Integrand size = 21, antiderivative size = 116 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{7/2} f}-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )} \]
-15/8*cos(f*x+e)/a^3/f+1/4*cos(f*x+e)^5/a/f/(b+a*cos(f*x+e)^2)^2+5/8*cos(f *x+e)^3/a^2/f/(b+a*cos(f*x+e)^2)+15/8*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b ^(1/2)/a^(7/2)/f
Result contains complex when optimal does not.
Time = 5.52 (sec) , antiderivative size = 656, normalized size of antiderivative = 5.66 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (15 \left (a^3+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (a^3+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+\frac {\sqrt {a} \left (24 a^4 \sqrt {b} \cos (e+f x)-24 a^3 b^{3/2} \cos (e+f x)-144 a^2 b^{5/2} \cos (e+f x)+512 b^{9/2} \cos (e+f x)-72 a^3 b^{3/2} \cos (e+f x) \cos (2 (e+f x))-24 a^3 \sqrt {b} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))+72 a^2 b^{3/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-1152 b^{7/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-15 a^{5/2} \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-15 a^{5/2} \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-512 b^{5/2} \cos (e) \cos (f x) (a+2 b+a \cos (2 (e+f x)))^2+512 b^{5/2} (a+2 b+a \cos (2 (e+f x)))^2 \sin (e) \sin (f x)+6 a^4 \sqrt {b} \csc (e+f x) \sin (4 (e+f x))\right )}{(a+2 b+a \cos (2 (e+f x)))^2}\right )}{4096 a^{7/2} b^{5/2} f \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(15*(a^3 + 64*b^3)*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]] + 15*(a^3 + 64*b^3)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(C os[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]] + (Sqrt[a]*(24*a^4*Sqr t[b]*Cos[e + f*x] - 24*a^3*b^(3/2)*Cos[e + f*x] - 144*a^2*b^(5/2)*Cos[e + f*x] + 512*b^(9/2)*Cos[e + f*x] - 72*a^3*b^(3/2)*Cos[e + f*x]*Cos[2*(e + f *x)] - 24*a^3*Sqrt[b]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 72*a^2 *b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 1152*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 15*a^(5/2)*ArcTan[(Sqrt[a] - Sqrt[ a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 15*a^ (5/2)*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 512*b^(5/2)*Cos[e]*Cos[f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 512*b^(5/2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sin[e]*Sin[f*x] + 6*a^4*Sqrt[b]*Csc[e + f*x]*Sin[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f* x)])^2))/(4096*a^(7/2)*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4621, 252, 252, 262, 218}
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.
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(e+f x)}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {5 \int \frac {\cos ^4(e+f x)}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {\frac {5 \left (\frac {3 \left (\frac {\cos (e+f x)}{a}-\frac {b \int \frac {1}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{a}\right )}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {5 \left (\frac {3 \left (\frac {\cos (e+f x)}{a}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
-((-1/4*Cos[e + f*x]^5/(a*(b + a*Cos[e + f*x]^2)^2) + (5*((3*(-((Sqrt[b]*A rcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/a^(3/2)) + Cos[e + f*x]/a))/(2*a) - Cos[e + f*x]^3/(2*a*(b + a*Cos[e + f*x]^2))))/(4*a))/f)
3.1.56.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, 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 = 2.95 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {7 b \sec \left (f x +e \right )^{3}}{8}+\frac {9 \sec \left (f x +e \right ) a}{8}}{\left (a +b \sec \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {\sec \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {1}{a^{3} \sec \left (f x +e \right )}}{f}\) | \(83\) |
| default | \(\frac {-\frac {b \left (\frac {\frac {7 b \sec \left (f x +e \right )^{3}}{8}+\frac {9 \sec \left (f x +e \right ) a}{8}}{\left (a +b \sec \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {\sec \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {1}{a^{3} \sec \left (f x +e \right )}}{f}\) | \(83\) |
| risch | \(-\frac {{\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}-\frac {{\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {b \left (9 a \,{\mathrm e}^{7 i \left (f x +e \right )}+27 a \,{\mathrm e}^{5 i \left (f x +e \right )}+28 b \,{\mathrm e}^{5 i \left (f x +e \right )}+27 a \,{\mathrm e}^{3 i \left (f x +e \right )}+28 b \,{\mathrm e}^{3 i \left (f x +e \right )}+9 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} f}-\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}+\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}\) | \(249\) |
1/f*(-1/a^3*b*((7/8*b*sec(f*x+e)^3+9/8*sec(f*x+e)*a)/(a+b*sec(f*x+e)^2)^2+ 15/8/(a*b)^(1/2)*arctan(sec(f*x+e)*b/(a*b)^(1/2)))-1/a^3/sec(f*x+e))
Time = 0.29 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.58 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {16 \, a^{2} \cos \left (f x + e\right )^{5} + 50 \, a b \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right )}{16 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {8 \, a^{2} \cos \left (f x + e\right )^{5} + 25 \, a b \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \]
[-1/16*(16*a^2*cos(f*x + e)^5 + 50*a*b*cos(f*x + e)^3 + 30*b^2*cos(f*x + e ) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-b/a)*log(-( a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b) ))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), -1/8*(8* a^2*cos(f*x + e)^5 + 25*a*b*cos(f*x + e)^3 + 15*b^2*cos(f*x + e) - 15*(a^2 *cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(b/a)*arctan(a*sqrt(b/a) *cos(f*x + e)/b))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b ^2*f)]
Timed out. \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {9 \, a b \cos \left (f x + e\right )^{3} + 7 \, b^{2} \cos \left (f x + e\right )}{a^{5} \cos \left (f x + e\right )^{4} + 2 \, a^{4} b \cos \left (f x + e\right )^{2} + a^{3} b^{2}} - \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {8 \, \cos \left (f x + e\right )}{a^{3}}}{8 \, f} \]
-1/8*((9*a*b*cos(f*x + e)^3 + 7*b^2*cos(f*x + e))/(a^5*cos(f*x + e)^4 + 2* a^4*b*cos(f*x + e)^2 + a^3*b^2) - 15*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(s qrt(a*b)*a^3) + 8*cos(f*x + e)/a^3)/f
Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} f} - \frac {\cos \left (f x + e\right )}{a^{3} f} - \frac {\frac {9 \, a b \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{3}} \]
15/8*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - cos(f*x + e)/( a^3*f) - 1/8*(9*a*b*cos(f*x + e)^3/f + 7*b^2*cos(f*x + e)/f)/((a*cos(f*x + e)^2 + b)^2*a^3)
Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\cos \left (e+f\,x\right )}{\sqrt {b}}\right )}{8\,a^{7/2}\,f}-\frac {\frac {7\,b^2\,\cos \left (e+f\,x\right )}{8}+\frac {9\,a\,b\,{\cos \left (e+f\,x\right )}^3}{8}}{f\,\left (a^5\,{\cos \left (e+f\,x\right )}^4+2\,a^4\,b\,{\cos \left (e+f\,x\right )}^2+a^3\,b^2\right )}-\frac {\cos \left (e+f\,x\right )}{a^3\,f} \]