3.356 \(\int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx\)
Optimal. Leaf size=50 \[ \frac {(d \cos (a+b x))^{n+3}}{b d^3 (n+3)}-\frac {(d \cos (a+b x))^{n+1}}{b d (n+1)} \]
[Out]
-(d*cos(b*x+a))^(1+n)/b/d/(1+n)+(d*cos(b*x+a))^(3+n)/b/d^3/(3+n)
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Rubi [A] time = 0.05, antiderivative size = 50, normalized size of antiderivative = 1.00,
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 =
{2565, 14} \[ \frac {(d \cos (a+b x))^{n+3}}{b d^3 (n+3)}-\frac {(d \cos (a+b x))^{n+1}}{b d (n+1)} \]
Antiderivative was successfully verified.
[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 2565
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*} \int (d \cos (a+b x))^n \sin ^3(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int x^n \left (1-\frac {x^2}{d^2}\right ) \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac {\operatorname {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*}
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Mathematica [A] time = 0.12, size = 50, normalized size = 1.00 \[ \frac {\cos (a+b x) ((n+1) \cos (2 (a+b x))-n-5) (d \cos (a+b x))^n}{2 b (n+1) (n+3)} \]
Antiderivative was successfully verified.
[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))
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fricas [A] time = 0.76, size = 50, normalized size = 1.00 \[ \frac {{\left ({\left (n + 1\right )} \cos \left (b x + a\right )^{3} - {\left (n + 3\right )} \cos \left (b x + a\right )\right )} \left (d \cos \left (b x + a\right )\right )^{n}}{b n^{2} + 4 \, b n + 3 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [A] time = 1.73, size = 100, normalized size = 2.00 \[ \frac {\frac {\left (d \cos \left (b x + a\right )\right )^{n} n \cos \left (b x + a\right )^{3}}{b} + \frac {\left (d \cos \left (b x + a\right )\right )^{n} \cos \left (b x + a\right )^{3}}{b} - \frac {\left (d \cos \left (b x + a\right )\right )^{n} n \cos \left (b x + a\right )}{b} - \frac {3 \, \left (d \cos \left (b x + a\right )\right )^{n} \cos \left (b x + a\right )}{b}}{n^{2} + 4 \, n + 3} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*cos(b*x+a))^n*sin(b*x+a)^3,x, algorithm="giac")
[Out]
((d*cos(b*x + a))^n*n*cos(b*x + a)^3/b + (d*cos(b*x + a))^n*cos(b*x + a)^3/b - (d*cos(b*x + a))^n*n*cos(b*x +
a)/b - 3*(d*cos(b*x + a))^n*cos(b*x + a)/b)/(n^2 + 4*n + 3)
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maple [C] time = 2.14, size = 1076, normalized size = 21.52 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*cos(b*x+a))^n*sin(b*x+a)^3,x)
[Out]
-1/8*exp(I*(b*x+a))^(-n)*d^n*(1/2)^n*(exp(2*I*(b*x+a))+1)^n/(3+n)/(1+n)/b*(n+9)*exp(-1/2*I*(Pi*n*csgn(I*cos(b*
x+a))^3-Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))^2-Pi*n*csgn(I*cos(b*x+a))^2*csgn(I*exp(-I*(b*x+a)
))+Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))*csgn(I*exp(-I*(b*x+a)))-Pi*n*csgn(I*cos(b*x+a))*csgn(I
*d*cos(b*x+a))^2+Pi*n*csgn(I*d)*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a))+Pi*n*csgn(I*d*cos(b*x+a))^3-Pi*n*csgn(
I*d)*csgn(I*d*cos(b*x+a))^2+2*b*x+2*a))-1/8*exp(I*(b*x+a))^(-n)*d^n*(1/2)^n*(exp(2*I*(b*x+a))+1)^n/(3+n)/(1+n)
/b*(n+9)*exp(1/2*I*(-Pi*n*csgn(I*cos(b*x+a))^3+Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))^2+Pi*n*csg
n(I*cos(b*x+a))^2*csgn(I*exp(-I*(b*x+a)))-Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))*csgn(I*exp(-I*(
b*x+a)))+Pi*n*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a))^2-Pi*n*csgn(I*d)*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a))
-Pi*n*csgn(I*d*cos(b*x+a))^3+Pi*n*csgn(I*d)*csgn(I*d*cos(b*x+a))^2+2*b*x+2*a))+1/8/(b*n+3*b)*(exp(2*I*(b*x+a))
+1)^n*(1/2)^n*d^n*exp(I*(b*x+a))^(-n)*exp(-1/2*I*(Pi*n*csgn(I*cos(b*x+a))^3-Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*
csgn(I*cos(b*x+a))^2-Pi*n*csgn(I*cos(b*x+a))^2*csgn(I*exp(-I*(b*x+a)))-Pi*n*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*
x+a))^2+Pi*n*csgn(I*d)*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a))+Pi*n*csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*
x+a))*csgn(I*exp(-I*(b*x+a)))+Pi*n*csgn(I*d*cos(b*x+a))^3-Pi*n*csgn(I*d)*csgn(I*d*cos(b*x+a))^2+6*b*x+6*a))+1/
8*exp(I*(b*x+a))^(-n)*d^n*(1/2)^n*(exp(2*I*(b*x+a))+1)^n/(b*n+3*b)*exp(1/2*I*(-Pi*n*csgn(I*cos(b*x+a))^3+Pi*n*
csgn(I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))^2+Pi*n*csgn(I*cos(b*x+a))^2*csgn(I*exp(-I*(b*x+a)))-Pi*n*csgn(
I*(exp(2*I*(b*x+a))+1))*csgn(I*cos(b*x+a))*csgn(I*exp(-I*(b*x+a)))+Pi*n*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a)
)^2-Pi*n*csgn(I*d)*csgn(I*cos(b*x+a))*csgn(I*d*cos(b*x+a))-Pi*n*csgn(I*d*cos(b*x+a))^3+Pi*n*csgn(I*d)*csgn(I*d
*cos(b*x+a))^2+6*b*x+6*a))
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maxima [A] time = 0.37, size = 52, normalized size = 1.04 \[ \frac {\frac {d^{n} \cos \left (b x + a\right )^{n} \cos \left (b x + a\right )^{3}}{n + 3} - \frac {\left (d \cos \left (b x + a\right )\right )^{n + 1}}{d {\left (n + 1\right )}}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
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mupad [B] time = 0.91, size = 65, normalized size = 1.30 \[ -\frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^n\,\left (9\,\cos \left (a+b\,x\right )-\cos \left (3\,a+3\,b\,x\right )+n\,\cos \left (a+b\,x\right )-n\,\cos \left (3\,a+3\,b\,x\right )\right )}{4\,b\,\left (n^2+4\,n+3\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[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))
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sympy [A] time = 12.78, size = 694, normalized size = 13.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)), (-
d**n*n*sin(a + b*x)**2*cos(a + b*x)*cos(a + b*x)**n/(b*n**2 + 4*b*n + 3*b) - 3*d**n*sin(a + b*x)**2*cos(a + b*
x)*cos(a + b*x)**n/(b*n**2 + 4*b*n + 3*b) - 2*d**n*cos(a + b*x)**3*cos(a + b*x)**n/(b*n**2 + 4*b*n + 3*b), Tru
e))
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