Integrand size = 25, antiderivative size = 65 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 b x}{2}+\frac {a \cos (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {3 b \tan (c+d x)}{2 d}-\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 d} \]
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 b (c+d x)}{2 d}+\frac {a \cos (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \tan (c+d x)}{d} \]
(-3*b*(c + d*x))/(2*d) + (a*Cos[c + d*x])/d + (a*Sec[c + d*x])/d + (b*Sin[ 2*(c + d*x)])/(4*d) + (b*Tan[c + d*x])/d
Time = 0.40 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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.
\(\displaystyle \int \sin (c+d x) \tan ^2(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 (a+b \sin (c+d x))}{\cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \sin (c+d x) \tan ^2(c+d x)dx+b \int \sin ^2(c+d x) \tan ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin (c+d x) \tan (c+d x)^2dx+b \int \sin (c+d x)^2 \tan (c+d x)^2dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle b \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {a \int \left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle b \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {a \int \left (\sec ^2(c+d x)-1\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b \int \sin (c+d x)^2 \tan (c+d x)^2dx-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d}\) |
\(\Big \downarrow \) 3071 |
\(\displaystyle \frac {b \int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{d}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {b \left (\frac {3}{2} \int \frac {\tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {b \left (\frac {3}{2} \left (\tan (c+d x)-\int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d}\) |
-((a*(-Cos[c + d*x] - Sec[c + d*x]))/d) + (b*((3*(-ArcTan[Tan[c + d*x]] + Tan[c + d*x]))/2 - Tan[c + d*x]^3/(2*(1 + Tan[c + d*x]^2))))/d
3.15.44.3.1 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 = 0.67 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02
method | result | size |
parallelrisch | \(\frac {4 a \cos \left (2 d x +2 c \right )+b \sin \left (3 d x +3 c \right )+\left (-12 b x d +16 a \right ) \cos \left (d x +c \right )+9 b \sin \left (d x +c \right )+12 a}{8 d \cos \left (d x +c \right )}\) | \(66\) |
derivativedivides | \(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(94\) |
default | \(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(94\) |
risch | \(-\frac {3 b x}{2}-\frac {i b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {i b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\) | \(102\) |
norman | \(\frac {\frac {3 b x}{2}-\frac {4 a}{d}-\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {3 b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {3 b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(154\) |
1/8*(4*a*cos(2*d*x+2*c)+b*sin(3*d*x+3*c)+(-12*b*d*x+16*a)*cos(d*x+c)+9*b*s in(d*x+c)+12*a)/d/cos(d*x+c)
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 \, b d x \cos \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right )^{2} - {\left (b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) - 2 \, a}{2 \, d \cos \left (d x + c\right )} \]
-1/2*(3*b*d*x*cos(d*x + c) - 2*a*cos(d*x + c)^2 - (b*cos(d*x + c)^2 + 2*b) *sin(d*x + c) - 2*a)/(d*cos(d*x + c))
\[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {{\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b - 2 \, a {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \]
-1/2*((3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*b - 2*a*(1/cos(d*x + c) + cos(d*x + c)))/d
Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3 \, {\left (d x + c\right )} b + \frac {4 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
-1/2*(3*(d*x + c)*b + 4*(b*tan(1/2*d*x + 1/2*c) + a)/(tan(1/2*d*x + 1/2*c) ^2 - 1) + 2*(b*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan (1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 15.89 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {3\,b\,x}{2}-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]