Optimal. Leaf size=134 \[ \frac{x \cos ^4(a+b x)}{8 b^2}+\frac{3 x \cos ^2(a+b x)}{8 b^2}-\frac{\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac{15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac{x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac{3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac{15 x}{64 b^2}+\frac{x^3}{8} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.108289, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3311, 30, 2635, 8} \[ \frac{x \cos ^4(a+b x)}{8 b^2}+\frac{3 x \cos ^2(a+b x)}{8 b^2}-\frac{\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac{15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac{x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac{3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac{15 x}{64 b^2}+\frac{x^3}{8} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3311
Rule 30
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int x^2 \cos ^4(a+b x) \, dx &=\frac{x \cos ^4(a+b x)}{8 b^2}+\frac{x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac{3}{4} \int x^2 \cos ^2(a+b x) \, dx-\frac{\int \cos ^4(a+b x) \, dx}{8 b^2}\\ &=\frac{3 x \cos ^2(a+b x)}{8 b^2}+\frac{x \cos ^4(a+b x)}{8 b^2}+\frac{3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac{x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac{3 \int x^2 \, dx}{8}-\frac{3 \int \cos ^2(a+b x) \, dx}{32 b^2}-\frac{3 \int \cos ^2(a+b x) \, dx}{8 b^2}\\ &=\frac{x^3}{8}+\frac{3 x \cos ^2(a+b x)}{8 b^2}+\frac{x \cos ^4(a+b x)}{8 b^2}-\frac{15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac{3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac{x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}-\frac{3 \int 1 \, dx}{64 b^2}-\frac{3 \int 1 \, dx}{16 b^2}\\ &=-\frac{15 x}{64 b^2}+\frac{x^3}{8}+\frac{3 x \cos ^2(a+b x)}{8 b^2}+\frac{x \cos ^4(a+b x)}{8 b^2}-\frac{15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac{3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac{x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}\\ \end{align*}
Mathematica [A] time = 0.176424, size = 92, normalized size = 0.69 \[ \frac{64 b^2 x^2 \sin (2 (a+b x))+8 b^2 x^2 \sin (4 (a+b x))-32 \sin (2 (a+b x))-\sin (4 (a+b x))+64 b x \cos (2 (a+b x))+4 b x \cos (4 (a+b x))+32 b^3 x^3}{256 b^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.029, size = 241, normalized size = 1.8 \begin{align*}{\frac{1}{{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \left ({\frac{\sin \left ( bx+a \right ) }{4} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\cos \left ( bx+a \right ) }{2}} \right ) }+{\frac{3\,bx}{8}}+{\frac{3\,a}{8}} \right ) +{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{8}}-{\frac{\sin \left ( bx+a \right ) }{32} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\cos \left ( bx+a \right ) }{2}} \right ) }-{\frac{15\,bx}{64}}-{\frac{15\,a}{64}}+{\frac{ \left ( 3\,bx+3\,a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{8}}-{\frac{3\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{16}}-{\frac{ \left ( bx+a \right ) ^{3}}{4}}-2\,a \left ( \left ( bx+a \right ) \left ( 1/4\, \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+3/2\,\cos \left ( bx+a \right ) \right ) \sin \left ( bx+a \right ) +3/8\,bx+3/8\,a \right ) -3/16\, \left ( bx+a \right ) ^{2}+1/16\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}+3/16\, \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) +{a}^{2} \left ({\frac{\sin \left ( bx+a \right ) }{4} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\cos \left ( bx+a \right ) }{2}} \right ) }+{\frac{3\,bx}{8}}+{\frac{3\,a}{8}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.06702, size = 254, normalized size = 1.9 \begin{align*} \frac{32 \,{\left (b x + a\right )}^{3} + 8 \,{\left (12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 4 \,{\left (24 \,{\left (b x + a\right )}^{2} + 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right ) + 16 \, \cos \left (2 \, b x + 2 \, a\right )\right )} a + 4 \,{\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 64 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (8 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{256 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.30974, size = 217, normalized size = 1.62 \begin{align*} \frac{8 \, b^{3} x^{3} + 8 \, b x \cos \left (b x + a\right )^{4} + 24 \, b x \cos \left (b x + a\right )^{2} - 15 \, b x +{\left (2 \,{\left (8 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right )^{3} + 3 \,{\left (8 \, b^{2} x^{2} - 5\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{64 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [A] time = 6.46099, size = 209, normalized size = 1.56 \begin{align*} \begin{cases} \frac{x^{3} \sin ^{4}{\left (a + b x \right )}}{8} + \frac{x^{3} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{x^{3} \cos ^{4}{\left (a + b x \right )}}{8} + \frac{3 x^{2} \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} + \frac{5 x^{2} \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac{15 x \sin ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{3 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{2}} + \frac{17 x \cos ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{15 \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{64 b^{3}} - \frac{17 \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \cos ^{4}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.12872, size = 113, normalized size = 0.84 \begin{align*} \frac{1}{8} \, x^{3} + \frac{x \cos \left (4 \, b x + 4 \, a\right )}{64 \, b^{2}} + \frac{x \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac{{\left (8 \, b^{2} x^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]