3.24 \(\int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx\)

Optimal. Leaf size=105 \[ \frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {13 a^3 x}{8} \]

[Out]

13/8*a^3*x+4*a^3*sin(d*x+c)/d+13/8*a^3*cos(d*x+c)*sin(d*x+c)/d+3/4*a^3*cos(d*x+c)^3*sin(d*x+c)/d-5/3*a^3*sin(d
*x+c)^3/d+1/5*a^3*sin(d*x+c)^5/d

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Rubi [A]  time = 0.12, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2635, 8, 2633} \[ \frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {13 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^3,x]

[Out]

(13*a^3*x)/8 + (4*a^3*Sin[c + d*x])/d + (13*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (3*a^3*Cos[c + d*x]^3*Sin[c
+ d*x])/(4*d) - (5*a^3*Sin[c + d*x]^3)/(3*d) + (a^3*Sin[c + d*x]^5)/(5*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 2757

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))^3 \, dx &=\int \left (a^3 \cos ^2(c+d x)+3 a^3 \cos ^3(c+d x)+3 a^3 \cos ^4(c+d x)+a^3 \cos ^5(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^2(c+d x) \, dx+a^3 \int \cos ^5(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \, dx\\ &=\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{2} a^3 \int 1 \, dx+\frac {1}{4} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {a^3 \sin ^5(c+d x)}{5 d}+\frac {1}{8} \left (9 a^3\right ) \int 1 \, dx\\ &=\frac {13 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {a^3 \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 63, normalized size = 0.60 \[ \frac {a^3 (1380 \sin (c+d x)+480 \sin (2 (c+d x))+170 \sin (3 (c+d x))+45 \sin (4 (c+d x))+6 \sin (5 (c+d x))+780 d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^3,x]

[Out]

(a^3*(780*d*x + 1380*Sin[c + d*x] + 480*Sin[2*(c + d*x)] + 170*Sin[3*(c + d*x)] + 45*Sin[4*(c + d*x)] + 6*Sin[
5*(c + d*x)]))/(480*d)

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fricas [A]  time = 2.16, size = 76, normalized size = 0.72 \[ \frac {195 \, a^{3} d x + {\left (24 \, a^{3} \cos \left (d x + c\right )^{4} + 90 \, a^{3} \cos \left (d x + c\right )^{3} + 152 \, a^{3} \cos \left (d x + c\right )^{2} + 195 \, a^{3} \cos \left (d x + c\right ) + 304 \, a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(195*a^3*d*x + (24*a^3*cos(d*x + c)^4 + 90*a^3*cos(d*x + c)^3 + 152*a^3*cos(d*x + c)^2 + 195*a^3*cos(d*x
+ c) + 304*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.79, size = 88, normalized size = 0.84 \[ \frac {13}{8} \, a^{3} x + \frac {a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {17 \, a^{3} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {23 \, a^{3} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

13/8*a^3*x + 1/80*a^3*sin(5*d*x + 5*c)/d + 3/32*a^3*sin(4*d*x + 4*c)/d + 17/48*a^3*sin(3*d*x + 3*c)/d + a^3*si
n(2*d*x + 2*c)/d + 23/8*a^3*sin(d*x + c)/d

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maple [A]  time = 0.05, size = 121, normalized size = 1.15 \[ \frac {\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*cos(d*x+c))^3,x)

[Out]

1/d*(1/5*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3*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*(2+cos(d*x+c)^2)*sin(d*x+c)+a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.63, size = 117, normalized size = 1.11 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))
*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a
^3)/d

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mupad [B]  time = 3.71, size = 105, normalized size = 1.00 \[ \frac {13\,a^3\,x}{8}+\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {91\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {416\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\frac {133\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {51\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + a*cos(c + d*x))^3,x)

[Out]

(13*a^3*x)/8 + ((133*a^3*tan(c/2 + (d*x)/2)^3)/6 + (416*a^3*tan(c/2 + (d*x)/2)^5)/15 + (91*a^3*tan(c/2 + (d*x)
/2)^7)/6 + (13*a^3*tan(c/2 + (d*x)/2)^9)/4 + (51*a^3*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)

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sympy [A]  time = 2.08, size = 272, normalized size = 2.59 \[ \begin {cases} \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{3} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((9*a**3*x*sin(c + d*x)**4/8 + 9*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**3*x*sin(c + d*x)**2/2
+ 9*a**3*x*cos(c + d*x)**4/8 + a**3*x*cos(c + d*x)**2/2 + 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) + 9*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*a**3*sin(c + d*x)**3/d + a**3*sin(c +
d*x)*cos(c + d*x)**4/d + 15*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 +
a**3*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(a*cos(c) + a)**3*cos(c)**2, True))

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