3.1093 \(\int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=127 \[ \frac {a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {b \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {b \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 b \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 b x}{128} \]

[Out]

3/128*b*x-1/5*a*cos(d*x+c)^5/d+1/7*a*cos(d*x+c)^7/d+3/128*b*cos(d*x+c)*sin(d*x+c)/d+1/64*b*cos(d*x+c)^3*sin(d*
x+c)/d-1/16*b*cos(d*x+c)^5*sin(d*x+c)/d-1/8*b*cos(d*x+c)^5*sin(d*x+c)^3/d

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Rubi [A]  time = 0.19, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2565, 14, 2568, 2635, 8} \[ \frac {a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {b \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {b \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 b \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 b x}{128} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x]),x]

[Out]

(3*b*x)/128 - (a*Cos[c + d*x]^5)/(5*d) + (a*Cos[c + d*x]^7)/(7*d) + (3*b*Cos[c + d*x]*Sin[c + d*x])/(128*d) +
(b*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) - (b*Cos[c + d*x]^5*Sin[c + d*x])/(16*d) - (b*Cos[c + d*x]^5*Sin[c + d*
x]^3)/(8*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 ^3(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx+b \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {b \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{8} (3 b) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {b \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {b \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{16} b \int \cos ^4(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^7(c+d x)}{7 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {b \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {b \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{64} (3 b) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^7(c+d x)}{7 d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{128 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {b \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {b \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{128} (3 b) \int 1 \, dx\\ &=\frac {3 b x}{128}-\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^7(c+d x)}{7 d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{128 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {b \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {b \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 77, normalized size = 0.61 \[ \frac {-1680 a \cos (c+d x)-560 a \cos (3 (c+d x))+112 a \cos (5 (c+d x))+80 a \cos (7 (c+d x))-280 b \sin (4 (c+d x))+35 b \sin (8 (c+d x))+840 b d x}{35840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x]),x]

[Out]

(840*b*d*x - 1680*a*Cos[c + d*x] - 560*a*Cos[3*(c + d*x)] + 112*a*Cos[5*(c + d*x)] + 80*a*Cos[7*(c + d*x)] - 2
80*b*Sin[4*(c + d*x)] + 35*b*Sin[8*(c + d*x)])/(35840*d)

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fricas [A]  time = 0.75, size = 84, normalized size = 0.66 \[ \frac {640 \, a \cos \left (d x + c\right )^{7} - 896 \, a \cos \left (d x + c\right )^{5} + 105 \, b d x + 35 \, {\left (16 \, b \cos \left (d x + c\right )^{7} - 24 \, b \cos \left (d x + c\right )^{5} + 2 \, b \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4480*(640*a*cos(d*x + c)^7 - 896*a*cos(d*x + c)^5 + 105*b*d*x + 35*(16*b*cos(d*x + c)^7 - 24*b*cos(d*x + c)^
5 + 2*b*cos(d*x + c)^3 + 3*b*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.22, size = 92, normalized size = 0.72 \[ \frac {3}{128} \, b x + \frac {a \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {3 \, a \cos \left (d x + c\right )}{64 \, d} + \frac {b \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {b \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

3/128*b*x + 1/448*a*cos(7*d*x + 7*c)/d + 1/320*a*cos(5*d*x + 5*c)/d - 1/64*a*cos(3*d*x + 3*c)/d - 3/64*a*cos(d
*x + c)/d + 1/1024*b*sin(8*d*x + 8*c)/d - 1/128*b*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.28, size = 106, normalized size = 0.83 \[ \frac {a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c)),x)

[Out]

1/d*(a*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+b*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*co
s(d*x+c)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128*c))

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maxima [A]  time = 0.38, size = 61, normalized size = 0.48 \[ \frac {1024 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a + 35 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b}{35840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/35840*(1024*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a + 35*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4
*c))*b)/d

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mupad [B]  time = 13.17, size = 209, normalized size = 1.65 \[ \frac {3\,b\,x}{128}-\frac {-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}-\frac {23\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {333\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}-\frac {671\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {671\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{64}+\frac {32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}-\frac {333\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{64}-\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {23\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{64}+\frac {32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {4\,a}{35}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*sin(c + d*x)^3*(a + b*sin(c + d*x)),x)

[Out]

(3*b*x)/128 - ((4*a)/35 + (3*b*tan(c/2 + (d*x)/2))/64 + (32*a*tan(c/2 + (d*x)/2)^2)/35 - (4*a*tan(c/2 + (d*x)/
2)^4)/5 + (32*a*tan(c/2 + (d*x)/2)^6)/5 + 4*a*tan(c/2 + (d*x)/2)^8 + 4*a*tan(c/2 + (d*x)/2)^12 + (23*b*tan(c/2
+ (d*x)/2)^3)/64 - (333*b*tan(c/2 + (d*x)/2)^5)/64 + (671*b*tan(c/2 + (d*x)/2)^7)/64 - (671*b*tan(c/2 + (d*x)
/2)^9)/64 + (333*b*tan(c/2 + (d*x)/2)^11)/64 - (23*b*tan(c/2 + (d*x)/2)^13)/64 - (3*b*tan(c/2 + (d*x)/2)^15)/6
4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^8)

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sympy [A]  time = 9.16, size = 248, normalized size = 1.95 \[ \begin {cases} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 a \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac {3 b x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 b x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {9 b x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 b x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 b \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 b \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} - \frac {11 b \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} - \frac {3 b \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right ) \sin ^{3}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**3*(a+b*sin(d*x+c)),x)

[Out]

Piecewise((-a*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*a*cos(c + d*x)**7/(35*d) + 3*b*x*sin(c + d*x)**8/128 +
3*b*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 9*b*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*b*x*sin(c + d*x)**2*c
os(c + d*x)**6/32 + 3*b*x*cos(c + d*x)**8/128 + 3*b*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 11*b*sin(c + d*x)**
5*cos(c + d*x)**3/(128*d) - 11*b*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) - 3*b*sin(c + d*x)*cos(c + d*x)**7/(1
28*d), Ne(d, 0)), (x*(a + b*sin(c))*sin(c)**3*cos(c)**4, True))

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