Integrand size = 19, antiderivative size = 57 \[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {\arctan (\cos (c+d x)+\sin (c+d x))}{a d}-\frac {\text {arctanh}\left (\frac {\cos (c+d x)-\sin (c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d} \] Output:
arctan(cos(d*x+c)+sin(d*x+c))/a/d-1/3*arctanh(1/3*(cos(d*x+c)-sin(d*x+c))* 3^(1/2))*3^(1/2)/a/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 3.03 (sec) , antiderivative size = 501, normalized size of antiderivative = 8.79 \[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {\left (\text {RootSum}\left [-1+4 i e^{2 i c} \text {$\#$1}^2+e^{4 i c} \text {$\#$1}^4\&,\frac {-\sqrt {3} d \sqrt {e^{-2 i c}} x-i \sqrt {3} \sqrt {e^{-2 i c}} \log \left (e^{i d x}-\text {$\#$1}\right )+3 d x \text {$\#$1}+3 i \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}-\frac {(3-2 i) \sqrt {3} d x \text {$\#$1}^2}{\sqrt {e^{-2 i c}}}-\frac {(2+3 i) \sqrt {3} \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-2 i c}}}+3 d e^{2 i c} x \text {$\#$1}^3+3 i e^{2 i c} \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}^3}{i+2 e^{2 i c} \text {$\#$1}^2}\&\right ]+\text {RootSum}\left [-1+4 i e^{2 i c} \text {$\#$1}^2+e^{4 i c} \text {$\#$1}^4\&,\frac {\sqrt {3} d \sqrt {e^{-2 i c}} x+i \sqrt {3} \sqrt {e^{-2 i c}} \log \left (e^{i d x}-\text {$\#$1}\right )+3 d x \text {$\#$1}+3 i \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}+\frac {(3-2 i) \sqrt {3} d x \text {$\#$1}^2}{\sqrt {e^{-2 i c}}}+\frac {(2+3 i) \sqrt {3} \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-2 i c}}}+3 d e^{2 i c} x \text {$\#$1}^3+3 i e^{2 i c} \log \left (e^{i d x}-\text {$\#$1}\right ) \text {$\#$1}^3}{i+2 e^{2 i c} \text {$\#$1}^2}\&\right ]\right ) (\cos (c)+i \sin (c))}{12 a d} \] Input:
Integrate[(a*Sec[c + d*x] + a*Sin[c + d*x])^(-1),x]
Output:
((RootSum[-1 + (4*I)*E^((2*I)*c)*#1^2 + E^((4*I)*c)*#1^4 & , (-(Sqrt[3]*d* Sqrt[E^((-2*I)*c)]*x) - I*Sqrt[3]*Sqrt[E^((-2*I)*c)]*Log[E^(I*d*x) - #1] + 3*d*x*#1 + (3*I)*Log[E^(I*d*x) - #1]*#1 - ((3 - 2*I)*Sqrt[3]*d*x*#1^2)/Sq rt[E^((-2*I)*c)] - ((2 + 3*I)*Sqrt[3]*Log[E^(I*d*x) - #1]*#1^2)/Sqrt[E^((- 2*I)*c)] + 3*d*E^((2*I)*c)*x*#1^3 + (3*I)*E^((2*I)*c)*Log[E^(I*d*x) - #1]* #1^3)/(I + 2*E^((2*I)*c)*#1^2) & ] + RootSum[-1 + (4*I)*E^((2*I)*c)*#1^2 + E^((4*I)*c)*#1^4 & , (Sqrt[3]*d*Sqrt[E^((-2*I)*c)]*x + I*Sqrt[3]*Sqrt[E^( (-2*I)*c)]*Log[E^(I*d*x) - #1] + 3*d*x*#1 + (3*I)*Log[E^(I*d*x) - #1]*#1 + ((3 - 2*I)*Sqrt[3]*d*x*#1^2)/Sqrt[E^((-2*I)*c)] + ((2 + 3*I)*Sqrt[3]*Log[ E^(I*d*x) - #1]*#1^2)/Sqrt[E^((-2*I)*c)] + 3*d*E^((2*I)*c)*x*#1^3 + (3*I)* E^((2*I)*c)*Log[E^(I*d*x) - #1]*#1^3)/(I + 2*E^((2*I)*c)*#1^2) & ])*(Cos[c ] + I*Sin[c]))/(12*a*d)
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.19, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4902, 27, 2492, 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.
| \(\displaystyle \int \frac {1}{a \sin (c+d x)+a \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a \sin (c+d x)+a \sec (c+d x)}dx\) |
| \(\Big \downarrow \) 4902 |
| \(\displaystyle \frac {2 \int \frac {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}{a \left (\tan ^4\left (\frac {1}{2} (c+d x)\right )-2 \tan ^3\left (\frac {1}{2} (c+d x)\right )+2 \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \frac {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}{\tan ^4\left (\frac {1}{2} (c+d x)\right )-2 \tan ^3\left (\frac {1}{2} (c+d x)\right )+2 \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 \tan \left (\frac {1}{2} (c+d x)\right )+1}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\) |
| \(\Big \downarrow \) 2492 |
| \(\displaystyle \frac {2 \int \left (\frac {3 i-\sqrt {3}}{6 \left (-i \tan ^2\left (\frac {1}{2} (c+d x)\right )+\left (i+\sqrt {3}\right ) \tan \left (\frac {1}{2} (c+d x)\right )+i\right )}+\frac {3 i+\sqrt {3}}{6 \left (-i \tan ^2\left (\frac {1}{2} (c+d x)\right )+\left (i-\sqrt {3}\right ) \tan \left (\frac {1}{2} (c+d x)\right )+i\right )}\right )d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {1}{6} \left (3+i \sqrt {3}\right ) \arctan \left (\frac {-2 i \tan \left (\frac {1}{2} (c+d x)\right )-\sqrt {3}+i}{\sqrt {2 \left (1+i \sqrt {3}\right )}}\right )-\frac {1}{6} \left (3-i \sqrt {3}\right ) \arctan \left (\frac {-2 i \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {3}+i}{\sqrt {2 \left (1-i \sqrt {3}\right )}}\right )\right )}{a d}\) |
Input:
Int[(a*Sec[c + d*x] + a*Sin[c + d*x])^(-1),x]
Output:
(2*(((3 + I*Sqrt[3])*ArcTan[(I - Sqrt[3] - (2*I)*Tan[(c + d*x)/2])/Sqrt[2* (1 + I*Sqrt[3])]])/6 - ((3 - I*Sqrt[3])*ArcTan[(I + Sqrt[3] - (2*I)*Tan[(c + d*x)/2])/Sqrt[2*(1 - I*Sqrt[3])]])/6))/(a*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Px_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2 + (d_.)*(x_)^3 + (e_.)*(x_)^4) ^(p_), x_Symbol] :> Simp[e^p Int[ExpandIntegrand[Px*(b/d + ((d + Sqrt[e*( (b^2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p*(b/d + ((d - Sqrt[e*((b^ 2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x] && ILtQ[p, 0] && EqQ[a*d^2 - b^2*e, 0]
Int[u_, x_Symbol] :> With[{w = Block[{$ShowSteps = False, $StepCounter = Nu ll}, Int[SubstFor[1/(1 + FreeFactors[Tan[FunctionOfTrig[u, x]/2], x]^2*x^2) , Tan[FunctionOfTrig[u, x]/2]/FreeFactors[Tan[FunctionOfTrig[u, x]/2], x], u, x], x]]}, Module[{v = FunctionOfTrig[u, x], d}, Simp[d = FreeFactors[Tan [v/2], x]; 2*(d/Coefficient[v, x, 1]) Subst[Int[SubstFor[1/(1 + d^2*x^2), Tan[v/2]/d, u, x], x], x, Tan[v/2]/d], x]] /; CalculusFreeQ[w, x]] /; Inve rseFunctionFreeQ[u, x] && !FalseQ[FunctionOfTrig[u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.38 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+2 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3}-3 \textit {\_R}^{2}+2 \textit {\_R} +1}}{d a}\) | \(69\) |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+2 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3}-3 \textit {\_R}^{2}+2 \textit {\_R} +1}}{d a}\) | \(69\) |
| risch | \(-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{2}-\frac {i}{2}+\frac {i \sqrt {3}}{2}-\frac {\sqrt {3}}{2}\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{2}-\frac {i}{2}+\frac {i \sqrt {3}}{2}-\frac {\sqrt {3}}{2}\right ) \sqrt {3}}{6 d a}-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{2}-\frac {i}{2}-\frac {i \sqrt {3}}{2}+\frac {\sqrt {3}}{2}\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{2}-\frac {i}{2}-\frac {i \sqrt {3}}{2}+\frac {\sqrt {3}}{2}\right ) \sqrt {3}}{6 d a}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{2}+\frac {i}{2}+\frac {i \sqrt {3}}{2}-\frac {\sqrt {3}}{2}\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{2}+\frac {i}{2}+\frac {i \sqrt {3}}{2}-\frac {\sqrt {3}}{2}\right ) \sqrt {3}}{6 d a}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{2}+\frac {i}{2}-\frac {i \sqrt {3}}{2}+\frac {\sqrt {3}}{2}\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{2}+\frac {i}{2}-\frac {i \sqrt {3}}{2}+\frac {\sqrt {3}}{2}\right ) \sqrt {3}}{6 d a}\) | \(282\) |
Input:
int(1/(a*sec(d*x+c)+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*sum((-_R^2+1)/(2*_R^3-3*_R^2+2*_R+1)*ln(tan(1/2*d*x+1/2*c)-_R),_R=Ro otOf(_Z^4-2*_Z^3+2*_Z^2+2*_Z+1))
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (54) = 108\).
Time = 0.09 (sec) , antiderivative size = 254, normalized size of antiderivative = 4.46 \[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=-\frac {\sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} {\left (a d \cos \left (d x + c\right ) - a d \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + \frac {1}{2} \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 1\right ) - \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} {\left (a d \cos \left (d x + c\right ) - a d \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - \frac {1}{2} \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 1\right ) + 2 \, \arctan \left (-\frac {3 \, \sqrt {\frac {1}{3}} {\left (a d \cos \left (d x + c\right ) - a d \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + 2}{\cos \left (d x + c\right ) + \sin \left (d x + c\right )}\right ) - 2 \, \arctan \left (-\frac {3 \, \sqrt {\frac {1}{3}} {\left (a d \cos \left (d x + c\right ) - a d \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 2}{\cos \left (d x + c\right ) + \sin \left (d x + c\right )}\right )}{4 \, a d} \] Input:
integrate(1/(a*sec(d*x+c)+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/4*(sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*log(-3/2*sqrt(1/3)*(a*d*cos(d*x + c) - a*d*sin(d*x + c))*sqrt(1/(a^2*d^2)) + 1/2*cos(d*x + c)*sin(d*x + c) - 1 ) - sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*log(-3/2*sqrt(1/3)*(a*d*cos(d*x + c) - a*d*sin(d*x + c))*sqrt(1/(a^2*d^2)) - 1/2*cos(d*x + c)*sin(d*x + c) + 1) + 2*arctan(-(3*sqrt(1/3)*(a*d*cos(d*x + c) - a*d*sin(d*x + c))*sqrt(1/(a^2 *d^2)) + 2)/(cos(d*x + c) + sin(d*x + c))) - 2*arctan(-(3*sqrt(1/3)*(a*d*c os(d*x + c) - a*d*sin(d*x + c))*sqrt(1/(a^2*d^2)) - 2)/(cos(d*x + c) + sin (d*x + c))))/(a*d)
\[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {\int \frac {1}{\sin {\left (c + d x \right )} + \sec {\left (c + d x \right )}}\, dx}{a} \] Input:
integrate(1/(a*sec(d*x+c)+a*sin(d*x+c)),x)
Output:
Integral(1/(sin(c + d*x) + sec(c + d*x)), x)/a
\[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\int { \frac {1}{a \sec \left (d x + c\right ) + a \sin \left (d x + c\right )} \,d x } \] Input:
integrate(1/(a*sec(d*x+c)+a*sin(d*x+c)),x, algorithm="maxima")
Output:
integrate(1/(a*sec(d*x + c) + a*sin(d*x + c)), x)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (54) = 108\).
Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.14 \[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {3 \, \pi + \sqrt {3} \log \left ({\left (\sqrt {3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right ) - \sqrt {3} \log \left ({\left (\sqrt {3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right ) + 6 \, \arctan \left ({\left (\sqrt {3} + 1\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 6 \, \arctan \left (-{\left (\sqrt {3} - 1\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{6 \, a d} \] Input:
integrate(1/(a*sec(d*x+c)+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*pi + sqrt(3)*log((sqrt(3) + tan(1/2*d*x + 1/2*c) - 1)^2 + tan(1/2*d *x + 1/2*c)^2) - sqrt(3)*log((sqrt(3) - tan(1/2*d*x + 1/2*c) + 1)^2 + tan( 1/2*d*x + 1/2*c)^2) + 6*arctan((sqrt(3) + 1)*tan(1/2*d*x + 1/2*c) + 1) + 6 *arctan(-(sqrt(3) - 1)*tan(1/2*d*x + 1/2*c) + 1))/(a*d)
Time = 16.10 (sec) , antiderivative size = 453, normalized size of antiderivative = 7.95 \[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {64\,\sqrt {3}\,\sqrt {-\frac {1}{6\,a^2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-\frac {1}{6\,a^2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}\,96{}\mathrm {i}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}+\frac {96\,\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-\frac {1}{6\,a^2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}\right )\,\sqrt {-\frac {1+\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}\,2{}\mathrm {i}}{d}-\frac {\mathrm {atan}\left (\frac {64\,\sqrt {3}\,\sqrt {-\frac {1}{6\,a^2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-\frac {1}{6\,a^2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}\,96{}\mathrm {i}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}+\frac {96\,\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-\frac {1}{6\,a^2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}}{\frac {64}{a}+\frac {64\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {\sqrt {3}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,64{}\mathrm {i}}{a}}\right )\,\sqrt {\frac {-1+\sqrt {3}\,1{}\mathrm {i}}{6\,a^2}}\,2{}\mathrm {i}}{d} \] Input:
int(1/(a*sin(c + d*x) + a/cos(c + d*x)),x)
Output:
(atan((64*3^(1/2)*(- (3^(1/2)*1i)/(6*a^2) - 1/(6*a^2))^(1/2))/(64/a + (64* tan(c/2 + (d*x)/2))/a + (3^(1/2)*tan(c/2 + (d*x)/2)*64i)/a) + (tan(c/2 + ( d*x)/2)*(- (3^(1/2)*1i)/(6*a^2) - 1/(6*a^2))^(1/2)*96i)/(64/a + (64*tan(c/ 2 + (d*x)/2))/a + (3^(1/2)*tan(c/2 + (d*x)/2)*64i)/a) + (96*3^(1/2)*tan(c/ 2 + (d*x)/2)*(- (3^(1/2)*1i)/(6*a^2) - 1/(6*a^2))^(1/2))/(64/a + (64*tan(c /2 + (d*x)/2))/a + (3^(1/2)*tan(c/2 + (d*x)/2)*64i)/a))*(-(3^(1/2)*1i + 1) /(6*a^2))^(1/2)*2i)/d - (atan((64*3^(1/2)*((3^(1/2)*1i)/(6*a^2) - 1/(6*a^2 ))^(1/2))/(64/a + (64*tan(c/2 + (d*x)/2))/a - (3^(1/2)*tan(c/2 + (d*x)/2)* 64i)/a) - (tan(c/2 + (d*x)/2)*((3^(1/2)*1i)/(6*a^2) - 1/(6*a^2))^(1/2)*96i )/(64/a + (64*tan(c/2 + (d*x)/2))/a - (3^(1/2)*tan(c/2 + (d*x)/2)*64i)/a) + (96*3^(1/2)*tan(c/2 + (d*x)/2)*((3^(1/2)*1i)/(6*a^2) - 1/(6*a^2))^(1/2)) /(64/a + (64*tan(c/2 + (d*x)/2))/a - (3^(1/2)*tan(c/2 + (d*x)/2)*64i)/a))* ((3^(1/2)*1i - 1)/(6*a^2))^(1/2)*2i)/d
\[ \int \frac {1}{a \sec (c+d x)+a \sin (c+d x)} \, dx=\frac {\int \frac {1}{\sec \left (d x +c \right )+\sin \left (d x +c \right )}d x}{a} \] Input:
int(1/(a*sec(d*x+c)+a*sin(d*x+c)),x)
Output:
int(1/(sec(c + d*x) + sin(c + d*x)),x)/a