*x] ))/((c + d)*Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]))*Sin[(e + f*x)/2])/( d*(c + d)*Sqrt[Cos[e + f*x]]*(d + c*Cos[e + f*x]))))/(8*c^2*f*(c + d*Sec[e + f*x])^2)
Time = 0.53 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 4428, 27, 198, 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.
\(\displaystyle \int \frac {(a \sec (e+f x)+a)^{5/2}}{(c+d \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4428 |
\(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {a^2 \cos (e+f x) (\sec (e+f x)+1)^2}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^4 \tan (e+f x) \int \frac {\cos (e+f x) (\sec (e+f x)+1)^2}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 198 |
\(\displaystyle -\frac {a^4 \tan (e+f x) \int \left (-\frac {(c-d)^2}{c d \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {\cos (e+f x)}{c^2 \sqrt {a-a \sec (e+f x)}}+\frac {c^2-d^2}{c^2 d \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 \tan (e+f x) \left (-\frac {2 \sqrt {c+d} (c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 d^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^2}+\frac {(c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c d^{3/2} (c+d)^{3/2}}+\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{a c d (c+d) (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^4*((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^2) + ((c - d)^2*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]) /(Sqrt[a]*c*d^(3/2)*(c + d)^(3/2)) - (2*(c - d)*Sqrt[c + d]*ArcTanh[(Sqrt[ d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^2*d^(3/2)) + ((c - d)^2*Sqrt[a - a*Sec[e + f*x]])/(a*c*d*(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]]))
3.2.64.3.1 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_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_))^(q_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && IntegersQ[p, q]
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. \(20137\) vs. \(2(281)=562\).
Time = 53.69 (sec) , antiderivative size = 20138, normalized size of antiderivative = 61.21
Time = 13.17 (sec) , antiderivative size = 2031, normalized size of antiderivative = 6.17 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\text {Too large to display} \]
[-1/2*(2*(a^2*c^3 - 2*a^2*c^2*d + a^2*c*d^2)*sqrt((a*cos(f*x + e) + a)/cos (f*x + e))*cos(f*x + e)*sin(f*x + e) + (a^2*c^3*d + 4*a^2*c^2*d^2 - 3*a^2* c*d^3 - 2*a^2*d^4 + (a^2*c^4 + 4*a^2*c^3*d - 3*a^2*c^2*d^2 - 2*a^2*c*d^3)* cos(f*x + e)^2 + (a^2*c^4 + 5*a^2*c^3*d + a^2*c^2*d^2 - 5*a^2*c*d^3 - 2*a^ 2*d^4)*cos(f*x + e))*sqrt(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*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*(a^2*c*d^2 + a^2*d^3 + (a^2*c^2* d + a^2*c*d^2)*cos(f*x + e)^2 + (a^2*c^2*d + 2*a^2*c*d^2 + a^2*d^3)*cos(f* x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-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*d + c^3*d^2)*f*cos(f*x + e)^2 + (c^4*d + 2*c^3*d^2 + c^2*d^3)*f*cos(f*x + e) + (c^3*d^2 + c^2*d^3)*f), -1/2*(2*(a^2*c^3 - 2*a^2 *c^2*d + a^2*c*d^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*s in(f*x + e) + 4*(a^2*c*d^2 + a^2*d^3 + (a^2*c^2*d + a^2*c*d^2)*cos(f*x + e )^2 + (a^2*c^2*d + 2*a^2*c*d^2 + a^2*d^3)*cos(f*x + e))*sqrt(a)*arctan(sqr t((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (a^2*c^3*d + 4*a^2*c^2*d^2 - 3*a^2*c*d^3 - 2*a^2*d^4 + (a^2*c^4 + 4*a^2* c^3*d - 3*a^2*c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e)^2 + (a^2*c^4 + 5*a^2*c^3 *d + a^2*c^2*d^2 - 5*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e))*sqrt(-a/(c*d ...
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}{\left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^2} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]