\(\int \cos (a+b x) \sin ^4(a+b x) \, dx\) [83]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 15 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b} \]

[Out]

1/5*sin(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2644, 30} \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b} \]

[In]

Int[Cos[a + b*x]*Sin[a + b*x]^4,x]

[Out]

Sin[a + b*x]^5/(5*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
 !(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^4 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\sin ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[a + b*x]^4,x]

[Out]

Sin[a + b*x]^5/(5*b)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method result size
derivativedivides \(\frac {\sin ^{5}\left (b x +a \right )}{5 b}\) \(14\)
default \(\frac {\sin ^{5}\left (b x +a \right )}{5 b}\) \(14\)
norman \(\frac {32 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{5}}\) \(32\)
parallelrisch \(\frac {10 \sin \left (b x +a \right )+\sin \left (5 b x +5 a \right )-5 \sin \left (3 b x +3 a \right )}{80 b}\) \(35\)
risch \(\frac {\sin \left (b x +a \right )}{8 b}+\frac {\sin \left (5 b x +5 a \right )}{80 b}-\frac {\sin \left (3 b x +3 a \right )}{16 b}\) \(41\)

[In]

int(cos(b*x+a)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/5*sin(b*x+a)^5/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (13) = 26\).

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{5 \, b} \]

[In]

integrate(cos(b*x+a)*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/5*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sin(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\begin {cases} \frac {\sin ^{5}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)*sin(b*x+a)**4,x)

[Out]

Piecewise((sin(a + b*x)**5/(5*b), Ne(b, 0)), (x*sin(a)**4*cos(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{5}}{5 \, b} \]

[In]

integrate(cos(b*x+a)*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/5*sin(b*x + a)^5/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{5}}{5 \, b} \]

[In]

integrate(cos(b*x+a)*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/5*sin(b*x + a)^5/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos (a+b x) \sin ^4(a+b x) \, dx=\frac {{\sin \left (a+b\,x\right )}^5}{5\,b} \]

[In]

int(cos(a + b*x)*sin(a + b*x)^4,x)

[Out]

sin(a + b*x)^5/(5*b)