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

Optimal. Leaf size=174 \[ \frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^2 x}{16}-\frac {a b \cos ^6(c+d x)}{3 d}-\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {b^2 x}{16} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3090, 2635, 8, 2565, 30, 2568} \[ \frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^2 x}{16}-\frac {a b \cos ^6(c+d x)}{3 d}-\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {b^2 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 30

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

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

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Mathematica [A]  time = 0.25, size = 147, normalized size = 0.84 \[ \frac {\left (5 a^2+b^2\right ) (c+d x)}{16 d}+\frac {\left (15 a^2+b^2\right ) \sin (2 (c+d x))}{64 d}+\frac {\left (3 a^2-b^2\right ) \sin (4 (c+d x))}{64 d}+\frac {\left (a^2-b^2\right ) \sin (6 (c+d x))}{192 d}-\frac {5 a b \cos (2 (c+d x))}{32 d}-\frac {a b \cos (4 (c+d x))}{16 d}-\frac {a b \cos (6 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^2 + b^2)*(c + d*x))/(16*d) - (5*a*b*Cos[2*(c + d*x)])/(32*d) - (a*b*Cos[4*(c + d*x)])/(16*d) - (a*b*Cos[
6*(c + d*x)])/(96*d) + ((15*a^2 + b^2)*Sin[2*(c + d*x)])/(64*d) + ((3*a^2 - b^2)*Sin[4*(c + d*x)])/(64*d) + ((
a^2 - b^2)*Sin[6*(c + d*x)])/(192*d)

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fricas [A]  time = 0.50, size = 95, normalized size = 0.55 \[ -\frac {16 \, a b \cos \left (d x + c\right )^{6} - 3 \, {\left (5 \, a^{2} + b^{2}\right )} d x - {\left (8 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 132, normalized size = 0.76 \[ \frac {1}{16} \, {\left (5 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {5 \, a b \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (3 \, a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (15 \, a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*a^2 + b^2)*x - 1/96*a*b*cos(6*d*x + 6*c)/d - 1/16*a*b*cos(4*d*x + 4*c)/d - 5/32*a*b*cos(2*d*x + 2*c)/d
+ 1/192*(a^2 - b^2)*sin(6*d*x + 6*c)/d + 1/64*(3*a^2 - b^2)*sin(4*d*x + 4*c)/d + 1/64*(15*a^2 + b^2)*sin(2*d*
x + 2*c)/d

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maple [A]  time = 11.25, size = 118, normalized size = 0.68 \[ \frac {b^{2} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a b \left (\cos ^{6}\left (d x +c \right )\right )}{3}+a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*(-1/6*cos(d*x+c)^5*sin(d*x+c)+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-1/3*a*b*
cos(d*x+c)^6+a^2*(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 = 102, normalized size = 0.59 \[ -\frac {64 \, a b \cos \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{192 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

-1/192*(64*a*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^2 - (4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*b^2)/d

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mupad [B]  time = 0.61, size = 156, normalized size = 0.90 \[ \frac {5\,a^2\,x}{16}+\frac {b^2\,x}{16}+\frac {5\,a^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}-\frac {b^2\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d}-\frac {a\,b\,{\cos \left (c+d\,x\right )}^6}{3\,d}+\frac {5\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d}+\frac {b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d} \]

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))^2,x)

[Out]

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

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sympy [A]  time = 3.48, size = 340, normalized size = 1.95 \[ \begin {cases} \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{2} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

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))**2,x)

[Out]

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

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