Integrand size = 21, antiderivative size = 108 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {b \left (a^2+b^2\right ) \log (\tan (c+d x))}{a^4 d}+\frac {b \left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{a^4 d} \]
-(a^2+b^2)*cot(d*x+c)/a^3/d+1/2*b*cot(d*x+c)^2/a^2/d-1/3*cot(d*x+c)^3/a/d- b*(a^2+b^2)*ln(tan(d*x+c))/a^4/d+b*(a^2+b^2)*ln(a+b*tan(d*x+c))/a^4/d
Time = 2.94 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {3 a^2 b \csc ^2(c+d x)-2 \cot (c+d x) \left (2 a^3+3 a b^2+a^3 \csc ^2(c+d x)\right )-6 b \left (a^2+b^2\right ) (\log (\sin (c+d x))-\log (a \cos (c+d x)+b \sin (c+d x)))}{6 a^4 d} \]
(3*a^2*b*Csc[c + d*x]^2 - 2*Cot[c + d*x]*(2*a^3 + 3*a*b^2 + a^3*Csc[c + d* x]^2) - 6*b*(a^2 + b^2)*(Log[Sin[c + d*x]] - Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(6*a^4*d)
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^4 (a+b \tan (c+d x))}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )}{b^4 (a+b \tan (c+d x))}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {b \int \left (\frac {\cot ^4(c+d x)}{a b^2}-\frac {\cot ^3(c+d x)}{a^2 b}+\frac {\left (a^2+b^2\right ) \cot ^2(c+d x)}{a^3 b^2}+\frac {\left (-a^2-b^2\right ) \cot (c+d x)}{a^4 b}+\frac {a^2+b^2}{a^4 (a+b \tan (c+d x))}\right )d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {\cot ^2(c+d x)}{2 a^2}-\frac {\left (a^2+b^2\right ) \log (b \tan (c+d x))}{a^4}+\frac {\left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{a^4}-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{a^3 b}-\frac {\cot ^3(c+d x)}{3 a b}\right )}{d}\) |
(b*(-(((a^2 + b^2)*Cot[c + d*x])/(a^3*b)) + Cot[c + d*x]^2/(2*a^2) - Cot[c + d*x]^3/(3*a*b) - ((a^2 + b^2)*Log[b*Tan[c + d*x]])/a^4 + ((a^2 + b^2)*L og[a + b*Tan[c + d*x]])/a^4))/d
3.1.59.3.1 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 = 1.42 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {\left (a^{2}+b^{2}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}}{d}\) | \(96\) |
default | \(\frac {\frac {\left (a^{2}+b^{2}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}}{d}\) | \(96\) |
risch | \(-\frac {2 \left (3 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{2}+3 i b^{2}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\) | \(227\) |
1/d*((a^2+b^2)/a^4*b*ln(a+b*tan(d*x+c))-1/3/a/tan(d*x+c)^3-(a^2+b^2)/a^3/t an(d*x+c)+1/2/a^2*b/tan(d*x+c)^2-(a^2+b^2)/a^4*b*ln(tan(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.93 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {2 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, a^{2} b \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )} \]
-1/6*(2*(2*a^3 + 3*a*b^2)*cos(d*x + c)^3 + 3*a^2*b*sin(d*x + c) + 3*(a^2*b + b^3 - (a^2*b + b^3)*cos(d*x + c)^2)*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)*sin(d*x + c) - 3*(a^2*b + b^3 - (a^2* b + b^3)*cos(d*x + c)^2)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 6*( a^3 + a*b^2)*cos(d*x + c))/((a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x + c))
\[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4}} - \frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {3 \, a b \tan \left (d x + c\right ) - 6 \, {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} - 2 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \]
1/6*(6*(a^2*b + b^3)*log(b*tan(d*x + c) + a)/a^4 - 6*(a^2*b + b^3)*log(tan (d*x + c))/a^4 + (3*a*b*tan(d*x + c) - 6*(a^2 + b^2)*tan(d*x + c)^2 - 2*a^ 2)/(a^3*tan(d*x + c)^3))/d
Time = 0.40 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.33 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {6 \, {\left (a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b} - \frac {11 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, b^{3} \tan \left (d x + c\right )^{3} - 6 \, a^{3} \tan \left (d x + c\right )^{2} - 6 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) - 2 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{3}}}{6 \, d} \]
-1/6*(6*(a^2*b + b^3)*log(abs(tan(d*x + c)))/a^4 - 6*(a^2*b^2 + b^4)*log(a bs(b*tan(d*x + c) + a))/(a^4*b) - (11*a^2*b*tan(d*x + c)^3 + 11*b^3*tan(d* x + c)^3 - 6*a^3*tan(d*x + c)^2 - 6*a*b^2*tan(d*x + c)^2 + 3*a^2*b*tan(d*x + c) - 2*a^3)/(a^4*tan(d*x + c)^3))/d
Time = 4.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,\left (a^2+b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (a^2\,b+b^3\right )}\right )\,\left (a^2+b^2\right )}{a^4\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+b^2\right )}{a^3}-\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \]