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

Optimal. Leaf size=301 \[ -\frac{a^2 b^2 \sin (c+d x) \cos ^5(c+d x)}{d}+\frac{a^2 b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a^2 b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a^2 b^2 x-\frac{2 a^3 b \cos ^6(c+d x)}{3 d}+\frac{a^4 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^4 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^4 x}{16}-\frac{2 a b^3 \sin ^6(c+d x)}{3 d}+\frac{a b^3 \sin ^4(c+d x)}{d}-\frac{b^4 \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac{b^4 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{b^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b^4 x}{16} \]

[Out]

(5*a^4*x)/16 + (3*a^2*b^2*x)/8 + (b^4*x)/16 - (2*a^3*b*Cos[c + d*x]^6)/(3*d) + (5*a^4*Cos[c + d*x]*Sin[c + d*x
])/(16*d) + (3*a^2*b^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a^4*Cos[
c + d*x]^3*Sin[c + d*x])/(24*d) + (a^2*b^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (b^4*Cos[c + d*x]^3*Sin[c + d*
x])/(8*d) + (a^4*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (a^2*b^2*Cos[c + d*x]^5*Sin[c + d*x])/d - (b^4*Cos[c + d
*x]^3*Sin[c + d*x]^3)/(6*d) + (a*b^3*Sin[c + d*x]^4)/d - (2*a*b^3*Sin[c + d*x]^6)/(3*d)

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Rubi [A]  time = 0.303332, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3090, 2635, 8, 2565, 30, 2568, 2564, 14} \[ -\frac{a^2 b^2 \sin (c+d x) \cos ^5(c+d x)}{d}+\frac{a^2 b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a^2 b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a^2 b^2 x-\frac{2 a^3 b \cos ^6(c+d x)}{3 d}+\frac{a^4 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^4 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^4 x}{16}-\frac{2 a b^3 \sin ^6(c+d x)}{3 d}+\frac{a b^3 \sin ^4(c+d x)}{d}-\frac{b^4 \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac{b^4 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{b^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{b^4 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^4*x)/16 + (3*a^2*b^2*x)/8 + (b^4*x)/16 - (2*a^3*b*Cos[c + d*x]^6)/(3*d) + (5*a^4*Cos[c + d*x]*Sin[c + d*x
])/(16*d) + (3*a^2*b^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a^4*Cos[
c + d*x]^3*Sin[c + d*x])/(24*d) + (a^2*b^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (b^4*Cos[c + d*x]^3*Sin[c + d*
x])/(8*d) + (a^4*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (a^2*b^2*Cos[c + d*x]^5*Sin[c + d*x])/d - (b^4*Cos[c + d
*x]^3*Sin[c + d*x]^3)/(6*d) + (a*b^3*Sin[c + d*x]^4)/d - (2*a*b^3*Sin[c + d*x]^6)/(3*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 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 8

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

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

Mathematica [C]  time = 0.427276, size = 178, normalized size = 0.59 \[ \frac{12 (a-i b) (a+i b) \left (5 a^2+b^2\right ) (c+d x)+3 \left (6 a^2 b^2+15 a^4-b^4\right ) \sin (2 (c+d x))+3 \left (-6 a^2 b^2+3 a^4-b^4\right ) \sin (4 (c+d x))+\left (-6 a^2 b^2+a^4+b^4\right ) \sin (6 (c+d x))-12 a b \left (5 a^2+3 b^2\right ) \cos (2 (c+d x))-4 a b \left (a^2-b^2\right ) \cos (6 (c+d x))-24 a^3 b \cos (4 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(a - I*b)*(a + I*b)*(5*a^2 + b^2)*(c + d*x) - 12*a*b*(5*a^2 + 3*b^2)*Cos[2*(c + d*x)] - 24*a^3*b*Cos[4*(c
+ d*x)] - 4*a*b*(a^2 - b^2)*Cos[6*(c + d*x)] + 3*(15*a^4 + 6*a^2*b^2 - b^4)*Sin[2*(c + d*x)] + 3*(3*a^4 - 6*a^
2*b^2 - b^4)*Sin[4*(c + d*x)] + (a^4 - 6*a^2*b^2 + b^4)*Sin[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.08, size = 219, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({b}^{4} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +4\,a{b}^{3} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-1/12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4} \right ) +6\,{a}^{2}{b}^{2} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{\frac{2\,{a}^{3}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+{a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*sin(d*x+c)*cos(d*x+c)^3+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*
c)+4*a*b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+6*a^2*b^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(co
s(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-2/3*a^3*b*cos(d*x+c)^6+a^4*(1/6*(cos(d*x+c)^5+5/4*cos(d
*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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Maxima [A]  time = 1.24684, size = 230, normalized size = 0.76 \begin{align*} -\frac{128 \, a^{3} b \cos \left (d x + c\right )^{6} +{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 6 \,{\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^{2} + 64 \,{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a b^{3} +{\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^{4}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(128*a^3*b*cos(d*x + c)^6 + (4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x +
2*c))*a^4 - 6*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2*b^2 + 64*(2*sin(d*x + c)^6 - 3*
sin(d*x + c)^4)*a*b^3 + (4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*b^4)/d

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Fricas [A]  time = 0.53733, size = 342, normalized size = 1.14 \begin{align*} -\frac{48 \, a b^{3} \cos \left (d x + c\right )^{4} + 32 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} d x -{\left (8 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(48*a*b^3*cos(d*x + c)^4 + 32*(a^3*b - a*b^3)*cos(d*x + c)^6 - 3*(5*a^4 + 6*a^2*b^2 + b^4)*d*x - (8*(a^4
- 6*a^2*b^2 + b^4)*cos(d*x + c)^5 + 2*(5*a^4 + 6*a^2*b^2 - 7*b^4)*cos(d*x + c)^3 + 3*(5*a^4 + 6*a^2*b^2 + b^4
)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 5.41183, size = 614, normalized size = 2.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*a**4*x*sin(c + d*x)**6/16 + 15*a**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**4*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**4*x*cos(c + d*x)**6/16 + 5*a**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**4*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 2*a**3*b*sin(c + d*x)**6/(3
*d) + 2*a**3*b*sin(c + d*x)**4*cos(c + d*x)**2/d + 2*a**3*b*sin(c + d*x)**2*cos(c + d*x)**4/d + 3*a**2*b**2*x*
sin(c + d*x)**6/8 + 9*a**2*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 9*a**2*b**2*x*sin(c + d*x)**2*cos(c + d*
x)**4/8 + 3*a**2*b**2*x*cos(c + d*x)**6/8 + 3*a**2*b**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + a**2*b**2*sin(c +
d*x)**3*cos(c + d*x)**3/d - 3*a**2*b**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) + a*b**3*sin(c + d*x)**6/(3*d) + a
*b**3*sin(c + d*x)**4*cos(c + d*x)**2/d + b**4*x*sin(c + d*x)**6/16 + 3*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2
/16 + 3*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + b**4*x*cos(c + d*x)**6/16 + b**4*sin(c + d*x)**5*cos(c + d
*x)/(16*d) - b**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)),
(x*(a*cos(c) + b*sin(c))**4*cos(c)**2, True))

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Giac [A]  time = 1.23918, size = 252, normalized size = 0.84 \begin{align*} -\frac{a^{3} b \cos \left (4 \, d x + 4 \, c\right )}{8 \, d} + \frac{1}{16} \,{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} x - \frac{{\left (a^{3} b - a b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{48 \, d} - \frac{{\left (5 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{16 \, d} + \frac{{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (3 \, a^{4} - 6 \, a^{2} b^{2} - b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (15 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^3*b*cos(4*d*x + 4*c)/d + 1/16*(5*a^4 + 6*a^2*b^2 + b^4)*x - 1/48*(a^3*b - a*b^3)*cos(6*d*x + 6*c)/d - 1
/16*(5*a^3*b + 3*a*b^3)*cos(2*d*x + 2*c)/d + 1/192*(a^4 - 6*a^2*b^2 + b^4)*sin(6*d*x + 6*c)/d + 1/64*(3*a^4 -
6*a^2*b^2 - b^4)*sin(4*d*x + 4*c)/d + 1/64*(15*a^4 + 6*a^2*b^2 - b^4)*sin(2*d*x + 2*c)/d