3.1.15 \(\int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx\) [15]

Optimal. Leaf size=87 \[ \frac {7 a^2 x}{8}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2836, 2715, 8, 2713} \begin {gather*} -\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^2 x}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^2,x]

[Out]

(7*a^2*x)/8 + (2*a^2*Sin[c + d*x])/d + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c + d
*x])/(4*d) - (2*a^2*Sin[c + d*x]^3)/(3*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 2836

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx &=\int \left (a^2 \cos ^2(c+d x)+2 a^2 \cos ^3(c+d x)+a^2 \cos ^4(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^2(c+d x) \, dx+a^2 \int \cos ^4(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{2} a^2 \int 1 \, dx+\frac {1}{4} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {a^2 x}{2}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {1}{8} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {7 a^2 x}{8}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 53, normalized size = 0.61 \begin {gather*} \frac {a^2 (84 d x+144 \sin (c+d x)+48 \sin (2 (c+d x))+16 \sin (3 (c+d x))+3 \sin (4 (c+d x)))}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^2,x]

[Out]

(a^2*(84*d*x + 144*Sin[c + d*x] + 48*Sin[2*(c + d*x)] + 16*Sin[3*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]
time = 0.10, size = 90, normalized size = 1.03

method	result	size
risch	\(\frac {7 a^{2} x}{8}+\frac {3 a^{2} \sin \left (d x +c \right )}{2 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\)	\(73\)
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(90\)
default	\(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(90\)
norman	\(\frac {\frac {7 a^{2} x}{8}+\frac {25 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {83 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {77 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {7 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {21 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {7 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*cos(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 83, normalized size = 0.95 \begin {gather*} -\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 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^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^
2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2)/d

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Fricas [A]
time = 0.40, size = 63, normalized size = 0.72 \begin {gather*} \frac {21 \, a^{2} d x + {\left (6 \, a^{2} \cos \left (d x + c\right )^{3} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 21 \, a^{2} \cos \left (d x + c\right ) + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(21*a^2*d*x + (6*a^2*cos(d*x + c)^3 + 16*a^2*cos(d*x + c)^2 + 21*a^2*cos(d*x + c) + 32*a^2)*sin(d*x + c))
/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (82) = 164\).
time = 0.20, size = 211, normalized size = 2.43 \begin {gather*} \begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{2} \cos ^{2}{\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)**2*(a+a*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]
time = 0.46, size = 72, normalized size = 0.83 \begin {gather*} \frac {7}{8} \, a^{2} x + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {3 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

7/8*a^2*x + 1/32*a^2*sin(4*d*x + 4*c)/d + 1/6*a^2*sin(3*d*x + 3*c)/d + 1/2*a^2*sin(2*d*x + 2*c)/d + 3/2*a^2*si
n(d*x + c)/d

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Mupad [B]
time = 3.50, size = 89, normalized size = 1.02 \begin {gather*} \frac {7\,a^2\,x}{8}+\frac {\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {77\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {25\,a^2\,\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)^2*(a + a*cos(c + d*x))^2,x)

[Out]

(7*a^2*x)/8 + ((83*a^2*tan(c/2 + (d*x)/2)^3)/12 + (77*a^2*tan(c/2 + (d*x)/2)^5)/12 + (7*a^2*tan(c/2 + (d*x)/2)
^7)/4 + (25*a^2*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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