Integrand size = 19, antiderivative size = 101 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cos (c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d} \]
b*arctanh(sin(d*x+c))/d-a*cos(d*x+c)/d+2/3*a*cos(d*x+c)^3/d-1/5*a*cos(d*x+ c)^5/d-b*sin(d*x+c)/d-1/3*b*sin(d*x+c)^3/d-1/5*b*sin(d*x+c)^5/d
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a \cos (c+d x)}{8 d}+\frac {5 a \cos (3 (c+d x))}{48 d}-\frac {a \cos (5 (c+d x))}{80 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d} \]
(b*ArcTanh[Sin[c + d*x]])/d - (5*a*Cos[c + d*x])/(8*d) + (5*a*Cos[3*(c + d *x)])/(48*d) - (a*Cos[5*(c + d*x)])/(80*d) - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (b*Sin[c + d*x]^5)/(5*d)
Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 4000, 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.
\(\displaystyle \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^5 (a+b \tan (c+d x))dx\) |
\(\Big \downarrow \) 4000 |
\(\displaystyle \int \left (a \sin ^5(c+d x)+b \sin ^5(c+d x) \tan (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}\) |
(b*ArcTanh[Sin[c + d*x]])/d - (a*Cos[c + d*x])/d + (2*a*Cos[c + d*x]^3)/(3 *d) - (a*Cos[c + d*x]^5)/(5*d) - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/( 3*d) - (b*Sin[c + d*x]^5)/(5*d)
3.1.11.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Int[Expand[Sin[e + f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Time = 3.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(80\) |
default | \(\frac {-\frac {a \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(80\) |
risch | \(\frac {11 i {\mathrm e}^{i \left (d x +c \right )} b}{16 d}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a}{16 d}-\frac {11 i {\mathrm e}^{-i \left (d x +c \right )} b}{16 d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a}{16 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {b \sin \left (5 d x +5 c \right )}{80 d}+\frac {5 a \cos \left (3 d x +3 c \right )}{48 d}+\frac {7 b \sin \left (3 d x +3 c \right )}{48 d}\) | \(161\) |
1/d*(-1/5*a*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)+b*(-1/5*sin(d*x +c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{5} - 20 \, a \cos \left (d x + c\right )^{3} + 30 \, a \cos \left (d x + c\right ) - 15 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{4} - 11 \, b \cos \left (d x + c\right )^{2} + 23 \, b\right )} \sin \left (d x + c\right )}{30 \, d} \]
-1/30*(6*a*cos(d*x + c)^5 - 20*a*cos(d*x + c)^3 + 30*a*cos(d*x + c) - 15*b *log(sin(d*x + c) + 1) + 15*b*log(-sin(d*x + c) + 1) + 2*(3*b*cos(d*x + c) ^4 - 11*b*cos(d*x + c)^2 + 23*b)*sin(d*x + c))/d
\[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sin ^{5}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a + {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} b}{30 \, d} \]
-1/30*(2*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*a + (6*s in(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin( d*x + c) - 1) + 30*sin(d*x + c))*b)/d
Leaf count of result is larger than twice the leaf count of optimal. 10412 vs. \(2 (93) = 186\).
Time = 1.73 (sec) , antiderivative size = 10412, normalized size of antiderivative = 103.09 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=\text {Too large to display} \]
-1/30*(15*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2* c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1 /2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^10*tan(1/2*c)^10 - 15*b*log(2*(tan(1/2*d *x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2* c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/ (tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2 *d*x)^10*tan(1/2*c)^10 + 16*a*tan(1/2*d*x)^10*tan(1/2*c)^10 + 75*b*log(2*( tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x) *tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2 *c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1) )*tan(1/2*d*x)^10*tan(1/2*c)^8 - 75*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^ 2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan( 1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^10*tan(1/2*c)^ 8 - 60*b*tan(1/2*d*x)^10*tan(1/2*c)^9 + 75*b*log(2*(tan(1/2*d*x)^2*tan(1/2 *c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/ 2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x) ^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^8*tan(1 /2*c)^10 - 75*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*t...
Time = 5.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.20 \[ \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a\,{\cos \left (c+d\,x\right )}^3}{3\,d}-\frac {a\,{\cos \left (c+d\,x\right )}^5}{5\,d}-\frac {a\,\cos \left (c+d\,x\right )}{d}-\frac {23\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {11\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}-\frac {b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]