3.4.51 \(\int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [351]

Optimal. Leaf size=65 \[ \frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 65, 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 = {2915, 12, 76} \begin {gather*} -\frac {a \sin ^6(c+d x)}{6 d}-\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

(a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^4)/(4*d) - (a*Sin[c + d*x]^5)/(5*d) - (a*Sin[c + d*x]^6)/(6*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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {\text {Subst}\left (\int \frac {(a-x) x^2 (a+x)^2}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int (a-x) x^2 (a+x)^2 \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\text {Subst}\left (\int \left (a^3 x^2+a^2 x^3-a x^4-x^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 51, normalized size = 0.78 \begin {gather*} \frac {a \left (-45 \cos (2 (c+d x))+5 \cos (6 (c+d x))+32 (7+3 \cos (2 (c+d x))) \sin ^3(c+d x)\right )}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

(a*(-45*Cos[2*(c + d*x)] + 5*Cos[6*(c + d*x)] + 32*(7 + 3*Cos[2*(c + d*x)])*Sin[c + d*x]^3))/(960*d)

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Maple [A]
time = 0.14, size = 74, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )}{d}\)	\(74\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )}{d}\)	\(74\)
risch	\(\frac {a \sin \left (d x +c \right )}{8 d}+\frac {a \cos \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (3 d x +3 c \right )}{48 d}-\frac {3 a \cos \left (2 d x +2 c \right )}{64 d}\)	\(74\)
norman	\(\frac {\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))+a*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*
cos(d*x+c)^4))

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Maxima [A]
time = 0.29, size = 50, normalized size = 0.77 \begin {gather*} -\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3)/d

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Fricas [A]
time = 0.38, size = 62, normalized size = 0.95 \begin {gather*} \frac {10 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(10*a*cos(d*x + c)^6 - 15*a*cos(d*x + c)^4 - 4*(3*a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - 2*a)*sin(d*x + c)
)/d

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Sympy [A]
time = 0.45, size = 90, normalized size = 1.38 \begin {gather*} \begin {cases} \frac {2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \cos ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((2*a*sin(c + d*x)**5/(15*d) + a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - a*sin(c + d*x)**2*cos(c + d*
x)**4/(4*d) - a*cos(c + d*x)**6/(12*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**2*cos(c)**3, True))

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Giac [A]
time = 0.51, size = 50, normalized size = 0.77 \begin {gather*} -\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3)/d

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Mupad [B]
time = 0.06, size = 49, normalized size = 0.75 \begin {gather*} \frac {-\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*sin(c + d*x)^2*(a + a*sin(c + d*x)),x)

[Out]

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

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