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

Optimal. Leaf size=175 \[ -\frac {a^3 \sin ^7(c+d x)}{7 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {a^3 \sin (c+d x)}{d}-\frac {3 a^2 b \cos ^7(c+d x)}{7 d}+\frac {3 a b^2 \sin ^7(c+d x)}{7 d}-\frac {6 a b^2 \sin ^5(c+d x)}{5 d}+\frac {a b^2 \sin ^3(c+d x)}{d}+\frac {b^3 \cos ^7(c+d x)}{7 d}-\frac {b^3 \cos ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3090, 2633, 2565, 30, 2564, 270, 14} \[ -\frac {3 a^2 b \cos ^7(c+d x)}{7 d}-\frac {a^3 \sin ^7(c+d x)}{7 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {a^3 \sin (c+d x)}{d}+\frac {3 a b^2 \sin ^7(c+d x)}{7 d}-\frac {6 a b^2 \sin ^5(c+d x)}{5 d}+\frac {a b^2 \sin ^3(c+d x)}{d}+\frac {b^3 \cos ^7(c+d x)}{7 d}-\frac {b^3 \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^3*Cos[c + d*x]^5)/(5*d) - (3*a^2*b*Cos[c + d*x]^7)/(7*d) + (b^3*Cos[c + d*x]^7)/(7*d) + (a^3*Sin[c + d*x])
/d - (a^3*Sin[c + d*x]^3)/d + (a*b^2*Sin[c + d*x]^3)/d + (3*a^3*Sin[c + d*x]^5)/(5*d) - (6*a*b^2*Sin[c + d*x]^
5)/(5*d) - (a^3*Sin[c + d*x]^7)/(7*d) + (3*a*b^2*Sin[c + d*x]^7)/(7*d)

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

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Mathematica [A]  time = 0.42, size = 204, normalized size = 1.17 \[ \frac {1225 a^3 \sin (c+d x)+245 a^3 \sin (3 (c+d x))+49 a^3 \sin (5 (c+d x))+5 a^3 \sin (7 (c+d x))-35 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-105 b \left (5 a^2+b^2\right ) \cos (c+d x)-105 a^2 b \cos (5 (c+d x))-15 a^2 b \cos (7 (c+d x))+525 a b^2 \sin (c+d x)-35 a b^2 \sin (3 (c+d x))-63 a b^2 \sin (5 (c+d x))-15 a b^2 \sin (7 (c+d x))+7 b^3 \cos (5 (c+d x))+5 b^3 \cos (7 (c+d x))}{2240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-105*b*(5*a^2 + b^2)*Cos[c + d*x] - 35*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 105*a^2*b*Cos[5*(c + d*x)] + 7*b^3*
Cos[5*(c + d*x)] - 15*a^2*b*Cos[7*(c + d*x)] + 5*b^3*Cos[7*(c + d*x)] + 1225*a^3*Sin[c + d*x] + 525*a*b^2*Sin[
c + d*x] + 245*a^3*Sin[3*(c + d*x)] - 35*a*b^2*Sin[3*(c + d*x)] + 49*a^3*Sin[5*(c + d*x)] - 63*a*b^2*Sin[5*(c
+ d*x)] + 5*a^3*Sin[7*(c + d*x)] - 15*a*b^2*Sin[7*(c + d*x)])/(2240*d)

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fricas [A]  time = 0.70, size = 123, normalized size = 0.70 \[ -\frac {7 \, b^{3} \cos \left (d x + c\right )^{5} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{7} - {\left (5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} + 16 \, a^{3} + 8 \, a b^{2} + 4 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{35 \, d} \]

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))^3,x, algorithm="fricas")

[Out]

-1/35*(7*b^3*cos(d*x + c)^5 + 5*(3*a^2*b - b^3)*cos(d*x + c)^7 - (5*(a^3 - 3*a*b^2)*cos(d*x + c)^6 + 3*(2*a^3
+ a*b^2)*cos(d*x + c)^4 + 16*a^3 + 8*a*b^2 + 4*(2*a^3 + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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giac [A]  time = 0.32, size = 197, normalized size = 1.13 \[ -\frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (15 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (9 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{64 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (7 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (7 \, a^{3} - a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} + \frac {5 \, {\left (7 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/448*(3*a^2*b - b^3)*cos(7*d*x + 7*c)/d - 1/320*(15*a^2*b - b^3)*cos(5*d*x + 5*c)/d - 1/64*(9*a^2*b + b^3)*c
os(3*d*x + 3*c)/d - 3/64*(5*a^2*b + b^3)*cos(d*x + c)/d + 1/448*(a^3 - 3*a*b^2)*sin(7*d*x + 7*c)/d + 1/320*(7*
a^3 - 9*a*b^2)*sin(5*d*x + 5*c)/d + 1/64*(7*a^3 - a*b^2)*sin(3*d*x + 3*c)/d + 5/64*(7*a^3 + 3*a*b^2)*sin(d*x +
c)/d

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maple [A]  time = 10.38, size = 145, normalized size = 0.83 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 126, normalized size = 0.72 \[ -\frac {15 \, a^{2} b \cos \left (d x + c\right )^{7} + {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{35 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/35*(15*a^2*b*cos(d*x + c)^7 + (5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*
a^3 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a*b^2 - (5*cos(d*x + c)^7 - 7*cos(d*x + c)^5
)*b^3)/d

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mupad [B]  time = 0.78, size = 214, normalized size = 1.22 \[ \frac {16\,a^3\,\sin \left (c+d\,x\right )}{35\,d}-\frac {b^3\,{\cos \left (c+d\,x\right )}^5}{5\,d}+\frac {b^3\,{\cos \left (c+d\,x\right )}^7}{7\,d}-\frac {3\,a^2\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d}+\frac {8\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {6\,a^3\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}+\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {4\,a\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*a^3*sin(c + d*x))/(35*d) - (b^3*cos(c + d*x)^5)/(5*d) + (b^3*cos(c + d*x)^7)/(7*d) - (3*a^2*b*cos(c + d*x)
^7)/(7*d) + (8*a^3*cos(c + d*x)^2*sin(c + d*x))/(35*d) + (6*a^3*cos(c + d*x)^4*sin(c + d*x))/(35*d) + (a^3*cos
(c + d*x)^6*sin(c + d*x))/(7*d) + (8*a*b^2*sin(c + d*x))/(35*d) + (4*a*b^2*cos(c + d*x)^2*sin(c + d*x))/(35*d)
+ (3*a*b^2*cos(c + d*x)^4*sin(c + d*x))/(35*d) - (3*a*b^2*cos(c + d*x)^6*sin(c + d*x))/(7*d)

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sympy [A]  time = 5.53, size = 233, normalized size = 1.33 \[ \begin {cases} \frac {16 a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \cos ^{7}{\left (c + d x \right )}}{7 d} + \frac {8 a b^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {4 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 b^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{3} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*a**3*sin(c + d*x)**7/(35*d) + 8*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*a**3*sin(c + d*x)
**3*cos(c + d*x)**4/d + a**3*sin(c + d*x)*cos(c + d*x)**6/d - 3*a**2*b*cos(c + d*x)**7/(7*d) + 8*a*b**2*sin(c
+ d*x)**7/(35*d) + 4*a*b**2*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + a*b**2*sin(c + d*x)**3*cos(c + d*x)**4/d -
b**3*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*b**3*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c
))**3*cos(c)**4, True))

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