Optimal. Leaf size=70 \[ -\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d} \]
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Rubi [A] time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3512, 723, 637} \[ -\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 637
Rule 723
Rule 3512
Rubi steps
\begin {align*} \int \cos ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d}+\frac {\left (2 \left (a^2+b^2\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {a+x}{\left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{3 b d}\\ &=-\frac {2 \left (a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{3 d}-\frac {\cos ^3(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.38, size = 81, normalized size = 1.16 \[ \frac {\left (b^3-3 a^2 b\right ) \cos (3 (c+d x))-9 b \left (a^2+b^2\right ) \cos (c+d x)+2 a \sin (c+d x) \left (\left (a^2-3 b^2\right ) \cos (2 (c+d x))+5 a^2+3 b^2\right )}{12 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.54, size = 77, normalized size = 1.10 \[ -\frac {3 \, b^{3} \cos \left (d x + c\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{3} + 3 \, a b^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.49, size = 75, normalized size = 1.07 \[ \frac {-\frac {b^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )-a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 77, normalized size = 1.10 \[ -\frac {3 \, a^{2} b \cos \left (d x + c\right )^{3} - 3 \, a b^{2} \sin \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.71, size = 104, normalized size = 1.49 \[ \frac {\frac {\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {2\,\sin \left (c+d\,x\right )\,a^3}{3}-a^2\,b\,{\cos \left (c+d\,x\right )}^3-\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\,a\,b^2+\frac {b^3\,{\cos \left (c+d\,x\right )}^3}{3}-b^3\,\cos \left (c+d\,x\right )}{d} \]
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 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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