Integrand size = 23, antiderivative size = 86 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{(a+b)^2 f}-\frac {(a-b) \text {arctanh}(\cos (e+f x))}{2 (a+b)^2 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 (a+b) f} \] Output:
a^(1/2)*b^(1/2)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/(a+b)^2/f-1/2*(a-b)*arc tanh(cos(f*x+e))/(a+b)^2/f-1/2*cot(f*x+e)*csc(f*x+e)/(a+b)/f
Result contains complex when optimal does not.
Time = 1.79 (sec) , antiderivative size = 371, normalized size of antiderivative = 4.31 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (-8 \sqrt {a} \sqrt {b} \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )-8 \sqrt {a} \sqrt {b} \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+a \csc ^2\left (\frac {1}{2} (e+f x)\right )+b \csc ^2\left (\frac {1}{2} (e+f x)\right )+4 a \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-4 b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-4 a \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-a \sec ^2\left (\frac {1}{2} (e+f x)\right )-b \sec ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sec ^2(e+f x)}{16 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )} \] Input:
Integrate[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
-1/16*((a + 2*b + a*Cos[2*(e + f*x)])*(-8*Sqrt[a]*Sqrt[b]*ArcTan[((-Sqrt[a ] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e ]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b ]] - 8*Sqrt[a]*Sqrt[b]*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I* Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[ e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] + a*Csc[(e + f*x)/2]^2 + b*Csc[( e + f*x)/2]^2 + 4*a*Log[Cos[(e + f*x)/2]] - 4*b*Log[Cos[(e + f*x)/2]] - 4* a*Log[Sin[(e + f*x)/2]] + 4*b*Log[Sin[(e + f*x)/2]] - a*Sec[(e + f*x)/2]^2 - b*Sec[(e + f*x)/2]^2)*Sec[e + f*x]^2)/((a + b)^2*f*(a + b*Sec[e + f*x]^ 2))
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4621, 373, 397, 218, 219}
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.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \sec (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle -\frac {\int \frac {\cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle -\frac {\frac {\cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right )}-\frac {\int \frac {b-a \cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {\cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right )}-\frac {\frac {2 a b \int \frac {1}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{a+b}-\frac {(a-b) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {\cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right )}-\frac {\frac {2 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a+b}-\frac {(a-b) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right )}-\frac {\frac {2 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a+b}-\frac {(a-b) \text {arctanh}(\cos (e+f x))}{a+b}}{2 (a+b)}}{f}\) |
Input:
Int[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
-((-1/2*((2*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(a + b ) - ((a - b)*ArcTanh[Cos[e + f*x]])/(a + b))/(a + b) + Cos[e + f*x]/(2*(a + b)*(1 - Cos[e + f*x]^2)))/f)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a +b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a -b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{2}}+\frac {a b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a +b \right )^{2} \sqrt {a b}}}{f}\) | \(115\) |
| default | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a +b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a -b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{2}}+\frac {a b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a +b \right )^{2} \sqrt {a b}}}{f}\) | \(115\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}}{f \left (a +b \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{2 f \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 f \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{2 f \left (a^{2}+2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 f \left (a^{2}+2 a b +b^{2}\right )}-\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{2 \left (a +b \right )^{2} f}+\frac {i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{2 \left (a +b \right )^{2} f}\) | \(263\) |
Input:
int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/(4*a+4*b)/(1+cos(f*x+e))+1/4/(a+b)^2*(-a+b)*ln(1+cos(f*x+e))+1/(4*a +4*b)/(-1+cos(f*x+e))+1/4*(a-b)/(a+b)^2*ln(-1+cos(f*x+e))+a*b/(a+b)^2/(a*b )^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2)))
Time = 0.12 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.80 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {2 \, \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a b} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (a + b\right )} \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}, \frac {4 \, \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {\sqrt {a b} \cos \left (f x + e\right )}{b}\right ) + 2 \, {\left (a + b\right )} \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}\right ] \] Input:
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[1/4*(2*sqrt(-a*b)*(cos(f*x + e)^2 - 1)*log(-(a*cos(f*x + e)^2 + 2*sqrt(-a *b)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 2*(a + b)*cos(f*x + e) - ( (a - b)*cos(f*x + e)^2 - a + b)*log(1/2*cos(f*x + e) + 1/2) + ((a - b)*cos (f*x + e)^2 - a + b)*log(-1/2*cos(f*x + e) + 1/2))/((a^2 + 2*a*b + b^2)*f* cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f), 1/4*(4*sqrt(a*b)*(cos(f*x + e)^2 - 1)*arctan(sqrt(a*b)*cos(f*x + e)/b) + 2*(a + b)*cos(f*x + e) - ((a - b)* cos(f*x + e)^2 - a + b)*log(1/2*cos(f*x + e) + 1/2) + ((a - b)*cos(f*x + e )^2 - a + b)*log(-1/2*cos(f*x + e) + 1/2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f)]
\[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(csc(f*x+e)**3/(a+b*sec(f*x+e)**2),x)
Output:
Integral(csc(e + f*x)**3/(a + b*sec(e + f*x)**2), x)
Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {4 \, a b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} - \frac {{\left (a - b\right )} \log \left (\cos \left (f x + e\right ) + 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {{\left (a - b\right )} \log \left (\cos \left (f x + e\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, \cos \left (f x + e\right )}{{\left (a + b\right )} \cos \left (f x + e\right )^{2} - a - b}}{4 \, f} \] Input:
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/4*(4*a*b*arctan(a*cos(f*x + e)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b) ) - (a - b)*log(cos(f*x + e) + 1)/(a^2 + 2*a*b + b^2) + (a - b)*log(cos(f* x + e) - 1)/(a^2 + 2*a*b + b^2) + 2*cos(f*x + e)/((a + b)*cos(f*x + e)^2 - a - b))/f
Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.63 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{2} f + 2 \, a b f + b^{2} f\right )} \sqrt {a b}} - \frac {{\left (a - b\right )} \log \left ({\left | \cos \left (f x + e\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} f + 2 \, a b f + b^{2} f\right )}} + \frac {{\left (a - b\right )} \log \left ({\left | \cos \left (f x + e\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} f + 2 \, a b f + b^{2} f\right )}} + \frac {\cos \left (f x + e\right )}{2 \, {\left (a f + b f\right )} {\left (\cos \left (f x + e\right )^{2} - 1\right )}} \] Input:
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
a*b*arctan(a*cos(f*x + e)/sqrt(a*b))/((a^2*f + 2*a*b*f + b^2*f)*sqrt(a*b)) - 1/4*(a - b)*log(abs(cos(f*x + e) + 1))/(a^2*f + 2*a*b*f + b^2*f) + 1/4* (a - b)*log(abs(cos(f*x + e) - 1))/(a^2*f + 2*a*b*f + b^2*f) + 1/2*cos(f*x + e)/((a*f + b*f)*(cos(f*x + e)^2 - 1))
Time = 12.99 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.56 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {2\,a\,\cos \left (e+f\,x\right )+2\,b\,\cos \left (e+f\,x\right )-a\,\ln \left (\cos \left (e+f\,x\right )-1\right )+a\,\ln \left (\cos \left (e+f\,x\right )+1\right )+b\,\ln \left (\cos \left (e+f\,x\right )-1\right )-b\,\ln \left (\cos \left (e+f\,x\right )+1\right )+a\,\ln \left (\cos \left (e+f\,x\right )-1\right )\,{\cos \left (e+f\,x\right )}^2-a\,\ln \left (\cos \left (e+f\,x\right )+1\right )\,{\cos \left (e+f\,x\right )}^2-b\,\ln \left (\cos \left (e+f\,x\right )-1\right )\,{\cos \left (e+f\,x\right )}^2+b\,\ln \left (\cos \left (e+f\,x\right )+1\right )\,{\cos \left (e+f\,x\right )}^2-\mathrm {atan}\left (\frac {a^3\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}+a\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}+a^2\,b\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,2{}\mathrm {i}}{a^3\,b+2\,a^2\,b^2+a\,b^3}\right )\,\sqrt {-a\,b}\,4{}\mathrm {i}+{\cos \left (e+f\,x\right )}^2\,\mathrm {atan}\left (\frac {a^3\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}+a\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}+a^2\,b\,\cos \left (e+f\,x\right )\,\sqrt {-a\,b}\,2{}\mathrm {i}}{a^3\,b+2\,a^2\,b^2+a\,b^3}\right )\,\sqrt {-a\,b}\,4{}\mathrm {i}}{-4\,f\,a^2\,{\cos \left (e+f\,x\right )}^2+4\,f\,a^2-8\,f\,a\,b\,{\cos \left (e+f\,x\right )}^2+8\,f\,a\,b-4\,f\,b^2\,{\cos \left (e+f\,x\right )}^2+4\,f\,b^2} \] Input:
int(1/(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)),x)
Output:
-(2*a*cos(e + f*x) + 2*b*cos(e + f*x) - atan((a^3*cos(e + f*x)*(-a*b)^(1/2 )*1i + a*b^2*cos(e + f*x)*(-a*b)^(1/2)*1i + a^2*b*cos(e + f*x)*(-a*b)^(1/2 )*2i)/(a*b^3 + a^3*b + 2*a^2*b^2))*(-a*b)^(1/2)*4i - a*log(cos(e + f*x) - 1) + a*log(cos(e + f*x) + 1) + b*log(cos(e + f*x) - 1) - b*log(cos(e + f*x ) + 1) + cos(e + f*x)^2*atan((a^3*cos(e + f*x)*(-a*b)^(1/2)*1i + a*b^2*cos (e + f*x)*(-a*b)^(1/2)*1i + a^2*b*cos(e + f*x)*(-a*b)^(1/2)*2i)/(a*b^3 + a ^3*b + 2*a^2*b^2))*(-a*b)^(1/2)*4i + a*log(cos(e + f*x) - 1)*cos(e + f*x)^ 2 - a*log(cos(e + f*x) + 1)*cos(e + f*x)^2 - b*log(cos(e + f*x) - 1)*cos(e + f*x)^2 + b*log(cos(e + f*x) + 1)*cos(e + f*x)^2)/(4*a^2*f + 4*b^2*f - 4 *a^2*f*cos(e + f*x)^2 - 4*b^2*f*cos(e + f*x)^2 + 8*a*b*f - 8*a*b*f*cos(e + f*x)^2)
Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.88 \[ \int \frac {\csc ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2}+2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2}-\cos \left (f x +e \right ) a -\cos \left (f x +e \right ) b +\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a -\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b}{2 \sin \left (f x +e \right )^{2} f \left (a^{2}+2 a b +b^{2}\right )} \] Input:
int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2),x)
Output:
( - 2*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b ))*sin(e + f*x)**2 + 2*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2 - cos(e + f*x)*a - cos(e + f*x)*b + lo g(tan((e + f*x)/2))*sin(e + f*x)**2*a - log(tan((e + f*x)/2))*sin(e + f*x) **2*b)/(2*sin(e + f*x)**2*f*(a**2 + 2*a*b + b**2))