Integrand size = 19, antiderivative size = 59 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {\left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*(2*a^2+b^2)*arctanh(sin(d*x+c))/d+2*a*b*tan(d*x+c)/d+1/2*b^2*sec(d*x+c )*tan(d*x+c)/d
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 a^2 \coth ^{-1}(\sin (c+d x))+b (b \text {arctanh}(\sin (c+d x))+(4 a+b \sec (c+d x)) \tan (c+d x))}{2 d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^2,x]
Output:
(2*a^2*ArcCoth[Sin[c + d*x]] + b*(b*ArcTanh[Sin[c + d*x]] + (4*a + b*Sec[c + d*x])*Tan[c + d*x]))/(2*d)
Time = 0.44 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4275, 3042, 4254, 24, 4534, 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.
\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 4275 |
\(\displaystyle \int \sec (c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right )dx+2 a b \int \sec ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \int 1d(-\tan (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a b \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{2} \left (2 a^2+b^2\right ) \int \sec (c+d x)dx+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (2 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\left (2 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^2,x]
Output:
((2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a*b*Tan[c + d*x])/d + (b^ 2*Sec[c + d*x]*Tan[c + d*x])/(2*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[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, 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 = 1.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(69\) |
default | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(69\) |
parts | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \tan \left (d x +c \right )}{d}\) | \(74\) |
parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 a b \sin \left (2 d x +2 c \right )+b^{2} \sin \left (d x +c \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(109\) |
norman | \(\frac {\frac {b \left (4 a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \left (4 a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(114\) |
risch | \(-\frac {i b \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )} a -b \,{\mathrm e}^{i \left (d x +c \right )}-4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\) | \(144\) |
Input:
int(sec(d*x+c)*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b*tan(d*x+c)+b^2*(1/2*sec(d*x+c)*ta n(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.58 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a b \cos \left (d x + c\right ) + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/4*((2*a^2 + b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^2 + b^2)*co s(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(4*a*b*cos(d*x + c) + b^2)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*sec(d*x+c))**2,x)
Output:
Integral((a + b*sec(c + d*x))**2*sec(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 8 \, a b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/4*(b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l og(sin(d*x + c) - 1)) - 4*a^2*log(sec(d*x + c) + tan(d*x + c)) - 8*a*b*tan (d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (55) = 110\).
Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.19 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/2*((2*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*a^2 + b^2)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(4*a*b*tan(1/2*d*x + 1/2*c)^3 - b^2*tan (1/2*d*x + 1/2*c)^3 - 4*a*b*tan(1/2*d*x + 1/2*c) - b^2*tan(1/2*d*x + 1/2*c ))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 11.56 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2+b^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b-b^2\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (b^2+4\,a\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((a + b/cos(c + d*x))^2/cos(c + d*x),x)
Output:
(atanh(tan(c/2 + (d*x)/2))*(2*a^2 + b^2))/d - (tan(c/2 + (d*x)/2)^3*(4*a*b - b^2) - tan(c/2 + (d*x)/2)*(4*a*b + b^2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*t an(c/2 + (d*x)/2)^2 + 1))
Time = 0.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 3.58 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\sin \left (d x +c \right ) b^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)*(a+b*sec(d*x+c))^2,x)
Output:
( - 4*cos(c + d*x)*sin(c + d*x)*a*b - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + 2*log(tan( (c + d*x)/2) - 1)*a**2 + log(tan((c + d*x)/2) - 1)*b**2 + 2*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**2*a**2 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)* *2*b**2 - 2*log(tan((c + d*x)/2) + 1)*a**2 - log(tan((c + d*x)/2) + 1)*b** 2 - sin(c + d*x)*b**2)/(2*d*(sin(c + d*x)**2 - 1))