Integrand size = 27, antiderivative size = 560 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=-\frac {\tan (e+f x)}{2 a (c-d)^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-3 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} \sqrt {a} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c (c-d)^2 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 (3 c-d) d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c^2 (c-d)^3 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {d^3 \tan (e+f x)}{a c (c-d)^2 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-1/2*tan(f*x+e)/a/(c-d)^2/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-d^3*tan( f*x+e)/a/c/(c-d)^2/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*arcta nh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c^2/f/a^(1/2)/(a-a*sec(f*x+e ))^(1/2)/(a+a*sec(f*x+e))^(1/2)-d^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^( 1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c/(c-d)^2/(c+d)^(3/2)/f/a^(1/2)/(a-a* sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-1/4*arctanh(1/2*(a-a*sec(f*x+e))^ (1/2)*2^(1/2)/a^(1/2))*tan(f*x+e)/(c-d)^2/f*2^(1/2)/a^(1/2)/(a-a*sec(f*x+e ))^(1/2)/(a+a*sec(f*x+e))^(1/2)-(c-3*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2) *2^(1/2)/a^(1/2))*2^(1/2)*tan(f*x+e)/(c-d)^3/f/a^(1/2)/(a-a*sec(f*x+e))^(1 /2)/(a+a*sec(f*x+e))^(1/2)-2*(3*c-d)*d^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+ e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c^2/(c-d)^3/f/a^(1/2)/(c+d)^(1/2 )/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)
Time = 11.85 (sec) , antiderivative size = 470, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\frac {\left ((-c-d)^{3/2} \left (-c^2 (5 c-13 d) \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )+4 \sqrt {2} (c-d)^3 \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right )+2 \sqrt {2} d^{5/2} \left (-7 c^2-3 c d+2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-c-d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) (d+c \cos (e+f x))^2 \sqrt {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )} \sec ^{\frac {7}{2}}(e+f x) \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)}}{c^2 (-c-d)^{3/2} (c-d)^3 f \sec ^2\left (\frac {1}{2} (e+f x)\right )^{3/2} (a (1+\sec (e+f x)))^{3/2} (c+d \sec (e+f x))^2}+\frac {\cos ^3\left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x))^2 \sec ^4(e+f x) \left (-\frac {2 \left (c^3+c^2 d+2 d^3\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{c^2 (-c+d)^2 (c+d)}+\frac {4 d^4 \sin \left (\frac {1}{2} (e+f x)\right )}{c^2 (-c+d)^2 (c+d) (d+c \cos (e+f x))}+\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{(-c+d)^2}\right )}{f (a (1+\sec (e+f x)))^{3/2} (c+d \sec (e+f x))^2} \]
(((-c - d)^(3/2)*(-(c^2*(5*c - 13*d)*ArcSin[Tan[(e + f*x)/2]]) + 4*Sqrt[2] *(c - d)^3*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]]) + 2*Sqrt[2]*d^(5/2)*(-7*c^2 - 3*c*d + 2*d^2)*ArcTanh[(Sqrt[d]*Tan[(e + f* x)/2])/(Sqrt[-c - d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])*(d + c*Cos[e + f*x])^2*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2]*Sec[e + f*x]^(7/2)*Sqrt[C os[(e + f*x)/2]^2*Sec[e + f*x]])/(c^2*(-c - d)^(3/2)*(c - d)^3*f*(Sec[(e + f*x)/2]^2)^(3/2)*(a*(1 + Sec[e + f*x]))^(3/2)*(c + d*Sec[e + f*x])^2) + ( Cos[(e + f*x)/2]^3*(d + c*Cos[e + f*x])^2*Sec[e + f*x]^4*((-2*(c^3 + c^2*d + 2*d^3)*Sin[(e + f*x)/2])/(c^2*(-c + d)^2*(c + d)) + (4*d^4*Sin[(e + f*x )/2])/(c^2*(-c + d)^2*(c + d)*(d + c*Cos[e + f*x])) + (Sec[(e + f*x)/2]*Ta n[(e + f*x)/2])/(-c + d)^2))/(f*(a*(1 + Sec[e + f*x]))^(3/2)*(c + d*Sec[e + f*x])^2)
Time = 0.66 (sec) , antiderivative size = 391, normalized size of antiderivative = 0.70, 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.
\(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{3/2} (c+d \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4428 |
\(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a^2 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 198 |
\(\displaystyle -\frac {\tan (e+f x) \int \left (-\frac {(3 c-d) d^3}{c^2 (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {d^3}{c (c-d)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {\cos (e+f x)}{c^2 \sqrt {a-a \sec (e+f x)}}+\frac {3 d-c}{(c-d)^3 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d)^2 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\tan (e+f x) \left (\frac {2 d^{5/2} (3 c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 (c-d)^3 \sqrt {c+d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^2}+\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c (c-d)^2 (c+d)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} (c-d)^2}+\frac {\sqrt {2} (c-3 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^3}+\frac {d^3 \sqrt {a-a \sec (e+f x)}}{a c (c-d)^2 (c+d) (c+d \sec (e+f x))}+\frac {\sqrt {a-a \sec (e+f x)}}{2 a (c-d)^2 (\sec (e+f x)+1)}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^2) + (Sqrt[2 ]*(c - 3*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]* (c - d)^3) + ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2 ]*Sqrt[a]*(c - d)^2) + (d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]]) /(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c*(c - d)^2*(c + d)^(3/2)) + (2*(3*c - d )*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d]) ])/(Sqrt[a]*c^2*(c - d)^3*Sqrt[c + d]) + Sqrt[a - a*Sec[e + f*x]]/(2*a*(c - d)^2*(1 + Sec[e + f*x])) + (d^3*Sqrt[a - a*Sec[e + f*x]])/(a*c*(c - d)^2 *(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]]))
3.2.76.3.1 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. \(69594\) vs. \(2(480)=960\).
Time = 17.96 (sec) , antiderivative size = 69595, normalized size of antiderivative = 124.28
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\int { \frac {1}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sec \left (f x + e\right ) + c\right )}^{2}} \,d x } \]
Exception generated. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]