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

Optimal. Leaf size=200 \[ -\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{112 d}+\frac {3 a^3 (8 A+3 B) \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {9 a^3 (8 A+3 B) \sin (c+d x) \cos (c+d x)}{128 d}+\frac {9}{128} a^3 x (8 A+3 B)-\frac {a (8 A+3 B) \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{56 d}-\frac {B \cos ^5(c+d x) (a \sin (c+d x)+a)^3}{8 d} \]

[Out]

9/128*a^3*(8*A+3*B)*x-3/80*a^3*(8*A+3*B)*cos(d*x+c)^5/d+9/128*a^3*(8*A+3*B)*cos(d*x+c)*sin(d*x+c)/d+3/64*a^3*(
8*A+3*B)*cos(d*x+c)^3*sin(d*x+c)/d-1/56*a*(8*A+3*B)*cos(d*x+c)^5*(a+a*sin(d*x+c))^2/d-1/8*B*cos(d*x+c)^5*(a+a*
sin(d*x+c))^3/d-3/112*(8*A+3*B)*cos(d*x+c)^5*(a^3+a^3*sin(d*x+c))/d

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Rubi [A]  time = 0.24, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{112 d}+\frac {3 a^3 (8 A+3 B) \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {9 a^3 (8 A+3 B) \sin (c+d x) \cos (c+d x)}{128 d}+\frac {9}{128} a^3 x (8 A+3 B)-\frac {a (8 A+3 B) \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{56 d}-\frac {B \cos ^5(c+d x) (a \sin (c+d x)+a)^3}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a^3*(8*A + 3*B)*x)/128 - (3*a^3*(8*A + 3*B)*Cos[c + d*x]^5)/(80*d) + (9*a^3*(8*A + 3*B)*Cos[c + d*x]*Sin[c
+ d*x])/(128*d) + (3*a^3*(8*A + 3*B)*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) - (a*(8*A + 3*B)*Cos[c + d*x]^5*(a +
a*Sin[c + d*x])^2)/(56*d) - (B*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3)/(8*d) - (3*(8*A + 3*B)*Cos[c + d*x]^5*(a
^3 + a^3*Sin[c + d*x]))/(112*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 ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}+\frac {1}{8} (8 A+3 B) \int \cos ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}+\frac {1}{56} (9 a (8 A+3 B)) \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac {1}{16} \left (3 a^2 (8 A+3 B)\right ) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac {1}{16} \left (3 a^3 (8 A+3 B)\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}+\frac {3 a^3 (8 A+3 B) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac {1}{64} \left (9 a^3 (8 A+3 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}+\frac {9 a^3 (8 A+3 B) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {3 a^3 (8 A+3 B) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac {1}{128} \left (9 a^3 (8 A+3 B)\right ) \int 1 \, dx\\ &=\frac {9}{128} a^3 (8 A+3 B) x-\frac {3 a^3 (8 A+3 B) \cos ^5(c+d x)}{80 d}+\frac {9 a^3 (8 A+3 B) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {3 a^3 (8 A+3 B) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a (8 A+3 B) \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac {3 (8 A+3 B) \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}\\ \end {align*}

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Mathematica [A]  time = 2.25, size = 183, normalized size = 0.92 \[ -\frac {a^3 \cos (c+d x) \left (16 (373 A+223 B) \cos (2 (c+d x))+32 (41 A+11 B) \cos (4 (c+d x))+\frac {2520 (8 A+3 B) \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}-10640 A \sin (c+d x)+560 A \sin (5 (c+d x))-80 A \cos (6 (c+d x))+4576 A-3045 B \sin (c+d x)+1365 B \sin (3 (c+d x))+595 B \sin (5 (c+d x))-35 B \sin (7 (c+d x))-240 B \cos (6 (c+d x))+2976 B\right )}{17920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/17920*(a^3*Cos[c + d*x]*(4576*A + 2976*B + (2520*(8*A + 3*B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[C
os[c + d*x]^2] + 16*(373*A + 223*B)*Cos[2*(c + d*x)] + 32*(41*A + 11*B)*Cos[4*(c + d*x)] - 80*A*Cos[6*(c + d*x
)] - 240*B*Cos[6*(c + d*x)] - 10640*A*Sin[c + d*x] - 3045*B*Sin[c + d*x] + 1365*B*Sin[3*(c + d*x)] + 560*A*Sin
[5*(c + d*x)] + 595*B*Sin[5*(c + d*x)] - 35*B*Sin[7*(c + d*x)]))/d

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fricas [A]  time = 0.71, size = 135, normalized size = 0.68 \[ \frac {640 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{7} - 3584 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{5} + 315 \, {\left (8 \, A + 3 \, B\right )} a^{3} d x + 35 \, {\left (16 \, B a^{3} \cos \left (d x + c\right )^{7} - 8 \, {\left (8 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} + 6 \, {\left (8 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (8 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(640*(A + 3*B)*a^3*cos(d*x + c)^7 - 3584*(A + B)*a^3*cos(d*x + c)^5 + 315*(8*A + 3*B)*a^3*d*x + 35*(16*
B*a^3*cos(d*x + c)^7 - 8*(8*A + 11*B)*a^3*cos(d*x + c)^5 + 6*(8*A + 3*B)*a^3*cos(d*x + c)^3 + 9*(8*A + 3*B)*a^
3*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.38, size = 217, normalized size = 1.08 \[ \frac {B a^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {9}{128} \, {\left (8 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (11 \, A a^{3} + B a^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (27 \, A a^{3} + 17 \, B a^{3}\right )} \cos \left (d x + c\right )}{64 \, d} - \frac {{\left (A a^{3} + B a^{3}\right )} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac {{\left (2 \, A a^{3} + 7 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {{\left (19 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*B*a^3*sin(8*d*x + 8*c)/d + 9/128*(8*A*a^3 + 3*B*a^3)*x + 1/448*(A*a^3 + 3*B*a^3)*cos(7*d*x + 7*c)/d - 1
/320*(11*A*a^3 + B*a^3)*cos(5*d*x + 5*c)/d - 1/64*(13*A*a^3 + 7*B*a^3)*cos(3*d*x + 3*c)/d - 1/64*(27*A*a^3 + 1
7*B*a^3)*cos(d*x + c)/d - 1/64*(A*a^3 + B*a^3)*sin(6*d*x + 6*c)/d - 1/128*(2*A*a^3 + 7*B*a^3)*sin(4*d*x + 4*c)
/d + 1/64*(19*A*a^3 + 3*B*a^3)*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.45, size = 323, normalized size = 1.62 \[ \frac {a^{3} A \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+3 a^{3} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+3 B \,a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )-\frac {3 a^{3} A \left (\cos ^{5}\left (d x +c \right )\right )}{5}+3 B \,a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+a^{3} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {B \,a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*A*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+B*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*A*(-1/6*sin(d*x+c)*c
os(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*B*a^3*(-1/7*sin(d*x+c)^2*cos(d*x+
c)^5-2/35*cos(d*x+c)^5)-3/5*a^3*A*cos(d*x+c)^5+3*B*a^3*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*co
s(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)+a^3*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)-1/5*B
*a^3*cos(d*x+c)^5)

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maxima [A]  time = 0.32, size = 232, normalized size = 1.16 \[ -\frac {21504 \, A a^{3} \cos \left (d x + c\right )^{5} + 7168 \, B a^{3} \cos \left (d x + c\right )^{5} - 1024 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} A a^{3} - 560 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{3} - 1120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 3072 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} B a^{3} - 560 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3} - 35 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3}}{35840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35840*(21504*A*a^3*cos(d*x + c)^5 + 7168*B*a^3*cos(d*x + c)^5 - 1024*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*
A*a^3 - 560*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*A*a^3 - 1120*(12*d*x + 12*c + sin(4*d*
x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 - 3072*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*B*a^3 - 560*(4*sin(2*d*x + 2
*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*B*a^3 - 35*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))
*B*a^3)/d

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mupad [B]  time = 10.69, size = 584, normalized size = 2.92 \[ \frac {9\,a^3\,\mathrm {atan}\left (\frac {9\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,A+3\,B\right )}{64\,\left (\frac {9\,A\,a^3}{8}+\frac {27\,B\,a^3}{64}\right )}\right )\,\left (8\,A+3\,B\right )}{64\,d}-\frac {9\,a^3\,\left (8\,A+3\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{64\,d}-\frac {\frac {46\,A\,a^3}{35}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,A\,a^3}{8}-\frac {27\,B\,a^3}{64}\right )+\frac {26\,B\,a^3}{35}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (6\,A\,a^3+2\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (30\,A\,a^3+10\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (22\,A\,a^3+18\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (46\,A\,a^3+26\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {74\,A\,a^3}{5}+\frac {14\,B\,a^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (\frac {7\,A\,a^3}{8}-\frac {27\,B\,a^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {158\,A\,a^3}{35}+\frac {138\,B\,a^3}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {218\,A\,a^3}{5}+\frac {158\,B\,a^3}{5}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {75\,A\,a^3}{8}+\frac {305\,B\,a^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {75\,A\,a^3}{8}+\frac {305\,B\,a^3}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {55\,A\,a^3}{8}+\frac {437\,B\,a^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {55\,A\,a^3}{8}+\frac {437\,B\,a^3}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {13\,A\,a^3}{8}+\frac {919\,B\,a^3}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {13\,A\,a^3}{8}+\frac {919\,B\,a^3}{64}\right )}{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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*a^3*atan((9*a^3*tan(c/2 + (d*x)/2)*(8*A + 3*B))/(64*((9*A*a^3)/8 + (27*B*a^3)/64)))*(8*A + 3*B))/(64*d) - (
9*a^3*(8*A + 3*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(64*d) - ((46*A*a^3)/35 - tan(c/2 + (d*x)/2)*((7*A*a^3
)/8 - (27*B*a^3)/64) + (26*B*a^3)/35 + tan(c/2 + (d*x)/2)^14*(6*A*a^3 + 2*B*a^3) + tan(c/2 + (d*x)/2)^10*(30*A
*a^3 + 10*B*a^3) + tan(c/2 + (d*x)/2)^12*(22*A*a^3 + 18*B*a^3) + tan(c/2 + (d*x)/2)^8*(46*A*a^3 + 26*B*a^3) +
tan(c/2 + (d*x)/2)^4*((74*A*a^3)/5 + (14*B*a^3)/5) + tan(c/2 + (d*x)/2)^15*((7*A*a^3)/8 - (27*B*a^3)/64) + tan
(c/2 + (d*x)/2)^2*((158*A*a^3)/35 + (138*B*a^3)/35) + tan(c/2 + (d*x)/2)^6*((218*A*a^3)/5 + (158*B*a^3)/5) - t
an(c/2 + (d*x)/2)^3*((75*A*a^3)/8 + (305*B*a^3)/64) + tan(c/2 + (d*x)/2)^13*((75*A*a^3)/8 + (305*B*a^3)/64) -
tan(c/2 + (d*x)/2)^5*((55*A*a^3)/8 + (437*B*a^3)/64) + tan(c/2 + (d*x)/2)^11*((55*A*a^3)/8 + (437*B*a^3)/64) +
tan(c/2 + (d*x)/2)^7*((13*A*a^3)/8 + (919*B*a^3)/64) - tan(c/2 + (d*x)/2)^9*((13*A*a^3)/8 + (919*B*a^3)/64))/
(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*tan(c/2 + (d*x)/2)^12 + 8*tan(c/2 + (d*x)/2)^14 + tan(c/2 + (d*x)/2)^16 + 1))

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sympy [A]  time = 11.67, size = 823, normalized size = 4.12 \[ \begin {cases} \frac {3 A a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 A a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {3 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 A a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {3 A a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 B a^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {3 B a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 B a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {9 B a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {9 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 B a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {3 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {11 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {6 B a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {B a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \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*(A+B*sin(d*x+c)),x)

[Out]

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

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