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

Optimal. Leaf size=158 \[ -\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac {a \left (2 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {3 a \left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a x \left (2 a^2+b^2\right )-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d} \]

[Out]

3/16*a*(2*a^2+b^2)*x-1/70*b*(17*a^2+4*b^2)*cos(d*x+c)^5/d+3/16*a*(2*a^2+b^2)*cos(d*x+c)*sin(d*x+c)/d+1/8*a*(2*
a^2+b^2)*cos(d*x+c)^3*sin(d*x+c)/d-3/14*a*b*cos(d*x+c)^5*(a+b*sin(d*x+c))/d-1/7*b*cos(d*x+c)^5*(a+b*sin(d*x+c)
)^2/d

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Rubi [A]  time = 0.22, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2692, 2862, 2669, 2635, 8} \[ -\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac {a \left (2 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {3 a \left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a x \left (2 a^2+b^2\right )-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(2*a^2 + b^2)*x)/16 - (b*(17*a^2 + 4*b^2)*Cos[c + d*x]^5)/(70*d) + (3*a*(2*a^2 + b^2)*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (a*(2*a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(8*d) - (3*a*b*Cos[c + d*x]^5*(a + b*Sin[c + d*
x]))/(14*d) - (b*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2)/(7*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 2692

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[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rule 2862

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[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{7} \int \cos ^4(c+d x) (a+b \sin (c+d x)) \left (7 a^2+2 b^2+9 a b \sin (c+d x)\right ) \, dx\\ &=-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{42} \int \cos ^4(c+d x) \left (21 a \left (2 a^2+b^2\right )+3 b \left (17 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{2} \left (a \left (2 a^2+b^2\right )\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac {a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{8} \left (3 a \left (2 a^2+b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac {3 a \left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{16} \left (3 a \left (2 a^2+b^2\right )\right ) \int 1 \, dx\\ &=\frac {3}{16} a \left (2 a^2+b^2\right ) x-\frac {b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac {3 a \left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 182, normalized size = 1.15 \[ \frac {560 a^3 \sin (2 (c+d x))+70 a^3 \sin (4 (c+d x))+840 a^3 c+840 a^3 d x-35 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))-105 b \left (8 a^2+b^2\right ) \cos (c+d x)-84 a^2 b \cos (5 (c+d x))+105 a b^2 \sin (2 (c+d x))-105 a b^2 \sin (4 (c+d x))-35 a b^2 \sin (6 (c+d x))+420 a b^2 c+420 a b^2 d x+7 b^3 \cos (5 (c+d x))+5 b^3 \cos (7 (c+d x))}{2240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(840*a^3*c + 420*a*b^2*c + 840*a^3*d*x + 420*a*b^2*d*x - 105*b*(8*a^2 + b^2)*Cos[c + d*x] - 35*(12*a^2*b + b^3
)*Cos[3*(c + d*x)] - 84*a^2*b*Cos[5*(c + d*x)] + 7*b^3*Cos[5*(c + d*x)] + 5*b^3*Cos[7*(c + d*x)] + 560*a^3*Sin
[2*(c + d*x)] + 105*a*b^2*Sin[2*(c + d*x)] + 70*a^3*Sin[4*(c + d*x)] - 105*a*b^2*Sin[4*(c + d*x)] - 35*a*b^2*S
in[6*(c + d*x)])/(2240*d)

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fricas [A]  time = 0.47, size = 117, normalized size = 0.74 \[ \frac {80 \, b^{3} \cos \left (d x + c\right )^{7} - 112 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{5} + 105 \, {\left (2 \, a^{3} + a b^{2}\right )} d x - 35 \, {\left (8 \, a b^{2} \cos \left (d x + c\right )^{5} - 2 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*(80*b^3*cos(d*x + c)^7 - 112*(3*a^2*b + b^3)*cos(d*x + c)^5 + 105*(2*a^3 + a*b^2)*d*x - 35*(8*a*b^2*cos(
d*x + c)^5 - 2*(2*a^3 + a*b^2)*cos(d*x + c)^3 - 3*(2*a^3 + a*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 1.26, size = 173, normalized size = 1.09 \[ \frac {b^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a b^{2} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} + \frac {3}{16} \, {\left (2 \, a^{3} + a b^{2}\right )} x - \frac {{\left (12 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {3 \, {\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{64 \, d} + \frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/448*b^3*cos(7*d*x + 7*c)/d - 1/64*a*b^2*sin(6*d*x + 6*c)/d + 3/16*(2*a^3 + a*b^2)*x - 1/320*(12*a^2*b - b^3)
*cos(5*d*x + 5*c)/d - 1/64*(12*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 3/64*(8*a^2*b + b^3)*cos(d*x + c)/d + 1/64*(2
*a^3 - 3*a*b^2)*sin(4*d*x + 4*c)/d + 1/64*(16*a^3 + 3*a*b^2)*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.28, size = 145, normalized size = 0.92 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {3 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 117, normalized size = 0.74 \[ -\frac {1344 \, a^{2} b \cos \left (d x + c\right )^{5} - 70 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 64 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{2240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.93, size = 474, normalized size = 3.00 \[ \frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )}{8\,\left (\frac {3\,a^3}{4}+\frac {3\,a\,b^2}{8}\right )}\right )\,\left (2\,a^2+b^2\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{8}-\frac {5\,a^3}{4}\right )+\frac {6\,a^2\,b}{5}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^3+\frac {11\,a\,b^2}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (3\,a^3+\frac {11\,a\,b^2}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {3\,a\,b^2}{8}-\frac {5\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {31\,a\,b^2}{8}-\frac {9\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {31\,a\,b^2}{8}-\frac {9\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (12\,a^2\,b+4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {12\,a^2\,b}{5}+\frac {4\,b^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (18\,a^2\,b-4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (24\,a^2\,b+8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {66\,a^2\,b}{5}-\frac {8\,b^3}{5}\right )+\frac {4\,b^3}{35}+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,a\,\left (2\,a^2+b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*atan((3*a*tan(c/2 + (d*x)/2)*(2*a^2 + b^2))/(8*((3*a*b^2)/8 + (3*a^3)/4)))*(2*a^2 + b^2))/(8*d) - (tan(c/
2 + (d*x)/2)*((3*a*b^2)/8 - (5*a^3)/4) + (6*a^2*b)/5 - tan(c/2 + (d*x)/2)^3*((11*a*b^2)/2 + 3*a^3) + tan(c/2 +
(d*x)/2)^11*((11*a*b^2)/2 + 3*a^3) - tan(c/2 + (d*x)/2)^13*((3*a*b^2)/8 - (5*a^3)/4) + tan(c/2 + (d*x)/2)^5*(
(31*a*b^2)/8 - (9*a^3)/4) - tan(c/2 + (d*x)/2)^9*((31*a*b^2)/8 - (9*a^3)/4) + tan(c/2 + (d*x)/2)^10*(12*a^2*b
+ 4*b^3) + tan(c/2 + (d*x)/2)^2*((12*a^2*b)/5 + (4*b^3)/5) + tan(c/2 + (d*x)/2)^8*(18*a^2*b - 4*b^3) + tan(c/2
+ (d*x)/2)^6*(24*a^2*b + 8*b^3) + tan(c/2 + (d*x)/2)^4*((66*a^2*b)/5 - (8*b^3)/5) + (4*b^3)/35 + 6*a^2*b*tan(
c/2 + (d*x)/2)^12)/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2
+ (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 + 1)) - (3*a*(2*a^2
+ b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*d)

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sympy [A]  time = 8.67, size = 348, normalized size = 2.20 \[ \begin {cases} \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 b^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{3} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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