3.1236 \(\int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=123 \[ \frac{a \sin ^{n+1}(c+d x)}{d (n+1)}-\frac{2 a \sin ^{n+3}(c+d x)}{d (n+3)}+\frac{a \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{b \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{2 b \sin ^{n+4}(c+d x)}{d (n+4)}+\frac{b \sin ^{n+6}(c+d x)}{d (n+6)} \]

[Out]

(a*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (b*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - (2*a*Sin[c + d*x]^(3 + n))/(d*(3
+ n)) - (2*b*Sin[c + d*x]^(4 + n))/(d*(4 + n)) + (a*Sin[c + d*x]^(5 + n))/(d*(5 + n)) + (b*Sin[c + d*x]^(6 +
n))/(d*(6 + n))

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Rubi [A]  time = 0.137182, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2837, 766} \[ \frac{a \sin ^{n+1}(c+d x)}{d (n+1)}-\frac{2 a \sin ^{n+3}(c+d x)}{d (n+3)}+\frac{a \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{b \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{2 b \sin ^{n+4}(c+d x)}{d (n+4)}+\frac{b \sin ^{n+6}(c+d x)}{d (n+6)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x]),x]

[Out]

(a*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (b*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - (2*a*Sin[c + d*x]^(3 + n))/(d*(3
+ n)) - (2*b*Sin[c + d*x]^(4 + n))/(d*(4 + n)) + (a*Sin[c + d*x]^(5 + n))/(d*(5 + n)) + (b*Sin[c + d*x]^(6 +
n))/(d*(6 + n))

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{x}{b}\right )^n (a+x) \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a b^4 \left (\frac{x}{b}\right )^n+b^5 \left (\frac{x}{b}\right )^{1+n}-2 a b^4 \left (\frac{x}{b}\right )^{2+n}-2 b^5 \left (\frac{x}{b}\right )^{3+n}+a b^4 \left (\frac{x}{b}\right )^{4+n}+b^5 \left (\frac{x}{b}\right )^{5+n}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{a \sin ^{1+n}(c+d x)}{d (1+n)}+\frac{b \sin ^{2+n}(c+d x)}{d (2+n)}-\frac{2 a \sin ^{3+n}(c+d x)}{d (3+n)}-\frac{2 b \sin ^{4+n}(c+d x)}{d (4+n)}+\frac{a \sin ^{5+n}(c+d x)}{d (5+n)}+\frac{b \sin ^{6+n}(c+d x)}{d (6+n)}\\ \end{align*}

Mathematica [A]  time = 0.195645, size = 97, normalized size = 0.79 \[ \frac{\sin ^{n+1}(c+d x) \left (\frac{a \sin ^4(c+d x)}{n+5}-\frac{2 a \sin ^2(c+d x)}{n+3}+\frac{a}{n+1}+\frac{b \sin ^5(c+d x)}{n+6}-\frac{2 b \sin ^3(c+d x)}{n+4}+\frac{b \sin (c+d x)}{n+2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]^(1 + n)*(a/(1 + n) + (b*Sin[c + d*x])/(2 + n) - (2*a*Sin[c + d*x]^2)/(3 + n) - (2*b*Sin[c + d*x]
^3)/(4 + n) + (a*Sin[c + d*x]^4)/(5 + n) + (b*Sin[c + d*x]^5)/(6 + n)))/d

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Maple [F]  time = 4.69, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+b\sin \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),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(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.95234, size = 713, normalized size = 5.8 \begin{align*} -\frac{{\left ({\left (b n^{5} + 15 \, b n^{4} + 85 \, b n^{3} + 225 \, b n^{2} + 274 \, b n + 120 \, b\right )} \cos \left (d x + c\right )^{6} -{\left (b n^{5} + 11 \, b n^{4} + 41 \, b n^{3} + 61 \, b n^{2} + 30 \, b n\right )} \cos \left (d x + c\right )^{4} - 8 \, b n^{3} - 72 \, b n^{2} - 4 \,{\left (b n^{4} + 9 \, b n^{3} + 23 \, b n^{2} + 15 \, b n\right )} \cos \left (d x + c\right )^{2} - 184 \, b n -{\left ({\left (a n^{5} + 16 \, a n^{4} + 95 \, a n^{3} + 260 \, a n^{2} + 324 \, a n + 144 \, a\right )} \cos \left (d x + c\right )^{4} + 8 \, a n^{3} + 96 \, a n^{2} + 4 \,{\left (a n^{4} + 13 \, a n^{3} + 56 \, a n^{2} + 92 \, a n + 48 \, a\right )} \cos \left (d x + c\right )^{2} + 352 \, a n + 384 \, a\right )} \sin \left (d x + c\right ) - 120 \, b\right )} \sin \left (d x + c\right )^{n}}{d n^{6} + 21 \, d n^{5} + 175 \, d n^{4} + 735 \, d n^{3} + 1624 \, d n^{2} + 1764 \, d n + 720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-((b*n^5 + 15*b*n^4 + 85*b*n^3 + 225*b*n^2 + 274*b*n + 120*b)*cos(d*x + c)^6 - (b*n^5 + 11*b*n^4 + 41*b*n^3 +
61*b*n^2 + 30*b*n)*cos(d*x + c)^4 - 8*b*n^3 - 72*b*n^2 - 4*(b*n^4 + 9*b*n^3 + 23*b*n^2 + 15*b*n)*cos(d*x + c)^
2 - 184*b*n - ((a*n^5 + 16*a*n^4 + 95*a*n^3 + 260*a*n^2 + 324*a*n + 144*a)*cos(d*x + c)^4 + 8*a*n^3 + 96*a*n^2
+ 4*(a*n^4 + 13*a*n^3 + 56*a*n^2 + 92*a*n + 48*a)*cos(d*x + c)^2 + 352*a*n + 384*a)*sin(d*x + c) - 120*b)*sin
(d*x + c)^n/(d*n^6 + 21*d*n^5 + 175*d*n^4 + 735*d*n^3 + 1624*d*n^2 + 1764*d*n + 720*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.17264, size = 512, normalized size = 4.16 \begin{align*} \frac{\frac{{\left (n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} + 4 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} - 2 \, n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + 3 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} - 12 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right ) - 10 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + 8 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right ) + 15 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )\right )} a}{n^{3} + 9 \, n^{2} + 23 \, n + 15} + \frac{{\left (n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} + 6 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} - 2 \, n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + 8 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} - 16 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + 10 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}\right )} b}{n^{3} + 12 \, n^{2} + 44 \, n + 48}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

((n^2*sin(d*x + c)^n*sin(d*x + c)^5 + 4*n*sin(d*x + c)^n*sin(d*x + c)^5 - 2*n^2*sin(d*x + c)^n*sin(d*x + c)^3
+ 3*sin(d*x + c)^n*sin(d*x + c)^5 - 12*n*sin(d*x + c)^n*sin(d*x + c)^3 + n^2*sin(d*x + c)^n*sin(d*x + c) - 10*
sin(d*x + c)^n*sin(d*x + c)^3 + 8*n*sin(d*x + c)^n*sin(d*x + c) + 15*sin(d*x + c)^n*sin(d*x + c))*a/(n^3 + 9*n
^2 + 23*n + 15) + (n^2*sin(d*x + c)^n*sin(d*x + c)^6 + 6*n*sin(d*x + c)^n*sin(d*x + c)^6 - 2*n^2*sin(d*x + c)^
n*sin(d*x + c)^4 + 8*sin(d*x + c)^n*sin(d*x + c)^6 - 16*n*sin(d*x + c)^n*sin(d*x + c)^4 + n^2*sin(d*x + c)^n*s
in(d*x + c)^2 - 24*sin(d*x + c)^n*sin(d*x + c)^4 + 10*n*sin(d*x + c)^n*sin(d*x + c)^2 + 24*sin(d*x + c)^n*sin(
d*x + c)^2)*b/(n^3 + 12*n^2 + 44*n + 48))/d