3.344 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=81 \[ \frac{4 (a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}-\frac{4 (a \sin (c+d x)+a)^{m+4}}{a^4 d (m+4)}+\frac{(a \sin (c+d x)+a)^{m+5}}{a^5 d (m+5)} \]

[Out]

(4*(a + a*Sin[c + d*x])^(3 + m))/(a^3*d*(3 + m)) - (4*(a + a*Sin[c + d*x])^(4 + m))/(a^4*d*(4 + m)) + (a + a*S
in[c + d*x])^(5 + m)/(a^5*d*(5 + m))

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Rubi [A]  time = 0.0690724, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac{4 (a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}-\frac{4 (a \sin (c+d x)+a)^{m+4}}{a^4 d (m+4)}+\frac{(a \sin (c+d x)+a)^{m+5}}{a^5 d (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + a*Sin[c + d*x])^(3 + m))/(a^3*d*(3 + m)) - (4*(a + a*Sin[c + d*x])^(4 + m))/(a^4*d*(4 + m)) + (a + a*S
in[c + d*x])^(5 + m)/(a^5*d*(5 + m))

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A]  time = 0.321718, size = 68, normalized size = 0.84 \[ \frac{(a (\sin (c+d x)+1))^{m+3} \left (-\frac{4 a^2 (\sin (c+d x)+1)}{m+4}+\frac{4 a^2}{m+3}+\frac{(a \sin (c+d x)+a)^2}{m+5}\right )}{a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.564, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{5} \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)^5*(a+a*sin(d*x+c))^m,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76582, size = 263, normalized size = 3.25 \begin{align*} \frac{{\left ({\left (m^{2} + 3 \, m\right )} \cos \left (d x + c\right )^{4} + 8 \, m \cos \left (d x + c\right )^{2} +{\left ({\left (m^{2} + 7 \, m + 12\right )} \cos \left (d x + c\right )^{4} + 8 \,{\left (m + 2\right )} \cos \left (d x + c\right )^{2} + 32\right )} \sin \left (d x + c\right ) + 32\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} + 12 \, d m^{2} + 47 \, d m + 60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((m^2 + 3*m)*cos(d*x + c)^4 + 8*m*cos(d*x + c)^2 + ((m^2 + 7*m + 12)*cos(d*x + c)^4 + 8*(m + 2)*cos(d*x + c)^2
+ 32)*sin(d*x + c) + 32)*(a*sin(d*x + c) + a)^m/(d*m^3 + 12*d*m^2 + 47*d*m + 60*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*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [B]  time = 1.09763, size = 397, normalized size = 4.9 \begin{align*} \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 4 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + 4 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} + 7 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{5}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 32 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 36 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m + 12 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{5}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 60 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 80 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2}}{{\left (a^{4} m^{3} + 12 \, a^{4} m^{2} + 47 \, a^{4} m + 60 \, a^{4}\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m*m^2 - 4*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m^2 + 4
*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2*m^2 + 7*(a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m*m - 3
2*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m + 36*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2*m +
12*(a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m - 60*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a + 80*(a
*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2)/((a^4*m^3 + 12*a^4*m^2 + 47*a^4*m + 60*a^4)*a*d)