3.1.25 \(\int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx\) [25]

Optimal. Leaf size=85 \[ \frac {15 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2830, 2724, 2717, 2715, 8, 2713} \begin {gather*} -\frac {a^3 \sin ^3(c+d x)}{4 d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {9 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {15 a^3 x}{8}+\frac {\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^3*x)/8 + (3*a^3*Sin[c + d*x])/d + (9*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + ((a + a*Cos[c + d*x])^3*Sin[
c + d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/(4*d)

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2717

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

Rule 2724

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]

Rule 2830

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx &=\frac {(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {3}{4} \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac {(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {3}{4} \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac {3 a^3 x}{4}+\frac {(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \left (3 a^3\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{4} \left (9 a^3\right ) \int \cos (c+d x) \, dx+\frac {1}{4} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {3 a^3 x}{4}+\frac {9 a^3 \sin (c+d x)}{4 d}+\frac {9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{8} \left (9 a^3\right ) \int 1 \, dx-\frac {\left (3 a^3\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac {15 a^3 x}{8}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 51, normalized size = 0.60 \begin {gather*} \frac {a^3 (60 d x+104 \sin (c+d x)+32 \sin (2 (c+d x))+8 \sin (3 (c+d x))+\sin (4 (c+d x)))}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(60*d*x + 104*Sin[c + d*x] + 32*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d)

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Maple [A]
time = 0.10, size = 100, normalized size = 1.18

method	result	size
risch	\(\frac {15 a^{3} x}{8}+\frac {13 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{4 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{d}\)	\(72\)
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+3 a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \sin \left (d x +c \right )}{d}\)	\(100\)
default	\(\frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+3 a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \sin \left (d x +c \right )}{d}\)	\(100\)
norman	\(\frac {\frac {15 a^{3} x}{8}+\frac {49 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {73 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {55 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {45 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 94, normalized size = 1.11 \begin {gather*} -\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - {\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^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, a^{3} \sin \left (d x + c\right )}{32 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3
- 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 32*a^3*sin(d*x + c))/d

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Fricas [A]
time = 0.41, size = 63, normalized size = 0.74 \begin {gather*} \frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(15*a^3*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^3*cos(d*x + c)^2 + 15*a^3*cos(d*x + c) + 24*a^3)*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (78) = 156\).
time = 0.21, size = 224, normalized size = 2.64 \begin {gather*} \begin {cases} \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 71, normalized size = 0.84 \begin {gather*} \frac {15}{8} \, a^{3} x + \frac {a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{3} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {13 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*a^3*x + 1/32*a^3*sin(4*d*x + 4*c)/d + 1/4*a^3*sin(3*d*x + 3*c)/d + a^3*sin(2*d*x + 2*c)/d + 13/4*a^3*sin(
d*x + c)/d

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Mupad [B]
time = 3.49, size = 89, normalized size = 1.05 \begin {gather*} \frac {15\,a^3\,x}{8}+\frac {\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {55\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {49\,a^3\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*a^3*x)/8 + ((73*a^3*tan(c/2 + (d*x)/2)^3)/4 + (55*a^3*tan(c/2 + (d*x)/2)^5)/4 + (15*a^3*tan(c/2 + (d*x)/2)
^7)/4 + (49*a^3*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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