Optimal. Leaf size=72 \[ \frac {3 \sin (a-c) \tanh ^{-1}(\sin (b x+c))}{2 b}+\frac {\cos (a-c) \sec (b x+c)}{b}-\frac {\sin (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac {\cos (a+b x)}{b} \]
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Rubi [A] time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4579, 4576, 2638, 3770, 2606, 8, 2611} \[ \frac {3 \sin (a-c) \tanh ^{-1}(\sin (b x+c))}{2 b}+\frac {\cos (a-c) \sec (b x+c)}{b}-\frac {\sin (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac {\cos (a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2606
Rule 2611
Rule 2638
Rule 3770
Rule 4576
Rule 4579
Rubi steps
\begin {align*} \int \cos (a+b x) \tan ^3(c+b x) \, dx &=-\left (\sin (a-c) \int \sec (c+b x) \tan ^2(c+b x) \, dx\right )+\int \sin (a+b x) \tan ^2(c+b x) \, dx\\ &=-\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b}+\cos (a-c) \int \sec (c+b x) \tan (c+b x) \, dx+\frac {1}{2} \sin (a-c) \int \sec (c+b x) \, dx-\int \cos (a+b x) \tan (c+b x) \, dx\\ &=\frac {\tanh ^{-1}(\sin (c+b x)) \sin (a-c)}{2 b}-\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b}+\frac {\cos (a-c) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+b x))}{b}+\sin (a-c) \int \sec (c+b x) \, dx-\int \sin (a+b x) \, dx\\ &=\frac {\cos (a+b x)}{b}+\frac {\cos (a-c) \sec (c+b x)}{b}+\frac {3 \tanh ^{-1}(\sin (c+b x)) \sin (a-c)}{2 b}-\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 70, normalized size = 0.97 \[ \frac {\sec ^2(b x+c) (2 \cos (a-b x-2 c)+\cos (a+3 b x+2 c)+5 \cos (a+b x))+12 \sin (a-c) \tanh ^{-1}\left (\cos (c) \tan \left (\frac {b x}{2}\right )+\sin (c)\right )}{4 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.49, size = 366, normalized size = 5.08 \[ \frac {16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \cos \left (b x + a\right ) + \frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (b x + a\right ) \tan \left (b x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.96, size = 181, normalized size = 2.51 \[ \frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {3 \,{\mathrm e}^{i \left (3 b x +5 a +2 c \right )}+{\mathrm e}^{i \left (3 b x +3 a +4 c \right )}+{\mathrm e}^{i \left (b x +5 a \right )}+3 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}}{2 b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.54, size = 1027, normalized size = 14.26 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]
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 \[ \int \cos {\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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