Integrand size = 25, antiderivative size = 49 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}-\frac {(A+C) \tan (c+d x)}{a d (1+\sec (c+d x))} \] Output:
A*x/a+C*arctanh(sin(d*x+c))/a/d-(A+C)*tan(d*x+c)/a/d/(1+sec(d*x+c))
Time = 0.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A d x+C \coth ^{-1}(\sin (c+d x))-(A+C) \tan \left (\frac {1}{2} (c+d x)\right )}{a d} \] Input:
Integrate[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x]),x]
Output:
(A*d*x + C*ArcCoth[Sin[c + d*x]] - (A + C)*Tan[(c + d*x)/2])/(a*d)
Time = 0.48 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4539, 3042, 4257, 4407, 3042, 4281}
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.
\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4539 |
\(\displaystyle \frac {\int \frac {a A-a C \sec (c+d x)}{\sec (c+d x) a+a}dx}{a}+\frac {C \int \sec (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a A-a C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\int \frac {a A-a C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {A x-a (A+C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A x-a (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {A x-\frac {a (A+C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}\) |
Input:
Int[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x]),x]
Output:
(C*ArcTanh[Sin[c + d*x]])/(a*d) + (A*x - (a*(A + C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/a
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_)), x_Symbol] :> Simp[C/b Int[Csc[e + f*x], x], x] + Simp[1/b In t[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, A, C}, x]
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20
method | result | size |
parallelrisch | \(\frac {-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-A -C \right )+d x A}{a d}\) | \(59\) |
derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(75\) |
default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(75\) |
risch | \(\frac {A x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\) | \(97\) |
norman | \(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A x}{a}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\) | \(125\) |
Input:
int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
(-C*ln(tan(1/2*d*x+1/2*c)-1)+C*ln(tan(1/2*d*x+1/2*c)+1)+tan(1/2*d*x+1/2*c) *(-A-C)+d*x*A)/a/d
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.80 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, A d x \cos \left (d x + c\right ) + 2 \, A d x + {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A + C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/2*(2*A*d*x*cos(d*x + c) + 2*A*d*x + (C*cos(d*x + c) + C)*log(sin(d*x + c ) + 1) - (C*cos(d*x + c) + C)*log(-sin(d*x + c) + 1) - 2*(A + C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate((A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
Output:
(Integral(A/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d *x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (49) = 98\).
Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.55 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
(A*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))) + C*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.63 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
((d*x + c)*A/a + C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - C*log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/a - (A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c)) /a)/d
Time = 12.55 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.06 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2\,A\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int((A + C/cos(c + d*x)^2)/(a + a/cos(c + d*x)),x)
Output:
(2*A*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*C*atanh(sin(c/2 + (d* x)/2)/cos(c/2 + (d*x)/2)))/(a*d) - (A*sin(c/2 + (d*x)/2) + C*sin(c/2 + (d* x)/2))/(a*d*cos(c/2 + (d*x)/2))
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.33 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +a d x}{a d} \] Input:
int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*c + log(tan((c + d*x)/2) + 1)*c - tan((c + d *x)/2)*a - tan((c + d*x)/2)*c + a*d*x)/(a*d)