Integrand size = 19, antiderivative size = 203 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=-\frac {4 a^2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {2 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \] Output:
-4*a^2*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/( a+b)^(5/2)/d-2*b^3*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a- b)^(5/2)/(a+b)^(5/2)/d-1/2*sin(d*x+c)/(a+b)^2/d/(1-cos(d*x+c))+1/2*sin(d*x +c)/(a-b)^2/d/(1+cos(d*x+c))+a*b^2*sin(d*x+c)/(a^2-b^2)^2/d/(b+a*cos(d*x+c ))
Time = 1.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {\frac {4 b \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}+\frac {\frac {2 a b^2 \sin (c+d x)}{(a+b)^2 (b+a \cos (c+d x))}+\tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}}{2 d} \] Input:
Integrate[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-2),x]
Output:
((4*b*(2*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ (a^2 - b^2)^(5/2) - Cot[(c + d*x)/2]/(a + b)^2 + ((2*a*b^2*Sin[c + d*x])/( (a + b)^2*(b + a*Cos[c + d*x])) + Tan[(c + d*x)/2])/(a - b)^2)/(2*d)
Time = 0.63 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4897, 3042, 3210, 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.
\(\displaystyle \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a \cos (c+d x)+b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan \left (c+d x-\frac {\pi }{2}\right )^2}{\left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3210 |
\(\displaystyle \int \left (-\frac {2 a^2 b}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {b^2}{\left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}-\frac {1}{2 (a+b)^2 (\cos (c+d x)-1)}+\frac {1}{2 (a-b)^2 (\cos (c+d x)+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 a^2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac {a b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {2 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)}\) |
Input:
Int[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-2),x]
Output:
(-4*a^2*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5 /2)*(a + b)^(5/2)*d) - (2*b^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[ a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) - Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d*x])) + (a*b^ 2*Sin[c + d*x])/((a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ ), x_Symbol] :> Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^m/ (1 - Sin[e + f*x]^2)^(p/2)), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, p/2]
Time = 0.58 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-4 a b +2 b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b}-\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(162\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-4 a b +2 b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b}-\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(162\) |
risch | \(-\frac {2 i \left (-2 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{2} b -{\mathrm e}^{3 i \left (d x +c \right )} b^{3}+a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-4 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} b^{3}+a^{3}+2 a \,b^{2}\right )}{\left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(477\) |
Input:
int(1/(a*sin(d*x+c)+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*tan(1/2*d*x+1/2*c)/(a^2-2*a*b+b^2)-1/2/(a+b)^2/tan(1/2*d*x+1/2*c) +2*b/(a-b)^2/(a+b)^2*(-a*b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-b*ta n(1/2*d*x+1/2*c)^2-a-b)-(2*a^2+b^2)/((a+b)*(a-b))^(1/2)*arctanh(tan(1/2*d* x+1/2*c)*(a-b)/((a+b)*(a-b))^(1/2))))
Time = 0.11 (sec) , antiderivative size = 526, normalized size of antiderivative = 2.59 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\left [\frac {6 \, a^{3} b^{2} - 6 \, a b^{4} + {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, \frac {3 \, a^{3} b^{2} - 3 \, a b^{4} - {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}\right ] \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
[1/2*(6*a^3*b^2 - 6*a*b^4 + (2*a^2*b^2 + b^4 + (2*a^3*b + a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^ 2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^ 2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 2*(a^5 + a^3* b^2 - 2*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))/ (((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c) + (a^6*b - 3*a^4*b^ 3 + 3*a^2*b^5 - b^7)*d)*sin(d*x + c)), (3*a^3*b^2 - 3*a*b^4 - (2*a^2*b^2 + b^4 + (2*a^3*b + a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - (a^ 5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))/(((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c) + (a^6*b - 3* a^4*b^3 + 3*a^2*b^5 - b^7)*d)*sin(d*x + c))]
\[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\int \frac {1}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))**2,x)
Output:
Integral((a*sin(c + d*x) + b*tan(c + d*x))**(-2), x)
Exception generated. \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.17 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.42 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (2 \, a^{2} b + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{2 \, d} \] Input:
integrate(1/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
1/2*(4*(2*a^2*b + b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)) )/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + tan(1/2*d*x + 1/2*c)/(a^2 - 2*a*b + b^2) - (a^3*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c) ^2 + 7*a*b^2*tan(1/2*d*x + 1/2*c)^2 - b^3*tan(1/2*d*x + 1/2*c)^2 - a^3 + a ^2*b + a*b^2 - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^3 - b *tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c)) ))/d
Time = 16.89 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {\frac {a^2-2\,a\,b+b^2}{a+b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^3-3\,a^2\,b+7\,a\,b^2-b^3\right )}{{\left (a+b\right )}^2}}{d\,\left (\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3+2\,a^2\,b+2\,a\,b^2-2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4-2{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^4}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )\,2{}\mathrm {i}}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:
int(1/(a*sin(c + d*x) + b*tan(c + d*x))^2,x)
Output:
((a^2 - 2*a*b + b^2)/(a + b) - (tan(c/2 + (d*x)/2)^2*(7*a*b^2 - 3*a^2*b + a^3 - b^3))/(a + b)^2)/(d*(tan(c/2 + (d*x)/2)^3*(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b^3) + tan(c/2 + (d*x)/2)*(2*a*b^2 + 2*a^2*b - 2*a^3 - 2*b^3))) + tan (c/2 + (d*x)/2)/(2*d*(a - b)^2) + (b*atan((a^4*tan(c/2 + (d*x)/2)*1i + b^4 *tan(c/2 + (d*x)/2)*1i - a^2*b^2*tan(c/2 + (d*x)/2)*2i)/((a + b)^(5/2)*(a - b)^(3/2)))*(2*a^2 + b^2)*2i)/(d*(a + b)^(5/2)*(a - b)^(5/2))
Time = 0.17 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.17 \[ \int \frac {1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b -2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (d x +c \right ) a^{2} b^{2}-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (d x +c \right ) b^{4}+\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+\sin \left (d x +c \right )^{2} a^{5}+\sin \left (d x +c \right )^{2} a^{3} b^{2}-2 \sin \left (d x +c \right )^{2} a \,b^{4}-a^{5}+2 a^{3} b^{2}-a \,b^{4}}{\sin \left (d x +c \right ) d \left (\cos \left (d x +c \right ) a^{7}-3 \cos \left (d x +c \right ) a^{5} b^{2}+3 \cos \left (d x +c \right ) a^{3} b^{4}-\cos \left (d x +c \right ) a \,b^{6}+a^{6} b -3 a^{4} b^{3}+3 a^{2} b^{5}-b^{7}\right )} \] Input:
int(1/(a*sin(d*x+c)+b*tan(d*x+c))^2,x)
Output:
( - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ sqrt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)*a**3*b - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2)) *cos(c + d*x)*sin(c + d*x)*a*b**3 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)*a**2*b* *2 - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b) /sqrt( - a**2 + b**2))*sin(c + d*x)*b**4 + cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + sin(c + d*x)**2*a**5 + sin(c + d*x)* *2*a**3*b**2 - 2*sin(c + d*x)**2*a*b**4 - a**5 + 2*a**3*b**2 - a*b**4)/(si n(c + d*x)*d*(cos(c + d*x)*a**7 - 3*cos(c + d*x)*a**5*b**2 + 3*cos(c + d*x )*a**3*b**4 - cos(c + d*x)*a*b**6 + a**6*b - 3*a**4*b**3 + 3*a**2*b**5 - b **7))