Integrand size = 27, antiderivative size = 258 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\frac {2 a^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} c^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 a d (2 c+5 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt {a+a \sec (e+f x)}}-\frac {2 \left (c^2+8 c d+8 d^2\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}} \]
2*a^3*(c+2*d)*(3*c+2*d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2/5*a*d*(2*c+5 *d)*(a-a*sec(f*x+e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2/7*d^2*(a-a*se c(f*x+e))^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2/3*(c^2+8*c*d+8*d^2)*(a^3 -a^3*sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*c^2*arctanh ((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/f/(a-a*sec(f*x+e))^(1/2)/(a+a* sec(f*x+e))^(1/2)
Time = 2.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.74 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\frac {a^2 \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt {a (1+\sec (e+f x))} \left (420 \sqrt {2} c^2 \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {7}{2}}(e+f x)+4 \left (35 c^2+196 c d+145 d^2+\left (420 c^2+987 c d+465 d^2\right ) \cos (e+f x)+\left (35 c^2+196 c d+115 d^2\right ) \cos (2 (e+f x))+140 c^2 \cos (3 (e+f x))+301 c d \cos (3 (e+f x))+115 d^2 \cos (3 (e+f x))\right ) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{420 f} \]
(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*(420*Sqrt[ 2]*c^2*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]*Cos[e + f*x]^(7/2) + 4*(35*c^2 + 1 96*c*d + 145*d^2 + (420*c^2 + 987*c*d + 465*d^2)*Cos[e + f*x] + (35*c^2 + 196*c*d + 115*d^2)*Cos[2*(e + f*x)] + 140*c^2*Cos[3*(e + f*x)] + 301*c*d*C os[3*(e + f*x)] + 115*d^2*Cos[3*(e + f*x)])*Sin[(e + f*x)/2]))/(420*f)
Time = 0.42 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.77, 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.
| \(\displaystyle \int (a \sec (e+f x)+a)^{5/2} (c+d \sec (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {a^2 \cos (e+f x) (\sec (e+f x)+1)^2 (c+d \sec (e+f x))^2}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \int \frac {\cos (e+f x) (\sec (e+f x)+1)^2 (c+d \sec (e+f x))^2}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 198 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \int \left (-\frac {d^2 (a-a \sec (e+f x))^{5/2}}{a^3}+\frac {d (2 c+5 d) (a-a \sec (e+f x))^{3/2}}{a^2}+\frac {\left (-c^2-8 d c-8 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a}+\frac {3 c^2+8 d c+4 d^2}{\sqrt {a-a \sec (e+f x)}}+\frac {c^2 \cos (e+f x)}{\sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^4 \tan (e+f x) \left (\frac {2 d^2 (a-a \sec (e+f x))^{7/2}}{7 a^4}-\frac {2 d (2 c+5 d) (a-a \sec (e+f x))^{5/2}}{5 a^3}+\frac {2 \left (c^2+8 c d+8 d^2\right ) (a-a \sec (e+f x))^{3/2}}{3 a^2}-\frac {2 c^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 (c+2 d) (3 c+2 d) \sqrt {a-a \sec (e+f x)}}{a}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^4*((-2*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/Sqrt[a] - (2*(c + 2*d)*(3*c + 2*d)*Sqrt[a - a*Sec[e + f*x]])/a + (2*(c^2 + 8*c*d + 8*d^2) *(a - a*Sec[e + f*x])^(3/2))/(3*a^2) - (2*d*(2*c + 5*d)*(a - a*Sec[e + f*x ])^(5/2))/(5*a^3) + (2*d^2*(a - a*Sec[e + f*x])^(7/2))/(7*a^4))*Tan[e + f* x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.61.3.1 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]
Time = 21.68 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.09
| method | result | size |
| default | \(\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (105 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c^{2} \cos \left (f x +e \right )+105 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c^{2}+280 \sin \left (f x +e \right ) c^{2}+602 \sin \left (f x +e \right ) c d +230 \sin \left (f x +e \right ) d^{2}+35 c^{2} \tan \left (f x +e \right )+196 c d \tan \left (f x +e \right )+115 d^{2} \tan \left (f x +e \right )+42 c d \tan \left (f x +e \right ) \sec \left (f x +e \right )+60 d^{2} \tan \left (f x +e \right ) \sec \left (f x +e \right )+15 d^{2} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )}{105 f \left (\cos \left (f x +e \right )+1\right )}\) | \(280\) |
| parts | \(\frac {2 c^{2} a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+8 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}+\frac {2 d^{2} a^{2} \left (46 \cos \left (f x +e \right )^{3}+23 \cos \left (f x +e \right )^{2}+12 \cos \left (f x +e \right )+3\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{21 f \left (\cos \left (f x +e \right )+1\right )}+\frac {4 c d \,a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (43 \sin \left (f x +e \right )+14 \tan \left (f x +e \right )+3 \sec \left (f x +e \right ) \tan \left (f x +e \right )\right )}{15 f \left (\cos \left (f x +e \right )+1\right )}\) | \(316\) |
2/105*a^2/f*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1)*(105*(-cos(f*x+e)/(cos (f*x+e)+1))^(1/2)*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+ e)+1))^(1/2))*c^2*cos(f*x+e)+105*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctan h(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*c^2+280*si n(f*x+e)*c^2+602*sin(f*x+e)*c*d+230*sin(f*x+e)*d^2+35*c^2*tan(f*x+e)+196*c *d*tan(f*x+e)+115*d^2*tan(f*x+e)+42*c*d*tan(f*x+e)*sec(f*x+e)+60*d^2*tan(f *x+e)*sec(f*x+e)+15*d^2*tan(f*x+e)*sec(f*x+e)^2)
Time = 0.29 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.94 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\left [\frac {105 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (15 \, a^{2} d^{2} + 2 \, {\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (15 \, a^{2} d^{2} + 2 \, {\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}\right ] \]
[1/105*(105*(a^2*c^2*cos(f*x + e)^4 + a^2*c^2*cos(f*x + e)^3)*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)) + 2*(1 5*a^2*d^2 + 2*(140*a^2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + ( 35*a^2*c^2 + 196*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^2 + 6*(7*a^2*c*d + 10 *a^2*d^2)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3), -2/105*(105*(a^2*c^2*cos(f*x + e)^4 + a^2*c^2*cos(f*x + e)^3)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/co s(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (15*a^2*d^2 + 2*(140*a^ 2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + (35*a^2*c^2 + 196*a^2* c*d + 115*a^2*d^2)*cos(f*x + e)^2 + 6*(7*a^2*c*d + 10*a^2*d^2)*cos(f*x + e ))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3)]
\[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{2}\, dx \]
\[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sec \left (f x + e\right ) + c\right )}^{2} \,d x } \]
-1/210*(105*((a^2*c^2*cos(2*f*x + 2*e)^2 + a^2*c^2*sin(2*f*x + 2*e)^2 + 2* a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f* x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e ), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*co s(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2* e) + 1)) + 1) - (a^2*c^2*cos(2*f*x + 2*e)^2 + a^2*c^2*sin(2*f*x + 2*e)^2 + 2*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2 *f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2 *cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 1) - 2*(a^2*c^2*f*cos(2*f*x + 2*e)^2 + a^2*c^2*f*sin(2*f*x + 2*e)^2 + 2*a^2*c^2*f*cos(2*f*x + 2*e) + a^2*c^2*f)*integrate((((cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f *x + 2*e) + sin(2*f*x + 2*e)^2)*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f* x + 2*e))) + (cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4 *f*x + 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2 *f*x + 2*e))*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(5/2 *arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - ((cos(2*f*x + 2*e)*sin (6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6*e)*...
\[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sec \left (f x + e\right ) + c\right )}^{2} \,d x } \]
Timed out. \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]