3.394 \(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=159 \[ -\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a^3 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {17 a^3 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {17 a^3 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {17 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {17 a^3 x}{128} \]

[Out]

17/128*a^3*x-4/5*a^3*cos(d*x+c)^5/d+5/7*a^3*cos(d*x+c)^7/d-1/9*a^3*cos(d*x+c)^9/d+17/128*a^3*cos(d*x+c)*sin(d*
x+c)/d+17/192*a^3*cos(d*x+c)^3*sin(d*x+c)/d-17/48*a^3*cos(d*x+c)^5*sin(d*x+c)/d-3/8*a^3*cos(d*x+c)^5*sin(d*x+c
)^3/d

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Rubi [A]  time = 0.32, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2568, 2635, 8, 2565, 14, 270} \[ -\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a^3 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {17 a^3 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {17 a^3 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {17 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {17 a^3 x}{128} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(17*a^3*x)/128 - (4*a^3*Cos[c + d*x]^5)/(5*d) + (5*a^3*Cos[c + d*x]^7)/(7*d) - (a^3*Cos[c + d*x]^9)/(9*d) + (1
7*a^3*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (17*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(192*d) - (17*a^3*Cos[c + d*x]
^5*Sin[c + d*x])/(48*d) - (3*a^3*Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

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

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 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^4(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^4(c+d x) \sin ^3(c+d x)+3 a^3 \cos ^4(c+d x) \sin ^4(c+d x)+a^3 \cos ^4(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+a^3 \int \cos ^4(c+d x) \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{6} a^3 \int \cos ^4(c+d x) \, dx+\frac {1}{8} \left (9 a^3\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {17 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{8} a^3 \int \cos ^2(c+d x) \, dx+\frac {1}{16} \left (3 a^3\right ) \int \cos ^4(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {4 a^3 \cos ^5(c+d x)}{5 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {17 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{16} a^3 \int 1 \, dx+\frac {1}{64} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a^3 x}{16}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {17 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {17 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{128} \left (9 a^3\right ) \int 1 \, dx\\ &=\frac {17 a^3 x}{128}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {17 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {17 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 106, normalized size = 0.67 \[ \frac {a^3 (5040 \sin (2 (c+d x))-12600 \sin (4 (c+d x))-1680 \sin (6 (c+d x))+945 \sin (8 (c+d x))-52920 \cos (c+d x)-16800 \cos (3 (c+d x))+4032 \cos (5 (c+d x))+2340 \cos (7 (c+d x))-140 \cos (9 (c+d x))+30240 c+42840 d x)}{322560 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(30240*c + 42840*d*x - 52920*Cos[c + d*x] - 16800*Cos[3*(c + d*x)] + 4032*Cos[5*(c + d*x)] + 2340*Cos[7*(
c + d*x)] - 140*Cos[9*(c + d*x)] + 5040*Sin[2*(c + d*x)] - 12600*Sin[4*(c + d*x)] - 1680*Sin[6*(c + d*x)] + 94
5*Sin[8*(c + d*x)]))/(322560*d)

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fricas [A]  time = 0.51, size = 111, normalized size = 0.70 \[ -\frac {4480 \, a^{3} \cos \left (d x + c\right )^{9} - 28800 \, a^{3} \cos \left (d x + c\right )^{7} + 32256 \, a^{3} \cos \left (d x + c\right )^{5} - 5355 \, a^{3} d x - 105 \, {\left (144 \, a^{3} \cos \left (d x + c\right )^{7} - 280 \, a^{3} \cos \left (d x + c\right )^{5} + 34 \, a^{3} \cos \left (d x + c\right )^{3} + 51 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/40320*(4480*a^3*cos(d*x + c)^9 - 28800*a^3*cos(d*x + c)^7 + 32256*a^3*cos(d*x + c)^5 - 5355*a^3*d*x - 105*(
144*a^3*cos(d*x + c)^7 - 280*a^3*cos(d*x + c)^5 + 34*a^3*cos(d*x + c)^3 + 51*a^3*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.33, size = 157, normalized size = 0.99 \[ \frac {17}{128} \, a^{3} x - \frac {a^{3} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {13 \, a^{3} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{96 \, d} - \frac {21 \, a^{3} \cos \left (d x + c\right )}{128 \, d} + \frac {3 \, a^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

17/128*a^3*x - 1/2304*a^3*cos(9*d*x + 9*c)/d + 13/1792*a^3*cos(7*d*x + 7*c)/d + 1/80*a^3*cos(5*d*x + 5*c)/d -
5/96*a^3*cos(3*d*x + 3*c)/d - 21/128*a^3*cos(d*x + c)/d + 3/1024*a^3*sin(8*d*x + 8*c)/d - 1/192*a^3*sin(6*d*x
+ 6*c)/d - 5/128*a^3*sin(4*d*x + 4*c)/d + 1/64*a^3*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.28, size = 216, normalized size = 1.36 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{9}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{63}-\frac {8 \left (\cos ^{5}\left (d x +c \right )\right )}{315}\right )+3 a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/9*sin(d*x+c)^4*cos(d*x+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos(d*x+c)^5)+3*a^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)+3*a^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+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))

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maxima [A]  time = 0.32, size = 138, normalized size = 0.87 \[ -\frac {1024 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 27648 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 1680 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 945 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{322560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/322560*(1024*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*a^3 - 27648*(5*cos(d*x + c)^7 - 7*
cos(d*x + c)^5)*a^3 - 1680*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3 - 945*(24*d*x + 24*
c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a^3)/d

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mupad [B]  time = 12.28, size = 437, normalized size = 2.75 \[ \frac {17\,a^3\,x}{128}-\frac {\frac {17\,a^3\,\left (c+d\,x\right )}{128}-\frac {35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}-\frac {537\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{32}+\frac {531\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{32}-\frac {531\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{32}+\frac {537\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{32}+\frac {35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{96}-\frac {17\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{64}-\frac {a^3\,\left (5355\,c+5355\,d\,x-15872\right )}{40320}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{128}-\frac {a^3\,\left (48195\,c+48195\,d\,x-142848\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{32}-\frac {a^3\,\left (192780\,c+192780\,d\,x-87552\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{32}-\frac {a^3\,\left (192780\,c+192780\,d\,x-483840\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {357\,a^3\,\left (c+d\,x\right )}{32}-\frac {a^3\,\left (449820\,c+449820\,d\,x-419328\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {1071\,a^3\,\left (c+d\,x\right )}{64}-\frac {a^3\,\left (674730\,c+674730\,d\,x+161280\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {357\,a^3\,\left (c+d\,x\right )}{32}-\frac {a^3\,\left (449820\,c+449820\,d\,x-913920\right )}{40320}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {1071\,a^3\,\left (c+d\,x\right )}{64}-\frac {a^3\,\left (674730\,c+674730\,d\,x-2161152\right )}{40320}\right )+\frac {17\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^3,x)

[Out]

(17*a^3*x)/128 - ((17*a^3*(c + d*x))/128 - (35*a^3*tan(c/2 + (d*x)/2)^3)/96 - (537*a^3*tan(c/2 + (d*x)/2)^5)/3
2 + (531*a^3*tan(c/2 + (d*x)/2)^7)/32 - (531*a^3*tan(c/2 + (d*x)/2)^11)/32 + (537*a^3*tan(c/2 + (d*x)/2)^13)/3
2 + (35*a^3*tan(c/2 + (d*x)/2)^15)/96 - (17*a^3*tan(c/2 + (d*x)/2)^17)/64 - (a^3*(5355*c + 5355*d*x - 15872))/
40320 + tan(c/2 + (d*x)/2)^2*((153*a^3*(c + d*x))/128 - (a^3*(48195*c + 48195*d*x - 142848))/40320) + tan(c/2
+ (d*x)/2)^4*((153*a^3*(c + d*x))/32 - (a^3*(192780*c + 192780*d*x - 87552))/40320) + tan(c/2 + (d*x)/2)^14*((
153*a^3*(c + d*x))/32 - (a^3*(192780*c + 192780*d*x - 483840))/40320) + tan(c/2 + (d*x)/2)^6*((357*a^3*(c + d*
x))/32 - (a^3*(449820*c + 449820*d*x - 419328))/40320) + tan(c/2 + (d*x)/2)^10*((1071*a^3*(c + d*x))/64 - (a^3
*(674730*c + 674730*d*x + 161280))/40320) + tan(c/2 + (d*x)/2)^12*((357*a^3*(c + d*x))/32 - (a^3*(449820*c + 4
49820*d*x - 913920))/40320) + tan(c/2 + (d*x)/2)^8*((1071*a^3*(c + d*x))/64 - (a^3*(674730*c + 674730*d*x - 21
61152))/40320) + (17*a^3*tan(c/2 + (d*x)/2))/64)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^9)

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sympy [A]  time = 21.62, size = 486, normalized size = 3.06 \[ \begin {cases} \frac {9 a^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {9 a^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {27 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {33 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {33 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {4 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {3 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {9 a^{3} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {8 a^{3} \cos ^{9}{\left (c + d x \right )}}{315 d} - \frac {6 a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \sin ^{2}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((9*a**3*x*sin(c + d*x)**8/128 + 9*a**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + a**3*x*sin(c + d*x)**6
/16 + 27*a**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a**3*x*si
n(c + d*x)**2*cos(c + d*x)**6/32 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 9*a**3*x*cos(c + d*x)**8/128
+ a**3*x*cos(c + d*x)**6/16 + 9*a**3*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 33*a**3*sin(c + d*x)**5*cos(c + d*
x)**3/(128*d) + a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**3*sin(c + d*x)**4*cos(c + d*x)**5/(5*d) - 33*a**
3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 4*a**3*sin(c + d*x)**
2*cos(c + d*x)**7/(35*d) - 3*a**3*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 9*a**3*sin(c + d*x)*cos(c + d*x)**7/
(128*d) - a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 8*a**3*cos(c + d*x)**9/(315*d) - 6*a**3*cos(c + d*x)**7/(
35*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**4, True))

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