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

Optimal. Leaf size=87 \[ -\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8} \]

[Out]

35/8*a^4*x+8*a^4*sin(d*x+c)/d+27/8*a^4*cos(d*x+c)*sin(d*x+c)/d+1/4*a^4*cos(d*x+c)^3*sin(d*x+c)/d-4/3*a^4*sin(d
*x+c)^3/d

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2645, 2637, 2635, 8, 2633} \[ -\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4,x]

[Out]

(35*a^4*x)/8 + (8*a^4*Sin[c + d*x])/d + (27*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]^3*Sin[c +
d*x])/(4*d) - (4*a^4*Sin[c + d*x]^3)/(3*d)

Rule 8

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

Rule 2633

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2645

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

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^4 \, dx &=\int \left (a^4+4 a^4 \cos (c+d x)+6 a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+a^4 \cos ^4(c+d x)\right ) \, dx\\ &=a^4 x+a^4 \int \cos ^4(c+d x) \, dx+\left (4 a^4\right ) \int \cos (c+d x) \, dx+\left (4 a^4\right ) \int \cos ^3(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^2(c+d x) \, dx\\ &=a^4 x+\frac {4 a^4 \sin (c+d x)}{d}+\frac {3 a^4 \cos (c+d x) \sin (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \left (3 a^4\right ) \int \cos ^2(c+d x) \, dx+\left (3 a^4\right ) \int 1 \, dx-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=4 a^4 x+\frac {8 a^4 \sin (c+d x)}{d}+\frac {27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=\frac {35 a^4 x}{8}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 56, normalized size = 0.64 \[ \frac {a^4 (672 \sin (c+d x)+168 \sin (2 (c+d x))+32 \sin (3 (c+d x))+3 \sin (4 (c+d x))+420 c+420 d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4,x]

[Out]

(a^4*(420*c + 420*d*x + 672*Sin[c + d*x] + 168*Sin[2*(c + d*x)] + 32*Sin[3*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(
96*d)

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fricas [A]  time = 1.03, size = 63, normalized size = 0.72 \[ \frac {105 \, a^{4} d x + {\left (6 \, a^{4} \cos \left (d x + c\right )^{3} + 32 \, a^{4} \cos \left (d x + c\right )^{2} + 81 \, a^{4} \cos \left (d x + c\right ) + 160 \, a^{4}\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(105*a^4*d*x + (6*a^4*cos(d*x + c)^3 + 32*a^4*cos(d*x + c)^2 + 81*a^4*cos(d*x + c) + 160*a^4)*sin(d*x + c
))/d

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giac [A]  time = 0.71, size = 72, normalized size = 0.83 \[ \frac {35}{8} \, a^{4} x + \frac {a^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{4} \sin \left (3 \, d x + 3 \, c\right )}{3 \, d} + \frac {7 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {7 \, a^{4} \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*a^4*x + 1/32*a^4*sin(4*d*x + 4*c)/d + 1/3*a^4*sin(3*d*x + 3*c)/d + 7/4*a^4*sin(2*d*x + 2*c)/d + 7*a^4*sin
(d*x + c)/d

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maple [A]  time = 0.05, size = 111, normalized size = 1.28 \[ \frac {a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} \sin \left (d x +c \right )+a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.74, size = 106, normalized size = 1.22 \[ a^{4} x - \frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4}}{3 \, d} + \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{32 \, d} + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{2 \, d} + \frac {4 \, a^{4} \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.56, size = 89, normalized size = 1.02 \[ \frac {35\,a^4\,x}{8}+\frac {\frac {35\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {385\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {511\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {93\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^4,x)

[Out]

(35*a^4*x)/8 + ((511*a^4*tan(c/2 + (d*x)/2)^3)/12 + (385*a^4*tan(c/2 + (d*x)/2)^5)/12 + (35*a^4*tan(c/2 + (d*x
)/2)^7)/4 + (93*a^4*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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sympy [A]  time = 1.08, size = 224, normalized size = 2.57 \[ \begin {cases} \frac {3 a^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + 3 a^{4} x \sin ^{2}{\left (c + d x \right )} + \frac {3 a^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + 3 a^{4} x \cos ^{2}{\left (c + d x \right )} + a^{4} x + \frac {3 a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {8 a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {4 a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {4 a^{4} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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