3.441 \(\int (a+b \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=137 \[ \frac {a b \left (19 a^2+16 b^2\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (8 a^4+24 a^2 b^2+3 b^4\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac {7 a b \sin (c+d x) (a+b \cos (c+d x))^2}{12 d} \]

[Out]

1/8*(8*a^4+24*a^2*b^2+3*b^4)*x+1/6*a*b*(19*a^2+16*b^2)*sin(d*x+c)/d+1/24*b^2*(26*a^2+9*b^2)*cos(d*x+c)*sin(d*x
+c)/d+7/12*a*b*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*b*(a+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.15, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2656, 2753, 2734} \[ \frac {a b \left (19 a^2+16 b^2\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (24 a^2 b^2+8 a^4+3 b^4\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac {7 a b \sin (c+d x) (a+b \cos (c+d x))^2}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4,x]

[Out]

((8*a^4 + 24*a^2*b^2 + 3*b^4)*x)/8 + (a*b*(19*a^2 + 16*b^2)*Sin[c + d*x])/(6*d) + (b^2*(26*a^2 + 9*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(24*d) + (7*a*b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (b*(a + b*Cos[c + d*x])^3*S
in[c + d*x])/(4*d)

Rule 2656

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^4 \, dx &=\frac {b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (4 a^2+3 b^2+7 a b \cos (c+d x)\right ) \, dx\\ &=\frac {7 a b (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a^4+24 a^2 b^2+3 b^4\right ) x+\frac {a b \left (19 a^2+16 b^2\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {7 a b (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 104, normalized size = 0.76 \[ \frac {24 b^2 \left (6 a^2+b^2\right ) \sin (2 (c+d x))+96 a b \left (4 a^2+3 b^2\right ) \sin (c+d x)+12 \left (8 a^4+24 a^2 b^2+3 b^4\right ) (c+d x)+32 a b^3 \sin (3 (c+d x))+3 b^4 \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4,x]

[Out]

(12*(8*a^4 + 24*a^2*b^2 + 3*b^4)*(c + d*x) + 96*a*b*(4*a^2 + 3*b^2)*Sin[c + d*x] + 24*b^2*(6*a^2 + b^2)*Sin[2*
(c + d*x)] + 32*a*b^3*Sin[3*(c + d*x)] + 3*b^4*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.77, size = 96, normalized size = 0.70 \[ \frac {3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} d x + {\left (6 \, b^{4} \cos \left (d x + c\right )^{3} + 32 \, a b^{3} \cos \left (d x + c\right )^{2} + 96 \, a^{3} b + 64 \, a b^{3} + 9 \, {\left (8 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*d*x + (6*b^4*cos(d*x + c)^3 + 32*a*b^3*cos(d*x + c)^2 + 96*a^3*b + 64*a*b
^3 + 9*(8*a^2*b^2 + b^4)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.54, size = 107, normalized size = 0.78 \[ \frac {b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a b^{3} \sin \left (3 \, d x + 3 \, c\right )}{3 \, d} + \frac {1}{8} \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} x + \frac {{\left (6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*b^4*sin(4*d*x + 4*c)/d + 1/3*a*b^3*sin(3*d*x + 3*c)/d + 1/8*(8*a^4 + 24*a^2*b^2 + 3*b^4)*x + 1/4*(6*a^2*b
^2 + b^4)*sin(2*d*x + 2*c)/d + (4*a^3*b + 3*a*b^3)*sin(d*x + c)/d

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maple [A]  time = 0.04, size = 116, normalized size = 0.85 \[ \frac {b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b \sin \left (d x +c \right )+a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.70, size = 111, normalized size = 0.81 \[ a^{4} x + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2}}{2 \, d} - \frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3}}{3 \, d} + \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{32 \, d} + \frac {4 \, a^{3} b \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*x + 3/2*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2*b^2/d - 4/3*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^3/d + 1/32*
(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*b^4/d + 4*a^3*b*sin(d*x + c)/d

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mupad [B]  time = 0.66, size = 123, normalized size = 0.90 \[ a^4\,x+\frac {3\,b^4\,x}{8}+3\,a^2\,b^2\,x+\frac {b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4,x)

[Out]

a^4*x + (3*b^4*x)/8 + 3*a^2*b^2*x + (b^4*sin(2*c + 2*d*x))/(4*d) + (b^4*sin(4*c + 4*d*x))/(32*d) + (a*b^3*sin(
3*c + 3*d*x))/(3*d) + (3*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (3*a*b^3*sin(c + d*x))/d + (4*a^3*b*sin(c + d*x))/d

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sympy [A]  time = 1.11, size = 240, normalized size = 1.75 \[ \begin {cases} a^{4} x + \frac {4 a^{3} b \sin {\left (c + d x \right )}}{d} + 3 a^{2} b^{2} x \sin ^{2}{\left (c + d x \right )} + 3 a^{2} b^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {3 a^{2} b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {8 a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {4 a b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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