3.341 \(\int \cos ^3(a+b x) (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=50 \[ \frac{(c \sin (a+b x))^{m+1}}{b c (m+1)}-\frac{(c \sin (a+b x))^{m+3}}{b c^3 (m+3)} \]

[Out]

(c*Sin[a + b*x])^(1 + m)/(b*c*(1 + m)) - (c*Sin[a + b*x])^(3 + m)/(b*c^3*(3 + m))

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Rubi [A]  time = 0.0506607, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2564, 14} \[ \frac{(c \sin (a+b x))^{m+1}}{b c (m+1)}-\frac{(c \sin (a+b x))^{m+3}}{b c^3 (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*(c*Sin[a + b*x])^m,x]

[Out]

(c*Sin[a + b*x])^(1 + m)/(b*c*(1 + m)) - (c*Sin[a + b*x])^(3 + m)/(b*c^3*(3 + m))

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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) (c \sin (a+b x))^m \, dx &=\frac{\operatorname{Subst}\left (\int x^m \left (1-\frac{x^2}{c^2}\right ) \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^m-\frac{x^{2+m}}{c^2}\right ) \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac{(c \sin (a+b x))^{1+m}}{b c (1+m)}-\frac{(c \sin (a+b x))^{3+m}}{b c^3 (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.0845519, size = 48, normalized size = 0.96 \[ \frac{\sin (a+b x) ((m+1) \cos (2 (a+b x))+m+5) (c \sin (a+b x))^m}{2 b (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*(c*Sin[a + b*x])^m,x]

[Out]

((5 + m + (1 + m)*Cos[2*(a + b*x)])*Sin[a + b*x]*(c*Sin[a + b*x])^m)/(2*b*(1 + m)*(3 + m))

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Maple [F]  time = 0.801, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{3} \left ( c\sin \left ( bx+a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*(c*sin(b*x+a))^m,x)

[Out]

int(cos(b*x+a)^3*(c*sin(b*x+a))^m,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*(c*sin(b*x+a))^m,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.45199, size = 113, normalized size = 2.26 \begin{align*} \frac{{\left ({\left (m + 1\right )} \cos \left (b x + a\right )^{2} + 2\right )} \left (c \sin \left (b x + a\right )\right )^{m} \sin \left (b x + a\right )}{b m^{2} + 4 \, b m + 3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*(c*sin(b*x+a))^m,x, algorithm="fricas")

[Out]

((m + 1)*cos(b*x + a)^2 + 2)*(c*sin(b*x + a))^m*sin(b*x + a)/(b*m^2 + 4*b*m + 3*b)

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Sympy [A]  time = 13.1929, size = 530, normalized size = 10.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*(c*sin(b*x+a))**m,x)

[Out]

Piecewise((x*(c*sin(a))**m*cos(a)**3, Eq(b, 0)), ((-log(sin(a + b*x))/b - cos(a + b*x)**2/(2*b*sin(a + b*x)**2
))/c**3, Eq(m, -3)), ((-log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2
+ b*x/2)**2 + b) - 2*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b
*x/2)**2 + b) - log(tan(a/2 + b*x/2)**2 + 1)/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a
/2 + b*x/2))*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + 2*log(tan(a/2 + b*x/2
))*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a/2 + b*x/2))/(b*tan(a/
2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) - 2*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*
x/2)**2 + b))/c, Eq(m, -1)), (c**m*m*sin(a + b*x)*sin(a + b*x)**m*cos(a + b*x)**2/(b*m**2 + 4*b*m + 3*b) + 2*c
**m*sin(a + b*x)**3*sin(a + b*x)**m/(b*m**2 + 4*b*m + 3*b) + 3*c**m*sin(a + b*x)*sin(a + b*x)**m*cos(a + b*x)*
*2/(b*m**2 + 4*b*m + 3*b), True))

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Giac [B]  time = 1.13056, size = 159, normalized size = 3.18 \begin{align*} -\frac{\left (c \sin \left (b x + a\right )\right )^{m} c^{3} m \sin \left (b x + a\right )^{3} + \left (c \sin \left (b x + a\right )\right )^{m} c^{3} \sin \left (b x + a\right )^{3} - \left (c \sin \left (b x + a\right )\right )^{m} c^{3} m \sin \left (b x + a\right ) - 3 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{3} \sin \left (b x + a\right )}{{\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*(c*sin(b*x+a))^m,x, algorithm="giac")

[Out]

-((c*sin(b*x + a))^m*c^3*m*sin(b*x + a)^3 + (c*sin(b*x + a))^m*c^3*sin(b*x + a)^3 - (c*sin(b*x + a))^m*c^3*m*s
in(b*x + a) - 3*(c*sin(b*x + a))^m*c^3*sin(b*x + a))/((c^2*m^2 + 4*c^2*m + 3*c^2)*b*c)