Integrand size = 21, antiderivative size = 167 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {a \left (a^2+6 b^2\right ) \cot (c+d x)}{d}-\frac {b \left (6 a^2+b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {a \left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {3 a^2 b \cot ^4(c+d x)}{4 d}-\frac {a^3 \cot ^5(c+d x)}{5 d}+\frac {b \left (3 a^2+2 b^2\right ) \log (\tan (c+d x))}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \tan ^2(c+d x)}{2 d} \]
-a*(a^2+6*b^2)*cot(d*x+c)/d-1/2*b*(6*a^2+b^2)*cot(d*x+c)^2/d-1/3*a*(2*a^2+ 3*b^2)*cot(d*x+c)^3/d-3/4*a^2*b*cot(d*x+c)^4/d-1/5*a^3*cot(d*x+c)^5/d+b*(3 *a^2+2*b^2)*ln(tan(d*x+c))/d+3*a*b^2*tan(d*x+c)/d+1/2*b^3*tan(d*x+c)^2/d
Leaf count is larger than twice the leaf count of optimal. \(515\) vs. \(2(167)=334\).
Time = 3.55 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.08 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {\csc ^5(c+d x) \sec ^2(c+d x) \left (40 a \left (5 a^2+3 b^2\right ) \cos (c+d x)+8 \left (a^3+15 a b^2\right ) \cos (3 (c+d x))-24 a^3 \cos (5 (c+d x))-360 a b^2 \cos (5 (c+d x))+8 a^3 \cos (7 (c+d x))+120 a b^2 \cos (7 (c+d x))+360 a^2 b \sin (c+d x)-240 b^3 \sin (c+d x)+225 a^2 b \log (\cos (c+d x)) \sin (c+d x)+150 b^3 \log (\cos (c+d x)) \sin (c+d x)-225 a^2 b \log (\sin (c+d x)) \sin (c+d x)-150 b^3 \log (\sin (c+d x)) \sin (c+d x)+270 a^2 b \sin (3 (c+d x))+180 b^3 \sin (3 (c+d x))+45 a^2 b \log (\cos (c+d x)) \sin (3 (c+d x))+30 b^3 \log (\cos (c+d x)) \sin (3 (c+d x))-45 a^2 b \log (\sin (c+d x)) \sin (3 (c+d x))-30 b^3 \log (\sin (c+d x)) \sin (3 (c+d x))-90 a^2 b \sin (5 (c+d x))-60 b^3 \sin (5 (c+d x))-135 a^2 b \log (\cos (c+d x)) \sin (5 (c+d x))-90 b^3 \log (\cos (c+d x)) \sin (5 (c+d x))+135 a^2 b \log (\sin (c+d x)) \sin (5 (c+d x))+90 b^3 \log (\sin (c+d x)) \sin (5 (c+d x))+45 a^2 b \log (\cos (c+d x)) \sin (7 (c+d x))+30 b^3 \log (\cos (c+d x)) \sin (7 (c+d x))-45 a^2 b \log (\sin (c+d x)) \sin (7 (c+d x))-30 b^3 \log (\sin (c+d x)) \sin (7 (c+d x))\right )}{960 d} \]
-1/960*(Csc[c + d*x]^5*Sec[c + d*x]^2*(40*a*(5*a^2 + 3*b^2)*Cos[c + d*x] + 8*(a^3 + 15*a*b^2)*Cos[3*(c + d*x)] - 24*a^3*Cos[5*(c + d*x)] - 360*a*b^2 *Cos[5*(c + d*x)] + 8*a^3*Cos[7*(c + d*x)] + 120*a*b^2*Cos[7*(c + d*x)] + 360*a^2*b*Sin[c + d*x] - 240*b^3*Sin[c + d*x] + 225*a^2*b*Log[Cos[c + d*x] ]*Sin[c + d*x] + 150*b^3*Log[Cos[c + d*x]]*Sin[c + d*x] - 225*a^2*b*Log[Si n[c + d*x]]*Sin[c + d*x] - 150*b^3*Log[Sin[c + d*x]]*Sin[c + d*x] + 270*a^ 2*b*Sin[3*(c + d*x)] + 180*b^3*Sin[3*(c + d*x)] + 45*a^2*b*Log[Cos[c + d*x ]]*Sin[3*(c + d*x)] + 30*b^3*Log[Cos[c + d*x]]*Sin[3*(c + d*x)] - 45*a^2*b *Log[Sin[c + d*x]]*Sin[3*(c + d*x)] - 30*b^3*Log[Sin[c + d*x]]*Sin[3*(c + d*x)] - 90*a^2*b*Sin[5*(c + d*x)] - 60*b^3*Sin[5*(c + d*x)] - 135*a^2*b*Lo g[Cos[c + d*x]]*Sin[5*(c + d*x)] - 90*b^3*Log[Cos[c + d*x]]*Sin[5*(c + d*x )] + 135*a^2*b*Log[Sin[c + d*x]]*Sin[5*(c + d*x)] + 90*b^3*Log[Sin[c + d*x ]]*Sin[5*(c + d*x)] + 45*a^2*b*Log[Cos[c + d*x]]*Sin[7*(c + d*x)] + 30*b^3 *Log[Cos[c + d*x]]*Sin[7*(c + d*x)] - 45*a^2*b*Log[Sin[c + d*x]]*Sin[7*(c + d*x)] - 30*b^3*Log[Sin[c + d*x]]*Sin[7*(c + d*x)]))/d
Time = 0.35 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3999, 522, 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.
| \(\displaystyle \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {\cot ^6(c+d x) (a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^6}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {b \int \left (\frac {a^3 \cot ^6(c+d x)}{b^2}+\frac {3 a^2 \cot ^5(c+d x)}{b}+\frac {\left (3 a b^4+2 a^3 b^2\right ) \cot ^4(c+d x)}{b^4}+\frac {\left (b^4+6 a^2 b^2\right ) \cot ^3(c+d x)}{b^3}+\frac {\left (a^3+6 b^2 a\right ) \cot ^2(c+d x)}{b^2}+\frac {\left (3 a^2+2 b^2\right ) \cot (c+d x)}{b}+3 a+b \tan (c+d x)\right )d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (-\frac {a^3 \cot ^5(c+d x)}{5 b}-\frac {a \left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 b}-\frac {1}{2} \left (6 a^2+b^2\right ) \cot ^2(c+d x)-\frac {a \left (a^2+6 b^2\right ) \cot (c+d x)}{b}+\left (3 a^2+2 b^2\right ) \log (b \tan (c+d x))-\frac {3}{4} a^2 \cot ^4(c+d x)+3 a b \tan (c+d x)+\frac {1}{2} b^2 \tan ^2(c+d x)\right )}{d}\) |
(b*(-((a*(a^2 + 6*b^2)*Cot[c + d*x])/b) - ((6*a^2 + b^2)*Cot[c + d*x]^2)/2 - (a*(2*a^2 + 3*b^2)*Cot[c + d*x]^3)/(3*b) - (3*a^2*Cot[c + d*x]^4)/4 - ( a^3*Cot[c + d*x]^5)/(5*b) + (3*a^2 + 2*b^2)*Log[b*Tan[c + d*x]] + 3*a*b*Ta n[c + d*x] + (b^2*Tan[c + d*x]^2)/2))/d
3.1.40.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Time = 19.52 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+3 a^{2} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(165\) |
| default | \(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+3 a^{2} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(165\) |
| risch | \(\frac {6 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}-18 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}-12 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+\frac {16 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{5}+32 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-16 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-16 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-\frac {16 i a^{3}}{15}-16 i a \,b^{2}+18 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+12 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-48 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{15}-\frac {32 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(427\) |
1/d*(b^3*(1/2/sin(d*x+c)^2/cos(d*x+c)^2-1/sin(d*x+c)^2+2*ln(tan(d*x+c)))+3 *a*b^2*(-1/3/sin(d*x+c)^3/cos(d*x+c)+4/3/sin(d*x+c)/cos(d*x+c)-8/3*cot(d*x +c))+3*a^2*b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c)))+a^3*(-8/1 5-1/5*csc(d*x+c)^4-4/15*csc(d*x+c)^2)*cot(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (157) = 314\).
Time = 0.28 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.05 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {32 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} - 80 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 180 \, a b^{2} \cos \left (d x + c\right ) + 60 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, b^{3} - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]
-1/60*(32*(a^3 + 15*a*b^2)*cos(d*x + c)^7 - 80*(a^3 + 15*a*b^2)*cos(d*x + c)^5 - 180*a*b^2*cos(d*x + c) + 60*(a^3 + 15*a*b^2)*cos(d*x + c)^3 + 30*(( 3*a^2*b + 2*b^3)*cos(d*x + c)^6 - 2*(3*a^2*b + 2*b^3)*cos(d*x + c)^4 + (3* a^2*b + 2*b^3)*cos(d*x + c)^2)*log(cos(d*x + c)^2)*sin(d*x + c) - 30*((3*a ^2*b + 2*b^3)*cos(d*x + c)^6 - 2*(3*a^2*b + 2*b^3)*cos(d*x + c)^4 + (3*a^2 *b + 2*b^3)*cos(d*x + c)^2)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 15*(2*(3*a^2*b + 2*b^3)*cos(d*x + c)^4 + 2*b^3 - 3*(3*a^2*b + 2*b^3)*cos(d *x + c)^2)*sin(d*x + c))/((d*cos(d*x + c)^6 - 2*d*cos(d*x + c)^4 + d*cos(d *x + c)^2)*sin(d*x + c))
\[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \csc ^{6}{\left (c + d x \right )}\, dx \]
Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.85 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {30 \, b^{3} \tan \left (d x + c\right )^{2} + 180 \, a b^{2} \tan \left (d x + c\right ) + 60 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {60 \, {\left (a^{3} + 6 \, a b^{2}\right )} \tan \left (d x + c\right )^{4} + 45 \, a^{2} b \tan \left (d x + c\right ) + 30 \, {\left (6 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{3} + 12 \, a^{3} + 20 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]
1/60*(30*b^3*tan(d*x + c)^2 + 180*a*b^2*tan(d*x + c) + 60*(3*a^2*b + 2*b^3 )*log(tan(d*x + c)) - (60*(a^3 + 6*a*b^2)*tan(d*x + c)^4 + 45*a^2*b*tan(d* x + c) + 30*(6*a^2*b + b^3)*tan(d*x + c)^3 + 12*a^3 + 20*(2*a^3 + 3*a*b^2) *tan(d*x + c)^2)/tan(d*x + c)^5)/d
Time = 0.77 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.13 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {30 \, b^{3} \tan \left (d x + c\right )^{2} + 180 \, a b^{2} \tan \left (d x + c\right ) + 60 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {411 \, a^{2} b \tan \left (d x + c\right )^{5} + 274 \, b^{3} \tan \left (d x + c\right )^{5} + 60 \, a^{3} \tan \left (d x + c\right )^{4} + 360 \, a b^{2} \tan \left (d x + c\right )^{4} + 180 \, a^{2} b \tan \left (d x + c\right )^{3} + 30 \, b^{3} \tan \left (d x + c\right )^{3} + 40 \, a^{3} \tan \left (d x + c\right )^{2} + 60 \, a b^{2} \tan \left (d x + c\right )^{2} + 45 \, a^{2} b \tan \left (d x + c\right ) + 12 \, a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]
1/60*(30*b^3*tan(d*x + c)^2 + 180*a*b^2*tan(d*x + c) + 60*(3*a^2*b + 2*b^3 )*log(abs(tan(d*x + c))) - (411*a^2*b*tan(d*x + c)^5 + 274*b^3*tan(d*x + c )^5 + 60*a^3*tan(d*x + c)^4 + 360*a*b^2*tan(d*x + c)^4 + 180*a^2*b*tan(d*x + c)^3 + 30*b^3*tan(d*x + c)^3 + 40*a^3*tan(d*x + c)^2 + 60*a*b^2*tan(d*x + c)^2 + 45*a^2*b*tan(d*x + c) + 12*a^3)/tan(d*x + c)^5)/d
Time = 4.32 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2\,b+\frac {b^3}{2}\right )+\frac {a^3}{5}+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^3+6\,a\,b^2\right )+\frac {3\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )}{4}\right )}{d}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}+\frac {3\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{d} \]