Integrand size = 20, antiderivative size = 115 \[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=\frac {d x}{4 b}+\frac {i (c+d x)^2}{2 d}-\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac {(c+d x) \sin ^2(a+b x)}{2 b} \]
1/4*d*x/b+1/2*I*(d*x+c)^2/d-(d*x+c)*ln(1+exp(2*I*(b*x+a)))/b+1/2*I*d*polyl og(2,-exp(2*I*(b*x+a)))/b^2-1/4*d*cos(b*x+a)*sin(b*x+a)/b^2-1/2*(d*x+c)*si n(b*x+a)^2/b
Time = 0.71 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=\frac {d x \cos (2 (a+b x))}{4 b}+\frac {a d \log (\cos (a+b x))}{b^2}-\frac {c \left (-\frac {1}{2} \cos ^2(a+b x)+\log (\cos (a+b x))\right )}{b}+\frac {d \left (\frac {1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )\right )}{b^2}-\frac {d \sin (2 (a+b x))}{8 b^2} \]
(d*x*Cos[2*(a + b*x)])/(4*b) + (a*d*Log[Cos[a + b*x]])/b^2 - (c*(-1/2*Cos[ a + b*x]^2 + Log[Cos[a + b*x]]))/b + (d*((I/2)*(a + b*x)^2 - (a + b*x)*Log [1 + E^((2*I)*(a + b*x))] + (I/2)*PolyLog[2, -E^((2*I)*(a + b*x))]))/b^2 - (d*Sin[2*(a + b*x)])/(8*b^2)
Time = 0.59 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4907, 3042, 4202, 2620, 2715, 2838, 4904, 3042, 3115, 24}
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.
| \(\displaystyle \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx\) |
| \(\Big \downarrow \) 4907 |
| \(\displaystyle \int (c+d x) \tan (a+b x)dx-\int (c+d x) \cos (a+b x) \sin (a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x) \tan (a+b x)dx-\int (c+d x) \cos (a+b x) \sin (a+b x)dx\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle -2 i \int \frac {e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}}dx-\int (c+d x) \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -2 i \left (\frac {i d \int \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\int (c+d x) \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -2 i \left (\frac {d \int e^{-2 i (a+b x)} \log \left (1+e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\int (c+d x) \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle -\int (c+d x) \cos (a+b x) \sin (a+b x)dx-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 4904 |
| \(\displaystyle \frac {d \int \sin ^2(a+b x)dx}{2 b}-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d \int \sin (a+b x)^2dx}{2 b}-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {d \left (\frac {\int 1dx}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )}{2 b}-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^2}{2 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d \left (\frac {x}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )}{2 b}+\frac {i (c+d x)^2}{2 d}\) |
((I/2)*(c + d*x)^2)/d - (2*I)*(((-1/2*I)*(c + d*x)*Log[1 + E^((2*I)*(a + b *x))])/b - (d*PolyLog[2, -E^((2*I)*(a + b*x))])/(4*b^2)) - ((c + d*x)*Sin[ a + b*x]^2)/(2*b) + (d*(x/2 - (Cos[a + b*x]*Sin[a + b*x])/(2*b)))/(2*b)
3.3.24.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x _)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) , x] - Simp[d*(m/(b*(n + 1))) Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b _.)*(x_)]^(p_.), x_Symbol] :> -Int[(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^ (p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x] /; Fr eeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
Time = 2.57 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.56
| method | result | size |
| risch | \(\frac {i d \,x^{2}}{2}-i c x +\frac {\left (2 d x b +2 c b +i d \right ) {\mathrm e}^{2 i \left (x b +a \right )}}{16 b^{2}}+\frac {\left (2 d x b +2 c b -i d \right ) {\mathrm e}^{-2 i \left (x b +a \right )}}{16 b^{2}}-\frac {c \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}+\frac {2 c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}+\frac {2 i d x a}{b}+\frac {i d \,a^{2}}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}+\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{2}}-\frac {2 d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}\) | \(179\) |
1/2*I*d*x^2-I*c*x+1/16*(2*d*x*b+I*d+2*c*b)/b^2*exp(2*I*(b*x+a))+1/16*(2*d* x*b-I*d+2*c*b)/b^2*exp(-2*I*(b*x+a))-1/b*c*ln(exp(2*I*(b*x+a))+1)+2/b*c*ln (exp(I*(b*x+a)))+2*I/b*d*x*a+I/b^2*d*a^2-1/b*d*ln(exp(2*I*(b*x+a))+1)*x+1/ 2*I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2-2/b^2*d*a*ln(exp(I*(b*x+a)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (96) = 192\).
Time = 0.28 (sec) , antiderivative size = 346, normalized size of antiderivative = 3.01 \[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=-\frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{4 \, b^{2}} \]
-1/4*(b*d*x - 2*(b*d*x + b*c)*cos(b*x + a)^2 + d*cos(b*x + a)*sin(b*x + a) + 2*I*d*dilog(I*cos(b*x + a) + sin(b*x + a)) - 2*I*d*dilog(I*cos(b*x + a) - sin(b*x + a)) - 2*I*d*dilog(-I*cos(b*x + a) + sin(b*x + a)) + 2*I*d*dil og(-I*cos(b*x + a) - sin(b*x + a)) + 2*(b*c - a*d)*log(cos(b*x + a) + I*si n(b*x + a) + I) + 2*(b*c - a*d)*log(cos(b*x + a) - I*sin(b*x + a) + I) + 2 *(b*d*x + a*d)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 2*(b*d*x + a*d)*lo g(I*cos(b*x + a) - sin(b*x + a) + 1) + 2*(b*d*x + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 2*(b*d*x + a*d)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + 2*(b*c - a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + 2*(b*c - a* d)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/b^2
\[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right ) \sin ^{3}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Time = 0.34 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.27 \[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=-\frac {-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x - 8 \, {\left (-i \, b d x - i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 i \, d {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]
-1/8*(-4*I*b^2*d*x^2 - 8*I*b^2*c*x - 8*(-I*b*d*x - I*b*c)*arctan2(sin(2*b* x + 2*a), cos(2*b*x + 2*a) + 1) - 2*(b*d*x + b*c)*cos(2*b*x + 2*a) - 4*I*d *dilog(-e^(2*I*b*x + 2*I*a)) + 4*(b*d*x + b*c)*log(cos(2*b*x + 2*a)^2 + si n(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + d*sin(2*b*x + 2*a))/b^2
\[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3\,\left (c+d\,x\right )}{\cos \left (a+b\,x\right )} \,d x \]