Optimal. Leaf size=94 \[ \frac {2 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac {\left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x))}{d}-\frac {(b \cos (c+d x)-a \sin (c+d x))^5}{5 d} \]
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Rubi [A] time = 0.05, antiderivative size = 94, 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 = {3072, 194} \[ \frac {2 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac {\left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x))}{d}-\frac {(b \cos (c+d x)-a \sin (c+d x))^5}{5 d} \]
Antiderivative was successfully verified.
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Rule 194
Rule 3072
Rubi steps
\begin {align*} \int (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=-\frac {\operatorname {Subst}\left (\int \left (a^2+b^2-x^2\right )^2 \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (a^4 \left (1+\frac {2 a^2 b^2+b^4}{a^4}\right )-2 a^2 \left (1+\frac {b^2}{a^2}\right ) x^2+x^4\right ) \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x))}{d}+\frac {2 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac {(b \cos (c+d x)-a \sin (c+d x))^5}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 156, normalized size = 1.66 \[ \frac {150 a \left (a^2+b^2\right )^2 \sin (c+d x)-150 b \left (a^2+b^2\right )^2 \cos (c+d x)+25 a \left (a^4-2 a^2 b^2-3 b^4\right ) \sin (3 (c+d x))+3 a \left (a^4-10 a^2 b^2+5 b^4\right ) \sin (5 (c+d x))+25 b \left (-3 a^4-2 a^2 b^2+b^4\right ) \cos (3 (c+d x))-3 b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos (5 (c+d x))}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 155, normalized size = 1.65 \[ -\frac {15 \, b^{5} \cos \left (d x + c\right ) + 3 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - {\left (8 \, a^{5} + 20 \, a^{3} b^{2} + 15 \, a b^{4} + 3 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{5} + 5 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.33, size = 187, normalized size = 1.99 \[ -\frac {{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, {\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - b^{5}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {5 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac {{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, {\left (a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 175, normalized size = 1.86 \[ \frac {-\frac {b^{5} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )+10 a^{2} b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+10 a^{3} b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-a^{4} b \left (\cos ^{5}\left (d x +c \right )\right )+\frac {a^{5} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 172, normalized size = 1.83 \[ -\frac {a^{4} b \cos \left (d x + c\right )^{5}}{d} + \frac {a b^{4} \sin \left (d x + c\right )^{5}}{d} + \frac {{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{5}}{15 \, d} - \frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{3} b^{2}}{3 \, d} + \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b^{3}}{3 \, d} - \frac {{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} b^{5}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.94, size = 248, normalized size = 2.64 \[ \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^5\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^5\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^5-\frac {15\,a^4\,b\,{\cos \left (c+d\,x\right )}^5}{2}-15\,\sin \left (c+d\,x\right )\,a^3\,b^2\,{\cos \left (c+d\,x\right )}^4+5\,\sin \left (c+d\,x\right )\,a^3\,b^2\,{\cos \left (c+d\,x\right )}^2+10\,\sin \left (c+d\,x\right )\,a^3\,b^2+15\,a^2\,b^3\,{\cos \left (c+d\,x\right )}^5-25\,a^2\,b^3\,{\cos \left (c+d\,x\right )}^3+\frac {15\,\sin \left (c+d\,x\right )\,a\,b^4\,{\cos \left (c+d\,x\right )}^4}{2}-15\,\sin \left (c+d\,x\right )\,a\,b^4\,{\cos \left (c+d\,x\right )}^2+\frac {15\,\sin \left (c+d\,x\right )\,a\,b^4}{2}-\frac {3\,b^5\,{\cos \left (c+d\,x\right )}^5}{2}+5\,b^5\,{\cos \left (c+d\,x\right )}^3-\frac {15\,b^5\,\cos \left (c+d\,x\right )}{2}\right )}{15\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.14, size = 267, normalized size = 2.84 \[ \begin {cases} \frac {8 a^{5} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{5} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{4} b \cos ^{5}{\left (c + d x \right )}}{d} + \frac {4 a^{3} b^{2} \sin ^{5}{\left (c + d x \right )}}{3 d} + \frac {10 a^{3} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {10 a^{2} b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a^{2} b^{3} \cos ^{5}{\left (c + d x \right )}}{3 d} + \frac {a b^{4} \sin ^{5}{\left (c + d x \right )}}{d} - \frac {b^{5} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {4 b^{5} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 b^{5} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{5} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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