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

Optimal. Leaf size=45 \[ \frac {(a \sin (c+d x)+a)^5}{5 a^2 d}-\frac {(a \sin (c+d x)+a)^4}{4 a d} \]

[Out]

-1/4*(a+a*sin(d*x+c))^4/a/d+1/5*(a+a*sin(d*x+c))^5/a^2/d

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Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2833, 12, 43} \[ \frac {(a \sin (c+d x)+a)^5}{5 a^2 d}-\frac {(a \sin (c+d x)+a)^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 30, normalized size = 0.67 \[ \frac {a^3 (\sin (c+d x)+1)^4 (4 \sin (c+d x)-1)}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[c + d*x])^4*(-1 + 4*Sin[c + d*x]))/(20*d)

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fricas [A]  time = 0.46, size = 71, normalized size = 1.58 \[ \frac {15 \, a^{3} \cos \left (d x + c\right )^{4} - 40 \, a^{3} \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 7 \, a^{3} \cos \left (d x + c\right )^{2} + 6 \, a^{3}\right )} \sin \left (d x + c\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(15*a^3*cos(d*x + c)^4 - 40*a^3*cos(d*x + c)^2 + 4*(a^3*cos(d*x + c)^4 - 7*a^3*cos(d*x + c)^2 + 6*a^3)*si
n(d*x + c))/d

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giac [A]  time = 0.16, size = 58, normalized size = 1.29 \[ \frac {4 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} + 10 \, a^{3} \sin \left (d x + c\right )^{2}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 + 10*a^3*sin(d*x + c)^2)/d

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maple [A]  time = 0.10, size = 57, normalized size = 1.27 \[ \frac {\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+a^{3} \left (\sin ^{3}\left (d x +c \right )\right )+\frac {a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*a^3*sin(d*x+c)^5+3/4*a^3*sin(d*x+c)^4+a^3*sin(d*x+c)^3+1/2*a^3*sin(d*x+c)^2)

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maxima [A]  time = 0.33, size = 58, normalized size = 1.29 \[ \frac {4 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} + 10 \, a^{3} \sin \left (d x + c\right )^{2}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 + 10*a^3*sin(d*x + c)^2)/d

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mupad [B]  time = 0.06, size = 56, normalized size = 1.24 \[ \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^4}{4}+a^3\,{\sin \left (c+d\,x\right )}^3+\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.55, size = 76, normalized size = 1.69 \[ \begin {cases} \frac {a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \sin {\relax (c )} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*sin(c + d*x)**5/(5*d) + 3*a**3*sin(c + d*x)**4/(4*d) + a**3*sin(c + d*x)**3/d + a**3*sin(c + d
*x)**2/(2*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)*cos(c), True))

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