Integrand size = 21, antiderivative size = 87 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(7 A+6 C) \tan (c+d x)}{7 d}+\frac {C \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {2 (7 A+6 C) \tan ^3(c+d x)}{21 d}+\frac {(7 A+6 C) \tan ^5(c+d x)}{35 d} \] Output:
1/7*(7*A+6*C)*tan(d*x+c)/d+1/7*C*sec(d*x+c)^6*tan(d*x+c)/d+2/21*(7*A+6*C)* tan(d*x+c)^3/d+1/35*(7*A+6*C)*tan(d*x+c)^5/d
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d}+\frac {C \left (\tan (c+d x)+\tan ^3(c+d x)+\frac {3}{5} \tan ^5(c+d x)+\frac {1}{7} \tan ^7(c+d x)\right )}{d} \] Input:
Integrate[Sec[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]
Output:
(A*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d + (C*(Tan[c + d*x] + Tan[c + d*x]^3 + (3*Tan[c + d*x]^5)/5 + Tan[c + d*x]^7/7))/d
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.
\(\displaystyle \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^6 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{7} (7 A+6 C) \int \sec ^6(c+d x)dx+\frac {C \tan (c+d x) \sec ^6(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} (7 A+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx+\frac {C \tan (c+d x) \sec ^6(c+d x)}{7 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {C \tan (c+d x) \sec ^6(c+d x)}{7 d}-\frac {(7 A+6 C) \int \left (\tan ^4(c+d x)+2 \tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {C \tan (c+d x) \sec ^6(c+d x)}{7 d}-\frac {(7 A+6 C) \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{7 d}\) |
Input:
Int[Sec[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]
Output:
(C*Sec[c + d*x]^6*Tan[c + d*x])/(7*d) - ((7*A + 6*C)*(-Tan[c + d*x] - (2*T an[c + d*x]^3)/3 - Tan[c + d*x]^5/5))/(7*d)
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 = 0.54 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-C \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(78\) |
default | \(\frac {-A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-C \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(78\) |
parts | \(-\frac {A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {C \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(80\) |
risch | \(\frac {16 i \left (70 A \,{\mathrm e}^{8 i \left (d x +c \right )}+175 A \,{\mathrm e}^{6 i \left (d x +c \right )}+210 C \,{\mathrm e}^{6 i \left (d x +c \right )}+147 A \,{\mathrm e}^{4 i \left (d x +c \right )}+126 C \,{\mathrm e}^{4 i \left (d x +c \right )}+49 A \,{\mathrm e}^{2 i \left (d x +c \right )}+42 C \,{\mathrm e}^{2 i \left (d x +c \right )}+7 A +6 C \right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(111\) |
parallelrisch | \(\frac {\left (1176 A +1008 C \right ) \sin \left (3 d x +3 c \right )+\left (392 A +336 C \right ) \sin \left (5 d x +5 c \right )+\left (56 A +48 C \right ) \sin \left (7 d x +7 c \right )+840 \sin \left (d x +c \right ) \left (A +2 C \right )}{105 d \left (\cos \left (7 d x +7 c \right )+7 \cos \left (5 d x +5 c \right )+21 \cos \left (3 d x +3 c \right )+35 \cos \left (d x +c \right )\right )}\) | \(113\) |
norman | \(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 \left (5 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 \left (5 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {8 \left (91 A +53 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}-\frac {2 \left (113 A +129 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {2 \left (113 A +129 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{7}}\) | \(169\) |
Input:
int(sec(d*x+c)^6*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(-A*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-C*(-16/35-1/ 7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (8 \, {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, C\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \] Input:
integrate(sec(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/105*(8*(7*A + 6*C)*cos(d*x + c)^6 + 4*(7*A + 6*C)*cos(d*x + c)^4 + 3*(7* A + 6*C)*cos(d*x + c)^2 + 15*C)*sin(d*x + c)/(d*cos(d*x + c)^7)
\[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**6*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**6, x)
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, C \tan \left (d x + c\right )^{7} + 21 \, {\left (A + 3 \, C\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \tan \left (d x + c\right )^{3} + 105 \, {\left (A + C\right )} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/105*(15*C*tan(d*x + c)^7 + 21*(A + 3*C)*tan(d*x + c)^5 + 35*(2*A + 3*C)* tan(d*x + c)^3 + 105*(A + C)*tan(d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, C \tan \left (d x + c\right )^{7} + 21 \, A \tan \left (d x + c\right )^{5} + 63 \, C \tan \left (d x + c\right )^{5} + 70 \, A \tan \left (d x + c\right )^{3} + 105 \, C \tan \left (d x + c\right )^{3} + 105 \, A \tan \left (d x + c\right ) + 105 \, C \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/105*(15*C*tan(d*x + c)^7 + 21*A*tan(d*x + c)^5 + 63*C*tan(d*x + c)^5 + 7 0*A*tan(d*x + c)^3 + 105*C*tan(d*x + c)^3 + 105*A*tan(d*x + c) + 105*C*tan (d*x + c))/d
Time = 11.36 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {C\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\left (\frac {A}{5}+\frac {3\,C}{5}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (\frac {2\,A}{3}+C\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\right )}{d} \] Input:
int((A + C/cos(c + d*x)^2)/cos(c + d*x)^6,x)
Output:
(tan(c + d*x)^3*((2*A)/3 + C) + (C*tan(c + d*x)^7)/7 + tan(c + d*x)*(A + C ) + tan(c + d*x)^5*(A/5 + (3*C)/5))/d
Time = 0.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.43 \[ \int \sec ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin \left (d x +c \right ) \left (56 \sin \left (d x +c \right )^{6} a +48 \sin \left (d x +c \right )^{6} c -196 \sin \left (d x +c \right )^{4} a -168 \sin \left (d x +c \right )^{4} c +245 \sin \left (d x +c \right )^{2} a +210 \sin \left (d x +c \right )^{2} c -105 a -105 c \right )}{105 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^6*(A+C*sec(d*x+c)^2),x)
Output:
(sin(c + d*x)*(56*sin(c + d*x)**6*a + 48*sin(c + d*x)**6*c - 196*sin(c + d *x)**4*a - 168*sin(c + d*x)**4*c + 245*sin(c + d*x)**2*a + 210*sin(c + d*x )**2*c - 105*a - 105*c))/(105*cos(c + d*x)*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))