3.45 \(\int \cos (c+d x) (a+a \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=22 \[ \frac {(a \sin (c+d x)+a)^9}{9 a d} \]

[Out]

1/9*(a+a*sin(d*x+c))^9/a/d

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2667, 32} \[ \frac {(a \sin (c+d x)+a)^9}{9 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sin[c + d*x])^8,x]

[Out]

(a + a*Sin[c + d*x])^9/(9*a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 (c+d x) (a+a \sin (c+d x))^8 \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^8 \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {(a+a \sin (c+d x))^9}{9 a d}\\ \end {align*}

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Mathematica [B]  time = 0.09, size = 147, normalized size = 6.68 \[ \frac {a^8 \sin ^9(c+d x)}{9 d}+\frac {a^8 \sin ^8(c+d x)}{d}+\frac {4 a^8 \sin ^7(c+d x)}{d}+\frac {28 a^8 \sin ^6(c+d x)}{3 d}+\frac {14 a^8 \sin ^5(c+d x)}{d}+\frac {14 a^8 \sin ^4(c+d x)}{d}+\frac {28 a^8 \sin ^3(c+d x)}{3 d}+\frac {4 a^8 \sin ^2(c+d x)}{d}+\frac {a^8 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Sin[c + d*x])^8,x]

[Out]

(a^8*Sin[c + d*x])/d + (4*a^8*Sin[c + d*x]^2)/d + (28*a^8*Sin[c + d*x]^3)/(3*d) + (14*a^8*Sin[c + d*x]^4)/d +
(14*a^8*Sin[c + d*x]^5)/d + (28*a^8*Sin[c + d*x]^6)/(3*d) + (4*a^8*Sin[c + d*x]^7)/d + (a^8*Sin[c + d*x]^8)/d
+ (a^8*Sin[c + d*x]^9)/(9*d)

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fricas [B]  time = 0.70, size = 122, normalized size = 5.55 \[ \frac {9 \, a^{8} \cos \left (d x + c\right )^{8} - 120 \, a^{8} \cos \left (d x + c\right )^{6} + 432 \, a^{8} \cos \left (d x + c\right )^{4} - 576 \, a^{8} \cos \left (d x + c\right )^{2} + {\left (a^{8} \cos \left (d x + c\right )^{8} - 40 \, a^{8} \cos \left (d x + c\right )^{6} + 240 \, a^{8} \cos \left (d x + c\right )^{4} - 448 \, a^{8} \cos \left (d x + c\right )^{2} + 256 \, a^{8}\right )} \sin \left (d x + c\right )}{9 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/9*(9*a^8*cos(d*x + c)^8 - 120*a^8*cos(d*x + c)^6 + 432*a^8*cos(d*x + c)^4 - 576*a^8*cos(d*x + c)^2 + (a^8*co
s(d*x + c)^8 - 40*a^8*cos(d*x + c)^6 + 240*a^8*cos(d*x + c)^4 - 448*a^8*cos(d*x + c)^2 + 256*a^8)*sin(d*x + c)
)/d

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giac [A]  time = 0.90, size = 20, normalized size = 0.91 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{9}}{9 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

1/9*(a*sin(d*x + c) + a)^9/(a*d)

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maple [A]  time = 0.08, size = 21, normalized size = 0.95 \[ \frac {\left (a +a \sin \left (d x +c \right )\right )^{9}}{9 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*sin(d*x+c))^8,x)

[Out]

1/9*(a+a*sin(d*x+c))^9/d/a

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maxima [A]  time = 0.66, size = 20, normalized size = 0.91 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{9}}{9 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/9*(a*sin(d*x + c) + a)^9/(a*d)

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mupad [B]  time = 4.75, size = 118, normalized size = 5.36 \[ \frac {\frac {a^8\,{\sin \left (c+d\,x\right )}^9}{9}+a^8\,{\sin \left (c+d\,x\right )}^8+4\,a^8\,{\sin \left (c+d\,x\right )}^7+\frac {28\,a^8\,{\sin \left (c+d\,x\right )}^6}{3}+14\,a^8\,{\sin \left (c+d\,x\right )}^5+14\,a^8\,{\sin \left (c+d\,x\right )}^4+\frac {28\,a^8\,{\sin \left (c+d\,x\right )}^3}{3}+4\,a^8\,{\sin \left (c+d\,x\right )}^2+a^8\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(a + a*sin(c + d*x))^8,x)

[Out]

(a^8*sin(c + d*x) + 4*a^8*sin(c + d*x)^2 + (28*a^8*sin(c + d*x)^3)/3 + 14*a^8*sin(c + d*x)^4 + 14*a^8*sin(c +
d*x)^5 + (28*a^8*sin(c + d*x)^6)/3 + 4*a^8*sin(c + d*x)^7 + a^8*sin(c + d*x)^8 + (a^8*sin(c + d*x)^9)/9)/d

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sympy [A]  time = 19.84, size = 148, normalized size = 6.73 \[ \begin {cases} \frac {a^{8} \sin ^{9}{\left (c + d x \right )}}{9 d} + \frac {a^{8} \sin ^{8}{\left (c + d x \right )}}{d} + \frac {4 a^{8} \sin ^{7}{\left (c + d x \right )}}{d} + \frac {28 a^{8} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {14 a^{8} \sin ^{5}{\left (c + d x \right )}}{d} + \frac {14 a^{8} \sin ^{4}{\left (c + d x \right )}}{d} + \frac {28 a^{8} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {4 a^{8} \sin ^{2}{\left (c + d x \right )}}{d} + \frac {a^{8} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{8} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))**8,x)

[Out]

Piecewise((a**8*sin(c + d*x)**9/(9*d) + a**8*sin(c + d*x)**8/d + 4*a**8*sin(c + d*x)**7/d + 28*a**8*sin(c + d*
x)**6/(3*d) + 14*a**8*sin(c + d*x)**5/d + 14*a**8*sin(c + d*x)**4/d + 28*a**8*sin(c + d*x)**3/(3*d) + 4*a**8*s
in(c + d*x)**2/d + a**8*sin(c + d*x)/d, Ne(d, 0)), (x*(a*sin(c) + a)**8*cos(c), True))

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