3.400 \(\int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=144 \[ \frac{\left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6}{3 b^5 d}-\frac{4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^5}{5 b^5 d}+\frac{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4}{4 b^5 d}+\frac{(a+b \sin (c+d x))^8}{8 b^5 d}-\frac{4 a (a+b \sin (c+d x))^7}{7 b^5 d} \]

[Out]

((a^2 - b^2)^2*(a + b*Sin[c + d*x])^4)/(4*b^5*d) - (4*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^5)/(5*b^5*d) + ((3*a^
2 - b^2)*(a + b*Sin[c + d*x])^6)/(3*b^5*d) - (4*a*(a + b*Sin[c + d*x])^7)/(7*b^5*d) + (a + b*Sin[c + d*x])^8/(
8*b^5*d)

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Rubi [A]  time = 0.132702, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2668, 697} \[ \frac{\left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6}{3 b^5 d}-\frac{4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^5}{5 b^5 d}+\frac{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4}{4 b^5 d}+\frac{(a+b \sin (c+d x))^8}{8 b^5 d}-\frac{4 a (a+b \sin (c+d x))^7}{7 b^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

((a^2 - b^2)^2*(a + b*Sin[c + d*x])^4)/(4*b^5*d) - (4*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^5)/(5*b^5*d) + ((3*a^
2 - b^2)*(a + b*Sin[c + d*x])^6)/(3*b^5*d) - (4*a*(a + b*Sin[c + d*x])^7)/(7*b^5*d) + (a + b*Sin[c + d*x])^8/(
8*b^5*d)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^3 \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (a^2-b^2\right )^2 (a+x)^3-4 \left (a^3-a b^2\right ) (a+x)^4+2 \left (3 a^2-b^2\right ) (a+x)^5-4 a (a+x)^6+(a+x)^7\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4}{4 b^5 d}-\frac{4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^5}{5 b^5 d}+\frac{\left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6}{3 b^5 d}-\frac{4 a (a+b \sin (c+d x))^7}{7 b^5 d}+\frac{(a+b \sin (c+d x))^8}{8 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.544047, size = 120, normalized size = 0.83 \[ \frac{\frac{1}{3} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6+\frac{1}{4} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4+\frac{1}{8} (a+b \sin (c+d x))^8-\frac{4}{7} a (a+b \sin (c+d x))^7-\frac{4}{5} a (a-b) (a+b) (a+b \sin (c+d x))^5}{b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

(((a^2 - b^2)^2*(a + b*Sin[c + d*x])^4)/4 - (4*a*(a - b)*(a + b)*(a + b*Sin[c + d*x])^5)/5 + ((3*a^2 - b^2)*(a
+ b*Sin[c + d*x])^6)/3 - (4*a*(a + b*Sin[c + d*x])^7)/7 + (a + b*Sin[c + d*x])^8/8)/(b^5*d)

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Maple [A]  time = 0.057, size = 135, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24}} \right ) +3\,a{b}^{2} \left ( -1/7\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2}}+{\frac{{a}^{3}\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(cos(d*x+c)^5*(a+b*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.94948, size = 194, normalized size = 1.35 \begin{align*} \frac{105 \, b^{3} \sin \left (d x + c\right )^{8} + 360 \, a b^{2} \sin \left (d x + c\right )^{7} + 140 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{6} + 168 \,{\left (a^{3} - 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{5} + 1260 \, a^{2} b \sin \left (d x + c\right )^{2} - 210 \,{\left (6 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{4} + 840 \, a^{3} \sin \left (d x + c\right ) - 280 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(105*b^3*sin(d*x + c)^8 + 360*a*b^2*sin(d*x + c)^7 + 140*(3*a^2*b - 2*b^3)*sin(d*x + c)^6 + 168*(a^3 - 6
*a*b^2)*sin(d*x + c)^5 + 1260*a^2*b*sin(d*x + c)^2 - 210*(6*a^2*b - b^3)*sin(d*x + c)^4 + 840*a^3*sin(d*x + c)
- 280*(2*a^3 - 3*a*b^2)*sin(d*x + c)^3)/d

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Fricas [A]  time = 2.37953, size = 281, normalized size = 1.95 \begin{align*} \frac{105 \, b^{3} \cos \left (d x + c\right )^{8} - 140 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 8 \,{\left (45 \, a b^{2} \cos \left (d x + c\right )^{6} - 3 \,{\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 56 \, a^{3} - 24 \, a b^{2} - 4 \,{\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/840*(105*b^3*cos(d*x + c)^8 - 140*(3*a^2*b + b^3)*cos(d*x + c)^6 - 8*(45*a*b^2*cos(d*x + c)^6 - 3*(7*a^3 + 3
*a*b^2)*cos(d*x + c)^4 - 56*a^3 - 24*a*b^2 - 4*(7*a^3 + 3*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [A]  time = 13.3257, size = 280, normalized size = 1.94 \begin{align*} \begin{cases} \frac{8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{a^{2} b \sin ^{6}{\left (c + d x \right )}}{2 d} + \frac{3 a^{2} b \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{3 a^{2} b \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 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 ^{8}{\left (c + d x \right )}}{24 d} + \frac{b^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \cos ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Piecewise((8*a**3*sin(c + d*x)**5/(15*d) + 4*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**3*sin(c + d*x)*co
s(c + d*x)**4/d + a**2*b*sin(c + d*x)**6/(2*d) + 3*a**2*b*sin(c + d*x)**4*cos(c + d*x)**2/(2*d) + 3*a**2*b*sin
(c + d*x)**2*cos(c + d*x)**4/(2*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)**8/(24*d) + b**3*sin(c + d*x)**6*cos(c
+ d*x)**2/(6*d) + b**3*sin(c + d*x)**4*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(a + b*sin(c))**3*cos(c)**5, True)
)

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Giac [A]  time = 1.11447, size = 250, normalized size = 1.74 \begin{align*} \frac{b^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{3 \, a b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac{{\left (24 \, a^{2} b + b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{3 \,{\left (10 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} + \frac{{\left (4 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (20 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \,{\left (8 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/1024*b^3*cos(8*d*x + 8*c)/d - 3/448*a*b^2*sin(7*d*x + 7*c)/d - 1/384*(6*a^2*b - b^3)*cos(6*d*x + 6*c)/d - 1/
256*(24*a^2*b + b^3)*cos(4*d*x + 4*c)/d - 3/128*(10*a^2*b + b^3)*cos(2*d*x + 2*c)/d + 1/320*(4*a^3 - 9*a*b^2)*
sin(5*d*x + 5*c)/d + 1/192*(20*a^3 - 3*a*b^2)*sin(3*d*x + 3*c)/d + 5/64*(8*a^3 + 3*a*b^2)*sin(d*x + c)/d