3.1117 \(\int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=194 \[ -\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}+\frac {b \left (8 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 b \left (8 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3}{128} b x \left (8 a^2+b^2\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d} \]

[Out]

3/128*b*(8*a^2+b^2)*x-1/560*a*(2*a^2+61*b^2)*cos(d*x+c)^5/d+3/128*b*(8*a^2+b^2)*cos(d*x+c)*sin(d*x+c)/d+1/64*b
*(8*a^2+b^2)*cos(d*x+c)^3*sin(d*x+c)/d-1/112*(2*a^2+7*b^2)*cos(d*x+c)^5*(a+b*sin(d*x+c))/d-3/56*a*cos(d*x+c)^5
*(a+b*sin(d*x+c))^2/d-1/8*cos(d*x+c)^5*(a+b*sin(d*x+c))^3/d

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Rubi [A]  time = 0.33, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}+\frac {b \left (8 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 b \left (8 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3}{128} b x \left (8 a^2+b^2\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(8*a^2 + b^2)*x)/128 - (a*(2*a^2 + 61*b^2)*Cos[c + d*x]^5)/(560*d) + (3*b*(8*a^2 + b^2)*Cos[c + d*x]*Sin[
c + d*x])/(128*d) + (b*(8*a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) - ((2*a^2 + 7*b^2)*Cos[c + d*x]^5*(a
+ b*Sin[c + d*x]))/(112*d) - (3*a*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2)/(56*d) - (Cos[c + d*x]^5*(a + b*Sin[c
+ d*x])^3)/(8*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 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) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{8} \int \cos ^4(c+d x) (3 b+3 a \sin (c+d x)) (a+b \sin (c+d x))^2 \, dx\\ &=-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{56} \int \cos ^4(c+d x) (a+b \sin (c+d x)) \left (27 a b+3 \left (2 a^2+7 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{336} \int \cos ^4(c+d x) \left (21 b \left (8 a^2+b^2\right )+3 a \left (2 a^2+61 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{16} \left (b \left (8 a^2+b^2\right )\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}+\frac {b \left (8 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{64} \left (3 b \left (8 a^2+b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}+\frac {3 b \left (8 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {b \left (8 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}+\frac {1}{128} \left (3 b \left (8 a^2+b^2\right )\right ) \int 1 \, dx\\ &=\frac {3}{128} b \left (8 a^2+b^2\right ) x-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}+\frac {3 b \left (8 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {b \left (8 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.87, size = 189, normalized size = 0.97 \[ \frac {-280 \left (4 a^3+3 a b^2\right ) \cos (3 (c+d x))-224 a^3 \cos (5 (c+d x))-280 a \left (8 a^2+9 b^2\right ) \cos (c+d x)+840 a^2 b \sin (2 (c+d x))-840 a^2 b \sin (4 (c+d x))-280 a^2 b \sin (6 (c+d x))+3360 a^2 b c+3360 a^2 b d x+168 a b^2 \cos (5 (c+d x))+120 a b^2 \cos (7 (c+d x))-140 b^3 \sin (4 (c+d x))+\frac {35}{2} b^3 \sin (8 (c+d x))+840 b^3 c+420 b^3 d x}{17920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3360*a^2*b*c + 840*b^3*c + 3360*a^2*b*d*x + 420*b^3*d*x - 280*a*(8*a^2 + 9*b^2)*Cos[c + d*x] - 280*(4*a^3 + 3
*a*b^2)*Cos[3*(c + d*x)] - 224*a^3*Cos[5*(c + d*x)] + 168*a*b^2*Cos[5*(c + d*x)] + 120*a*b^2*Cos[7*(c + d*x)]
+ 840*a^2*b*Sin[2*(c + d*x)] - 840*a^2*b*Sin[4*(c + d*x)] - 140*b^3*Sin[4*(c + d*x)] - 280*a^2*b*Sin[6*(c + d*
x)] + (35*b^3*Sin[8*(c + d*x)])/2)/(17920*d)

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fricas [A]  time = 0.86, size = 136, normalized size = 0.70 \[ \frac {1920 \, a b^{2} \cos \left (d x + c\right )^{7} - 896 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, {\left (8 \, a^{2} b + b^{3}\right )} d x + 35 \, {\left (16 \, b^{3} \cos \left (d x + c\right )^{7} - 8 \, {\left (8 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(1920*a*b^2*cos(d*x + c)^7 - 896*(a^3 + 3*a*b^2)*cos(d*x + c)^5 + 105*(8*a^2*b + b^3)*d*x + 35*(16*b^3*
cos(d*x + c)^7 - 8*(8*a^2*b + 3*b^3)*cos(d*x + c)^5 + 2*(8*a^2*b + b^3)*cos(d*x + c)^3 + 3*(8*a^2*b + b^3)*cos
(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.36, size = 184, normalized size = 0.95 \[ \frac {3 \, a b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {b^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} b \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} + \frac {3 \, a^{2} b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {3}{128} \, {\left (8 \, a^{2} b + b^{3}\right )} x - \frac {{\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (8 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} - \frac {{\left (6 \, a^{2} b + b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/448*a*b^2*cos(7*d*x + 7*c)/d + 1/1024*b^3*sin(8*d*x + 8*c)/d - 1/64*a^2*b*sin(6*d*x + 6*c)/d + 3/64*a^2*b*si
n(2*d*x + 2*c)/d + 3/128*(8*a^2*b + b^3)*x - 1/320*(4*a^3 - 3*a*b^2)*cos(5*d*x + 5*c)/d - 1/64*(4*a^3 + 3*a*b^
2)*cos(3*d*x + 3*c)/d - 1/64*(8*a^3 + 9*a*b^2)*cos(d*x + c)/d - 1/128*(6*a^2*b + b^3)*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.33, size = 180, normalized size = 0.93 \[ \frac {-\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+3 a^{2} b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+3 a \,b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+b^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/5*a^3*cos(d*x+c)^5+3*a^2*b*(-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*a*b^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+b^3*(-1/8*sin(d*x+c)^3*cos(d*x+c)
^5-1/16*sin(d*x+c)*cos(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 = 0.40, size = 117, normalized size = 0.60 \[ -\frac {7168 \, a^{3} \cos \left (d x + c\right )^{5} - 560 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b - 3072 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a b^{2} - 35 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3}}{35840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35840*(7168*a^3*cos(d*x + c)^5 - 560*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2*b - 30
72*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a*b^2 - 35*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*b^
3)/d

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mupad [B]  time = 10.92, size = 552, normalized size = 2.85 \[ \frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^2+b^2\right )}{64\,\left (\frac {3\,a^2\,b}{8}+\frac {3\,b^3}{64}\right )}\right )\,\left (8\,a^2+b^2\right )}{64\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^2\,b}{8}+\frac {3\,b^3}{64}\right )+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+\frac {12\,a\,b^2}{35}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (6\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (14\,a^3+12\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {12\,a\,b^2}{5}-\frac {26\,a^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {6\,a^3}{5}+\frac {96\,a\,b^2}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {62\,a^3}{5}+\frac {96\,a\,b^2}{5}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (\frac {3\,a^2\,b}{8}+\frac {3\,b^3}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {41\,a^2\,b}{8}-\frac {23\,b^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {41\,a^2\,b}{8}-\frac {23\,b^3}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {13\,a^2\,b}{8}+\frac {333\,b^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {13\,a^2\,b}{8}+\frac {333\,b^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {31\,a^2\,b}{8}+\frac {671\,b^3}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {31\,a^2\,b}{8}+\frac {671\,b^3}{64}\right )+\frac {2\,a^3}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\left (8\,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 )}{64\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*b*atan((3*b*tan(c/2 + (d*x)/2)*(8*a^2 + b^2))/(64*((3*a^2*b)/8 + (3*b^3)/64)))*(8*a^2 + b^2))/(64*d) - (tan
(c/2 + (d*x)/2)*((3*a^2*b)/8 + (3*b^3)/64) + 10*a^3*tan(c/2 + (d*x)/2)^10 + 2*a^3*tan(c/2 + (d*x)/2)^14 + (12*
a*b^2)/35 + tan(c/2 + (d*x)/2)^12*(12*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^8*(12*a*b^2 + 14*a^3) - tan(c/2 + (d
*x)/2)^4*((12*a*b^2)/5 - (26*a^3)/5) + tan(c/2 + (d*x)/2)^2*((96*a*b^2)/35 + (6*a^3)/5) + tan(c/2 + (d*x)/2)^6
*((96*a*b^2)/5 + (62*a^3)/5) - tan(c/2 + (d*x)/2)^15*((3*a^2*b)/8 + (3*b^3)/64) - tan(c/2 + (d*x)/2)^3*((41*a^
2*b)/8 - (23*b^3)/64) + tan(c/2 + (d*x)/2)^13*((41*a^2*b)/8 - (23*b^3)/64) - tan(c/2 + (d*x)/2)^5*((13*a^2*b)/
8 + (333*b^3)/64) + tan(c/2 + (d*x)/2)^11*((13*a^2*b)/8 + (333*b^3)/64) + tan(c/2 + (d*x)/2)^7*((31*a^2*b)/8 +
(671*b^3)/64) - tan(c/2 + (d*x)/2)^9*((31*a^2*b)/8 + (671*b^3)/64) + (2*a^3)/5)/(d*(8*tan(c/2 + (d*x)/2)^2 +
28*tan(c/2 + (d*x)/2)^4 + 56*tan(c/2 + (d*x)/2)^6 + 70*tan(c/2 + (d*x)/2)^8 + 56*tan(c/2 + (d*x)/2)^10 + 28*ta
n(c/2 + (d*x)/2)^12 + 8*tan(c/2 + (d*x)/2)^14 + tan(c/2 + (d*x)/2)^16 + 1)) - (3*b*(8*a^2 + b^2)*(atan(tan(c/2
+ (d*x)/2)) - (d*x)/2))/(64*d)

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sympy [A]  time = 10.06, size = 456, normalized size = 2.35 \[ \begin {cases} - \frac {a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a^{2} b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{2} b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{2} b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a^{2} b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {3 a b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {6 a b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac {3 b^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 b^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {9 b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 b^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 b^{3} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 b^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} - \frac {11 b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} - \frac {3 b^{3} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{3} \sin {\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**3*cos(c + d*x)**5/(5*d) + 3*a**2*b*x*sin(c + d*x)**6/16 + 9*a**2*b*x*sin(c + d*x)**4*cos(c + d*
x)**2/16 + 9*a**2*b*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**2*b*x*cos(c + d*x)**6/16 + 3*a**2*b*sin(c + d*
x)**5*cos(c + d*x)/(16*d) + a**2*b*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3*a**2*b*sin(c + d*x)*cos(c + d*x)*
*5/(16*d) - 3*a*b**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 6*a*b**2*cos(c + d*x)**7/(35*d) + 3*b**3*x*sin(c
+ d*x)**8/128 + 3*b**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 9*b**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*
b**3*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 3*b**3*x*cos(c + d*x)**8/128 + 3*b**3*sin(c + d*x)**7*cos(c + d*x)
/(128*d) + 11*b**3*sin(c + d*x)**5*cos(c + d*x)**3/(128*d) - 11*b**3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) -
3*b**3*sin(c + d*x)*cos(c + d*x)**7/(128*d), Ne(d, 0)), (x*(a + b*sin(c))**3*sin(c)*cos(c)**4, True))

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