Integrand size = 27, antiderivative size = 592 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=-\frac {\tan (e+f x)}{4 a^2 (c-d) f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)}}-\frac {(c-2 d) \tan (e+f x)}{2 a^2 (c-d)^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}-\frac {3 \tan (e+f x)}{16 a^2 (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {(c-2 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} a^{3/2} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{16 \sqrt {2} a^{3/2} (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} \left (c^2-3 c d+3 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} (c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{a^{3/2} c (c-d)^3 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]
-1/4*tan(f*x+e)/a^2/(c-d)/f/(1+sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)-1/2*(c -2*d)*tan(f*x+e)/a^2/(c-d)^2/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-3/16* tan(f*x+e)/a^2/(c-d)/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*arctanh((a- a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/a^(3/2)/c/f/(a-a*sec(f*x+e))^(1/2) /(a+a*sec(f*x+e))^(1/2)-1/4*(c-2*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)^2/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)/f*2^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f *x+e))^(1/2)-(c^2-3*c*d+3*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)^3/f/(a-a*sec(f*x+e))^(1/2)/(a+a* sec(f*x+e))^(1/2)+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/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^ (1/2)/(a+a*sec(f*x+e))^(1/2)
Time = 11.41 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\frac {\cos ^4\left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x)) \sec ^{\frac {7}{2}}(e+f x) \left (-\frac {4 \left (\sqrt {-c-d} \left (c \left (43 c^2-126 c d+115 d^2\right ) \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )-32 \sqrt {2} (c-d)^3 \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right )+32 \sqrt {2} d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-c-d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {1+\sec (e+f x)}}{c \sqrt {-c-d} \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}+(c-d) (11 c-19 d+(15 c-23 d) \cos (e+f x)) \sec ^3\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {3}{2} (e+f x)\right )\right )\right )}{16 (c-d)^3 f (a (1+\sec (e+f x)))^{5/2} (c+d \sec (e+f x))} \]
(Cos[(e + f*x)/2]^4*(d + c*Cos[e + f*x])*Sec[e + f*x]^(7/2)*((-4*(Sqrt[-c - d]*(c*(43*c^2 - 126*c*d + 115*d^2)*ArcSin[Tan[(e + f*x)/2]] - 32*Sqrt[2] *(c - d)^3*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]]) + 32*Sqrt[2]*d^(7/2)*ArcTanh[(Sqrt[d]*Tan[(e + f*x)/2])/(Sqrt[-c - d]*Sqr t[Cos[e + f*x]/(1 + Cos[e + f*x])])])*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x]) ]*Sqrt[1 + Sec[e + f*x]])/(c*Sqrt[-c - d]*Sqrt[Sec[(e + f*x)/2]^2]) + (c - d)*(11*c - 19*d + (15*c - 23*d)*Cos[e + f*x])*Sec[(e + f*x)/2]^3*Sqrt[Sec [e + f*x]]*(Sin[(e + f*x)/2] - Sin[(3*(e + f*x))/2])))/(16*(c - d)^3*f*(a* (1 + Sec[e + f*x]))^(5/2)*(c + d*Sec[e + f*x]))
Time = 0.63 (sec) , antiderivative size = 420, 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.
| \(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{5/2} (c+d \sec (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a^3 (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 198 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\frac {d^4}{c (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {\cos (e+f x)}{c \sqrt {a-a \sec (e+f x)}}+\frac {-c^2+3 d c-3 d^2}{(c-d)^3 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}+\frac {2 d-c}{(c-d)^2 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d) (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {\sqrt {2} \left (c^2-3 c d+3 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^3}-\frac {2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c (c-d)^3 \sqrt {c+d}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} \sqrt {a} (c-d)}+\frac {(c-2 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} (c-d)^2}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c}+\frac {3 \sqrt {a-a \sec (e+f x)}}{16 a (c-d) (\sec (e+f x)+1)}+\frac {(c-2 d) \sqrt {a-a \sec (e+f x)}}{2 a (c-d)^2 (\sec (e+f x)+1)}+\frac {\sqrt {a-a \sec (e+f x)}}{4 a (c-d) (\sec (e+f x)+1)^2}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c) + ((c - 2*d )*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*Sqrt[a]* (c - d)^2) + (3*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(16*S qrt[2]*Sqrt[a]*(c - d)) + (Sqrt[2]*(c^2 - 3*c*d + 3*d^2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]*(c - d)^3) - (2*d^(7/2)*ArcTa nh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c*( c - d)^3*Sqrt[c + d]) + Sqrt[a - a*Sec[e + f*x]]/(4*a*(c - d)*(1 + Sec[e + f*x])^2) + ((c - 2*d)*Sqrt[a - a*Sec[e + f*x]])/(2*a*(c - d)^2*(1 + Sec[e + f*x])) + (3*Sqrt[a - a*Sec[e + f*x]])/(16*a*(c - d)*(1 + Sec[e + f*x])) )*Tan[e + f*x])/(a*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.81.3.1 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. \(2344\) vs. \(2(509)=1018\).
Time = 16.76 (sec) , antiderivative size = 2345, normalized size of antiderivative = 3.96
1/192/f/(d/(c-d))^(1/2)/(c-d)^3/c/((c+d)*(c-d))^(1/2)/a^3*(576*((c+d)*(c-d ))^(1/2)*2^(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*( -cot(f*x+e)+csc(f*x+e)))*(d/(c-d))^(1/2)*c*d^2+12*((c+d)*(c-d))^(1/2)*((1- cos(f*x+e))^2*csc(f*x+e)^2-1)^(5/2)*(d/(c-d))^(1/2)*c^2*d*(-cot(f*x+e)+csc (f*x+e))-576*((c+d)*(c-d))^(1/2)*2^(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2 *csc(f*x+e)^2-1)^(1/2)*(-cot(f*x+e)+csc(f*x+e)))*(d/(c-d))^(1/2)*c^2*d-96* 2^(1/2)*ln(-2*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^ (1/2)*c+2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+ ((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))+c-d)/(c*(-cot(f*x+e)+csc(f*x+ e))-(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*d^4+96*2^(1/2)*ln(-2* (((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c-2^(1/2) *(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^( 1/2)*(-cot(f*x+e)+csc(f*x+e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot(f*x+ e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*d^4-12*((c+d)*(c-d))^(1/2)*((1-cos( f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(d/(c-d))^(1/2)*c^2*d*(1-cos(f*x+e))^5*csc (f*x+e)^5+5*((c+d)*(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(3/2)*(d /(c-d))^(1/2)*c^3*(-cot(f*x+e)+csc(f*x+e))-17*((c+d)*(c-d))^(1/2)*((1-cos( f*x+e))^2*csc(f*x+e)^2-1)^(3/2)*(d/(c-d))^(1/2)*d^3*(-cot(f*x+e)+csc(f*x+e ))+87*((c+d)*(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(d/(c-d) )^(1/2)*c^3*(-cot(f*x+e)+csc(f*x+e))+192*((c+d)*(c-d))^(1/2)*2^(1/2)*ar...
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\int { \frac {1}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \]
Exception generated. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \]