Integrand size = 19, antiderivative size = 87 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {b \log (\tan (c+d x))}{d} \] Output:
-a*cot(d*x+c)/d-b*cot(d*x+c)^2/d-2/3*a*cot(d*x+c)^3/d-1/4*b*cot(d*x+c)^4/d -1/5*a*cot(d*x+c)^5/d+b*ln(tan(d*x+c))/d
Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.33 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {8 a \cot (c+d x)}{15 d}-\frac {b \csc ^2(c+d x)}{2 d}-\frac {4 a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}-\frac {b \log (\cos (c+d x))}{d}+\frac {b \log (\sin (c+d x))}{d} \] Input:
Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x]),x]
Output:
(-8*a*Cot[c + d*x])/(15*d) - (b*Csc[c + d*x]^2)/(2*d) - (4*a*Cot[c + d*x]* Csc[c + d*x]^2)/(15*d) - (b*Csc[c + d*x]^4)/(4*d) - (a*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) - (b*Log[Cos[c + d*x]])/d + (b*Log[Sin[c + d*x]])/d
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 4889, 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.
\(\displaystyle \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (c+d x)}{\sin (c+d x)^6}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \frac {\int \cot ^6(c+d x) (a+b \tan (c+d x)) \left (\tan ^2(c+d x)+1\right )^2d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (a \cot ^6(c+d x)+b \cot ^5(c+d x)+2 a \cot ^4(c+d x)+2 b \cot ^3(c+d x)+a \cot ^2(c+d x)+b \cot (c+d x)\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{5} a \cot ^5(c+d x)-\frac {2}{3} a \cot ^3(c+d x)-a \cot (c+d x)-\frac {1}{4} b \cot ^4(c+d x)-b \cot ^2(c+d x)+b \log (\tan (c+d x))}{d}\) |
Input:
Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x]),x]
Output:
(-(a*Cot[c + d*x]) - b*Cot[c + d*x]^2 - (2*a*Cot[c + d*x]^3)/3 - (b*Cot[c + d*x]^4)/4 - (a*Cot[c + d*x]^5)/5 + b*Log[Tan[c + d*x]])/d
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[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 9.49 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(66\) |
default | \(\frac {b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(66\) |
risch | \(\frac {2 b \,{\mathrm e}^{8 i \left (d x +c \right )}-10 b \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}}{3}+10 b \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {16 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a}{15}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(134\) |
Input:
int(csc(d*x+c)^6*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c)))+a*(-8/15-1/5*cs c(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. 174 vs. \(2 (81) = 162\).
Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.00 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {32 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 30 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) + 60 \, a \cos \left (d x + c\right ) - 15 \, {\left (2 \, b \cos \left (d x + c\right )^{2} - 3 \, b\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(32*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 30*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(cos(d*x + c)^2)*sin(d*x + c) - 30*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c ) + 60*a*cos(d*x + c) - 15*(2*b*cos(d*x + c)^2 - 3*b)*sin(d*x + c))/((d*co s(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
\[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \csc ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate(csc(d*x+c)**6*(a+b*tan(d*x+c)),x)
Output:
Integral((a + b*tan(c + d*x))*csc(c + d*x)**6, x)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {60 \, b \log \left (\tan \left (d x + c\right )\right ) - \frac {60 \, a \tan \left (d x + c\right )^{4} + 60 \, b \tan \left (d x + c\right )^{3} + 40 \, a \tan \left (d x + c\right )^{2} + 15 \, b \tan \left (d x + c\right ) + 12 \, a}{\tan \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
1/60*(60*b*log(tan(d*x + c)) - (60*a*tan(d*x + c)^4 + 60*b*tan(d*x + c)^3 + 40*a*tan(d*x + c)^2 + 15*b*tan(d*x + c) + 12*a)/tan(d*x + c)^5)/d
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {60 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {60 \, a \tan \left (d x + c\right )^{4} + 60 \, b \tan \left (d x + c\right )^{3} + 40 \, a \tan \left (d x + c\right )^{2} + 15 \, b \tan \left (d x + c\right ) + 12 \, a}{\tan \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
1/60*(60*b*log(abs(tan(d*x + c))) - (60*a*tan(d*x + c)^4 + 60*b*tan(d*x + c)^3 + 40*a*tan(d*x + c)^2 + 15*b*tan(d*x + c) + 12*a)/tan(d*x + c)^5)/d
Time = 0.97 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4+b\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{4}+\frac {a}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \] Input:
int((a + b*tan(c + d*x))/sin(c + d*x)^6,x)
Output:
(b*log(tan(c + d*x)))/d - (a/5 + (b*tan(c + d*x))/4 + (2*a*tan(c + d*x)^2) /3 + a*tan(c + d*x)^4 + b*tan(c + d*x)^3)/(d*tan(c + d*x)^5)
Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.78 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a -128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -96 \cos \left (d x +c \right ) a -480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{5} b -480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5} b +480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} b +165 \sin \left (d x +c \right )^{5} b -240 \sin \left (d x +c \right )^{3} b -120 \sin \left (d x +c \right ) b}{480 \sin \left (d x +c \right )^{5} d} \] Input:
int(csc(d*x+c)^6*(a+b*tan(d*x+c)),x)
Output:
( - 256*cos(c + d*x)*sin(c + d*x)**4*a - 128*cos(c + d*x)*sin(c + d*x)**2* a - 96*cos(c + d*x)*a - 480*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*b - 480*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*b + 480*log(tan((c + d*x)/2) )*sin(c + d*x)**5*b + 165*sin(c + d*x)**5*b - 240*sin(c + d*x)**3*b - 120* sin(c + d*x)*b)/(480*sin(c + d*x)**5*d)