Integrand size = 21, antiderivative size = 70 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
1/8*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/8*(4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+ 1/4*C*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {3 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Input:
Integrate[Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]
Output:
(A*ArcTanh[Sin[c + d*x]])/(2*d) + (3*C*ArcTanh[Sin[c + d*x]])/(8*d) + (A*S ec[c + d*x]*Tan[c + d*x])/(2*d) + (3*C*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4534, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \int \sec ^3(c+d x)dx+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]
Output:
(C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*A + 3*C)*(ArcTanh[Sin[c + d*x] ]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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 = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
| default | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
| parts | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(87\) |
| parallelrisch | \(\frac {-8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (4 A +3 C \right ) \sin \left (3 d x +3 c \right )+4 \left (A +\frac {11 C}{4}\right ) \sin \left (d x +c \right )}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(143\) |
| norman | \(\frac {-\frac {\left (4 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {\left (4 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(157\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (4 A \,{\mathrm e}^{6 i \left (d x +c \right )}+3 C \,{\mathrm e}^{6 i \left (d x +c \right )}+4 A \,{\mathrm e}^{4 i \left (d x +c \right )}+11 C \,{\mathrm e}^{4 i \left (d x +c \right )}-4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-11 C \,{\mathrm e}^{2 i \left (d x +c \right )}-4 A -3 C \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(199\) |
Input:
int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*(-(-1/4 *sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/16*((4*A + 3*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (4*A + 3*C)*cos(d *x + c)^4*log(-sin(d*x + c) + 1) + 2*((4*A + 3*C)*cos(d*x + c)^2 + 2*C)*si n(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.39 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )^{3} - {\left (4 \, A + 5 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/16*((4*A + 3*C)*log(sin(d*x + c) + 1) - (4*A + 3*C)*log(sin(d*x + c) - 1 ) - 2*((4*A + 3*C)*sin(d*x + c)^3 - (4*A + 5*C)*sin(d*x + c))/(sin(d*x + c )^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.40 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, A \sin \left (d x + c\right )^{3} + 3 \, C \sin \left (d x + c\right )^{3} - 4 \, A \sin \left (d x + c\right ) - 5 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/16*((4*A + 3*C)*log(abs(sin(d*x + c) + 1)) - (4*A + 3*C)*log(abs(sin(d*x + c) - 1)) - 2*(4*A*sin(d*x + c)^3 + 3*C*sin(d*x + c)^3 - 4*A*sin(d*x + c ) - 5*C*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d
Time = 11.40 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {A}{2}+\frac {5\,C}{8}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d} \] Input:
int((A + C/cos(c + d*x)^2)/cos(c + d*x)^3,x)
Output:
(sin(c + d*x)*(A/2 + (5*C)/8) - sin(c + d*x)^3*(A/2 + (3*C)/8))/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) + (atanh(sin(c + d*x))*(A/2 + (3*C)/8))/ d
Time = 0.17 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.46 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -4 \sin \left (d x +c \right )^{3} a -3 \sin \left (d x +c \right )^{3} c +4 \sin \left (d x +c \right ) a +5 \sin \left (d x +c \right ) c}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x)
Output:
( - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 4*log(tan((c + d*x)/2) - 1 )*a - 3*log(tan((c + d*x)/2) - 1)*c + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **2*c + 4*log(tan((c + d*x)/2) + 1)*a + 3*log(tan((c + d*x)/2) + 1)*c - 4* sin(c + d*x)**3*a - 3*sin(c + d*x)**3*c + 4*sin(c + d*x)*a + 5*sin(c + d*x )*c)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))