Integrand size = 23, antiderivative size = 107 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}+\frac {2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}-\frac {(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac {2 (a-2 b) b \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]
-(a^2-6*a*b+6*b^2)*cos(f*x+e)/f+2/3*(a-2*b)*(a-b)*cos(f*x+e)^3/f-1/5*(a-b) ^2*cos(f*x+e)^5/f+2*(a-2*b)*b*sec(f*x+e)/f+1/3*b^2*sec(f*x+e)^3/f
Time = 1.50 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.91 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {-30 \left (5 a^2-38 a b+41 b^2\right ) \cos (e+f x)+5 (5 a-13 b) (a-b) \cos (3 (e+f x))-3 (a-b)^2 \cos (5 (e+f x))+480 (a-2 b) b \sec (e+f x)+80 b^2 \sec ^3(e+f x)}{240 f} \]
(-30*(5*a^2 - 38*a*b + 41*b^2)*Cos[e + f*x] + 5*(5*a - 13*b)*(a - b)*Cos[3 *(e + f*x)] - 3*(a - b)^2*Cos[5*(e + f*x)] + 480*(a - 2*b)*b*Sec[e + f*x] + 80*b^2*Sec[e + f*x]^3)/(240*f)
Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4147, 355, 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.
\(\displaystyle \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^2dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \frac {\int \cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^2d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {\int \left ((a-b)^2 \cos ^6(e+f x)+2 (a-2 b) (b-a) \cos ^4(e+f x)+\left (a^2-6 b a+6 b^2\right ) \cos ^2(e+f x)+b^2 \sec ^2(e+f x)+2 (a-2 b) b\right )d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)-\frac {1}{5} (a-b)^2 \cos ^5(e+f x)+\frac {2}{3} (a-2 b) (a-b) \cos ^3(e+f x)+2 b (a-2 b) \sec (e+f x)+\frac {1}{3} b^2 \sec ^3(e+f x)}{f}\) |
(-((a^2 - 6*a*b + 6*b^2)*Cos[e + f*x]) + (2*(a - 2*b)*(a - b)*Cos[e + f*x] ^3)/3 - ((a - b)^2*Cos[e + f*x]^5)/5 + 2*(a - 2*b)*b*Sec[e + f*x] + (b^2*S ec[e + f*x]^3)/3)/f
3.1.43.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 2.63 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.73
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (\frac {\sin \left (f x +e \right )^{8}}{\cos \left (f x +e \right )}+\left (\frac {16}{5}+\sin \left (f x +e \right )^{6}+\frac {6 \sin \left (f x +e \right )^{4}}{5}+\frac {8 \sin \left (f x +e \right )^{2}}{5}\right ) \cos \left (f x +e \right )\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{10}}{3 \cos \left (f x +e \right )^{3}}-\frac {7 \sin \left (f x +e \right )^{10}}{3 \cos \left (f x +e \right )}-\frac {7 \left (\frac {128}{35}+\sin \left (f x +e \right )^{8}+\frac {8 \sin \left (f x +e \right )^{6}}{7}+\frac {48 \sin \left (f x +e \right )^{4}}{35}+\frac {64 \sin \left (f x +e \right )^{2}}{35}\right ) \cos \left (f x +e \right )}{3}\right )}{f}\) | \(185\) |
default | \(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (\frac {\sin \left (f x +e \right )^{8}}{\cos \left (f x +e \right )}+\left (\frac {16}{5}+\sin \left (f x +e \right )^{6}+\frac {6 \sin \left (f x +e \right )^{4}}{5}+\frac {8 \sin \left (f x +e \right )^{2}}{5}\right ) \cos \left (f x +e \right )\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{10}}{3 \cos \left (f x +e \right )^{3}}-\frac {7 \sin \left (f x +e \right )^{10}}{3 \cos \left (f x +e \right )}-\frac {7 \left (\frac {128}{35}+\sin \left (f x +e \right )^{8}+\frac {8 \sin \left (f x +e \right )^{6}}{7}+\frac {48 \sin \left (f x +e \right )^{4}}{35}+\frac {64 \sin \left (f x +e \right )^{2}}{35}\right ) \cos \left (f x +e \right )}{3}\right )}{f}\) | \(185\) |
risch | \(-\frac {{\mathrm e}^{5 i \left (f x +e \right )} a^{2}}{160 f}+\frac {{\mathrm e}^{5 i \left (f x +e \right )} a b}{80 f}-\frac {{\mathrm e}^{5 i \left (f x +e \right )} b^{2}}{160 f}+\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )} a^{2}}{96 f}-\frac {3 \,{\mathrm e}^{3 i \left (f x +e \right )} a b}{16 f}+\frac {13 \,{\mathrm e}^{3 i \left (f x +e \right )} b^{2}}{96 f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a^{2}}{16 f}+\frac {19 \,{\mathrm e}^{i \left (f x +e \right )} a b}{8 f}-\frac {41 \,{\mathrm e}^{i \left (f x +e \right )} b^{2}}{16 f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{16 f}+\frac {19 \,{\mathrm e}^{-i \left (f x +e \right )} a b}{8 f}-\frac {41 \,{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{16 f}+\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )} a^{2}}{96 f}-\frac {3 \,{\mathrm e}^{-3 i \left (f x +e \right )} a b}{16 f}+\frac {13 \,{\mathrm e}^{-3 i \left (f x +e \right )} b^{2}}{96 f}-\frac {{\mathrm e}^{-5 i \left (f x +e \right )} a^{2}}{160 f}+\frac {{\mathrm e}^{-5 i \left (f x +e \right )} a b}{80 f}-\frac {{\mathrm e}^{-5 i \left (f x +e \right )} b^{2}}{160 f}-\frac {4 b \,{\mathrm e}^{i \left (f x +e \right )} \left (-3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+6 b \,{\mathrm e}^{4 i \left (f x +e \right )}-6 a \,{\mathrm e}^{2 i \left (f x +e \right )}+10 b \,{\mathrm e}^{2 i \left (f x +e \right )}-3 a +6 b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\) | \(425\) |
1/f*(-1/5*a^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*a*b*(sin(f* x+e)^8/cos(f*x+e)+(16/5+sin(f*x+e)^6+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*co s(f*x+e))+b^2*(1/3*sin(f*x+e)^10/cos(f*x+e)^3-7/3*sin(f*x+e)^10/cos(f*x+e) -7/3*(128/35+sin(f*x+e)^8+8/7*sin(f*x+e)^6+48/35*sin(f*x+e)^4+64/35*sin(f* x+e)^2)*cos(f*x+e)))
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} - 10 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 15 \, {\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 30 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 5 \, b^{2}}{15 \, f \cos \left (f x + e\right )^{3}} \]
-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^8 - 10*(a^2 - 3*a*b + 2*b^2)*cos (f*x + e)^6 + 15*(a^2 - 6*a*b + 6*b^2)*cos(f*x + e)^4 - 30*(a*b - 2*b^2)*c os(f*x + e)^2 - 5*b^2)/(f*cos(f*x + e)^3)
\[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2} \sin ^{5}{\left (e + f x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right ) - \frac {5 \, {\left (6 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{15 \, f} \]
-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 10*(a^2 - 3*a*b + 2*b^2)*cos (f*x + e)^3 + 15*(a^2 - 6*a*b + 6*b^2)*cos(f*x + e) - 5*(6*(a*b - 2*b^2)*c os(f*x + e)^2 + b^2)/cos(f*x + e)^3)/f
Timed out. \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]
Time = 11.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.71 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {2\,a^2\,{\cos \left (e+f\,x\right )}^3}{3\,f}-\frac {6\,b^2\,\cos \left (e+f\,x\right )}{f}-\frac {a^2\,\cos \left (e+f\,x\right )}{f}-\frac {a^2\,{\cos \left (e+f\,x\right )}^5}{5\,f}-\frac {4\,b^2}{f\,\cos \left (e+f\,x\right )}+\frac {b^2}{3\,f\,{\cos \left (e+f\,x\right )}^3}+\frac {4\,b^2\,{\cos \left (e+f\,x\right )}^3}{3\,f}-\frac {b^2\,{\cos \left (e+f\,x\right )}^5}{5\,f}+\frac {6\,a\,b\,\cos \left (e+f\,x\right )}{f}+\frac {2\,a\,b}{f\,\cos \left (e+f\,x\right )}-\frac {2\,a\,b\,{\cos \left (e+f\,x\right )}^3}{f}+\frac {2\,a\,b\,{\cos \left (e+f\,x\right )}^5}{5\,f} \]
(2*a^2*cos(e + f*x)^3)/(3*f) - (6*b^2*cos(e + f*x))/f - (a^2*cos(e + f*x)) /f - (a^2*cos(e + f*x)^5)/(5*f) - (4*b^2)/(f*cos(e + f*x)) + b^2/(3*f*cos( e + f*x)^3) + (4*b^2*cos(e + f*x)^3)/(3*f) - (b^2*cos(e + f*x)^5)/(5*f) + (6*a*b*cos(e + f*x))/f + (2*a*b)/(f*cos(e + f*x)) - (2*a*b*cos(e + f*x)^3) /f + (2*a*b*cos(e + f*x)^5)/(5*f)