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

Optimal. Leaf size=55 \[ \frac {2 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac {(a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, 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} \[ \frac {2 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac {(a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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 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])

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])

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 52, normalized size = 0.95 \[ -\frac {(\sin (c+d x)+1)^2 ((m+2) \sin (c+d x)-m-4) (a (\sin (c+d x)+1))^m}{d (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 61, normalized size = 1.11 \[ \frac {{\left (m \cos \left (d x + c\right )^{2} + {\left ({\left (m + 2\right )} \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) + 4\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} + 5 \, d m + 6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(m*cos(d*x + c)^2 + ((m + 2)*cos(d*x + c)^2 + 4)*sin(d*x + c) + 4)*(a*sin(d*x + c) + a)^m/(d*m^2 + 5*d*m + 6*d
)

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giac [B]  time = 1.47, size = 152, normalized size = 2.76 \[ -\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{3} + {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{2} + 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right )^{3} - {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right ) - {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right ) - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{2} + 5 \, m + 6\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*sin(d*x + c) + a)^m*m*sin(d*x + c)^3 + (a*sin(d*x + c) + a)^m*m*sin(d*x + c)^2 + 2*(a*sin(d*x + c) + a)^m
*sin(d*x + c)^3 - (a*sin(d*x + c) + a)^m*m*sin(d*x + c) - (a*sin(d*x + c) + a)^m*m - 6*(a*sin(d*x + c) + a)^m*
sin(d*x + c) - 4*(a*sin(d*x + c) + a)^m)/((m^2 + 5*m + 6)*d)

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maple [F]  time = 1.74, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.93, size = 111, normalized size = 2.02 \[ -\frac {\frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 2 \, a^{m} m \sin \left (d x + c\right ) + 2 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} - \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m + 1}}{a {\left (m + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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) - (a*sin(d*x + c) + a)^(m + 1)/(a*(m + 1)))/d

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mupad [B]  time = 0.73, size = 85, normalized size = 1.55 \[ \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (2\,m+18\,\sin \left (c+d\,x\right )+2\,\sin \left (3\,c+3\,d\,x\right )+m\,\sin \left (c+d\,x\right )-2\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+m\,\sin \left (3\,c+3\,d\,x\right )+16\right )}{4\,d\,\left (m^2+5\,m+6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*(sin(c + d*x) + 1))^m*(2*m + 18*sin(c + d*x) + 2*sin(3*c + 3*d*x) + m*sin(c + d*x) - 2*m*(2*sin(c + d*x)^2
- 1) + m*sin(3*c + 3*d*x) + 16))/(4*d*(5*m + m^2 + 6))

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sympy [A]  time = 21.47, size = 1114, normalized size = 20.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*(a*sin(c) + a)**m*cos(c)**3, 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) - 2*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) - cos(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) - 2/(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) - 2*sin(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)), (2*tan(c/2 + d*x/2)**3
/(a*d*tan(c/2 + d*x/2)**4 + 2*a*d*tan(c/2 + d*x/2)**2 + a*d) - 2*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**4
+ 2*a*d*tan(c/2 + d*x/2)**2 + a*d) + 2*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**4 + 2*a*d*tan(c/2 + d*x/2)**2 +
a*d), Eq(m, -1)), (m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)*cos(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*
d) + m**2*(a*sin(c + d*x) + a)**m*cos(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)**3/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 4*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) + 5*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)*cos(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) + 5*m*(a*sin(c + d*x)
+ a)**m*cos(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 4*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**
3 + 6*d*m**2 + 11*d*m + 6*d) + 6*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) +
6*(a*sin(c + d*x) + a)**m*sin(c + d*x)*cos(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 6*(a*sin(c + d*x)
+ a)**m*cos(c + d*x)**2/(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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