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

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

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 301, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.43, size = 178, normalized size = 0.59 \[ \frac {-24 a^3 b \cos (4 (c+d x))+12 (a-i b) (a+i b) \left (5 a^2+b^2\right ) (c+d x)-12 a b \left (5 a^2+3 b^2\right ) \cos (2 (c+d x))-4 a b \left (a^2-b^2\right ) \cos (6 (c+d x))+3 \left (15 a^4+6 a^2 b^2-b^4\right ) \sin (2 (c+d x))+3 \left (3 a^4-6 a^2 b^2-b^4\right ) \sin (4 (c+d x))+\left (a^4-6 a^2 b^2+b^4\right ) \sin (6 (c+d x))}{192 d} \]

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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fricas [A]  time = 0.69, size = 151, normalized size = 0.50 \[ -\frac {48 \, a b^{3} \cos \left (d x + c\right )^{4} + 32 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} d x - {\left (8 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

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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giac [A]  time = 2.02, size = 187, normalized size = 0.62 \[ -\frac {a^{3} b \cos \left (4 \, d x + 4 \, c\right )}{8 \, d} + \frac {1}{16} \, {\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} x - \frac {{\left (a^{3} b - a b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{48 \, d} - \frac {{\left (5 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{16 \, d} + \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (3 \, a^{4} - 6 \, a^{2} b^{2} - b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (15 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

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

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maple [A]  time = 28.97, size = 219, normalized size = 0.73 \[ \frac {b^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+4 a \,b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+6 a^{2} b^{2} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a^{3} b \left (\cos ^{6}\left (d x +c \right )\right )}{3}+a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

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*cos(d*x+c)^3*sin(d*x+c)+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*cos(d*x+c)^5*sin(d*x+c)+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 = 0.34, size = 170, normalized size = 0.56 \[ -\frac {128 \, a^{3} b \cos \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 6 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b^{2} + 64 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a b^{3} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{4}}{192 \, d} \]

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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mupad [B]  time = 2.31, size = 471, normalized size = 1.56 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (-\frac {11\,a^4}{8}+\frac {3\,a^2\,b^2}{4}+\frac {b^4}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {15\,a^4}{4}-\frac {39\,a^2\,b^2}{2}+\frac {19\,b^4}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {15\,a^4}{4}-\frac {39\,a^2\,b^2}{2}+\frac {19\,b^4}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,a^4}{24}-\frac {47\,a^2\,b^2}{4}+\frac {17\,b^4}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {5\,a^4}{24}-\frac {47\,a^2\,b^2}{4}+\frac {17\,b^4}{24}\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-\frac {11\,a^4}{8}+\frac {3\,a^2\,b^2}{4}+\frac {b^4}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {32\,a\,b^3}{3}-\frac {80\,a^3\,b}{3}\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\,a\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (5\,a^4+6\,a^2\,b^2+b^4\right )}{8\,d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^2+b^2\right )\,\left (a^2+b^2\right )}{8\,\left (\frac {5\,a^4}{8}+\frac {3\,a^2\,b^2}{4}+\frac {b^4}{8}\right )}\right )\,\left (5\,a^2+b^2\right )\,\left (a^2+b^2\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^11*(b^4/8 - (11*a^4)/8 + (3*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^5*((15*a^4)/4 + (19*b^4)/4 -
(39*a^2*b^2)/2) - tan(c/2 + (d*x)/2)^7*((15*a^4)/4 + (19*b^4)/4 - (39*a^2*b^2)/2) - tan(c/2 + (d*x)/2)^3*((5*a
^4)/24 + (17*b^4)/24 - (47*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^9*((5*a^4)/24 + (17*b^4)/24 - (47*a^2*b^2)/4) - ta
n(c/2 + (d*x)/2)*(b^4/8 - (11*a^4)/8 + (3*a^2*b^2)/4) - tan(c/2 + (d*x)/2)^6*((32*a*b^3)/3 - (80*a^3*b)/3) + 8
*a^3*b*tan(c/2 + (d*x)/2)^2 + 16*a*b^3*tan(c/2 + (d*x)/2)^4 + 16*a*b^3*tan(c/2 + (d*x)/2)^8 + 8*a^3*b*tan(c/2
+ (d*x)/2)^10)/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (
d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) - ((atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(5*a
^4 + b^4 + 6*a^2*b^2))/(8*d) + (atan((tan(c/2 + (d*x)/2)*(5*a^2 + b^2)*(a^2 + b^2))/(8*((5*a^4)/8 + b^4/8 + (3
*a^2*b^2)/4)))*(5*a^2 + b^2)*(a^2 + b^2))/(8*d)

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sympy [A]  time = 3.99, size = 563, normalized size = 1.87 \[ \begin {cases} \frac {5 a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {2 a^{3} b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {3 a^{2} b^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {9 a^{2} b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {9 a^{2} b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} + \frac {a b^{3} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {a b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {b^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{4} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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*cos(c + d*x)**6/(3
*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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