3.93 \(\int \cos ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=337 \[ -\frac{10 a^3 b^2 \sin ^9(c+d x)}{9 d}+\frac{30 a^3 b^2 \sin ^7(c+d x)}{7 d}-\frac{6 a^3 b^2 \sin ^5(c+d x)}{d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^9(c+d x)}{9 d}-\frac{10 a^2 b^3 \cos ^7(c+d x)}{7 d}-\frac{5 a^4 b \cos ^9(c+d x)}{9 d}+\frac{a^5 \sin ^9(c+d x)}{9 d}-\frac{4 a^5 \sin ^7(c+d x)}{7 d}+\frac{6 a^5 \sin ^5(c+d x)}{5 d}-\frac{4 a^5 \sin ^3(c+d x)}{3 d}+\frac{a^5 \sin (c+d x)}{d}+\frac{5 a b^4 \sin ^9(c+d x)}{9 d}-\frac{10 a b^4 \sin ^7(c+d x)}{7 d}+\frac{a b^4 \sin ^5(c+d x)}{d}-\frac{b^5 \cos ^9(c+d x)}{9 d}+\frac{2 b^5 \cos ^7(c+d x)}{7 d}-\frac{b^5 \cos ^5(c+d x)}{5 d} \]

[Out]

-(b^5*Cos[c + d*x]^5)/(5*d) - (10*a^2*b^3*Cos[c + d*x]^7)/(7*d) + (2*b^5*Cos[c + d*x]^7)/(7*d) - (5*a^4*b*Cos[
c + d*x]^9)/(9*d) + (10*a^2*b^3*Cos[c + d*x]^9)/(9*d) - (b^5*Cos[c + d*x]^9)/(9*d) + (a^5*Sin[c + d*x])/d - (4
*a^5*Sin[c + d*x]^3)/(3*d) + (10*a^3*b^2*Sin[c + d*x]^3)/(3*d) + (6*a^5*Sin[c + d*x]^5)/(5*d) - (6*a^3*b^2*Sin
[c + d*x]^5)/d + (a*b^4*Sin[c + d*x]^5)/d - (4*a^5*Sin[c + d*x]^7)/(7*d) + (30*a^3*b^2*Sin[c + d*x]^7)/(7*d) -
(10*a*b^4*Sin[c + d*x]^7)/(7*d) + (a^5*Sin[c + d*x]^9)/(9*d) - (10*a^3*b^2*Sin[c + d*x]^9)/(9*d) + (5*a*b^4*S
in[c + d*x]^9)/(9*d)

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Rubi [A]  time = 0.299991, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3090, 2633, 2565, 30, 2564, 270, 14} \[ -\frac{10 a^3 b^2 \sin ^9(c+d x)}{9 d}+\frac{30 a^3 b^2 \sin ^7(c+d x)}{7 d}-\frac{6 a^3 b^2 \sin ^5(c+d x)}{d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^9(c+d x)}{9 d}-\frac{10 a^2 b^3 \cos ^7(c+d x)}{7 d}-\frac{5 a^4 b \cos ^9(c+d x)}{9 d}+\frac{a^5 \sin ^9(c+d x)}{9 d}-\frac{4 a^5 \sin ^7(c+d x)}{7 d}+\frac{6 a^5 \sin ^5(c+d x)}{5 d}-\frac{4 a^5 \sin ^3(c+d x)}{3 d}+\frac{a^5 \sin (c+d x)}{d}+\frac{5 a b^4 \sin ^9(c+d x)}{9 d}-\frac{10 a b^4 \sin ^7(c+d x)}{7 d}+\frac{a b^4 \sin ^5(c+d x)}{d}-\frac{b^5 \cos ^9(c+d x)}{9 d}+\frac{2 b^5 \cos ^7(c+d x)}{7 d}-\frac{b^5 \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

-(b^5*Cos[c + d*x]^5)/(5*d) - (10*a^2*b^3*Cos[c + d*x]^7)/(7*d) + (2*b^5*Cos[c + d*x]^7)/(7*d) - (5*a^4*b*Cos[
c + d*x]^9)/(9*d) + (10*a^2*b^3*Cos[c + d*x]^9)/(9*d) - (b^5*Cos[c + d*x]^9)/(9*d) + (a^5*Sin[c + d*x])/d - (4
*a^5*Sin[c + d*x]^3)/(3*d) + (10*a^3*b^2*Sin[c + d*x]^3)/(3*d) + (6*a^5*Sin[c + d*x]^5)/(5*d) - (6*a^3*b^2*Sin
[c + d*x]^5)/d + (a*b^4*Sin[c + d*x]^5)/d - (4*a^5*Sin[c + d*x]^7)/(7*d) + (30*a^3*b^2*Sin[c + d*x]^7)/(7*d) -
(10*a*b^4*Sin[c + d*x]^7)/(7*d) + (a^5*Sin[c + d*x]^9)/(9*d) - (10*a^3*b^2*Sin[c + d*x]^9)/(9*d) + (5*a*b^4*S
in[c + d*x]^9)/(9*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

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

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 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]]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \cos ^9(c+d x)+5 a^4 b \cos ^8(c+d x) \sin (c+d x)+10 a^3 b^2 \cos ^7(c+d x) \sin ^2(c+d x)+10 a^2 b^3 \cos ^6(c+d x) \sin ^3(c+d x)+5 a b^4 \cos ^5(c+d x) \sin ^4(c+d x)+b^5 \cos ^4(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^5 \int \cos ^9(c+d x) \, dx+\left (5 a^4 b\right ) \int \cos ^8(c+d x) \sin (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \cos ^7(c+d x) \sin ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \cos ^6(c+d x) \sin ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \cos ^5(c+d x) \sin ^4(c+d x) \, dx+b^5 \int \cos ^4(c+d x) \sin ^5(c+d x) \, dx\\ &=-\frac{a^5 \operatorname{Subst}\left (\int \left (1-4 x^2+6 x^4-4 x^6+x^8\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (5 a^4 b\right ) \operatorname{Subst}\left (\int x^8 \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (10 a^3 b^2\right ) \operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^3 \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (5 a b^4\right ) \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^5 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{5 a^4 b \cos ^9(c+d x)}{9 d}+\frac{a^5 \sin (c+d x)}{d}-\frac{4 a^5 \sin ^3(c+d x)}{3 d}+\frac{6 a^5 \sin ^5(c+d x)}{5 d}-\frac{4 a^5 \sin ^7(c+d x)}{7 d}+\frac{a^5 \sin ^9(c+d x)}{9 d}+\frac{\left (10 a^3 b^2\right ) \operatorname{Subst}\left (\int \left (x^2-3 x^4+3 x^6-x^8\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (5 a b^4\right ) \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^5 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b^5 \cos ^5(c+d x)}{5 d}-\frac{10 a^2 b^3 \cos ^7(c+d x)}{7 d}+\frac{2 b^5 \cos ^7(c+d x)}{7 d}-\frac{5 a^4 b \cos ^9(c+d x)}{9 d}+\frac{10 a^2 b^3 \cos ^9(c+d x)}{9 d}-\frac{b^5 \cos ^9(c+d x)}{9 d}+\frac{a^5 \sin (c+d x)}{d}-\frac{4 a^5 \sin ^3(c+d x)}{3 d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{6 a^5 \sin ^5(c+d x)}{5 d}-\frac{6 a^3 b^2 \sin ^5(c+d x)}{d}+\frac{a b^4 \sin ^5(c+d x)}{d}-\frac{4 a^5 \sin ^7(c+d x)}{7 d}+\frac{30 a^3 b^2 \sin ^7(c+d x)}{7 d}-\frac{10 a b^4 \sin ^7(c+d x)}{7 d}+\frac{a^5 \sin ^9(c+d x)}{9 d}-\frac{10 a^3 b^2 \sin ^9(c+d x)}{9 d}+\frac{5 a b^4 \sin ^9(c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 1.02615, size = 278, normalized size = 0.82 \[ \frac{630 a \left (70 a^2 b^2+63 a^4+15 b^4\right ) \sin (c+d x)+420 a \left (21 a^4-5 b^4\right ) \sin (3 (c+d x))+252 a \left (-20 a^2 b^2+9 a^4-5 b^4\right ) \sin (5 (c+d x))+45 a \left (-50 a^2 b^2+9 a^4+5 b^4\right ) \sin (7 (c+d x))+35 a \left (-10 a^2 b^2+a^4+5 b^4\right ) \sin (9 (c+d x))-630 b \left (30 a^2 b^2+35 a^4+3 b^4\right ) \cos (c+d x)-420 b \left (20 a^2 b^2+35 a^4+b^4\right ) \cos (3 (c+d x))+252 b \left (b^4-25 a^4\right ) \cos (5 (c+d x))+45 b \left (30 a^2 b^2-35 a^4+b^4\right ) \cos (7 (c+d x))-35 b \left (-10 a^2 b^2+5 a^4+b^4\right ) \cos (9 (c+d x))}{80640 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(-630*b*(35*a^4 + 30*a^2*b^2 + 3*b^4)*Cos[c + d*x] - 420*b*(35*a^4 + 20*a^2*b^2 + b^4)*Cos[3*(c + d*x)] + 252*
b*(-25*a^4 + b^4)*Cos[5*(c + d*x)] + 45*b*(-35*a^4 + 30*a^2*b^2 + b^4)*Cos[7*(c + d*x)] - 35*b*(5*a^4 - 10*a^2
*b^2 + b^4)*Cos[9*(c + d*x)] + 630*a*(63*a^4 + 70*a^2*b^2 + 15*b^4)*Sin[c + d*x] + 420*a*(21*a^4 - 5*b^4)*Sin[
3*(c + d*x)] + 252*a*(9*a^4 - 20*a^2*b^2 - 5*b^4)*Sin[5*(c + d*x)] + 45*a*(9*a^4 - 50*a^2*b^2 + 5*b^4)*Sin[7*(
c + d*x)] + 35*a*(a^4 - 10*a^2*b^2 + 5*b^4)*Sin[9*(c + d*x)])/(80640*d)

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Maple [A]  time = 0.213, size = 291, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{5} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{9}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{63}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{315}} \right ) +5\,a{b}^{4} \left ( -1/9\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1/21\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{ \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{105}} \right ) +10\,{a}^{2}{b}^{3} \left ( -1/9\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{63}} \right ) +10\,{a}^{3}{b}^{2} \left ( -1/9\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{8}+{\frac{\sin \left ( dx+c \right ) }{63} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) } \right ) -{\frac{5\,{a}^{4}b \left ( \cos \left ( dx+c \right ) \right ) ^{9}}{9}}+{\frac{{a}^{5}\sin \left ( dx+c \right ) }{9} \left ({\frac{128}{35}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{8}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{48\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{64\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

1/d*(b^5*(-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)+5*a*b^4*(-1/9*sin(
d*x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+10*a^
2*b^3*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)+10*a^3*b^2*(-1/9*sin(d*x+c)*cos(d*x+c)^8+1/63*(16/5+c
os(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))-5/9*a^4*b*cos(d*x+c)^9+1/9*a^5*(128/35+cos(d*x+c)^8
+8/7*cos(d*x+c)^6+48/35*cos(d*x+c)^4+64/35*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.20285, size = 302, normalized size = 0.9 \begin{align*} -\frac{175 \, a^{4} b \cos \left (d x + c\right )^{9} -{\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{5} + 10 \,{\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{3} b^{2} - 50 \,{\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} b^{3} - 5 \,{\left (35 \, \sin \left (d x + c\right )^{9} - 90 \, \sin \left (d x + c\right )^{7} + 63 \, \sin \left (d x + c\right )^{5}\right )} a b^{4} +{\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 )} b^{5}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/315*(175*a^4*b*cos(d*x + c)^9 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 378*sin(d*x + c)^5 - 420*sin(d*x
+ c)^3 + 315*sin(d*x + c))*a^5 + 10*(35*sin(d*x + c)^9 - 135*sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin(d*x
+ c)^3)*a^3*b^2 - 50*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^2*b^3 - 5*(35*sin(d*x + c)^9 - 90*sin(d*x + c)^7
+ 63*sin(d*x + c)^5)*a*b^4 + (35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*b^5)/d

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Fricas [A]  time = 0.561943, size = 504, normalized size = 1.5 \begin{align*} -\frac{63 \, b^{5} \cos \left (d x + c\right )^{5} + 35 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{9} + 90 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{7} -{\left (35 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{8} + 10 \,{\left (4 \, a^{5} + 5 \, a^{3} b^{2} - 25 \, a b^{4}\right )} \cos \left (d x + c\right )^{6} + 128 \, a^{5} + 160 \, a^{3} b^{2} + 40 \, a b^{4} + 3 \,{\left (16 \, a^{5} + 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (16 \, a^{5} + 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

-1/315*(63*b^5*cos(d*x + c)^5 + 35*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^9 + 90*(5*a^2*b^3 - b^5)*cos(d*x
+ c)^7 - (35*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^8 + 10*(4*a^5 + 5*a^3*b^2 - 25*a*b^4)*cos(d*x + c)^6 +
128*a^5 + 160*a^3*b^2 + 40*a*b^4 + 3*(16*a^5 + 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 + 4*(16*a^5 + 20*a^3*b^2 +
5*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [A]  time = 32.2598, size = 440, normalized size = 1.31 \begin{align*} \begin{cases} \frac{128 a^{5} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac{64 a^{5} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{16 a^{5} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{8 a^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac{a^{5} \sin{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{d} - \frac{5 a^{4} b \cos ^{9}{\left (c + d x \right )}}{9 d} + \frac{32 a^{3} b^{2} \sin ^{9}{\left (c + d x \right )}}{63 d} + \frac{16 a^{3} b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac{4 a^{3} b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{10 a^{3} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac{10 a^{2} b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac{20 a^{2} b^{3} \cos ^{9}{\left (c + d x \right )}}{63 d} + \frac{8 a b^{4} \sin ^{9}{\left (c + d x \right )}}{63 d} + \frac{4 a b^{4} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac{a b^{4} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{b^{5} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{4 b^{5} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{8 b^{5} \cos ^{9}{\left (c + d x \right )}}{315 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{5} \cos ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Piecewise((128*a**5*sin(c + d*x)**9/(315*d) + 64*a**5*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 16*a**5*sin(c +
d*x)**5*cos(c + d*x)**4/(5*d) + 8*a**5*sin(c + d*x)**3*cos(c + d*x)**6/(3*d) + a**5*sin(c + d*x)*cos(c + d*x)
**8/d - 5*a**4*b*cos(c + d*x)**9/(9*d) + 32*a**3*b**2*sin(c + d*x)**9/(63*d) + 16*a**3*b**2*sin(c + d*x)**7*co
s(c + d*x)**2/(7*d) + 4*a**3*b**2*sin(c + d*x)**5*cos(c + d*x)**4/d + 10*a**3*b**2*sin(c + d*x)**3*cos(c + d*x
)**6/(3*d) - 10*a**2*b**3*sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 20*a**2*b**3*cos(c + d*x)**9/(63*d) + 8*a*b*
*4*sin(c + d*x)**9/(63*d) + 4*a*b**4*sin(c + d*x)**7*cos(c + d*x)**2/(7*d) + a*b**4*sin(c + d*x)**5*cos(c + d*
x)**4/d - b**5*sin(c + d*x)**4*cos(c + d*x)**5/(5*d) - 4*b**5*sin(c + d*x)**2*cos(c + d*x)**7/(35*d) - 8*b**5*
cos(c + d*x)**9/(315*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**5*cos(c)**4, True))

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Giac [A]  time = 1.43427, size = 423, normalized size = 1.26 \begin{align*} -\frac{{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac{{\left (35 \, a^{4} b - 30 \, a^{2} b^{3} - b^{5}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac{{\left (25 \, a^{4} b - b^{5}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (35 \, a^{4} b + 20 \, a^{2} b^{3} + b^{5}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{{\left (35 \, a^{4} b + 30 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )}{128 \, d} + \frac{{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac{{\left (9 \, a^{5} - 50 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac{{\left (9 \, a^{5} - 20 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (21 \, a^{5} - 5 \, a b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{{\left (63 \, a^{5} + 70 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \sin \left (d x + c\right )}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

-1/2304*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(9*d*x + 9*c)/d - 1/1792*(35*a^4*b - 30*a^2*b^3 - b^5)*cos(7*d*x + 7*c
)/d - 1/320*(25*a^4*b - b^5)*cos(5*d*x + 5*c)/d - 1/192*(35*a^4*b + 20*a^2*b^3 + b^5)*cos(3*d*x + 3*c)/d - 1/1
28*(35*a^4*b + 30*a^2*b^3 + 3*b^5)*cos(d*x + c)/d + 1/2304*(a^5 - 10*a^3*b^2 + 5*a*b^4)*sin(9*d*x + 9*c)/d + 1
/1792*(9*a^5 - 50*a^3*b^2 + 5*a*b^4)*sin(7*d*x + 7*c)/d + 1/320*(9*a^5 - 20*a^3*b^2 - 5*a*b^4)*sin(5*d*x + 5*c
)/d + 1/192*(21*a^5 - 5*a*b^4)*sin(3*d*x + 3*c)/d + 1/128*(63*a^5 + 70*a^3*b^2 + 15*a*b^4)*sin(d*x + c)/d