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

Optimal. Leaf size=130 \[ -\frac {3 a^3 \cos ^5(c+d x)}{10 d}-\frac {3 \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{14 d}+\frac {3 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {9 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {9 a^3 x}{16}-\frac {a \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

9/16*a^3*x-3/10*a^3*cos(d*x+c)^5/d+9/16*a^3*cos(d*x+c)*sin(d*x+c)/d+3/8*a^3*cos(d*x+c)^3*sin(d*x+c)/d-1/7*a*co
s(d*x+c)^5*(a+a*sin(d*x+c))^2/d-3/14*cos(d*x+c)^5*(a^3+a^3*sin(d*x+c))/d

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2678, 2669, 2635, 8} \[ -\frac {3 a^3 \cos ^5(c+d x)}{10 d}-\frac {3 \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{14 d}+\frac {3 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {9 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {9 a^3 x}{16}-\frac {a \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a^3*x)/16 - (3*a^3*Cos[c + d*x]^5)/(10*d) + (9*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (3*a^3*Cos[c + d*x]^
3*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*d) - (3*Cos[c + d*x]^5*(a^3 + a^3*Sin[c +
d*x]))/(14*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]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac {1}{7} (9 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {3 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{14 d}+\frac {1}{2} \left (3 a^2\right ) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {3 a^3 \cos ^5(c+d x)}{10 d}-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {3 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{14 d}+\frac {1}{2} \left (3 a^3\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {3 a^3 \cos ^5(c+d x)}{10 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {3 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{14 d}+\frac {1}{8} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {3 a^3 \cos ^5(c+d x)}{10 d}+\frac {9 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {3 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{14 d}+\frac {1}{16} \left (9 a^3\right ) \int 1 \, dx\\ &=\frac {9 a^3 x}{16}-\frac {3 a^3 \cos ^5(c+d x)}{10 d}+\frac {9 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {3 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{14 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.78, size = 161, normalized size = 1.24 \[ -\frac {a^3 \left (\sqrt {\sin (c+d x)+1} \left (80 \sin ^7(c+d x)+200 \sin ^6(c+d x)-72 \sin ^5(c+d x)-558 \sin ^4(c+d x)-306 \sin ^3(c+d x)+411 \sin ^2(c+d x)+613 \sin (c+d x)-368\right )-630 \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}\right ) \cos ^5(c+d x)}{560 d (\sin (c+d x)-1)^3 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/560*(a^3*Cos[c + d*x]^5*(-630*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[
c + d*x]]*(-368 + 613*Sin[c + d*x] + 411*Sin[c + d*x]^2 - 306*Sin[c + d*x]^3 - 558*Sin[c + d*x]^4 - 72*Sin[c +
d*x]^5 + 200*Sin[c + d*x]^6 + 80*Sin[c + d*x]^7)))/(d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 85, normalized size = 0.65 \[ \frac {80 \, a^{3} \cos \left (d x + c\right )^{7} - 448 \, a^{3} \cos \left (d x + c\right )^{5} + 315 \, a^{3} d x - 35 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 6 \, a^{3} \cos \left (d x + c\right )^{3} - 9 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*(80*a^3*cos(d*x + c)^7 - 448*a^3*cos(d*x + c)^5 + 315*a^3*d*x - 35*(8*a^3*cos(d*x + c)^5 - 6*a^3*cos(d*x
+ c)^3 - 9*a^3*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.84, size = 123, normalized size = 0.95 \[ \frac {9}{16} \, a^{3} x + \frac {a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {11 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {27 \, a^{3} \cos \left (d x + c\right )}{64 \, d} - \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac {a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {19 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/16*a^3*x + 1/448*a^3*cos(7*d*x + 7*c)/d - 11/320*a^3*cos(5*d*x + 5*c)/d - 13/64*a^3*cos(3*d*x + 3*c)/d - 27/
64*a^3*cos(d*x + c)/d - 1/64*a^3*sin(6*d*x + 6*c)/d - 1/64*a^3*sin(4*d*x + 4*c)/d + 19/64*a^3*sin(2*d*x + 2*c)
/d

________________________________________________________________________________________

maple [A]  time = 0.18, size = 143, normalized size = 1.10 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {3 \left (\cos ^{5}\left (d x +c \right )\right ) a^{3}}{5}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+3*a^3*(-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)-3/5*cos(d*x+c)^5*a^3+a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*s
in(d*x+c)+3/8*d*x+3/8*c))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 115, normalized size = 0.88 \[ -\frac {1344 \, a^{3} \cos \left (d x + c\right )^{5} - 64 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 70 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{2240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2240*(1344*a^3*cos(d*x + c)^5 - 64*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^3 - 35*(4*sin(2*d*x + 2*c)^3 + 1
2*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3 - 70*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3)/d

________________________________________________________________________________________

mupad [B]  time = 6.71, size = 389, normalized size = 2.99 \[ \frac {9\,a^3\,x}{16}-\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}-\frac {17\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {17\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{2}+\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {a^3\,\left (315\,c+315\,d\,x\right )}{560}-\frac {a^3\,\left (315\,c+315\,d\,x-736\right )}{560}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (315\,c+315\,d\,x\right )}{80}-\frac {a^3\,\left (2205\,c+2205\,d\,x-1792\right )}{560}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {a^3\,\left (315\,c+315\,d\,x\right )}{80}-\frac {a^3\,\left (2205\,c+2205\,d\,x-3360\right )}{560}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {3\,a^3\,\left (315\,c+315\,d\,x\right )}{80}-\frac {a^3\,\left (6615\,c+6615\,d\,x-6496\right )}{560}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {3\,a^3\,\left (315\,c+315\,d\,x\right )}{80}-\frac {a^3\,\left (6615\,c+6615\,d\,x-8960\right )}{560}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^3\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^3\,\left (11025\,c+11025\,d\,x-7840\right )}{560}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^3\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^3\,\left (11025\,c+11025\,d\,x-17920\right )}{560}\right )-\frac {7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*a^3*x)/16 - ((13*a^3*tan(c/2 + (d*x)/2)^5)/8 - (17*a^3*tan(c/2 + (d*x)/2)^3)/2 - (13*a^3*tan(c/2 + (d*x)/2)
^9)/8 + (17*a^3*tan(c/2 + (d*x)/2)^11)/2 + (7*a^3*tan(c/2 + (d*x)/2)^13)/8 + (a^3*(315*c + 315*d*x))/560 - (a^
3*(315*c + 315*d*x - 736))/560 + tan(c/2 + (d*x)/2)^2*((a^3*(315*c + 315*d*x))/80 - (a^3*(2205*c + 2205*d*x -
1792))/560) + tan(c/2 + (d*x)/2)^12*((a^3*(315*c + 315*d*x))/80 - (a^3*(2205*c + 2205*d*x - 3360))/560) + tan(
c/2 + (d*x)/2)^4*((3*a^3*(315*c + 315*d*x))/80 - (a^3*(6615*c + 6615*d*x - 6496))/560) + tan(c/2 + (d*x)/2)^10
*((3*a^3*(315*c + 315*d*x))/80 - (a^3*(6615*c + 6615*d*x - 8960))/560) + tan(c/2 + (d*x)/2)^8*((a^3*(315*c + 3
15*d*x))/16 - (a^3*(11025*c + 11025*d*x - 7840))/560) + tan(c/2 + (d*x)/2)^6*((a^3*(315*c + 315*d*x))/16 - (a^
3*(11025*c + 11025*d*x - 17920))/560) - (7*a^3*tan(c/2 + (d*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

________________________________________________________________________________________

sympy [A]  time = 10.00, size = 335, normalized size = 2.58 \[ \begin {cases} \frac {3 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {5 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {3 a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**6/16 + 9*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**3*x*sin(c + d*x)**
4/8 + 9*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 + 3*a**3*x*cos(
c + d*x)**6/16 + 3*a**3*x*cos(c + d*x)**4/8 + 3*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/(2*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)**5/(5*
d) - 3*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 5*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a**3*cos(c + d
*x)**7/(35*d) - 3*a**3*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**4, True))

________________________________________________________________________________________