3.929 \(\int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=108 \[ \frac{3 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac{3 (a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}+\frac{(a \sin (c+d x)+a)^{m+4}}{a^4 d (m+4)}-\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[Out]

-((a + a*Sin[c + d*x])^(1 + m)/(a*d*(1 + m))) + (3*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d*(2 + m)) - (3*(a + a*S
in[c + d*x])^(3 + m))/(a^3*d*(3 + m)) + (a + a*Sin[c + d*x])^(4 + m)/(a^4*d*(4 + m))

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Rubi [A]  time = 0.0969453, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ \frac{3 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac{3 (a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}+\frac{(a \sin (c+d x)+a)^{m+4}}{a^4 d (m+4)}-\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + a*Sin[c + d*x])^(1 + m)/(a*d*(1 + m))) + (3*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d*(2 + m)) - (3*(a + a*S
in[c + d*x])^(3 + m))/(a^3*d*(3 + m)) + (a + a*Sin[c + d*x])^(4 + m)/(a^4*d*(4 + m))

Rule 2833

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.571176, size = 94, normalized size = 0.87 \[ \frac{\left (\left (m^3+6 m^2+11 m+6\right ) \sin ^3(c+d x)-3 \left (m^2+3 m+2\right ) \sin ^2(c+d x)+6 (m+1) \sin (c+d x)-6\right ) (a (\sin (c+d x)+1))^{m+1}}{a d (m+1) (m+2) (m+3) (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(1 + m)*(-6 + 6*(1 + m)*Sin[c + d*x] - 3*(2 + 3*m + m^2)*Sin[c + d*x]^2 + (6 + 11*m +
6*m^2 + m^3)*Sin[c + d*x]^3))/(a*d*(1 + m)*(2 + m)*(3 + m)*(4 + m))

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Maple [F]  time = 1.652, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03799, size = 161, normalized size = 1.49 \begin{align*} \frac{{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} a^{m} \sin \left (d x + c\right )^{4} +{\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (d x + c\right )^{3} - 3 \,{\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} + 6 \, a^{m} m \sin \left (d x + c\right ) - 6 \, a^{m}\right )}{\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((m^3 + 6*m^2 + 11*m + 6)*a^m*sin(d*x + c)^4 + (m^3 + 3*m^2 + 2*m)*a^m*sin(d*x + c)^3 - 3*(m^2 + m)*a^m*sin(d*
x + c)^2 + 6*a^m*m*sin(d*x + c) - 6*a^m)*(sin(d*x + c) + 1)^m/((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*d)

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Fricas [A]  time = 1.80017, size = 336, normalized size = 3.11 \begin{align*} \frac{{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} \cos \left (d x + c\right )^{4} + m^{3} -{\left (2 \, m^{3} + 9 \, m^{2} + 19 \, m + 12\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} +{\left (m^{3} -{\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} + 8 \, m\right )} \sin \left (d x + c\right ) + 8 \, m\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} + 10 \, d m^{3} + 35 \, d m^{2} + 50 \, d m + 24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((m^3 + 6*m^2 + 11*m + 6)*cos(d*x + c)^4 + m^3 - (2*m^3 + 9*m^2 + 19*m + 12)*cos(d*x + c)^2 + 3*m^2 + (m^3 - (
m^3 + 3*m^2 + 2*m)*cos(d*x + c)^2 + 3*m^2 + 8*m)*sin(d*x + c) + 8*m)*(a*sin(d*x + c) + a)^m/(d*m^4 + 10*d*m^3
+ 35*d*m^2 + 50*d*m + 24*d)

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Sympy [A]  time = 70.7406, size = 1547, normalized size = 14.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*(a*sin(c) + a)**m*sin(c)**3*cos(c), Eq(d, 0)), (6*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(6*a**4*d
*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 18*log(sin(c + d*x) + 1)*s
in(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 18
*log(sin(c + d*x) + 1)*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c +
d*x) + 6*a**4*d) + 6*log(sin(c + d*x) + 1)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*s
in(c + d*x) + 6*a**4*d) - 6*sin(c + d*x)**3/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*
sin(c + d*x) + 6*a**4*d) + 9*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*si
n(c + d*x) + 6*a**4*d) + 5/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*
a**4*d), Eq(m, -4)), (-6*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*
x) + 2*a**3*d) - 12*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a
**3*d) - 6*log(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 2*sin(c + d*x
)**3/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 12*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2
+ 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 9/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Eq(m,
-3)), (6*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 6*log(sin(c + d*x) + 1)/(2*a*
*2*d*sin(c + d*x) + 2*a**2*d) + 4*sin(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 3*sin(c + d*x)*cos(c +
d*x)**2/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 3*cos(c + d*x)**2/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 6/(2*a**2*
d*sin(c + d*x) + 2*a**2*d), Eq(m, -2)), (-log(sin(c + d*x) + 1)/(a*d) + sin(c + d*x)**3/(3*a*d) + sin(c + d*x)
/(a*d) + cos(c + d*x)**2/(2*a*d), Eq(m, -1)), (m**3*(a*sin(c + d*x) + a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**
3 + 35*d*m**2 + 50*d*m + 24*d) + m**3*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**4 + 10*d*m**3 + 35*d*m**2
+ 50*d*m + 24*d) + 6*m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 2
4*d) + 3*m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) - 3*m**
2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 11*m*(a*sin(c + d
*x) + a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 2*m*(a*sin(c + d*x) + a)**m*sin
(c + d*x)**3/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) - 3*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d
*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 6*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)/(d*m**4 + 10*d*m**3
+ 35*d*m**2 + 50*d*m + 24*d) + 6*(a*sin(c + d*x) + a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*
d*m + 24*d) - 6*(a*sin(c + d*x) + a)**m/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d), True))

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Giac [B]  time = 1.24764, size = 686, normalized size = 6.35 \begin{align*} \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{3} - 3 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{3} + 3 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{3} -{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m^{3} + 6 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 21 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + 24 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} - 9 \,{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m^{2} + 11 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 42 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 57 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m - 26 \,{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m + 6 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 24 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 36 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} - 24 \,{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3}}{{\left (a^{3} m^{4} + 10 \, a^{3} m^{3} + 35 \, a^{3} m^{2} + 50 \, a^{3} m + 24 \, a^{3}\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m^3 - 3*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a*m^3 + 3
*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m^3 - (a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3*m^3 +
6*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m^2 - 21*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a*m^2
+ 24*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m^2 - 9*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3
*m^2 + 11*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m - 42*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a
*m + 57*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m - 26*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a
^3*m + 6*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m - 24*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a +
36*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2 - 24*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3)/((a
^3*m^4 + 10*a^3*m^3 + 35*a^3*m^2 + 50*a^3*m + 24*a^3)*a*d)