Integrand size = 21, antiderivative size = 65 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \] Output:
1/5*(5*A+4*C)*tan(d*x+c)/d+1/5*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*A+4*C)* tan(d*x+c)^3/d
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d}+\frac {C \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \] Input:
Integrate[Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]
Output:
(A*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d + (C*(Tan[c + d*x] + (2*Tan[c + d* x]^3)/3 + Tan[c + d*x]^5/5))/d
Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{5} (5 A+4 C) \int \sec ^4(c+d x)dx+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}-\frac {(5 A+4 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}-\frac {(5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 d}\) |
Input:
Int[Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]
Output:
(C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) - ((5*A + 4*C)*(-Tan[c + d*x] - Tan[ c + d*x]^3/3))/(5*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 = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
default | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
parts | \(-\frac {A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(60\) |
parallelrisch | \(\frac {\left (50 A +40 C \right ) \sin \left (3 d x +3 c \right )+\left (10 A +8 C \right ) \sin \left (5 d x +5 c \right )+40 \sin \left (d x +c \right ) \left (A +2 C \right )}{15 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(85\) |
risch | \(\frac {4 i \left (15 A \,{\mathrm e}^{6 i \left (d x +c \right )}+35 A \,{\mathrm e}^{4 i \left (d x +c \right )}+40 C \,{\mathrm e}^{4 i \left (d x +c \right )}+25 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 C \,{\mathrm e}^{2 i \left (d x +c \right )}+5 A +4 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(87\) |
norman | \(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}\) | \(119\) |
Input:
int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(-A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-C*(-8/15-1/5*sec(d*x+c)^4-4/15* sec(d*x+c)^2)*tan(d*x+c))
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/15*(2*(5*A + 4*C)*cos(d*x + c)^4 + (5*A + 4*C)*cos(d*x + c)^2 + 3*C)*sin (d*x + c)/(d*cos(d*x + c)^5)
\[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, {\left (A + 2 \, C\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/15*(3*C*tan(d*x + c)^5 + 5*(A + 2*C)*tan(d*x + c)^3 + 15*(A + C)*tan(d*x + c))/d
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, A \tan \left (d x + c\right )^{3} + 10 \, C \tan \left (d x + c\right )^{3} + 15 \, A \tan \left (d x + c\right ) + 15 \, C \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/15*(3*C*tan(d*x + c)^5 + 5*A*tan(d*x + c)^3 + 10*C*tan(d*x + c)^3 + 15*A *tan(d*x + c) + 15*C*tan(d*x + c))/d
Time = 11.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {C\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\left (\frac {A}{3}+\frac {2\,C}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\right )}{d} \] Input:
int((A + C/cos(c + d*x)^2)/cos(c + d*x)^4,x)
Output:
((C*tan(c + d*x)^5)/5 + tan(c + d*x)*(A + C) + tan(c + d*x)^3*(A/3 + (2*C) /3))/d
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin \left (d x +c \right ) \left (10 \sin \left (d x +c \right )^{4} a +8 \sin \left (d x +c \right )^{4} c -25 \sin \left (d x +c \right )^{2} a -20 \sin \left (d x +c \right )^{2} c +15 a +15 c \right )}{15 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x)
Output:
(sin(c + d*x)*(10*sin(c + d*x)**4*a + 8*sin(c + d*x)**4*c - 25*sin(c + d*x )**2*a - 20*sin(c + d*x)**2*c + 15*a + 15*c))/(15*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))