3.52 \(\int \cos (a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ \frac {\sin ^3(a+b x)}{3 b} \]

[Out]

1/3*sin(b*x+a)^3/b

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2564, 30} \[ \frac {\sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Sin[a + b*x]^2,x]

[Out]

Sin[a + b*x]^3/(3*b)

Rule 30

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

Rule 2564

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*} \int \cos (a+b x) \sin ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {\sin ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {\sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Sin[a + b*x]^2,x]

[Out]

Sin[a + b*x]^3/(3*b)

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fricas [A]  time = 0.44, size = 21, normalized size = 1.40 \[ -\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/3*(cos(b*x + a)^2 - 1)*sin(b*x + a)/b

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giac [A]  time = 0.23, size = 13, normalized size = 0.87 \[ \frac {\sin \left (b x + a\right )^{3}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/3*sin(b*x + a)^3/b

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maple [A]  time = 0.00, size = 14, normalized size = 0.93 \[ \frac {\sin ^{3}\left (b x +a \right )}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*sin(b*x+a)^2,x)

[Out]

1/3*sin(b*x+a)^3/b

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maxima [A]  time = 0.31, size = 13, normalized size = 0.87 \[ \frac {\sin \left (b x + a\right )^{3}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*sin(b*x + a)^3/b

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mupad [B]  time = 0.02, size = 13, normalized size = 0.87 \[ \frac {{\sin \left (a+b\,x\right )}^3}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*sin(a + b*x)^2,x)

[Out]

sin(a + b*x)^3/(3*b)

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sympy [A]  time = 0.71, size = 20, normalized size = 1.33 \[ \begin {cases} \frac {\sin ^{3}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\relax (a )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)**2,x)

[Out]

Piecewise((sin(a + b*x)**3/(3*b), Ne(b, 0)), (x*sin(a)**2*cos(a), True))

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