Integrand size = 27, antiderivative size = 82 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 a x}{2}+\frac {2 b \cos (c+d x)}{d}-\frac {b \cos ^3(c+d x)}{3 d}+\frac {b \sec (c+d x)}{d}+\frac {3 a \tan (c+d x)}{2 d}-\frac {a \sin ^2(c+d x) \tan (c+d x)}{2 d} \]
-3/2*a*x+2*b*cos(d*x+c)/d-1/3*b*cos(d*x+c)^3/d+b*sec(d*x+c)/d+3/2*a*tan(d* x+c)/d-1/2*a*sin(d*x+c)^2*tan(d*x+c)/d
Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 a (c+d x)}{2 d}+\frac {7 b \cos (c+d x)}{4 d}-\frac {b \cos (3 (c+d x))}{12 d}+\frac {b \sec (c+d x)}{d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \tan (c+d x)}{d} \]
(-3*a*(c + d*x))/(2*d) + (7*b*Cos[c + d*x])/(4*d) - (b*Cos[3*(c + d*x)])/( 12*d) + (b*Sec[c + d*x])/d + (a*Sin[2*(c + d*x)])/(4*d) + (a*Tan[c + d*x]) /d
Time = 0.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3317, 3042, 3070, 244, 2009, 3071, 252, 262, 216}
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.
| \(\displaystyle \int \sin ^2(c+d x) \tan ^2(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a+b \sin (c+d x))}{\cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \sin ^2(c+d x) \tan ^2(c+d x)dx+b \int \sin ^3(c+d x) \tan ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^2dx+b \int \sin (c+d x)^3 \tan (c+d x)^2dx\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {b \int \left (1-\cos ^2(c+d x)\right )^2 \sec ^2(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {b \int \left (\cos ^2(c+d x)+\sec ^2(c+d x)-2\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {b \left (\frac {1}{3} \cos ^3(c+d x)-2 \cos (c+d x)-\sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle \frac {a \int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{d}-\frac {b \left (\frac {1}{3} \cos ^3(c+d x)-2 \cos (c+d x)-\sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {a \left (\frac {3}{2} \int \frac {\tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {b \left (\frac {1}{3} \cos ^3(c+d x)-2 \cos (c+d x)-\sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {a \left (\frac {3}{2} \left (\tan (c+d x)-\int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {b \left (\frac {1}{3} \cos ^3(c+d x)-2 \cos (c+d x)-\sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {a \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {b \left (\frac {1}{3} \cos ^3(c+d x)-2 \cos (c+d x)-\sec (c+d x)\right )}{d}\) |
-((b*(-2*Cos[c + d*x] + Cos[c + d*x]^3/3 - Sec[c + d*x]))/d) + (a*((3*(-Ar cTan[Tan[c + d*x]] + Tan[c + d*x]))/2 - Tan[c + d*x]^3/(2*(1 + Tan[c + d*x ]^2))))/d
3.15.43.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m + n - 1)/2)/x^n, x], x, Cos[e + f *x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Time = 0.68 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\frac {20 b \cos \left (2 d x +2 c \right )-\cos \left (4 d x +4 c \right ) b +3 a \sin \left (3 d x +3 c \right )+\left (-36 a x d +64 b \right ) \cos \left (d x +c \right )+27 a \sin \left (d x +c \right )+45 b}{24 d \cos \left (d x +c \right )}\) | \(79\) |
| derivativedivides | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(104\) |
| default | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(104\) |
| risch | \(-\frac {3 a x}{2}-\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {7 b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {7 b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a +2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b \cos \left (3 d x +3 c \right )}{12 d}\) | \(117\) |
| norman | \(\frac {\frac {3 a x}{2}-\frac {16 b}{3 d}-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+3 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {32 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(171\) |
1/24*(20*b*cos(2*d*x+2*c)-cos(4*d*x+4*c)*b+3*a*sin(3*d*x+3*c)+(-36*a*d*x+6 4*b)*cos(d*x+c)+27*a*sin(d*x+c)+45*b)/d/cos(d*x+c)
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {2 \, b \cos \left (d x + c\right )^{4} + 9 \, a d x \cos \left (d x + c\right ) - 12 \, b \cos \left (d x + c\right )^{2} - 3 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 6 \, b}{6 \, d \cos \left (d x + c\right )} \]
-1/6*(2*b*cos(d*x + c)^4 + 9*a*d*x*cos(d*x + c) - 12*b*cos(d*x + c)^2 - 3* (a*cos(d*x + c)^2 + 2*a)*sin(d*x + c) - 6*b)/(d*cos(d*x + c))
\[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b}{6 \, d} \]
-1/6*(3*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c)) *a + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b)/d
Time = 0.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {9 \, {\left (d x + c\right )} a + \frac {12 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
-1/6*(9*(d*x + c)*a + 12*(a*tan(1/2*d*x + 1/2*c) + b)/(tan(1/2*d*x + 1/2*c )^2 - 1) + 2*(3*a*tan(1/2*d*x + 1/2*c)^5 - 6*b*tan(1/2*d*x + 1/2*c)^4 - 24 *b*tan(1/2*d*x + 1/2*c)^2 - 3*a*tan(1/2*d*x + 1/2*c) - 10*b)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 18.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3\,a\,x}{2}-\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {32\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {16\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]