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

Optimal. Leaf size=94 \[ \frac{2 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac{\left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x))}{d}-\frac{(b \cos (c+d x)-a \sin (c+d x))^5}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0452859, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3072, 194} \[ \frac{2 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac{\left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x))}{d}-\frac{(b \cos (c+d x)-a \sin (c+d x))^5}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3072

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int
[(a^2 + b^2 - x^2)^((n - 1)/2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 + b^2, 0] && IGtQ[(n - 1)/2, 0]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.466708, size = 156, normalized size = 1.66 \[ \frac{150 a \left (a^2+b^2\right )^2 \sin (c+d x)+25 a \left (-2 a^2 b^2+a^4-3 b^4\right ) \sin (3 (c+d x))+3 a \left (-10 a^2 b^2+a^4+5 b^4\right ) \sin (5 (c+d x))-150 b \left (a^2+b^2\right )^2 \cos (c+d x)+25 b \left (-2 a^2 b^2-3 a^4+b^4\right ) \cos (3 (c+d x))-3 b \left (-10 a^2 b^2+5 a^4+b^4\right ) \cos (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-150*b*(a^2 + b^2)^2*Cos[c + d*x] + 25*b*(-3*a^4 - 2*a^2*b^2 + b^4)*Cos[3*(c + d*x)] - 3*b*(5*a^4 - 10*a^2*b^
2 + b^4)*Cos[5*(c + d*x)] + 150*a*(a^2 + b^2)^2*Sin[c + d*x] + 25*a*(a^4 - 2*a^2*b^2 - 3*b^4)*Sin[3*(c + d*x)]
+ 3*a*(a^4 - 10*a^2*b^2 + 5*b^4)*Sin[5*(c + d*x)])/(240*d)

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Maple [A]  time = 0.105, size = 175, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{{b}^{5}\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}+10\,{a}^{2}{b}^{3} \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +10\,{a}^{3}{b}^{2} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{a}^{4}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{{a}^{5}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/5*b^5*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)+a*b^4*sin(d*x+c)^5+10*a^2*b^3*(-1/5*sin(d*x+c)^2*
cos(d*x+c)^3-2/15*cos(d*x+c)^3)+10*a^3*b^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-a^4
*b*cos(d*x+c)^5+1/5*a^5*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 0.997787, size = 232, normalized size = 2.47 \begin{align*} -\frac{a^{4} b \cos \left (d x + c\right )^{5}}{d} + \frac{a b^{4} \sin \left (d x + c\right )^{5}}{d} + \frac{{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{5}}{15 \, d} - \frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{3} b^{2}}{3 \, d} + \frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b^{3}}{3 \, d} - \frac{{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} b^{5}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^4*b*cos(d*x + c)^5/d + a*b^4*sin(d*x + c)^5/d + 1/15*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c
))*a^5/d - 2/3*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^3*b^2/d + 2/3*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2
*b^3/d - 1/15*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*b^5/d

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Fricas [A]  time = 2.49648, size = 354, normalized size = 3.77 \begin{align*} -\frac{15 \, b^{5} \cos \left (d x + c\right ) + 3 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} + 10 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} -{\left (8 \, a^{5} + 20 \, a^{3} b^{2} + 15 \, a b^{4} + 3 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, a^{5} + 5 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*b^5*cos(d*x + c) + 3*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^5 + 10*(5*a^2*b^3 - b^5)*cos(d*x + c)
^3 - (8*a^5 + 20*a^3*b^2 + 15*a*b^4 + 3*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 + 2*(2*a^5 + 5*a^3*b^2 - 1
5*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [A]  time = 2.97084, size = 267, normalized size = 2.84 \begin{align*} \begin{cases} \frac{8 a^{5} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{5} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{a^{4} b \cos ^{5}{\left (c + d x \right )}}{d} + \frac{4 a^{3} b^{2} \sin ^{5}{\left (c + d x \right )}}{3 d} + \frac{10 a^{3} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac{10 a^{2} b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{4 a^{2} b^{3} \cos ^{5}{\left (c + d x \right )}}{3 d} + \frac{a b^{4} \sin ^{5}{\left (c + d x \right )}}{d} - \frac{b^{5} \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{4 b^{5} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{8 b^{5} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a**5*sin(c + d*x)**5/(15*d) + 4*a**5*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**5*sin(c + d*x)*co
s(c + d*x)**4/d - a**4*b*cos(c + d*x)**5/d + 4*a**3*b**2*sin(c + d*x)**5/(3*d) + 10*a**3*b**2*sin(c + d*x)**3*
cos(c + d*x)**2/(3*d) - 10*a**2*b**3*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 4*a**2*b**3*cos(c + d*x)**5/(3*d)
+ a*b**4*sin(c + d*x)**5/d - b**5*sin(c + d*x)**4*cos(c + d*x)/d - 4*b**5*sin(c + d*x)**2*cos(c + d*x)**3/(3*
d) - 8*b**5*cos(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**5, True))

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Giac [B]  time = 1.12192, size = 252, normalized size = 2.68 \begin{align*} -\frac{{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{5 \,{\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - b^{5}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{5 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac{{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{5 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{5 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(5*d*x + 5*c)/d - 5/48*(3*a^4*b + 2*a^2*b^3 - b^5)*cos(3*d*x + 3*c)/d -
5/8*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)/d + 1/80*(a^5 - 10*a^3*b^2 + 5*a*b^4)*sin(5*d*x + 5*c)/d + 5/48*(a^
5 - 2*a^3*b^2 - 3*a*b^4)*sin(3*d*x + 3*c)/d + 5/8*(a^5 + 2*a^3*b^2 + a*b^4)*sin(d*x + c)/d