3.1094 \(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=103 \[ -\frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {a x}{16}+\frac {b \cos ^7(c+d x)}{7 d}-\frac {b \cos ^5(c+d x)}{5 d} \]

[Out]

1/16*a*x-1/5*b*cos(d*x+c)^5/d+1/7*b*cos(d*x+c)^7/d+1/16*a*cos(d*x+c)*sin(d*x+c)/d+1/24*a*cos(d*x+c)^3*sin(d*x+
c)/d-1/6*a*cos(d*x+c)^5*sin(d*x+c)/d

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Rubi [A]  time = 0.15, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2568, 2635, 8, 2565, 14} \[ -\frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {a x}{16}+\frac {b \cos ^7(c+d x)}{7 d}-\frac {b \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + b*Sin[c + d*x]),x]

[Out]

(a*x)/16 - (b*Cos[c + d*x]^5)/(5*d) + (b*Cos[c + d*x]^7)/(7*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*Cos
[c + d*x]^3*Sin[c + d*x])/(24*d) - (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+b \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} a \int \cos ^4(c+d x) \, dx-\frac {b \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{8} a \int \cos ^2(c+d x) \, dx-\frac {b \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {b \cos ^5(c+d x)}{5 d}+\frac {b \cos ^7(c+d x)}{7 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{16} a \int 1 \, dx\\ &=\frac {a x}{16}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {b \cos ^7(c+d x)}{7 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 88, normalized size = 0.85 \[ \frac {105 a \sin (2 (c+d x))-105 a \sin (4 (c+d x))-35 a \sin (6 (c+d x))+420 a d x-315 b \cos (c+d x)-105 b \cos (3 (c+d x))+21 b \cos (5 (c+d x))+15 b \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + b*Sin[c + d*x]),x]

[Out]

(420*a*d*x - 315*b*Cos[c + d*x] - 105*b*Cos[3*(c + d*x)] + 21*b*Cos[5*(c + d*x)] + 15*b*Cos[7*(c + d*x)] + 105
*a*Sin[2*(c + d*x)] - 105*a*Sin[4*(c + d*x)] - 35*a*Sin[6*(c + d*x)])/(6720*d)

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fricas [A]  time = 0.83, size = 73, normalized size = 0.71 \[ \frac {240 \, b \cos \left (d x + c\right )^{7} - 336 \, b \cos \left (d x + c\right )^{5} + 105 \, a d x - 35 \, {\left (8 \, a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/1680*(240*b*cos(d*x + c)^7 - 336*b*cos(d*x + c)^5 + 105*a*d*x - 35*(8*a*cos(d*x + c)^5 - 2*a*cos(d*x + c)^3
- 3*a*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.20, size = 107, normalized size = 1.04 \[ \frac {1}{16} \, a x + \frac {b \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {b \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {b \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {3 \, b \cos \left (d x + c\right )}{64 \, d} - \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*a*x + 1/448*b*cos(7*d*x + 7*c)/d + 1/320*b*cos(5*d*x + 5*c)/d - 1/64*b*cos(3*d*x + 3*c)/d - 3/64*b*cos(d*
x + c)/d - 1/192*a*sin(6*d*x + 6*c)/d - 1/64*a*sin(4*d*x + 4*c)/d + 1/64*a*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.29, size = 88, normalized size = 0.85 \[ \frac {a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c)),x)

[Out]

1/d*(a*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)+b*(-1/7*si
n(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5))

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maxima [A]  time = 0.39, size = 65, normalized size = 0.63 \[ \frac {35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a + 192 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6720*(35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a + 192*(5*cos(d*x + c)^7 - 7*cos(d*x +
c)^5)*b)/d

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mupad [B]  time = 13.02, size = 181, normalized size = 1.76 \[ \frac {a\,x}{16}-\frac {-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {31\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}-4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {31\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}-\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {4\,b}{35}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + b*sin(c + d*x)),x)

[Out]

(a*x)/16 - ((4*b)/35 + (a*tan(c/2 + (d*x)/2))/8 - (11*a*tan(c/2 + (d*x)/2)^3)/6 + (31*a*tan(c/2 + (d*x)/2)^5)/
24 - (31*a*tan(c/2 + (d*x)/2)^9)/24 + (11*a*tan(c/2 + (d*x)/2)^11)/6 - (a*tan(c/2 + (d*x)/2)^13)/8 + (4*b*tan(
c/2 + (d*x)/2)^2)/5 - (8*b*tan(c/2 + (d*x)/2)^4)/5 + 8*b*tan(c/2 + (d*x)/2)^6 - 4*b*tan(c/2 + (d*x)/2)^8 + 4*b
*tan(c/2 + (d*x)/2)^10)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 5.49, size = 192, normalized size = 1.86 \[ \begin {cases} \frac {a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {b \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 b \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right ) \sin ^{2}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+b*sin(d*x+c)),x)

[Out]

Piecewise((a*x*sin(c + d*x)**6/16 + 3*a*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a*x*sin(c + d*x)**2*cos(c + d
*x)**4/16 + a*x*cos(c + d*x)**6/16 + a*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a*sin(c + d*x)**3*cos(c + d*x)**3
/(6*d) - a*sin(c + d*x)*cos(c + d*x)**5/(16*d) - b*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*b*cos(c + d*x)**7
/(35*d), Ne(d, 0)), (x*(a + b*sin(c))*sin(c)**2*cos(c)**4, True))

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