Integrand size = 23, antiderivative size = 114 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{7/2} f}-\frac {(a+2 b) \cos (e+f x)}{a^3 f}+\frac {\cos ^3(e+f x)}{3 a^2 f}-\frac {b (a+b) \cos (e+f x)}{2 a^3 f \left (b+a \cos ^2(e+f x)\right )} \]
-(a+2*b)*cos(f*x+e)/a^3/f+1/3*cos(f*x+e)^3/a^2/f-1/2*b*(a+b)*cos(f*x+e)/a^ 3/f/(b+a*cos(f*x+e)^2)+1/2*(3*a+5*b)*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 = 2.91 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.54 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}+\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {9 a^3 \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {9 a^3 \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {32 \sqrt {a} \cos (e+f x) \left (9 a^2+56 a b+60 b^2+4 a (2 a+5 b) \cos (2 (e+f x))-a^2 \cos (4 (e+f x))\right )}{a+2 b+a \cos (2 (e+f x))}}{384 a^{7/2} f} \]
((3*(3*a^3 + 192*a*b^2 + 320*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]])/b^(3/2) + (3*(3*a^3 + 192*a*b^2 + 320*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]])/b^(3/2) - (9*a^3*ArcTan[(Sqrt[a ] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (9*a^3*ArcTan[(Sqrt[ a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (32*Sqrt[a]*Cos[e + f*x]*(9*a^2 + 56*a*b + 60*b^2 + 4*a*(2*a + 5*b)*Cos[2*(e + f*x)] - a^2*Co s[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)]))/(384*a^(7/2)*f)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4621, 360, 1467, 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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\cos ^2(e+f x)\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \frac {2 a^2 \cos ^4(e+f x)-2 a (a+b) \cos ^2(e+f x)+b (a+b)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^3}}{f}\) |
\(\Big \downarrow \) 1467 |
\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \left (2 a \cos ^2(e+f x)-2 (a+2 b)+\frac {5 b^2+3 a b}{a \cos ^2(e+f x)+b}\right )d\cos (e+f x)}{2 a^3}}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{\sqrt {a}}-2 (a+2 b) \cos (e+f x)+\frac {2}{3} a \cos ^3(e+f x)}{2 a^3}}{f}\) |
-(((b*(a + b)*Cos[e + f*x])/(2*a^3*(b + a*Cos[e + f*x]^2)) - ((Sqrt[b]*(3* a + 5*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/Sqrt[a] - 2*(a + 2*b)*Cos [e + f*x] + (2*a*Cos[e + f*x]^3)/3)/(2*a^3))/f)
3.1.42.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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 = 2.46 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 b \cos \left (f x +e \right )}{a^{3}}+\frac {b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\) | \(102\) |
default | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 b \cos \left (f x +e \right )}{a^{3}}+\frac {b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\) | \(102\) |
risch | \(\frac {{\mathrm e}^{3 i \left (f x +e \right )}}{24 a^{2} f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )}}{8 f \,a^{2}}-\frac {{\mathrm e}^{i \left (f x +e \right )} b}{f \,a^{3}}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )}}{8 f \,a^{2}}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b}{f \,a^{3}}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )}}{24 a^{2} f}-\frac {\left (a +b \right ) b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}+\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 a^{4} f}-\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}-\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 a^{4} f}\) | \(362\) |
1/f*(1/a^3*(1/3*a*cos(f*x+e)^3-cos(f*x+e)*a-2*b*cos(f*x+e))+b/a^3*((-1/2*a -1/2*b)*cos(f*x+e)/(b+a*cos(f*x+e)^2)+1/2*(3*a+5*b)/(a*b)^(1/2)*arctan(a*c os(f*x+e)/(a*b)^(1/2))))
Time = 0.28 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.61 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac {2 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) - 3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \]
[1/12*(4*a^2*cos(f*x + e)^5 - 4*(3*a^2 + 5*a*b)*cos(f*x + e)^3 + 3*((3*a^2 + 5*a*b)*cos(f*x + e)^2 + 3*a*b + 5*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)) - 6*(3*a*b + 5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f), 1/6*(2*a^2*cos(f*x + e)^5 - 2*(3*a^2 + 5*a*b)*cos(f*x + e)^3 + 3*((3*a^2 + 5*a*b)*cos(f*x + e )^2 + 3*a*b + 5*b^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) - 3*(3*a *b + 5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{4} \cos \left (f x + e\right )^{2} + a^{3} b} - \frac {3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (a \cos \left (f x + e\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \cos \left (f x + e\right )\right )}}{a^{3}}}{6 \, f} \]
-1/6*(3*(a*b + b^2)*cos(f*x + e)/(a^4*cos(f*x + e)^2 + a^3*b) - 3*(3*a*b + 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2*(a*cos(f*x + e)^3 - 3*(a + 2*b)*cos(f*x + e))/a^3)/f
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} f} - \frac {\frac {a b \cos \left (f x + e\right )}{f} + \frac {b^{2} \cos \left (f x + e\right )}{f}}{2 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} a^{3}} + \frac {a^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) - 6 \, a^{3} b f^{11} \cos \left (f x + e\right )}{3 \, a^{6} f^{12}} \]
1/2*(3*a*b + 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - 1 /2*(a*b*cos(f*x + e)/f + b^2*cos(f*x + e)/f)/((a*cos(f*x + e)^2 + b)*a^3) + 1/3*(a^4*f^11*cos(f*x + e)^3 - 3*a^4*f^11*cos(f*x + e) - 6*a^3*b*f^11*co s(f*x + e))/(a^6*f^12)
Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.14 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\cos \left (e+f\,x\right )}^3}{3\,a^2\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {2\,b}{a^3}+\frac {1}{a^2}\right )}{f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {b^2}{2}+\frac {a\,b}{2}\right )}{f\,\left (a^4\,{\cos \left (e+f\,x\right )}^2+b\,a^3\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (3\,a+5\,b\right )}{5\,b^2+3\,a\,b}\right )\,\left (3\,a+5\,b\right )}{2\,a^{7/2}\,f} \]