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]

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

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Rubi [A]  time = 0.13, antiderivative size = 144, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.52, 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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fricas [A]  time = 0.49, size = 117, normalized size = 0.81 \[ \frac {105 \, b^{3} \cos \left (d x + c\right )^{8} - 140 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 8 \, {\left (45 \, a b^{2} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 56 \, a^{3} - 24 \, a b^{2} - 4 \, {\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]

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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giac [A]  time = 1.00, size = 185, normalized size = 1.28 \[ \frac {b^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {3 \, a b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (24 \, a^{2} b + b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {3 \, {\left (10 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} + \frac {{\left (4 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

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

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maple [A]  time = 0.23, size = 135, normalized size = 0.94 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )+3 a \,b^{2} \left (-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {a^{2} b \left (\cos ^{6}\left (d x +c \right )\right )}{2}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

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*cos(d*x+c)^6*sin(d*x+c)+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 = 0.34, size = 144, normalized size = 1.00 \[ \frac {105 \, b^{3} \sin \left (d x + c\right )^{8} + 360 \, a b^{2} \sin \left (d x + c\right )^{7} + 140 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{6} + 168 \, {\left (a^{3} - 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{5} + 1260 \, a^{2} b \sin \left (d x + c\right )^{2} - 210 \, {\left (6 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{4} + 840 \, a^{3} \sin \left (d x + c\right ) - 280 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3}}{840 \, d} \]

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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mupad [B]  time = 0.09, size = 141, normalized size = 0.98 \[ \frac {{\sin \left (c+d\,x\right )}^3\,\left (a\,b^2-\frac {2\,a^3}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {6\,a\,b^2}{5}-\frac {a^3}{5}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {a^2\,b}{2}-\frac {b^3}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{4}\right )+a^3\,\sin \left (c+d\,x\right )+\frac {b^3\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\sin \left (c+d\,x\right )}^7}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 12.92, size = 202, normalized size = 1.40 \[ \begin {cases} \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{2} b \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {8 a b^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {4 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{3} \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{3} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

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*cos(c + d*x)**6/(2*d) + 8*a*b**2*sin(c + d*x)**7/(35*d) + 4*a*b**2*sin(c + d*x)**5*co
s(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)**6/(6*d) -
b**3*cos(c + d*x)**8/(24*d), Ne(d, 0)), (x*(a + b*sin(c))**3*cos(c)**5, True))

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