3.286 \(\int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=117 \[ \frac {3 a^3 \cos ^5(c+d x)}{5 d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac {7 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7 a^3 x}{16} \]

[Out]

7/16*a^3*x-4/3*a^3*cos(d*x+c)^3/d+3/5*a^3*cos(d*x+c)^5/d+7/16*a^3*cos(d*x+c)*sin(d*x+c)/d-7/8*a^3*cos(d*x+c)^3
*sin(d*x+c)/d-1/6*a^3*cos(d*x+c)^3*sin(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 133, normalized size of antiderivative = 1.14, 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} \[ -\frac {7 a^3 \cos ^3(c+d x)}{24 d}-\frac {7 \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{40 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7 a^3 x}{16}-\frac {a \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{10 d}-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^3*x)/16 - (7*a^3*Cos[c + d*x]^3)/(24*d) + (7*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Cos[c + d*x]^3*(a
+ a*Sin[c + d*x])^2)/(10*d) - (Cos[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(6*d) - (7*Cos[c + d*x]^3*(a^3 + a^3*Si
n[c + d*x]))/(40*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 ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx &=-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac {1}{2} \int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac {1}{10} (7 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (7 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {7 a^3 \cos ^3(c+d x)}{24 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (7 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {7 a^3 \cos ^3(c+d x)}{24 d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{16} \left (7 a^3\right ) \int 1 \, dx\\ &=\frac {7 a^3 x}{16}-\frac {7 a^3 \cos ^3(c+d x)}{24 d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.38, size = 76, normalized size = 0.65 \[ \frac {a^3 (-15 \sin (2 (c+d x))-105 \sin (4 (c+d x))+5 \sin (6 (c+d x))-600 \cos (c+d x)-140 \cos (3 (c+d x))+36 \cos (5 (c+d x))+450 c+420 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(450*c + 420*d*x - 600*Cos[c + d*x] - 140*Cos[3*(c + d*x)] + 36*Cos[5*(c + d*x)] - 15*Sin[2*(c + d*x)] -
105*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 85, normalized size = 0.73 \[ \frac {144 \, a^{3} \cos \left (d x + c\right )^{5} - 320 \, a^{3} \cos \left (d x + c\right )^{3} + 105 \, a^{3} d x + 5 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 50 \, a^{3} \cos \left (d x + c\right )^{3} + 21 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(144*a^3*cos(d*x + c)^5 - 320*a^3*cos(d*x + c)^3 + 105*a^3*d*x + 5*(8*a^3*cos(d*x + c)^5 - 50*a^3*cos(d*
x + c)^3 + 21*a^3*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.20, size = 106, normalized size = 0.91 \[ \frac {7}{16} \, a^{3} x + \frac {3 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {7 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {5 \, a^{3} \cos \left (d x + c\right )}{8 \, d} + \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {7 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a^3*x + 3/80*a^3*cos(5*d*x + 5*c)/d - 7/48*a^3*cos(3*d*x + 3*c)/d - 5/8*a^3*cos(d*x + c)/d + 1/192*a^3*si
n(6*d*x + 6*c)/d - 7/64*a^3*sin(4*d*x + 4*c)/d - 1/64*a^3*sin(2*d*x + 2*c)/d

________________________________________________________________________________________

maple [A]  time = 0.17, size = 156, normalized size = 1.33 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+3 a^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*
c)+3*a^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+3*a^3*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)
*sin(d*x+c)+1/8*d*x+1/8*c)-1/3*a^3*cos(d*x+c)^3)

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 106, normalized size = 0.91 \[ -\frac {320 \, a^{3} \cos \left (d x + c\right )^{3} - 192 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 90 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(320*a^3*cos(d*x + c)^3 - 192*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 + 5*(4*sin(2*d*x + 2*c)^3 - 12*
d*x - 12*c + 3*sin(4*d*x + 4*c))*a^3 - 90*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^3)/d

________________________________________________________________________________________

mupad [B]  time = 10.72, size = 349, normalized size = 2.98 \[ \frac {7\,a^3\,x}{16}-\frac {\frac {37\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {37\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}-\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {a^3\,\left (105\,c+105\,d\,x\right )}{240}-\frac {a^3\,\left (105\,c+105\,d\,x-352\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^3\,\left (630\,c+630\,d\,x-480\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^3\,\left (630\,c+630\,d\,x-1632\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^3\,\left (1575\,c+1575\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^3\,\left (1575\,c+1575\,d\,x-4320\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^3\,\left (2100\,c+2100\,d\,x-3520\right )}{240}\right )+\frac {7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a^3*x)/16 - ((37*a^3*tan(c/2 + (d*x)/2)^7)/4 - (37*a^3*tan(c/2 + (d*x)/2)^5)/4 - (73*a^3*tan(c/2 + (d*x)/2)
^3)/24 + (73*a^3*tan(c/2 + (d*x)/2)^9)/24 - (7*a^3*tan(c/2 + (d*x)/2)^11)/8 + (a^3*(105*c + 105*d*x))/240 - (a
^3*(105*c + 105*d*x - 352))/240 + tan(c/2 + (d*x)/2)^10*((a^3*(105*c + 105*d*x))/40 - (a^3*(630*c + 630*d*x -
480))/240) + tan(c/2 + (d*x)/2)^2*((a^3*(105*c + 105*d*x))/40 - (a^3*(630*c + 630*d*x - 1632))/240) + tan(c/2
+ (d*x)/2)^4*((a^3*(105*c + 105*d*x))/16 - (a^3*(1575*c + 1575*d*x - 960))/240) + tan(c/2 + (d*x)/2)^8*((a^3*(
105*c + 105*d*x))/16 - (a^3*(1575*c + 1575*d*x - 4320))/240) + tan(c/2 + (d*x)/2)^6*((a^3*(105*c + 105*d*x))/1
2 - (a^3*(2100*c + 2100*d*x - 3520))/240) + (7*a^3*tan(c/2 + (d*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

________________________________________________________________________________________

sympy [A]  time = 4.88, size = 328, normalized size = 2.80 \[ \begin {cases} \frac {a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x*sin(c + d*x)**6/16 + 3*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**3*x*sin(c + d*x)**4/
8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**3*x*cos(c +
d*x)**6/16 + 3*a**3*x*cos(c + d*x)**4/8 + a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**3*sin(c + d*x)**3*cos(
c + d*x)**3/(6*d) + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**3/d - a**3*
sin(c + d*x)*cos(c + d*x)**5/(16*d) - 3*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a**3*cos(c + d*x)**5/(5*d)
- a**3*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)*cos(c)**2, True))

________________________________________________________________________________________