3.28 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=67 \[ \frac {(a \sin (c+d x)+a)^8}{8 a^5 d}-\frac {4 (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac {2 (a \sin (c+d x)+a)^6}{3 a^3 d} \]

[Out]

2/3*(a+a*sin(d*x+c))^6/a^3/d-4/7*(a+a*sin(d*x+c))^7/a^4/d+1/8*(a+a*sin(d*x+c))^8/a^5/d

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Rubi [A]  time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac {(a \sin (c+d x)+a)^8}{8 a^5 d}-\frac {4 (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac {2 (a \sin (c+d x)+a)^6}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

(2*(a + a*Sin[c + d*x])^6)/(3*a^3*d) - (4*(a + a*Sin[c + d*x])^7)/(7*a^4*d) + (a + a*Sin[c + d*x])^8/(8*a^5*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^2 (a+x)^5 \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (4 a^2 (a+x)^5-4 a (a+x)^6+(a+x)^7\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {2 (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {4 (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {(a+a \sin (c+d x))^8}{8 a^5 d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 58, normalized size = 0.87 \[ -\frac {a^3 (\sin (c+d x)+1)^3 \left (21 \sin ^2(c+d x)-54 \sin (c+d x)+37\right ) \cos ^6(c+d x)}{168 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/168*(a^3*Cos[c + d*x]^6*(1 + Sin[c + d*x])^3*(37 - 54*Sin[c + d*x] + 21*Sin[c + d*x]^2))/(d*(-1 + Sin[c + d
*x])^3)

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fricas [A]  time = 0.64, size = 85, normalized size = 1.27 \[ \frac {21 \, a^{3} \cos \left (d x + c\right )^{8} - 112 \, a^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (9 \, a^{3} \cos \left (d x + c\right )^{6} - 6 \, a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/168*(21*a^3*cos(d*x + c)^8 - 112*a^3*cos(d*x + c)^6 - 8*(9*a^3*cos(d*x + c)^6 - 6*a^3*cos(d*x + c)^4 - 8*a^3
*cos(d*x + c)^2 - 16*a^3)*sin(d*x + c))/d

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giac [B]  time = 0.98, size = 134, normalized size = 2.00 \[ \frac {a^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {5 \, a^{3} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {25 \, a^{3} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {33 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {3 \, a^{3} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a^{3} \sin \left (5 \, d x + 5 \, c\right )}{64 \, d} + \frac {17 \, a^{3} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {55 \, a^{3} \sin \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/1024*a^3*cos(8*d*x + 8*c)/d - 5/384*a^3*cos(6*d*x + 6*c)/d - 25/256*a^3*cos(4*d*x + 4*c)/d - 33/128*a^3*cos(
2*d*x + 2*c)/d - 3/448*a^3*sin(7*d*x + 7*c)/d - 1/64*a^3*sin(5*d*x + 5*c)/d + 17/192*a^3*sin(3*d*x + 3*c)/d +
55/64*a^3*sin(d*x + c)/d

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maple [B]  time = 0.17, size = 133, normalized size = 1.99 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) a^{3}}{2}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)+3*a^3*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(
d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/2*cos(d*x+c)^6*a^3+1/5*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*
x+c))

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maxima [A]  time = 0.37, size = 108, normalized size = 1.61 \[ \frac {21 \, a^{3} \sin \left (d x + c\right )^{8} + 72 \, a^{3} \sin \left (d x + c\right )^{7} + 28 \, a^{3} \sin \left (d x + c\right )^{6} - 168 \, a^{3} \sin \left (d x + c\right )^{5} - 210 \, a^{3} \sin \left (d x + c\right )^{4} + 56 \, a^{3} \sin \left (d x + c\right )^{3} + 252 \, a^{3} \sin \left (d x + c\right )^{2} + 168 \, a^{3} \sin \left (d x + c\right )}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/168*(21*a^3*sin(d*x + c)^8 + 72*a^3*sin(d*x + c)^7 + 28*a^3*sin(d*x + c)^6 - 168*a^3*sin(d*x + c)^5 - 210*a^
3*sin(d*x + c)^4 + 56*a^3*sin(d*x + c)^3 + 252*a^3*sin(d*x + c)^2 + 168*a^3*sin(d*x + c))/d

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mupad [B]  time = 4.54, size = 106, normalized size = 1.58 \[ \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6}{6}-a^3\,{\sin \left (c+d\,x\right )}^5-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^2}{2}+a^3\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*sin(c + d*x))^3,x)

[Out]

(a^3*sin(c + d*x) + (3*a^3*sin(c + d*x)^2)/2 + (a^3*sin(c + d*x)^3)/3 - (5*a^3*sin(c + d*x)^4)/4 - a^3*sin(c +
d*x)^5 + (a^3*sin(c + d*x)^6)/6 + (3*a^3*sin(c + d*x)^7)/7 + (a^3*sin(c + d*x)^8)/8)/d

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sympy [A]  time = 13.40, size = 196, normalized size = 2.93 \[ \begin {cases} \frac {8 a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {4 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{3} \cos ^{8}{\left (c + d x \right )}}{24 d} - \frac {a^{3} \cos ^{6}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((8*a**3*sin(c + d*x)**7/(35*d) + 4*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 8*a**3*sin(c + d*x)*
*5/(15*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**4/d + 4*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - a**3*sin(c
+ d*x)**2*cos(c + d*x)**6/(6*d) + a**3*sin(c + d*x)*cos(c + d*x)**4/d - a**3*cos(c + d*x)**8/(24*d) - a**3*co
s(c + d*x)**6/(2*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**5, True))

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