\(\int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx\) [356]

Optimal result

Integrand size = 19, antiderivative size = 50 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=-\frac {(d \cos (a+b x))^{1+n}}{b d (1+n)}+\frac {(d \cos (a+b x))^{3+n}}{b d^3 (3+n)} \]

[Out]

-(d*cos(b*x+a))^(1+n)/b/d/(1+n)+(d*cos(b*x+a))^(3+n)/b/d^3/(3+n)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2645, 14} \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\frac {(d \cos (a+b x))^{n+3}}{b d^3 (n+3)}-\frac {(d \cos (a+b x))^{n+1}}{b d (n+1)} \]

[In]

Int[(d*Cos[a + b*x])^n*Sin[a + b*x]^3,x]

[Out]

-((d*Cos[a + b*x])^(1 + n)/(b*d*(1 + n))) + (d*Cos[a + b*x])^(3 + n)/(b*d^3*(3 + 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]]

Rule 2645

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int x^n \left (1-\frac {x^2}{d^2}\right ) \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {\text {Subst}\left (\int \left (x^n-\frac {x^{2+n}}{d^2}\right ) \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {(d \cos (a+b x))^{1+n}}{b d (1+n)}+\frac {(d \cos (a+b x))^{3+n}}{b d^3 (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\frac {\cos (a+b x) (d \cos (a+b x))^n (-5-n+(1+n) \cos (2 (a+b x)))}{2 b (1+n) (3+n)} \]

[In]

Integrate[(d*Cos[a + b*x])^n*Sin[a + b*x]^3,x]

[Out]

(Cos[a + b*x]*(d*Cos[a + b*x])^n*(-5 - n + (1 + n)*Cos[2*(a + b*x)]))/(2*b*(1 + n)*(3 + n))

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(-\frac {\left (\left (-n -1\right ) \cos \left (3 b x +3 a \right )+\cos \left (b x +a \right ) \left (n +9\right )\right ) \left (d \cos \left (b x +a \right )\right )^{n}}{4 b \left (3+n \right ) \left (1+n \right )}\)	\(52\)
derivativedivides	\(\frac {\left (\cos ^{3}\left (b x +a \right )\right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (3+n \right )}-\frac {\cos \left (b x +a \right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (1+n \right )}\)	\(59\)
default	\(\frac {\left (\cos ^{3}\left (b x +a \right )\right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (3+n \right )}-\frac {\cos \left (b x +a \right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (1+n \right )}\)	\(59\)

[In]

int((d*cos(b*x+a))^n*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*((-n-1)*cos(3*b*x+3*a)+cos(b*x+a)*(n+9))*(d*cos(b*x+a))^n/b/(3+n)/(1+n)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\frac {{\left ({\left (n + 1\right )} \cos \left (b x + a\right )^{3} - {\left (n + 3\right )} \cos \left (b x + a\right )\right )} \left (d \cos \left (b x + a\right )\right )^{n}}{b n^{2} + 4 \, b n + 3 \, b} \]

[In]

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

[Out]

((n + 1)*cos(b*x + a)^3 - (n + 3)*cos(b*x + a))*(d*cos(b*x + a))^n/(b*n^2 + 4*b*n + 3*b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 688 vs. \(2 (37) = 74\).

Time = 1.17 (sec) , antiderivative size = 688, normalized size of antiderivative = 13.76 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*cos(b*x+a))**n*sin(b*x+a)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\frac {\frac {d^{n} \cos \left (b x + a\right )^{n} \cos \left (b x + a\right )^{3}}{n + 3} - \frac {\left (d \cos \left (b x + a\right )\right )^{n + 1}}{d {\left (n + 1\right )}}}{b} \]

[In]

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

[Out]

(d^n*cos(b*x + a)^n*cos(b*x + a)^3/(n + 3) - (d*cos(b*x + a))^(n + 1)/(d*(n + 1)))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (50) = 100\).

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.34 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=\frac {\left (d \cos \left (b x + a\right )\right )^{n} d^{3} n \cos \left (b x + a\right )^{3} + \left (d \cos \left (b x + a\right )\right )^{n} d^{3} \cos \left (b x + a\right )^{3} - \left (d \cos \left (b x + a\right )\right )^{n} d^{3} n \cos \left (b x + a\right ) - 3 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{3} \cos \left (b x + a\right )}{{\left (d^{2} n^{2} + 4 \, d^{2} n + 3 \, d^{2}\right )} b d} \]

[In]

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

[Out]

((d*cos(b*x + a))^n*d^3*n*cos(b*x + a)^3 + (d*cos(b*x + a))^n*d^3*cos(b*x + a)^3 - (d*cos(b*x + a))^n*d^3*n*co
s(b*x + a) - 3*(d*cos(b*x + a))^n*d^3*cos(b*x + a))/((d^2*n^2 + 4*d^2*n + 3*d^2)*b*d)

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx=-\frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^n\,\left (9\,\cos \left (a+b\,x\right )-\cos \left (3\,a+3\,b\,x\right )+n\,\cos \left (a+b\,x\right )-n\,\cos \left (3\,a+3\,b\,x\right )\right )}{4\,b\,\left (n^2+4\,n+3\right )} \]

[In]

int(sin(a + b*x)^3*(d*cos(a + b*x))^n,x)

[Out]

-((d*cos(a + b*x))^n*(9*cos(a + b*x) - cos(3*a + 3*b*x) + n*cos(a + b*x) - n*cos(3*a + 3*b*x)))/(4*b*(4*n + n^
2 + 3))