Integrand size = 27, antiderivative size = 536 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\frac {2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {3 a^{7/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 c d^{3/2} (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^{7/2} (c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 d^{3/2} \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 a^{7/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^3 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^3 (c-d)^2 \tan (e+f x)}{2 c d (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {a^3 (c-d) \tan (e+f x)}{c^2 d f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {3 a^3 (c-d)^2 \tan (e+f x)}{4 c d (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-1/2*a^3*(c-d)^2*tan(f*x+e)/c/d/(c+d)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e) )^(1/2)+a^3*(c-d)*tan(f*x+e)/c^2/d/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/ 2)-3/4*a^3*(c-d)^2*tan(f*x+e)/c/d/(c+d)^2/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^3 /f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-3/4*a^(7/2)*(c-d)^2*arcta nh(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)^(5/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+a^(7/2)*(c-d )*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c ^2/d^(3/2)/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-2*a ^(7/2)*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^3/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2 )
Time = 9.97 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\frac {\left (8 d^{3/2} (c+d)^2 \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )-\frac {\left (c^4+10 c^3 d-15 c^2 d^2-20 c d^3-8 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-c-d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )}{\sqrt {-c-d}}\right ) (d+c \cos (e+f x))^3 \sec ^6\left (\frac {1}{2} (e+f x)\right ) \sqrt {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\sec (e+f x)} \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)} (a (1+\sec (e+f x)))^{5/2}}{16 \sqrt {2} c^3 d^{3/2} (c+d)^2 f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} (c+d \sec (e+f x))^3}+\frac {(d+c \cos (e+f x))^3 \sec ^5\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) (a (1+\sec (e+f x)))^{5/2} \left (-\frac {\left (c^3-12 c^2 d+5 c d^2+6 d^3\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{16 c^3 d (c+d)^2}+\frac {-c^2 d \sin \left (\frac {1}{2} (e+f x)\right )+2 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )-d^3 \sin \left (\frac {1}{2} (e+f x)\right )}{8 c^3 (c+d) (d+c \cos (e+f x))^2}+\frac {3 c^3 \sin \left (\frac {1}{2} (e+f x)\right )-14 c^2 d \sin \left (\frac {1}{2} (e+f x)\right )+3 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+8 d^3 \sin \left (\frac {1}{2} (e+f x)\right )}{16 c^3 (c+d)^2 (d+c \cos (e+f x))}\right )}{f (c+d \sec (e+f x))^3} \]
((8*d^(3/2)*(c + d)^2*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] - ((c^4 + 10*c^3*d - 15*c^2*d^2 - 20*c*d^3 - 8*d^4)*ArcTanh[(Sq rt[d]*Tan[(e + f*x)/2])/(Sqrt[-c - d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x]) ])])/Sqrt[-c - d])*(d + c*Cos[e + f*x])^3*Sec[(e + f*x)/2]^6*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]]*(a*(1 + Sec[e + f*x]))^(5/2))/(16*Sqrt[2]*c^3*d^(3/2)*(c + d)^2*f* Sqrt[Sec[(e + f*x)/2]^2]*(c + d*Sec[e + f*x])^3) + ((d + c*Cos[e + f*x])^3 *Sec[(e + f*x)/2]^5*Sec[e + f*x]*(a*(1 + Sec[e + f*x]))^(5/2)*(-1/16*((c^3 - 12*c^2*d + 5*c*d^2 + 6*d^3)*Sin[(e + f*x)/2])/(c^3*d*(c + d)^2) + (-(c^ 2*d*Sin[(e + f*x)/2]) + 2*c*d^2*Sin[(e + f*x)/2] - d^3*Sin[(e + f*x)/2])/( 8*c^3*(c + d)*(d + c*Cos[e + f*x])^2) + (3*c^3*Sin[(e + f*x)/2] - 14*c^2*d *Sin[(e + f*x)/2] + 3*c*d^2*Sin[(e + f*x)/2] + 8*d^3*Sin[(e + f*x)/2])/(16 *c^3*(c + d)^2*(d + c*Cos[e + f*x]))))/(f*(c + d*Sec[e + f*x])^3)
Time = 0.66 (sec) , antiderivative size = 405, normalized size of antiderivative = 0.76, 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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{5/2}}{(c+d \sec (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {a^2 \cos (e+f x) (\sec (e+f x)+1)^2}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \int \frac {\cos (e+f x) (\sec (e+f x)+1)^2}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 198 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \int \left (-\frac {(c-d)^2}{c d \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}+\frac {\cos (e+f x)}{c^3 \sqrt {a-a \sec (e+f x)}}-\frac {d}{c^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {c^2-d^2}{c^2 d \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \left (\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^3 \sqrt {c+d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^3}-\frac {(c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 d^{3/2} \sqrt {c+d}}+\frac {3 (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{4 \sqrt {a} c d^{3/2} (c+d)^{5/2}}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{a c^2 d (c+d \sec (e+f x))}+\frac {3 (c-d)^2 \sqrt {a-a \sec (e+f x)}}{4 a c d (c+d)^2 (c+d \sec (e+f x))}+\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{2 a c d (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^4*((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^3) + (3* (c - d)^2*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d]) ])/(4*Sqrt[a]*c*d^(3/2)*(c + d)^(5/2)) - ((c - d)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^2*d^(3/2)*Sqrt[c + d ]) + (2*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^3*Sqrt[c + d]) + ((c - d)^2*Sqrt[a - a*Sec[e + f*x]])/ (2*a*c*d*(c + d)*(c + d*Sec[e + f*x])^2) - ((c - d)*Sqrt[a - a*Sec[e + f*x ]])/(a*c^2*d*(c + d*Sec[e + f*x])) + (3*(c - d)^2*Sqrt[a - a*Sec[e + f*x]] )/(4*a*c*d*(c + d)^2*(c + d*Sec[e + f*x])))*Tan[e + f*x])/(f*Sqrt[a - a*Se c[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.65.3.1 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. \(42562\) vs. \(2(460)=920\).
Time = 235.82 (sec) , antiderivative size = 42563, normalized size of antiderivative = 79.41
Time = 22.58 (sec) , antiderivative size = 3351, normalized size of antiderivative = 6.25 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\text {Too large to display} \]
[-1/8*((a^2*c^4*d^2 + 10*a^2*c^3*d^3 - 15*a^2*c^2*d^4 - 20*a^2*c*d^5 - 8*a ^2*d^6 + (a^2*c^6 + 10*a^2*c^5*d - 15*a^2*c^4*d^2 - 20*a^2*c^3*d^3 - 8*a^2 *c^2*d^4)*cos(f*x + e)^3 + (a^2*c^6 + 12*a^2*c^5*d + 5*a^2*c^4*d^2 - 50*a^ 2*c^3*d^3 - 48*a^2*c^2*d^4 - 16*a^2*c*d^5)*cos(f*x + e)^2 + (2*a^2*c^5*d + 21*a^2*c^4*d^2 - 20*a^2*c^3*d^3 - 55*a^2*c^2*d^4 - 36*a^2*c*d^5 - 8*a^2*d ^6)*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)) - 8*(a^2*c^2*d^3 + 2*a^2*c*d^4 + a^2*d^ 5 + (a^2*c^4*d + 2*a^2*c^3*d^2 + a^2*c^2*d^3)*cos(f*x + e)^3 + (a^2*c^4*d + 4*a^2*c^3*d^2 + 5*a^2*c^2*d^3 + 2*a^2*c*d^4)*cos(f*x + e)^2 + (2*a^2*c^3 *d^2 + 5*a^2*c^2*d^3 + 4*a^2*c*d^4 + a^2*d^5)*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))*co s(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*((a^ 2*c^5 - 12*a^2*c^4*d + 5*a^2*c^3*d^2 + 6*a^2*c^2*d^3)*cos(f*x + e)^2 - (a^ 2*c^4*d + 10*a^2*c^3*d^2 - 7*a^2*c^2*d^3 - 4*a^2*c*d^4)*cos(f*x + e))*sqrt ((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/((c^7*d + 2*c^6*d^2 + c^ 5*d^3)*f*cos(f*x + e)^3 + (c^7*d + 4*c^6*d^2 + 5*c^5*d^3 + 2*c^4*d^4)*f*co s(f*x + e)^2 + (2*c^6*d^2 + 5*c^5*d^3 + 4*c^4*d^4 + c^3*d^5)*f*cos(f*x + e ) + (c^5*d^3 + 2*c^4*d^4 + c^3*d^5)*f), -1/8*(16*(a^2*c^2*d^3 + 2*a^2*c...
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}{\left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]