3.930 \(\int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=80 \[ -\frac{2 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}+\frac{(a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}+\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0842618, antiderivative size = 80, 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} \[ -\frac{2 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}+\frac{(a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}+\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

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

Mathematica [A]  time = 0.187009, size = 77, normalized size = 0.96 \[ -\frac{(a (\sin (c+d x)+1))^{m+1} \left (\left (m^2+3 m+2\right ) \cos (2 (c+d x))+4 (m+1) \sin (c+d x)-m^2-3 m-6\right )}{2 a d (m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

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

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Maple [F]  time = 1.47, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+a*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^m,x)

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Maxima [A]  time = 1.04014, size = 113, normalized size = 1.41 \begin{align*} \frac{{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (d x + c\right )^{3} +{\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 2 \, a^{m} m \sin \left (d x + c\right ) + 2 \, a^{m}\right )}{\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

((m^2 + 3*m + 2)*a^m*sin(d*x + c)^3 + (m^2 + m)*a^m*sin(d*x + c)^2 - 2*a^m*m*sin(d*x + c) + 2*a^m)*(sin(d*x +
c) + 1)^m/((m^3 + 6*m^2 + 11*m + 6)*d)

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Fricas [A]  time = 1.8318, size = 217, normalized size = 2.71 \begin{align*} -\frac{{\left ({\left (m^{2} + m\right )} \cos \left (d x + c\right )^{2} - m^{2} +{\left ({\left (m^{2} + 3 \, m + 2\right )} \cos \left (d x + c\right )^{2} - m^{2} - m - 2\right )} \sin \left (d x + c\right ) - m - 2\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} + 6 \, d m^{2} + 11 \, d m + 6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

-((m^2 + m)*cos(d*x + c)^2 - m^2 + ((m^2 + 3*m + 2)*cos(d*x + c)^2 - m^2 - m - 2)*sin(d*x + c) - m - 2)*(a*sin
(d*x + c) + a)^m/(d*m^3 + 6*d*m^2 + 11*d*m + 6*d)

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Sympy [A]  time = 21.4815, size = 756, normalized size = 9.45 \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)**2*(a+a*sin(d*x+c))**m,x)

[Out]

Piecewise((x*(a*sin(c) + a)**m*sin(c)**2*cos(c), Eq(d, 0)), (2*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) + 4*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) + 2*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) + 4*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) + 3/(2*a*
*3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Eq(m, -3)), (-2*log(sin(c + d*x) + 1)*sin(c + d*x)/(
a**2*d*sin(c + d*x) + a**2*d) - 2*log(sin(c + d*x) + 1)/(a**2*d*sin(c + d*x) + a**2*d) - sin(c + d*x)**3/(a**2
*d*sin(c + d*x) + a**2*d) - sin(c + d*x)*cos(c + d*x)**2/(a**2*d*sin(c + d*x) + a**2*d) - cos(c + d*x)**2/(a**
2*d*sin(c + d*x) + a**2*d) - 2/(a**2*d*sin(c + d*x) + a**2*d), Eq(m, -2)), (log(sin(c + d*x) + 1)/(a*d) - sin(
c + d*x)/(a*d) - cos(c + d*x)**2/(2*a*d), Eq(m, -1)), (m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 +
6*d*m**2 + 11*d*m + 6*d) + m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 3
*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + m*(a*sin(c + d*x) + a)**m*sin(
c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) - 2*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)/(d*m**3 + 6*d*m**2 +
11*d*m + 6*d) + 2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 2*(a*sin(c + d
*x) + a)**m/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d), True))

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Giac [B]  time = 1.25146, size = 387, normalized size = 4.84 \begin{align*} \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 2 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} +{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} + 3 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 8 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 5 \,{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m + 2 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 6 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 6 \,{\left (a \sin \left (d x + c\right ) + a\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2}}{{\left (a^{2} m^{3} + 6 \, a^{2} m^{2} + 11 \, a^{2} m + 6 \, a^{2}\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)^2*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

((a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*m^2 - 2*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a*m^2 + (
a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^2*m^2 + 3*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*m - 8*(a*
sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a*m + 5*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^2*m + 2*(a*si
n(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m - 6*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a + 6*(a*sin(d*x +
c) + a)*(a*sin(d*x + c) + a)^m*a^2)/((a^2*m^3 + 6*a^2*m^2 + 11*a^2*m + 6*a^2)*a*d)