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

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.57, size = 94, normalized size = 0.87 \[ \frac {\left (-3 \left (m^2+3 m+2\right ) \sin ^2(c+d x)+\left (m^3+6 m^2+11 m+6\right ) \sin ^3(c+d x)+6 (m+1) \sin (c+d x)-6\right ) (a (\sin (c+d x)+1))^{m+1}}{a d (m+1) (m+2) (m+3) (m+4)} \]

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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fricas [A]  time = 0.48, size = 140, normalized size = 1.30 \[ \frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} \cos \left (d x + c\right )^{4} + m^{3} - {\left (2 \, m^{3} + 9 \, m^{2} + 19 \, m + 12\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} + {\left (m^{3} - {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} + 8 \, m\right )} \sin \left (d x + c\right ) + 8 \, m\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} + 10 \, d m^{3} + 35 \, d m^{2} + 50 \, d m + 24 \, d} \]

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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giac [B]  time = 0.28, size = 274, normalized size = 2.54 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{3} \sin \left (d x + c\right )^{4} + {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{3} \sin \left (d x + c\right )^{3} + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} \sin \left (d x + c\right )^{4} + 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} \sin \left (d x + c\right )^{3} + 11 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{4} - 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} \sin \left (d x + c\right )^{2} + 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{3} + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right )^{4} - 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{2} + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right ) - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} d} \]

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

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maple [F]  time = 3.23, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*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 = 0.69, size = 119, normalized size = 1.10 \[ \frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} a^{m} \sin \left (d x + c\right )^{4} + {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (d x + c\right )^{3} - 3 \, {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} + 6 \, a^{m} m \sin \left (d x + c\right ) - 6 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} d} \]

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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mupad [B]  time = 10.74, size = 224, normalized size = 2.07 \[ \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (21\,m-24\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+60\,m\,\sin \left (c+d\,x\right )-32\,m\,\cos \left (2\,c+2\,d\,x\right )+11\,m\,\cos \left (4\,c+4\,d\,x\right )-4\,m\,\sin \left (3\,c+3\,d\,x\right )+18\,m^2\,\sin \left (c+d\,x\right )+6\,m^3\,\sin \left (c+d\,x\right )+6\,m^2+3\,m^3-12\,m^2\,\cos \left (2\,c+2\,d\,x\right )-4\,m^3\,\cos \left (2\,c+2\,d\,x\right )+6\,m^2\,\cos \left (4\,c+4\,d\,x\right )+m^3\,\cos \left (4\,c+4\,d\,x\right )-6\,m^2\,\sin \left (3\,c+3\,d\,x\right )-2\,m^3\,\sin \left (3\,c+3\,d\,x\right )-30\right )}{8\,d\,\left (m^4+10\,m^3+35\,m^2+50\,m+24\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*(sin(c + d*x) + 1))^m*(21*m - 24*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + 60*m*sin(c + d*x) - 32*m*cos(2*c
+ 2*d*x) + 11*m*cos(4*c + 4*d*x) - 4*m*sin(3*c + 3*d*x) + 18*m^2*sin(c + d*x) + 6*m^3*sin(c + d*x) + 6*m^2 + 3
*m^3 - 12*m^2*cos(2*c + 2*d*x) - 4*m^3*cos(2*c + 2*d*x) + 6*m^2*cos(4*c + 4*d*x) + m^3*cos(4*c + 4*d*x) - 6*m^
2*sin(3*c + 3*d*x) - 2*m^3*sin(3*c + 3*d*x) - 30))/(8*d*(50*m + 35*m^2 + 10*m^3 + m^4 + 24))

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sympy [A]  time = 28.58, size = 1508, normalized size = 13.96 \[ \text {result too large to display} \]

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) + 18*sin(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) + 27*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) + 11/(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) + sin(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) - 3*sin(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 + 24*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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