Optimal. Leaf size=201 \[ -\frac {2 a^5 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}+\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^2}+\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^3} \]
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Rubi [A] time = 0.52, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3872, 2866, 12, 2659, 208} \[ \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}+\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^2}+\frac {\csc (c+d x) \left (15 a^4 b-a \left (9 a^2 b^2+8 a^4-2 b^4\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^3}-\frac {2 a^5 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 208
Rule 2659
Rule 2866
Rule 3872
Rubi steps
\begin {align*} \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cot (c+d x) \csc ^5(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac {\int \frac {\left (a b-4 a^2 \cos (c+d x)\right ) \csc ^4(c+d x)}{-b-a \cos (c+d x)} \, dx}{5 \left (a^2-b^2\right )}\\ &=\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac {\int \frac {\left (a b \left (7 a^2-2 b^2\right )-2 a^2 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=\frac {\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac {\int \frac {15 a^5 b}{-b-a \cos (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac {\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac {\left (a^5 b\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=\frac {\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac {\left (2 a^5 b\right ) \operatorname {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 )^3 d}\\ &=-\frac {2 a^5 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 1.29, size = 277, normalized size = 1.38 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b) \left (\frac {2 \left (64 a^2-43 a b+9 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3}-\frac {2 \left (64 a^2+43 a b+9 b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3}+\frac {960 a^5 b \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}-\frac {3 \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )}{2 (a+b)}+\frac {96 \sin ^6\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)}{a-b}-\frac {(19 a+9 b) \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )}{2 (a+b)^2}+\frac {8 (19 a-9 b) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)}{(a-b)^2}\right )}{480 d (a+b \sec (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.89, size = 861, normalized size = 4.28 \[ \left [\frac {46 \, a^{6} b - 68 \, a^{4} b^{3} + 28 \, a^{2} b^{5} - 6 \, b^{7} - 2 \, {\left (8 \, a^{7} + a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 30 \, {\left (a^{6} b - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{7} - a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (a^{5} b \cos \left (d x + c\right )^{4} - 2 \, a^{5} b \cos \left (d x + c\right )^{2} + a^{5} 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 ) - 10 \, {\left (7 \, a^{6} b - 8 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 30 \, {\left (a^{7} - a^{5} b^{2}\right )} \cos \left (d x + c\right )}{30 \, {\left ({\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d\right )} \sin \left (d x + c\right )}, \frac {23 \, a^{6} b - 34 \, a^{4} b^{3} + 14 \, a^{2} b^{5} - 3 \, b^{7} - {\left (8 \, a^{7} + a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left (a^{6} b - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, a^{7} - a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (a^{5} b \cos \left (d x + c\right )^{4} - 2 \, a^{5} b \cos \left (d x + c\right )^{2} + a^{5} 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 ) - 5 \, {\left (7 \, a^{6} b - 8 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (a^{7} - a^{5} b^{2}\right )} \cos \left (d x + c\right )}{15 \, {\left ({\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.35, size = 541, normalized size = 2.69 \[ -\frac {\frac {960 \, {\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^{5} b}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 150 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 420 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 420 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 180 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} + \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.60, size = 282, normalized size = 1.40 \[ \frac {\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{5}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{5}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{3}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b}{3}+\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+10 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 \left (a -b \right )^{3}}-\frac {2 a^{5} b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{160 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5 a +3 b}{96 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {10 a^{2}+8 a b +2 b^{2}}{32 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.69, size = 387, normalized size = 1.93 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,d\,\left (32\,a-32\,b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {4}{3\,\left (32\,a-32\,b\right )}+\frac {32\,a+32\,b}{3\,{\left (32\,a-32\,b\right )}^2}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5}{32\,a-32\,b}+\frac {\left (\frac {4}{32\,a-32\,b}+\frac {32\,a+32\,b}{{\left (32\,a-32\,b\right )}^2}\right )\,\left (32\,a+32\,b\right )}{32\,a-32\,b}\right )}{d}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{5\,\left (a+b\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-5\,a^5+11\,a^4\,b-4\,a^3\,b^2-4\,a^2\,b^3+a\,b^4+b^5\right )}{{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^4-12\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3-3\,b^4\right )}{3\,{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (32\,a^3-96\,a^2\,b+96\,a\,b^2-32\,b^3\right )}-\frac {2\,a^5\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{5/2}}\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{6}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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