3.4.45 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^m \, dx\) [345]

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

[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 = 0.04, 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 = {2746, 45} \begin {gather*} \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)} \end {gather*}

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 45

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 2746

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 {\text {Subst}\left (\int (a-x) (a+x)^{1+m} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {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 = 0.21, size = 65, normalized size = 1.18 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 (a (1+\sin (c+d x)))^m (-4-m+(2+m) \sin (c+d x))}{d (2+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.13, 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] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (55) = 110\).
time = 0.28, size = 111, normalized size = 2.02 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.39, size = 61, normalized size = 1.11 \begin {gather*} \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} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1114 vs. \(2 (44) = 88\).
time = 3.81, size = 1114, normalized size = 20.25 \begin {gather*} \begin {cases} x \left (a \sin {\left (c \right )} + a\right )^{m} \cos ^{3}{\left (c \right )} & \text {for}\: d = 0 \\- \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {\cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: m = -3 \\\frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2 \sin ^{2}{\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {\cos ^{2}{\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {2}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: m = -2 \\\frac {2 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: m = -1 \\\frac {m^{2} \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {m^{2} \left (a \sin {\left (c + d x \right )} + a\right )^{m} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {2 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{3}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {4 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {5 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {2 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {5 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {4 \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{3}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {6 \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {6 \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {6 \left (a \sin {\left (c + d x \right )} + a\right )^{m} \cos ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} - \frac {2 \left (a \sin {\left (c + d x \right )} + a\right )^{m}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} & \text {otherwise} \end {cases} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (55) = 110\).
time = 5.65, size = 137, normalized size = 2.49 \begin {gather*} -\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 \, {\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 )}^{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}{{\left (a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}\right )} a d} \end {gather*}

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

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Mupad [B]
time = 0.73, size = 85, normalized size = 1.55 \begin {gather*} \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 )} \end {gather*}

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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