Integrand size = 21, antiderivative size = 87 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(7 a+6 b) \tan (e+f x)}{7 f}+\frac {b \sec ^6(e+f x) \tan (e+f x)}{7 f}+\frac {2 (7 a+6 b) \tan ^3(e+f x)}{21 f}+\frac {(7 a+6 b) \tan ^5(e+f x)}{35 f} \] Output:
1/7*(7*a+6*b)*tan(f*x+e)/f+1/7*b*sec(f*x+e)^6*tan(f*x+e)/f+2/21*(7*a+6*b)* tan(f*x+e)^3/f+1/35*(7*a+6*b)*tan(f*x+e)^5/f
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \left (\tan (e+f x)+\frac {2}{3} \tan ^3(e+f x)+\frac {1}{5} \tan ^5(e+f x)\right )}{f}+\frac {b \left (\tan (e+f x)+\tan ^3(e+f x)+\frac {3}{5} \tan ^5(e+f x)+\frac {1}{7} \tan ^7(e+f x)\right )}{f} \] Input:
Integrate[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
(a*(Tan[e + f*x] + (2*Tan[e + f*x]^3)/3 + Tan[e + f*x]^5/5))/f + (b*(Tan[e + f*x] + Tan[e + f*x]^3 + (3*Tan[e + f*x]^5)/5 + Tan[e + f*x]^7/7))/f
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4534, 3042, 4254, 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.
\(\displaystyle \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^6 \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{7} (7 a+6 b) \int \sec ^6(e+f x)dx+\frac {b \tan (e+f x) \sec ^6(e+f x)}{7 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} (7 a+6 b) \int \csc \left (e+f x+\frac {\pi }{2}\right )^6dx+\frac {b \tan (e+f x) \sec ^6(e+f x)}{7 f}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {b \tan (e+f x) \sec ^6(e+f x)}{7 f}-\frac {(7 a+6 b) \int \left (\tan ^4(e+f x)+2 \tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{7 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \tan (e+f x) \sec ^6(e+f x)}{7 f}-\frac {(7 a+6 b) \left (-\frac {1}{5} \tan ^5(e+f x)-\frac {2}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{7 f}\) |
Input:
Int[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]
Output:
(b*Sec[e + f*x]^6*Tan[e + f*x])/(7*f) - ((7*a + 6*b)*(-Tan[e + f*x] - (2*T an[e + f*x]^3)/3 - Tan[e + f*x]^5/5))/(7*f)
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 1.63 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-b \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}\) | \(78\) |
default | \(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-b \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}\) | \(78\) |
parts | \(-\frac {a \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}-\frac {b \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}\) | \(80\) |
risch | \(\frac {16 i \left (70 a \,{\mathrm e}^{8 i \left (f x +e \right )}+175 a \,{\mathrm e}^{6 i \left (f x +e \right )}+210 b \,{\mathrm e}^{6 i \left (f x +e \right )}+147 a \,{\mathrm e}^{4 i \left (f x +e \right )}+126 b \,{\mathrm e}^{4 i \left (f x +e \right )}+49 a \,{\mathrm e}^{2 i \left (f x +e \right )}+42 b \,{\mathrm e}^{2 i \left (f x +e \right )}+7 a +6 b \right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\) | \(111\) |
parallelrisch | \(\frac {\left (1176 a +1008 b \right ) \sin \left (3 f x +3 e \right )+\left (392 a +336 b \right ) \sin \left (5 f x +5 e \right )+\left (56 a +48 b \right ) \sin \left (7 f x +7 e \right )+840 \sin \left (f x +e \right ) \left (a +2 b \right )}{105 f \left (\cos \left (7 f x +7 e \right )+7 \cos \left (5 f x +5 e \right )+21 \cos \left (3 f x +3 e \right )+35 \cos \left (f x +e \right )\right )}\) | \(113\) |
norman | \(\frac {-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{f}+\frac {4 \left (5 a +3 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}+\frac {4 \left (5 a +3 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{3 f}+\frac {8 \left (91 a +53 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{35 f}-\frac {2 \left (113 a +129 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}-\frac {2 \left (113 a +129 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{15 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{7}}\) | \(169\) |
Input:
int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-a*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)-b*(-16/35-1/ 7*sec(f*x+e)^6-6/35*sec(f*x+e)^4-8/35*sec(f*x+e)^2)*tan(f*x+e))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (8 \, {\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{6} + 4 \, {\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{2} + 15 \, b\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \] Input:
integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/105*(8*(7*a + 6*b)*cos(f*x + e)^6 + 4*(7*a + 6*b)*cos(f*x + e)^4 + 3*(7* a + 6*b)*cos(f*x + e)^2 + 15*b)*sin(f*x + e)/(f*cos(f*x + e)^7)
\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{6}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*sec(e + f*x)**6, x)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {15 \, b \tan \left (f x + e\right )^{7} + 21 \, {\left (a + 3 \, b\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (2 \, a + 3 \, b\right )} \tan \left (f x + e\right )^{3} + 105 \, {\left (a + b\right )} \tan \left (f x + e\right )}{105 \, f} \] Input:
integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/105*(15*b*tan(f*x + e)^7 + 21*(a + 3*b)*tan(f*x + e)^5 + 35*(2*a + 3*b)* tan(f*x + e)^3 + 105*(a + b)*tan(f*x + e))/f
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {15 \, b \tan \left (f x + e\right )^{7} + 21 \, a \tan \left (f x + e\right )^{5} + 63 \, b \tan \left (f x + e\right )^{5} + 70 \, a \tan \left (f x + e\right )^{3} + 105 \, b \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right ) + 105 \, b \tan \left (f x + e\right )}{105 \, f} \] Input:
integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/105*(15*b*tan(f*x + e)^7 + 21*a*tan(f*x + e)^5 + 63*b*tan(f*x + e)^5 + 7 0*a*tan(f*x + e)^3 + 105*b*tan(f*x + e)^3 + 105*a*tan(f*x + e) + 105*b*tan (f*x + e))/f
Time = 15.69 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^7}{7}+\left (\frac {a}{5}+\frac {3\,b}{5}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {2\,a}{3}+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a+b\right )\,\mathrm {tan}\left (e+f\,x\right )}{f} \] Input:
int((a + b/cos(e + f*x)^2)/cos(e + f*x)^6,x)
Output:
(tan(e + f*x)^5*(a/5 + (3*b)/5) + (b*tan(e + f*x)^7)/7 + tan(e + f*x)^3*(( 2*a)/3 + b) + tan(e + f*x)*(a + b))/f
Time = 0.17 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.43 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\sin \left (f x +e \right ) \left (56 \sin \left (f x +e \right )^{6} a +48 \sin \left (f x +e \right )^{6} b -196 \sin \left (f x +e \right )^{4} a -168 \sin \left (f x +e \right )^{4} b +245 \sin \left (f x +e \right )^{2} a +210 \sin \left (f x +e \right )^{2} b -105 a -105 b \right )}{105 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x)
Output:
(sin(e + f*x)*(56*sin(e + f*x)**6*a + 48*sin(e + f*x)**6*b - 196*sin(e + f *x)**4*a - 168*sin(e + f*x)**4*b + 245*sin(e + f*x)**2*a + 210*sin(e + f*x )**2*b - 105*a - 105*b))/(105*cos(e + f*x)*f*(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin(e + f*x)**2 - 1))