Integrand size = 34, antiderivative size = 155 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=-\frac {64 c^4 \tan (e+f x)}{3 a^2 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{3 a^2 f}-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2} \] Output:
-64/3*c^4*tan(f*x+e)/a^2/f/(c-c*sec(f*x+e))^(1/2)-16/3*c^3*(c-c*sec(f*x+e) )^(1/2)*tan(f*x+e)/a^2/f-4*c^2*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(a^2+a^ 2*sec(f*x+e))+2/3*c*(c-c*sec(f*x+e))^(5/2)*tan(f*x+e)/f/(a+a*sec(f*x+e))^2
Time = 1.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.46 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=\frac {2 c^4 \left (-45-69 \sec (e+f x)-15 \sec ^2(e+f x)+\sec ^3(e+f x)\right ) \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \] Input:
Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^2 ,x]
Output:
(2*c^4*(-45 - 69*Sec[e + f*x] - 15*Sec[e + f*x]^2 + Sec[e + f*x]^3)*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])
Time = 0.82 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 4442, 3042, 4442, 3042, 4280, 3042, 4279}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a \sec (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{7/2}}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4442 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{\sec (e+f x) a+a}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}{\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}dx}{a}\) |
| \(\Big \downarrow \) 4442 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \left (\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)}-\frac {4 c \int \sec (e+f x) (c-c \sec (e+f x))^{3/2}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \left (\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)}-\frac {4 c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \left (\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)}-\frac {4 c \left (\frac {4}{3} c \int \sec (e+f x) \sqrt {c-c \sec (e+f x)}dx-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 f}\right )}{a}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \left (\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)}-\frac {4 c \left (\frac {4}{3} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 f}\right )}{a}\right )}{a}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2}-\frac {2 c \left (\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)}-\frac {4 c \left (-\frac {8 c^2 \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 f}\right )}{a}\right )}{a}\) |
Input:
Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^2,x]
Output:
(2*c*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (2*c*((2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(a + a*Sec[e + f *x])) - (4*c*((-8*c^2*Tan[e + f*x])/(3*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c* Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*f)))/a))/a
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[a*((2*m - 1)/m) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege rQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), x] - Simp[d*((2*n - 1)/(b*(2*m + 1))) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && LtQ[ m, -2^(-1)]
Time = 9.40 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.68
| method | result | size |
| default | \(\frac {\frac {4 \left (996 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-1245 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+465 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-35 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, c^{3} \sqrt {2}\, \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \csc \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}-\frac {8 \left (91 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-115 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+45 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, c^{3} \sqrt {2}\, \sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}}{a^{2}}\) | \(261\) |
Input:
int(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x,method=_RETURNV ERBOSE)
Output:
1/a^2*(4/15/f*(996*cos(1/2*f*x+1/2*e)^8-1245*cos(1/2*f*x+1/2*e)^6+465*cos( 1/2*f*x+1/2*e)^4-35*cos(1/2*f*x+1/2*e)^2-5)*(-c/(2*cos(1/2*f*x+1/2*e)^2-1) *sin(1/2*f*x+1/2*e)^2)^(1/2)*c^3*2^(1/2)/(2*cos(1/2*f*x+1/2*e)^2-1)^2*sec( 1/2*f*x+1/2*e)^3*csc(1/2*f*x+1/2*e)-8/5/f*(91*cos(1/2*f*x+1/2*e)^6-115*cos (1/2*f*x+1/2*e)^4+45*cos(1/2*f*x+1/2*e)^2-5)*(-c/(2*cos(1/2*f*x+1/2*e)^2-1 )*sin(1/2*f*x+1/2*e)^2)^(1/2)*c^3*2^(1/2)/(2*cos(1/2*f*x+1/2*e)^2-1)^2*sec (1/2*f*x+1/2*e)*csc(1/2*f*x+1/2*e))
Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=\frac {2 \, {\left (45 \, c^{3} \cos \left (f x + e\right )^{3} + 69 \, c^{3} \cos \left (f x + e\right )^{2} + 15 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorith m="fricas")
Output:
2/3*(45*c^3*cos(f*x + e)^3 + 69*c^3*cos(f*x + e)^2 + 15*c^3*cos(f*x + e) - c^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/((a^2*f*cos(f*x + e)^2 + a^2 *f*cos(f*x + e))*sin(f*x + e))
Timed out. \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=\text {Timed out} \] Input:
integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**2,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.21 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=-\frac {4 \, {\left (16 \, \sqrt {2} c^{\frac {7}{2}} - \frac {56 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {70 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {35 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {4 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {\sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{3 \, a^{2} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {7}{2}}} \] Input:
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorith m="maxima")
Output:
-4/3*(16*sqrt(2)*c^(7/2) - 56*sqrt(2)*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 70*sqrt(2)*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 35*sqrt (2)*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*sqrt(2)*c^(7/2)*sin(f* x + e)^8/(cos(f*x + e) + 1)^8 + sqrt(2)*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/(a^2*f*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)^(7/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(7/2))
Time = 0.46 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.78 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=-\frac {4 \, \sqrt {2} c^{3} {\left (\frac {9 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + c^{2}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a^{2}} - \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a^{4} c^{2} + 9 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} a^{4} c^{3}}{a^{6} c^{3}}\right )}}{3 \, f} \] Input:
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorith m="giac")
Output:
-4/3*sqrt(2)*c^3*((9*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + c^2)/((c*tan(1/2*f *x + 1/2*e)^2 - c)^(3/2)*a^2) - ((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*a^4* c^2 + 9*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*a^4*c^3)/(a^6*c^3))/f
Time = 15.46 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.21 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=\frac {2\,c^3\,\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,138{}\mathrm {i}+{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,195{}\mathrm {i}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,268{}\mathrm {i}+{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,195{}\mathrm {i}+{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,138{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,45{}\mathrm {i}+45{}\mathrm {i}\right )}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}-1\right )} \] Input:
int((c - c/cos(e + f*x))^(7/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^2),x)
Output:
(2*c^3*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*(exp( e*1i + f*x*1i)*138i + exp(e*2i + f*x*2i)*195i + exp(e*3i + f*x*3i)*268i + exp(e*4i + f*x*4i)*195i + exp(e*5i + f*x*5i)*138i + exp(e*6i + f*x*6i)*45i + 45i))/(3*a^2*f*(exp(e*1i + f*x*1i) + 1)^3*(exp(e*1i + f*x*1i) - exp(e*2 i + f*x*2i) + exp(e*3i + f*x*3i) - 1))
\[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx=\frac {\sqrt {c}\, c^{3} \left (-\left (\int \frac {\sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{2}+2 \sec \left (f x +e \right )+1}d x \right )+3 \left (\int \frac {\sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{2}+2 \sec \left (f x +e \right )+1}d x \right )-3 \left (\int \frac {\sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{2}+2 \sec \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )}{\sec \left (f x +e \right )^{2}+2 \sec \left (f x +e \right )+1}d x \right )}{a^{2}} \] Input:
int(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x)
Output:
(sqrt(c)*c**3*( - int((sqrt( - sec(e + f*x) + 1)*sec(e + f*x)**4)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1),x) + 3*int((sqrt( - sec(e + f*x) + 1)*sec(e + f*x)**3)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1),x) - 3*int((sqrt( - sec (e + f*x) + 1)*sec(e + f*x)**2)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1),x) + int((sqrt( - sec(e + f*x) + 1)*sec(e + f*x))/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1),x)))/a**2