Optimal. Leaf size=151 \[ \frac {8 \sin (c+d x)}{15 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac {4 (b \cos (c+d x)-a \sin (c+d x))}{15 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5} \]
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Rubi [A] time = 0.07, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3076, 3075} \[ \frac {8 \sin (c+d x)}{15 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac {4 (b \cos (c+d x)-a \sin (c+d x))}{15 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5} \]
Antiderivative was successfully verified.
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Rule 3075
Rule 3076
Rubi steps
\begin {align*} \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx &=-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {4 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx}{5 \left (a^2+b^2\right )}\\ &=-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {4 (b \cos (c+d x)-a \sin (c+d x))}{15 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {8 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{15 \left (a^2+b^2\right )^2}\\ &=-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {4 (b \cos (c+d x)-a \sin (c+d x))}{15 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {8 \sin (c+d x)}{15 a \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.52, size = 182, normalized size = 1.21 \[ \frac {10 a^4 \sin (c+d x)+5 a^4 \sin (3 (c+d x))+a^4 \sin (5 (c+d x))+\left (4 a b^3-4 a^3 b\right ) \cos (5 (c+d x))+20 a^2 b^2 \sin (c+d x)-6 a^2 b^2 \sin (5 (c+d x))-10 a b \left (a^2+b^2\right ) \cos (3 (c+d x))+10 b^4 \sin (c+d x)-5 b^4 \sin (3 (c+d x))+b^4 \sin (5 (c+d x))}{30 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 2.05, size = 441, normalized size = 2.92 \[ -\frac {8 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left ({\left (a^{11} - 7 \, a^{9} b^{2} - 22 \, a^{7} b^{4} - 14 \, a^{5} b^{6} + 5 \, a^{3} b^{8} + 5 \, a b^{10}\right )} d \cos \left (d x + c\right )^{5} + 10 \, {\left (a^{9} b^{2} + 2 \, a^{7} b^{4} - 2 \, a^{3} b^{8} - a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 5 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left ({\left (5 \, a^{10} b + 5 \, a^{8} b^{3} - 14 \, a^{6} b^{5} - 22 \, a^{4} b^{7} - 7 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 12 \, a^{4} b^{7} + 2 \, a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 118, normalized size = 0.78 \[ -\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 10 \, b^{4} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 5 \, a b^{3} \tan \left (d x + c\right ) + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4}}{15 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{5} b^{5} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 125, normalized size = 0.83 \[ \frac {-\frac {a^{4}+2 a^{2} b^{2}+b^{4}}{5 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{5}}+\frac {a \left (a^{2}+b^{2}\right )}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{4}}-\frac {6 a^{2}+2 b^{2}}{3 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{3}}+\frac {2 a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {1}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.79, size = 174, normalized size = 1.15 \[ -\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4} + 10 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} + 5 \, {\left (3 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{15 \, {\left (b^{10} \tan \left (d x + c\right )^{5} + 5 \, a b^{9} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{8} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{7} \tan \left (d x + c\right )^{2} + 5 \, a^{4} b^{6} \tan \left (d x + c\right ) + a^{5} b^{5}\right )} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.15, size = 470, normalized size = 3.11 \[ \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{a}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2-6\,b^2\right )}{3\,a^3}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2-6\,b^2\right )}{3\,a^3}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2}-\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (29\,a^4-112\,a^2\,b^2+24\,b^4\right )}{15\,a^5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (60\,a^4\,b-160\,a^2\,b^3+32\,b^5\right )+a^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^5-40\,a^3\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (5\,a^5-40\,a^3\,b^2\right )+10\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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