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

Optimal. Leaf size=129 \[ -\frac {a^2 \cos ^5(c+d x)}{15 d}-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{21 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 x}{8}-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

1/8*a^2*x-1/15*a^2*cos(d*x+c)^5/d+1/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/12*a^2*cos(d*x+c)^3*sin(d*x+c)/d-1/7*cos(d
*x+c)^5*(a+a*sin(d*x+c))^2/d-1/21*cos(d*x+c)^5*(a^2+a^2*sin(d*x+c))/d

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac {a^2 \cos ^5(c+d x)}{15 d}-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{21 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 x}{8}-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/8 - (a^2*Cos[c + d*x]^5)/(15*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c +
d*x])/(12*d) - (Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*d) - (Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(21*d
)

Rule 8

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

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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2678

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

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac {2}{7} \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac {1}{3} a \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {a^2 \cos ^5(c+d x)}{15 d}-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac {1}{3} a^2 \int \cos ^4(c+d x) \, dx\\ &=-\frac {a^2 \cos ^5(c+d x)}{15 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac {1}{4} a^2 \int \cos ^2(c+d x) \, dx\\ &=-\frac {a^2 \cos ^5(c+d x)}{15 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac {1}{8} a^2 \int 1 \, dx\\ &=\frac {a^2 x}{8}-\frac {a^2 \cos ^5(c+d x)}{15 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 86, normalized size = 0.67 \[ \frac {a^2 (210 \sin (2 (c+d x))-210 \sin (4 (c+d x))-70 \sin (6 (c+d x))-1155 \cos (c+d x)-525 \cos (3 (c+d x))-63 \cos (5 (c+d x))+15 \cos (7 (c+d x))+840 c+840 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(840*c + 840*d*x - 1155*Cos[c + d*x] - 525*Cos[3*(c + d*x)] - 63*Cos[5*(c + d*x)] + 15*Cos[7*(c + d*x)] +
210*Sin[2*(c + d*x)] - 210*Sin[4*(c + d*x)] - 70*Sin[6*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.49, size = 85, normalized size = 0.66 \[ \frac {120 \, a^{2} \cos \left (d x + c\right )^{7} - 336 \, a^{2} \cos \left (d x + c\right )^{5} + 105 \, a^{2} d x - 35 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(120*a^2*cos(d*x + c)^7 - 336*a^2*cos(d*x + c)^5 + 105*a^2*d*x - 35*(8*a^2*cos(d*x + c)^5 - 2*a^2*cos(d*
x + c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.26, size = 123, normalized size = 0.95 \[ \frac {1}{8} \, a^{2} x + \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {3 \, a^{2} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {11 \, a^{2} \cos \left (d x + c\right )}{64 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*x + 1/448*a^2*cos(7*d*x + 7*c)/d - 3/320*a^2*cos(5*d*x + 5*c)/d - 5/64*a^2*cos(3*d*x + 3*c)/d - 11/64*
a^2*cos(d*x + c)/d - 1/96*a^2*sin(6*d*x + 6*c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d + 1/32*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.27, size = 106, normalized size = 0.82 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+2*a^2*(-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)-1/5*a^2*cos(d*x+c)^5)

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maxima [A]  time = 0.35, size = 82, normalized size = 0.64 \[ -\frac {672 \, a^{2} \cos \left (d x + c\right )^{5} - 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360*(672*a^2*cos(d*x + c)^5 - 96*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^2 - 35*(4*sin(2*d*x + 2*c)^3 + 12
*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2)/d

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mupad [B]  time = 10.75, size = 388, normalized size = 3.01 \[ \frac {a^2\,x}{8}-\frac {\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}-\frac {18}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (7\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {7\,c}{8}+\frac {7\,d\,x}{8}-2\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (7\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {7\,c}{8}+\frac {7\,d\,x}{8}-\frac {8}{5}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (21\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {21\,c}{8}+\frac {21\,d\,x}{8}-8\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (21\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {21\,c}{8}+\frac {21\,d\,x}{8}-\frac {14}{5}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (35\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {35\,c}{8}+\frac {35\,d\,x}{8}-2\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (35\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {35\,c}{8}+\frac {35\,d\,x}{8}-16\right )\right )+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x)/8 - ((31*a^2*tan(c/2 + (d*x)/2)^5)/12 - (11*a^2*tan(c/2 + (d*x)/2)^3)/3 - (31*a^2*tan(c/2 + (d*x)/2)^9
)/12 + (11*a^2*tan(c/2 + (d*x)/2)^11)/3 - (a^2*tan(c/2 + (d*x)/2)^13)/4 + a^2*(c/8 + (d*x)/8) - a^2*(c/8 + (d*
x)/8 - 18/35) + tan(c/2 + (d*x)/2)^12*(7*a^2*(c/8 + (d*x)/8) - a^2*((7*c)/8 + (7*d*x)/8 - 2)) + tan(c/2 + (d*x
)/2)^2*(7*a^2*(c/8 + (d*x)/8) - a^2*((7*c)/8 + (7*d*x)/8 - 8/5)) + tan(c/2 + (d*x)/2)^10*(21*a^2*(c/8 + (d*x)/
8) - a^2*((21*c)/8 + (21*d*x)/8 - 8)) + tan(c/2 + (d*x)/2)^4*(21*a^2*(c/8 + (d*x)/8) - a^2*((21*c)/8 + (21*d*x
)/8 - 14/5)) + tan(c/2 + (d*x)/2)^8*(35*a^2*(c/8 + (d*x)/8) - a^2*((35*c)/8 + (35*d*x)/8 - 2)) + tan(c/2 + (d*
x)/2)^6*(35*a^2*(c/8 + (d*x)/8) - a^2*((35*c)/8 + (35*d*x)/8 - 16)) + (a^2*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2
+ (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 8.01, size = 223, normalized size = 1.73 \[ \begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin {\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x*sin(c + d*x)**6/8 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 3*a**2*x*sin(c + d*x)**2*co
s(c + d*x)**4/8 + a**2*x*cos(c + d*x)**6/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + a**2*sin(c + d*x)**3*co
s(c + d*x)**3/(3*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) - 2
*a**2*cos(c + d*x)**7/(35*d) - a**2*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos(c)**4, T
rue))

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