Optimal. Leaf size=122 \[ -\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)}{b^5}+\frac {a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac {\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac {a \csc ^4(x)}{4 b^2}-\frac {\csc ^5(x)}{5 b}+\frac {\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}-\frac {\log (\sin (x))}{a} \]
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Rubi [A]
time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3970, 908}
\begin {gather*} \frac {\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}+\frac {a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac {\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}-\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)}{b^5}+\frac {a \csc ^4(x)}{4 b^2}-\frac {\log (\sin (x))}{a}-\frac {\csc ^5(x)}{5 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 908
Rule 3970
Rubi steps
\begin {align*} \int \frac {\cot ^7(x)}{a+b \csc (x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^3}{x (a+x)} \, dx,x,b \csc (x)\right )}{b^6}\\ &=\frac {\text {Subst}\left (\int \left (-a^4 \left (1+\frac {3 b^2 \left (-a^2+b^2\right )}{a^4}\right )+\frac {b^6}{a x}+a \left (a^2-3 b^2\right ) x-\left (a^2-3 b^2\right ) x^2+a x^3-x^4+\frac {\left (a^2-b^2\right )^3}{a (a+x)}\right ) \, dx,x,b \csc (x)\right )}{b^6}\\ &=-\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)}{b^5}+\frac {a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac {\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac {a \csc ^4(x)}{4 b^2}-\frac {\csc ^5(x)}{5 b}+\frac {\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}-\frac {\log (\sin (x))}{a}\\ \end {align*}
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Mathematica [A]
time = 0.31, size = 132, normalized size = 1.08 \begin {gather*} \frac {-60 b \left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)+30 a b^2 \left (a^2-3 b^2\right ) \csc ^2(x)-20 b^3 \left (a^2-3 b^2\right ) \csc ^3(x)+15 a b^4 \csc ^4(x)-12 b^5 \csc ^5(x)-60 a \left (a^4-3 a^2 b^2+3 b^4\right ) \log (\sin (x))+\frac {60 \left (a^2-b^2\right )^3 \log (b+a \sin (x))}{a}}{60 b^6} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 149, normalized size = 1.22
method | result | size |
default | \(\frac {\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \ln \left (b +a \sin \left (x \right )\right )}{b^{6} a}-\frac {1}{5 b \sin \left (x \right )^{5}}-\frac {a^{2}-3 b^{2}}{3 b^{3} \sin \left (x \right )^{3}}-\frac {a^{4}-3 a^{2} b^{2}+3 b^{4}}{b^{5} \sin \left (x \right )}+\frac {a}{4 b^{2} \sin \left (x \right )^{4}}+\frac {\left (a^{2}-3 b^{2}\right ) a}{2 b^{4} \sin \left (x \right )^{2}}-\frac {\left (a^{4}-3 a^{2} b^{2}+3 b^{4}\right ) a \ln \left (\sin \left (x \right )\right )}{b^{6}}\) | \(149\) |
risch | \(\frac {i x}{a}-\frac {2 i \left (45 i a \,b^{3} {\mathrm e}^{8 i x}-15 i a^{3} b \,{\mathrm e}^{8 i x}+15 a^{4} {\mathrm e}^{9 i x}-45 a^{2} b^{2} {\mathrm e}^{9 i x}+45 b^{4} {\mathrm e}^{9 i x}-45 i a \,b^{3} {\mathrm e}^{2 i x}+15 i a^{3} b \,{\mathrm e}^{2 i x}-60 a^{4} {\mathrm e}^{7 i x}+160 a^{2} b^{2} {\mathrm e}^{7 i x}-120 b^{4} {\mathrm e}^{7 i x}+105 i a \,b^{3} {\mathrm e}^{4 i x}-45 i a^{3} b \,{\mathrm e}^{4 i x}+90 a^{4} {\mathrm e}^{5 i x}-230 a^{2} b^{2} {\mathrm e}^{5 i x}+198 b^{4} {\mathrm e}^{5 i x}-105 i a \,b^{3} {\mathrm e}^{6 i x}+45 i a^{3} b \,{\mathrm e}^{6 i x}-60 a^{4} {\mathrm e}^{3 i x}+160 a^{2} b^{2} {\mathrm e}^{3 i x}-120 b^{4} {\mathrm e}^{3 i x}+15 \,{\mathrm e}^{i x} a^{4}-45 \,{\mathrm e}^{i x} a^{2} b^{2}+45 \,{\mathrm e}^{i x} b^{4}\right )}{15 b^{5} \left ({\mathrm e}^{2 i x}-1\right )^{5}}-\frac {a^{5} \ln \left ({\mathrm e}^{2 i x}-1\right )}{b^{6}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i x}-1\right )}{b^{4}}-\frac {3 a \ln \left ({\mathrm e}^{2 i x}-1\right )}{b^{2}}+\frac {a^{5} \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{b^{6}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{b^{4}}+\frac {3 a \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a}\) | \(438\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 149, normalized size = 1.22 \begin {gather*} -\frac {{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (\sin \left (x\right )\right )}{b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \sin \left (x\right ) + b\right )}{a b^{6}} + \frac {15 \, a b^{3} \sin \left (x\right ) - 60 \, {\left (a^{4} - 3 \, a^{2} b^{2} + 3 \, b^{4}\right )} \sin \left (x\right )^{4} - 12 \, b^{4} + 30 \, {\left (a^{3} b - 3 \, a b^{3}\right )} \sin \left (x\right )^{3} - 20 \, {\left (a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (x\right )^{2}}{60 \, b^{5} \sin \left (x\right )^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 327 vs.
\(2 (114) = 228\).
time = 2.90, size = 327, normalized size = 2.68 \begin {gather*} -\frac {60 \, a^{5} b - 160 \, a^{3} b^{3} + 132 \, a b^{5} + 60 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (x\right )^{4} - 20 \, {\left (6 \, a^{5} b - 17 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (x\right )^{2} - 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} + {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{4} - 2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \log \left (a \sin \left (x\right ) + b\right ) \sin \left (x\right ) + 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{4} - 2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (x\right )\right ) \sin \left (x\right ) - 15 \, {\left (2 \, a^{4} b^{2} - 5 \, a^{2} b^{4} - 2 \, {\left (a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{60 \, {\left (a b^{6} \cos \left (x\right )^{4} - 2 \, a b^{6} \cos \left (x\right )^{2} + a b^{6}\right )} \sin \left (x\right )} \end {gather*}
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 \begin {gather*} \int \frac {\cot ^{7}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 155, normalized size = 1.27 \begin {gather*} -\frac {{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left ({\left | \sin \left (x\right ) \right |}\right )}{b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a b^{6}} + \frac {15 \, a b^{4} \sin \left (x\right ) - 12 \, b^{5} - 60 \, {\left (a^{4} b - 3 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \left (x\right )^{4} + 30 \, {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (x\right )^{3} - 20 \, {\left (a^{2} b^{3} - 3 \, b^{5}\right )} \sin \left (x\right )^{2}}{60 \, b^{6} \sin \left (x\right )^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.87, size = 284, normalized size = 2.33 \begin {gather*} {\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {3}{32\,b}-\frac {a^2}{24\,b^3}\right )-\frac {19\,\mathrm {tan}\left (\frac {x}{2}\right )}{16\,b}+\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{160\,b}-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (\frac {a}{32\,b^2}+\frac {a\,\left (\frac {9}{32\,b}-\frac {a^2}{8\,b^3}\right )}{b}\right )+\frac {11\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,b^3}+\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{64\,b^2}-\frac {a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{2\,b^5}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (a^5-3\,a^3\,b^2+3\,a\,b^4\right )}{b^6}-\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^5\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (10\,a\,b^3-4\,a^3\,b\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (3\,b^4-\frac {4\,a^2\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (16\,a^4-44\,a^2\,b^2+38\,b^4\right )+\frac {b^4}{5}-\frac {a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{2}\right )}{32\,b^5}+\frac {\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )\,{\left (a^2-b^2\right )}^3}{a\,b^6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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