Integrand size = 32, antiderivative size = 122 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=-\frac {64 c^3 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{35 f}-\frac {2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f} \]
-2/7*c*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f-64/105*c^3*(a+ a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-16/35*c^2*(a+a*sec(f*x+e ))*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f
Time = 0.68 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.49 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=-\frac {2 a c^3 (1+\sec (e+f x)) \left (71-54 \sec (e+f x)+15 \sec ^2(e+f x)\right ) \tan (e+f x)}{105 f \sqrt {c-c \sec (e+f x)}} \]
(-2*a*c^3*(1 + Sec[e + f*x])*(71 - 54*Sec[e + f*x] + 15*Sec[e + f*x]^2)*Ta n[e + f*x])/(105*f*Sqrt[c - c*Sec[e + f*x]])
Time = 0.60 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 4443, 3042, 4443, 3042, 4441}
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.
| \(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4443 |
| \(\displaystyle \frac {8}{7} c \int \sec (e+f x) (\sec (e+f x) a+a) (c-c \sec (e+f x))^{3/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{7} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\) |
| \(\Big \downarrow \) 4443 |
| \(\displaystyle \frac {8}{7} c \left (\frac {4}{5} c \int \sec (e+f x) (\sec (e+f x) a+a) \sqrt {c-c \sec (e+f x)}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{7} c \left (\frac {4}{5} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\) |
| \(\Big \downarrow \) 4441 |
| \(\displaystyle \frac {8}{7} c \left (-\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)}{15 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\) |
(-2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(7*f) + (8*c*((-8*c^2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(15*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x] )/(5*f)))/7
3.1.65.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*Sq rt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[2*a*c*Cot[e + f *x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1)*Sqrt[c + d*Csc[e + f*x]])), x] / ; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[(-d)*Cot[e + f *x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[c*((2*n - 1)/(m + n)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b *c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && !LtQ[m, -2^(-1)] && !(IGtQ[m - 1/2, 0] && LtQ[m, n])
Time = 7.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.55
| method | result | size |
| default | \(\frac {2 a \,c^{2} \left (71 \cos \left (f x +e \right )^{2}-54 \cos \left (f x +e \right )+15\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right )^{2} \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )}{105 f}\) | \(67\) |
| parts | \(\frac {2 a \left (\sec \left (f x +e \right )-1\right )^{2} \left (43 \cos \left (f x +e \right )^{2}-14 \cos \left (f x +e \right )+3\right ) c^{2} \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \csc \left (f x +e \right )}{15 f \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {2 a \left (46 \cos \left (f x +e \right )^{3}-23 \cos \left (f x +e \right )^{2}+12 \cos \left (f x +e \right )-3\right ) \left (\sec \left (f x +e \right )-1\right )^{2} c^{2} \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )}{21 f \left (\cos \left (f x +e \right )-1\right )^{2}}\) | \(170\) |
2/105*a*c^2/f*(71*cos(f*x+e)^2-54*cos(f*x+e)+15)*(-c*(sec(f*x+e)-1))^(1/2) *(cos(f*x+e)+1)^2*sec(f*x+e)^3*csc(f*x+e)
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.86 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=\frac {2 \, {\left (71 \, a c^{2} \cos \left (f x + e\right )^{4} + 88 \, a c^{2} \cos \left (f x + e\right )^{3} - 22 \, a c^{2} \cos \left (f x + e\right )^{2} - 24 \, a c^{2} \cos \left (f x + e\right ) + 15 \, a c^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )} \]
2/105*(71*a*c^2*cos(f*x + e)^4 + 88*a*c^2*cos(f*x + e)^3 - 22*a*c^2*cos(f* x + e)^2 - 24*a*c^2*cos(f*x + e) + 15*a*c^2)*sqrt((c*cos(f*x + e) - c)/cos (f*x + e))/(f*cos(f*x + e)^3*sin(f*x + e))
\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=a \left (\int c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}\, dx + \int \left (- c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{4}{\left (e + f x \right )}\, dx\right ) \]
a*(Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x), x) + Integral(-c* *2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2, x) + Integral(-c**2*sqrt(-c* sec(e + f*x) + c)*sec(e + f*x)**3, x) + Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**4, x))
\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \sec \left (f x + e\right ) \,d x } \]
-2/105*(105*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(3/4)*(3*(a*c^2*f*cos(2*f*x + 2*e)^2 + a*c^2*f*sin(2*f*x + 2*e)^2 + 2 *a*c^2*f*cos(2*f*x + 2*e) + a*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(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (c os(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - c os(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*si n(7/2*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)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*cos(7/2*arctan2(sin(2*f*x + 2*e), co s(2*f*x + 2*e))) - (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)*sin(7/2*arcta n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))/(((cos(2*f*x + 2*e)^4 + sin(2*f*x + 2*e)^4 + (cos (2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e)^2 + 4*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e)^2 + 2*cos(2*f*x + 2*e)^3 + (cos(2*f*x + 2*e)^2 + ...
Time = 0.83 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.68 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=\frac {16 \, \sqrt {2} {\left (35 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{2} + 42 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{3} + 15 \, c^{4}\right )} a c^{2}}{105 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {7}{2}} f} \]
16/105*sqrt(2)*(35*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^2 + 42*(c*tan(1/2*f* x + 1/2*e)^2 - c)*c^3 + 15*c^4)*a*c^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(7/2 )*f)
Time = 17.24 (sec) , antiderivative size = 384, normalized size of antiderivative = 3.15 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx=\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,2{}\mathrm {i}}{f}+\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,142{}\mathrm {i}}{105\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}+\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,16{}\mathrm {i}}{7\,f}-\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,16{}\mathrm {i}}{7\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,8{}\mathrm {i}}{5\,f}-\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,184{}\mathrm {i}}{35\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,4{}\mathrm {i}}{3\,f}+\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,244{}\mathrm {i}}{105\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )} \]
((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^2*2i) /f + (a*c^2*exp(e*1i + f*x*1i)*142i)/(105*f)))/(exp(e*1i + f*x*1i) - 1) + ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^2*16i )/(7*f) - (a*c^2*exp(e*1i + f*x*1i)*16i)/(7*f)))/((exp(e*1i + f*x*1i) - 1) *(exp(e*2i + f*x*2i) + 1)^3) - ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^2*8i)/(5*f) - (a*c^2*exp(e*1i + f*x*1i)*184i)/(3 5*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^2) - ((c - c/(ex p(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^2*4i)/(3*f) + (a *c^2*exp(e*1i + f*x*1i)*244i)/(105*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2 i + f*x*2i) + 1))