Integrand size = 27, antiderivative size = 756 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=-\frac {\tan (e+f x)}{4 a^2 (c-d)^2 f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)}}-\frac {(c-3 d) \tan (e+f x)}{2 a^2 (c-d)^3 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}-\frac {3 \tan (e+f x)}{16 a^2 (c-d)^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {(c-3 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} a^{3/2} (c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{16 \sqrt {2} a^{3/2} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} \left (c^2-4 c d+6 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} (c-d)^4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{a^{3/2} c (c-d)^3 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 (4 c-d) d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{a^{3/2} c^2 (c-d)^4 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {d^4 \tan (e+f x)}{a^2 c (c-d)^3 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-1/4*tan(f*x+e)/a^2/(c-d)^2/f/(1+sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)-1/2* (c-3*d)*tan(f*x+e)/a^2/(c-d)^3/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-3/1 6*tan(f*x+e)/a^2/(c-d)^2/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+d^4*tan(f *x+e)/a^2/c/(c-d)^3/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*arct anh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/a^(3/2)/c^2/f/(a-a*sec(f*x+ e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+d^(7/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^ (1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/a^(3/2)/c/(c-d)^3/(c+d)^(3/2)/f/(a-a *sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-1/4*(c-3*d)*arctanh(1/2*(a-a*sec (f*x+e))^(1/2)*2^(1/2)/a^(1/2))*tan(f*x+e)/a^(3/2)/(c-d)^3/f*2^(1/2)/(a-a* sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-3/32*arctanh(1/2*(a-a*sec(f*x+e)) ^(1/2)*2^(1/2)/a^(1/2))*tan(f*x+e)/a^(3/2)/(c-d)^2/f*2^(1/2)/(a-a*sec(f*x+ e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-(c^2-4*c*d+6*d^2)*arctanh(1/2*(a-a*sec(f* x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*tan(f*x+e)/a^(3/2)/(c-d)^4/f/(a-a*sec (f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+2*(4*c-d)*d^(7/2)*arctanh(d^(1/2)*(a -a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/a^(3/2)/c^2/(c-d)^4/f /(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)
Time = 13.65 (sec) , antiderivative size = 465, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x))^2 \sec ^{\frac {9}{2}}(e+f x) \left (-\frac {\left (c^2 \left (43 c^3-123 c^2 d+53 c d^2+219 d^3\right ) \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )-32 \sqrt {2} (c-d)^4 (c+d) \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )+\frac {16 \sqrt {2} d^{7/2} \left (9 c^2+5 c d-2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-c-d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )}{\sqrt {-c-d}}\right ) \sqrt {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)}}{c^2 (c-d)^4 (c+d) \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}+\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)} \left (\frac {2 \left (15 c^2-16 c d-31 d^2-\frac {16 d^4 \cos (e+f x)}{c (d+c \cos (e+f x))}\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{c+d}+32 (c-d) \csc ^4(e+f x) \sin ^5\left (\frac {1}{2} (e+f x)\right )+(-19 c+35 d) \sec \left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{(-c+d)^3}\right )}{4 f (a (1+\sec (e+f x)))^{5/2} (c+d \sec (e+f x))^2} \]
(Cos[(e + f*x)/2]^4*(d + c*Cos[e + f*x])^2*Sec[e + f*x]^(9/2)*(-(((c^2*(43 *c^3 - 123*c^2*d + 53*c*d^2 + 219*d^3)*ArcSin[Tan[(e + f*x)/2]] - 32*Sqrt[ 2]*(c - d)^4*(c + d)*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] + (16*Sqrt[2]*d^(7/2)*(9*c^2 + 5*c*d - 2*d^2)*ArcTanh[(Sqrt[d]*T an[(e + f*x)/2])/(Sqrt[-c - d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])/Sq rt[-c - d])*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2]*Sqrt[Cos[(e + f*x)/2]^2* Sec[e + f*x]])/(c^2*(c - d)^4*(c + d)*Sqrt[Sec[(e + f*x)/2]^2])) + (Cos[(e + f*x)/2]*Sqrt[Sec[e + f*x]]*((2*(15*c^2 - 16*c*d - 31*d^2 - (16*d^4*Cos[ e + f*x])/(c*(d + c*Cos[e + f*x])))*Sin[(e + f*x)/2])/(c + d) + 32*(c - d) *Csc[e + f*x]^4*Sin[(e + f*x)/2]^5 + (-19*c + 35*d)*Sec[(e + f*x)/2]*Tan[( e + f*x)/2]))/(-c + d)^3))/(4*f*(a*(1 + Sec[e + f*x]))^(5/2)*(c + d*Sec[e + f*x])^2)
Time = 0.76 (sec) , antiderivative size = 540, normalized size of antiderivative = 0.71, 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.
| \(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{5/2} (c+d \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a^3 (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 198 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\frac {(4 c-d) d^4}{c^2 (c-d)^4 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {d^4}{c (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {\cos (e+f x)}{c^2 \sqrt {a-a \sec (e+f x)}}+\frac {-c^2+4 d c-6 d^2}{(c-d)^4 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}+\frac {3 d-c}{(c-d)^3 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d)^2 (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (-\frac {2 d^{7/2} (4 c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 (c-d)^4 \sqrt {c+d}}+\frac {\sqrt {2} \left (c^2-4 c d+6 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^4}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^2}-\frac {d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c (c-d)^3 (c+d)^{3/2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} \sqrt {a} (c-d)^2}+\frac {(c-3 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} (c-d)^3}-\frac {d^4 \sqrt {a-a \sec (e+f x)}}{a c (c-d)^3 (c+d) (c+d \sec (e+f x))}+\frac {3 \sqrt {a-a \sec (e+f x)}}{16 a (c-d)^2 (\sec (e+f x)+1)}+\frac {(c-3 d) \sqrt {a-a \sec (e+f x)}}{2 a (c-d)^3 (\sec (e+f x)+1)}+\frac {\sqrt {a-a \sec (e+f x)}}{4 a (c-d)^2 (\sec (e+f x)+1)^2}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^2) + ((c - 3 *d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*Sqrt[a ]*(c - d)^3) + (3*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(16 *Sqrt[2]*Sqrt[a]*(c - d)^2) + (Sqrt[2]*(c^2 - 4*c*d + 6*d^2)*ArcTanh[Sqrt[ a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]*(c - d)^4) - (d^(7/2)*Arc Tanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c *(c - d)^3*(c + d)^(3/2)) - (2*(4*c - d)*d^(7/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^2*(c - d)^4*Sqrt[c + d]) + Sqrt[a - a*Sec[e + f*x]]/(4*a*(c - d)^2*(1 + Sec[e + f*x])^2) + ((c - 3*d)*Sqrt[a - a*Sec[e + f*x]])/(2*a*(c - d)^3*(1 + Sec[e + f*x])) + (3*S qrt[a - a*Sec[e + f*x]])/(16*a*(c - d)^2*(1 + Sec[e + f*x])) - (d^4*Sqrt[a - a*Sec[e + f*x]])/(a*c*(c - d)^3*(c + d)*(c + d*Sec[e + f*x])))*Tan[e + f*x])/(a*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.82.3.1 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. \(85543\) vs. \(2(653)=1306\).
Time = 20.18 (sec) , antiderivative size = 85544, normalized size of antiderivative = 113.15
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2} \, dx=\text {Hanged} \]