Integrand size = 21, antiderivative size = 194 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {\left (a^4+12 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-\frac {2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {a^3 b \cot ^4(c+d x)}{d}-\frac {a^4 \cot ^5(c+d x)}{5 d}+\frac {4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}+\frac {2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {2 a b^3 \tan ^2(c+d x)}{d}+\frac {b^4 \tan ^3(c+d x)}{3 d} \]
-(a^4+12*a^2*b^2+b^4)*cot(d*x+c)/d-2*a*b*(2*a^2+b^2)*cot(d*x+c)^2/d-2/3*a^ 2*(a^2+3*b^2)*cot(d*x+c)^3/d-a^3*b*cot(d*x+c)^4/d-1/5*a^4*cot(d*x+c)^5/d+4 *a*b*(a^2+2*b^2)*ln(tan(d*x+c))/d+2*b^2*(3*a^2+b^2)*tan(d*x+c)/d+2*a*b^3*t an(d*x+c)^2/d+1/3*b^4*tan(d*x+c)^3/d
Time = 6.10 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.20 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {\left (15 a^3 b \cot ^4(c+d x)+3 a^4 \cot ^5(c+d x)+2 a \cos ^2(c+d x) \left (-15 b^3+15 b \left (a^2+b^2\right ) \cot ^2(c+d x)+a \left (2 a^2+15 b^2\right ) \cot ^3(c+d x)\right )+\cos ^4(c+d x) \left (\left (8 a^4+150 a^2 b^2+15 b^4\right ) \cot (c+d x)+60 a b \left (a^2+2 b^2\right ) (\log (\cos (c+d x))-\log (\sin (c+d x)))\right )-5 b^2 \left (18 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)-\frac {5}{2} b^4 \sin (2 (c+d x))\right ) (a+b \tan (c+d x))^4}{15 d (a \cos (c+d x)+b \sin (c+d x))^4} \]
-1/15*((15*a^3*b*Cot[c + d*x]^4 + 3*a^4*Cot[c + d*x]^5 + 2*a*Cos[c + d*x]^ 2*(-15*b^3 + 15*b*(a^2 + b^2)*Cot[c + d*x]^2 + a*(2*a^2 + 15*b^2)*Cot[c + d*x]^3) + Cos[c + d*x]^4*((8*a^4 + 150*a^2*b^2 + 15*b^4)*Cot[c + d*x] + 60 *a*b*(a^2 + 2*b^2)*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]])) - 5*b^2*(18*a^ 2 + 5*b^2)*Cos[c + d*x]^3*Sin[c + d*x] - (5*b^4*Sin[2*(c + d*x)])/2)*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)
Time = 0.38 (sec) , antiderivative size = 178, 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.
| \(\displaystyle \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^4}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {\cot ^6(c+d x) (a+b \tan (c+d x))^4 \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^6}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {b \int \left (\frac {a^4 \cot ^6(c+d x)}{b^2}+\frac {4 a^3 \cot ^5(c+d x)}{b}+\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^4(c+d x)}{b^2}+\frac {4 a \left (2 a^2+b^2\right ) \cot ^3(c+d x)}{b}+\frac {\left (a^4+12 b^2 a^2+b^4\right ) \cot ^2(c+d x)}{b^2}+\frac {4 \left (a^3+2 b^2 a\right ) \cot (c+d x)}{b}+b^2 \tan ^2(c+d x)+2 \left (3 a^2+b^2\right )+4 a b \tan (c+d x)\right )d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (-\frac {a^4 \cot ^5(c+d x)}{5 b}-a^3 \cot ^4(c+d x)+2 b \left (3 a^2+b^2\right ) \tan (c+d x)-\frac {2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 b}-2 a \left (2 a^2+b^2\right ) \cot ^2(c+d x)+4 a \left (a^2+2 b^2\right ) \log (b \tan (c+d x))-\frac {\left (a^4+12 a^2 b^2+b^4\right ) \cot (c+d x)}{b}+2 a b^2 \tan ^2(c+d x)+\frac {1}{3} b^3 \tan ^3(c+d x)\right )}{d}\) |
(b*(-(((a^4 + 12*a^2*b^2 + b^4)*Cot[c + d*x])/b) - 2*a*(2*a^2 + b^2)*Cot[c + d*x]^2 - (2*a^2*(a^2 + 3*b^2)*Cot[c + d*x]^3)/(3*b) - a^3*Cot[c + d*x]^ 4 - (a^4*Cot[c + d*x]^5)/(5*b) + 4*a*(a^2 + 2*b^2)*Log[b*Tan[c + d*x]] + 2 *b*(3*a^2 + b^2)*Tan[c + d*x] + 2*a*b^2*Tan[c + d*x]^2 + (b^3*Tan[c + d*x] ^3)/3))/d
3.1.49.3.1 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 = 25.78 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {b^{4} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+4 a \,b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+4 a^{3} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{4} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(218\) |
| default | \(\frac {b^{4} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+4 a \,b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+4 a^{3} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{4} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(218\) |
| risch | \(\frac {64 i a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-32 i a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+64 i a^{2} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-128 i a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-56 a^{3} b \,{\mathrm e}^{10 i \left (d x +c \right )}+16 a \,b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+64 i a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a^{4}}{15}+16 a \,b^{3} {\mathrm e}^{14 i \left (d x +c \right )}-\frac {32 i a^{4} {\mathrm e}^{10 i \left (d x +c \right )}}{3}-32 a \,b^{3} {\mathrm e}^{12 i \left (d x +c \right )}+8 a^{3} b \,{\mathrm e}^{14 i \left (d x +c \right )}-\frac {32 i b^{4} {\mathrm e}^{10 i \left (d x +c \right )}}{3}-16 a^{3} b \,{\mathrm e}^{12 i \left (d x +c \right )}+\frac {112 i b^{4} {\mathrm e}^{8 i \left (d x +c \right )}}{3}+16 a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+32 a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-\frac {128 i b^{4} {\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {32 i b^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-32 i a^{2} b^{2}+\frac {32 i b^{4} {\mathrm e}^{4 i \left (d x +c \right )}}{3}+56 a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-16 a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+\frac {32 i a^{4} {\mathrm e}^{4 i \left (d x +c \right )}}{15}-\frac {256 i a^{4} {\mathrm e}^{6 i \left (d x +c \right )}}{15}-\frac {80 i a^{4} {\mathrm e}^{8 i \left (d x +c \right )}}{3}-8 a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-16 a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {32 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{15}-\frac {16 i b^{4}}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {8 a \,b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {8 a \,b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(559\) |
1/d*(b^4*(1/3/sin(d*x+c)/cos(d*x+c)^3+4/3/sin(d*x+c)/cos(d*x+c)-8/3*cot(d* x+c))+4*a*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 )))+6*a^2*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))+4*a^3*b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c)))+a^ 4*(-8/15-1/5*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. 386 vs. \(2 (188) = 376\).
Time = 0.29 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.99 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {8 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{8} - 20 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + 15 \, {\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - 5 \, b^{4} - 10 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 2 \, a b^{3} \cos \left (d x + c\right ) - 3 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )} \]
-1/15*(8*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^8 - 20*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^6 + 15*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^4 - 5 *b^4 - 10*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 + 30*((a^3*b + 2*a*b^3)*cos(d*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x + c)^3)*log(cos(d*x + c)^2)*sin(d*x + c) - 30*((a^3*b + 2*a*b^3)*cos(d*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x + c) ^3)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 15*(2*(a^3*b + 2*a*b^3)* cos(d*x + c)^5 + 2*a*b^3*cos(d*x + c) - 3*(a^3*b + 2*a*b^3)*cos(d*x + c)^3 )*sin(d*x + c))/((d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^5 + d*cos(d*x + c)^3 )*sin(d*x + c))
Timed out. \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=\text {Timed out} \]
Time = 0.56 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.88 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 60 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + 30 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right ) - \frac {15 \, a^{3} b \tan \left (d x + c\right ) + 15 \, {\left (a^{4} + 12 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + 3 \, a^{4} + 30 \, {\left (2 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )^{3} + 10 \, {\left (a^{4} + 3 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]
1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 60*(a^3*b + 2*a*b^3 )*log(tan(d*x + c)) + 30*(3*a^2*b^2 + b^4)*tan(d*x + c) - (15*a^3*b*tan(d* x + c) + 15*(a^4 + 12*a^2*b^2 + b^4)*tan(d*x + c)^4 + 3*a^4 + 30*(2*a^3*b + a*b^3)*tan(d*x + c)^3 + 10*(a^4 + 3*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c )^5)/d
Time = 1.18 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.21 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 90 \, a^{2} b^{2} \tan \left (d x + c\right ) + 30 \, b^{4} \tan \left (d x + c\right ) + 60 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {137 \, a^{3} b \tan \left (d x + c\right )^{5} + 274 \, a b^{3} \tan \left (d x + c\right )^{5} + 15 \, a^{4} \tan \left (d x + c\right )^{4} + 180 \, a^{2} b^{2} \tan \left (d x + c\right )^{4} + 15 \, b^{4} \tan \left (d x + c\right )^{4} + 60 \, a^{3} b \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 10 \, a^{4} \tan \left (d x + c\right )^{2} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]
1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 90*a^2*b^2*tan(d*x + c) + 30*b^4*tan(d*x + c) + 60*(a^3*b + 2*a*b^3)*log(abs(tan(d*x + c))) - (137*a^3*b*tan(d*x + c)^5 + 274*a*b^3*tan(d*x + c)^5 + 15*a^4*tan(d*x + c )^4 + 180*a^2*b^2*tan(d*x + c)^4 + 15*b^4*tan(d*x + c)^4 + 60*a^3*b*tan(d* x + c)^3 + 30*a*b^3*tan(d*x + c)^3 + 10*a^4*tan(d*x + c)^2 + 30*a^2*b^2*ta n(d*x + c)^2 + 15*a^3*b*tan(d*x + c) + 3*a^4)/tan(d*x + c)^5)/d
Time = 4.62 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.93 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (4\,a^3\,b+8\,a\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,a^4}{3}+2\,a^2\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,a^3\,b+2\,a\,b^3\right )+\frac {a^4}{5}+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+12\,a^2\,b^2+b^4\right )+a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (6\,a^2\,b^2+2\,b^4\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]
(log(tan(c + d*x))*(8*a*b^3 + 4*a^3*b))/d - (cot(c + d*x)^5*(tan(c + d*x)^ 2*((2*a^4)/3 + 2*a^2*b^2) + tan(c + d*x)^3*(2*a*b^3 + 4*a^3*b) + a^4/5 + t an(c + d*x)^4*(a^4 + b^4 + 12*a^2*b^2) + a^3*b*tan(c + d*x)))/d + (b^4*tan (c + d*x)^3)/(3*d) + (tan(c + d*x)*(2*b^4 + 6*a^2*b^2))/d + (2*a*b^3*tan(c + d*x)^2)/d