Integrand size = 21, antiderivative size = 140 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 a^3 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \]
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Time = 0.37 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2945, 12, 2738, 214} \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 a^3 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}+\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )^2} \]
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Rule 12
Rule 214
Rule 2738
Rule 2945
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cot (c+d x) \csc ^3(c+d x)}{-b-a \cos (c+d x)} \, dx \\ & = \frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\int \frac {\left (a b-2 a^2 \cos (c+d x)\right ) \csc ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\int \frac {3 a^3 b}{-b-a \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2} \\ & = \frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (a^3 b\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2} \\ & = \frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^3 b\right ) \text {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d} \\ & = -\frac {2 a^3 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \\ \end{align*}
Time = 1.07 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {24 a^3 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} \left (10 a^2 b-4 b^3+\left (-6 a^3+3 a b^2\right ) \cos (c+d x)-6 a^2 b \cos (2 (c+d x))+2 a^3 \cos (3 (c+d x))+a b^2 \cos (3 (c+d x))\right ) \csc ^3(c+d x)}{12 (a-b)^2 (a+b)^2 \sqrt {a^2-b^2} d} \]
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Time = 0.66 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a +b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(165\) |
default | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a +b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(165\) |
risch | \(\frac {2 i \left (3 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-10 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{3}-a \,b^{2}\right )}{3 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(299\) |
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Time = 0.31 (sec) , antiderivative size = 558, normalized size of antiderivative = 3.99 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\left [-\frac {8 \, a^{4} b - 10 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{3 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
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\[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (128) = 256\).
Time = 0.33 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.92 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]
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Time = 14.00 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.56 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2}{8\,a-8\,b}+\frac {8\,a+8\,b}{{\left (8\,a-8\,b\right )}^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\left (8\,a-8\,b\right )}-\frac {\frac {a^2-2\,a\,b+b^2}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3-5\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^2-16\,a\,b+8\,b^2\right )}-\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
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