3.43 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=45 \[ \frac {(a \sin (c+d x)+a)^{10}}{5 a^2 d}-\frac {(a \sin (c+d x)+a)^{11}}{11 a^3 d} \]

[Out]

1/5*(a+a*sin(d*x+c))^10/a^2/d-1/11*(a+a*sin(d*x+c))^11/a^3/d

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Rubi [A]  time = 0.05, antiderivative size = 45, 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} \[ \frac {(a \sin (c+d x)+a)^{10}}{5 a^2 d}-\frac {(a \sin (c+d x)+a)^{11}}{11 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^8,x]

[Out]

(a + a*Sin[c + d*x])^10/(5*a^2*d) - (a + a*Sin[c + d*x])^11/(11*a^3*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 ^3(c+d x) (a+a \sin (c+d x))^8 \, dx &=\frac {\operatorname {Subst}\left (\int (a-x) (a+x)^9 \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 a (a+x)^9-(a+x)^{10}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {(a+a \sin (c+d x))^{10}}{5 a^2 d}-\frac {(a+a \sin (c+d x))^{11}}{11 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 43, normalized size = 0.96 \[ -\frac {a^8 (5 \sin (c+d x)-6) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^{20}}{55 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^8,x]

[Out]

-1/55*(a^8*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^20*(-6 + 5*Sin[c + d*x]))/d

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fricas [B]  time = 0.75, size = 136, normalized size = 3.02 \[ \frac {44 \, a^{8} \cos \left (d x + c\right )^{10} - 550 \, a^{8} \cos \left (d x + c\right )^{8} + 1760 \, a^{8} \cos \left (d x + c\right )^{6} - 1760 \, a^{8} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{8} \cos \left (d x + c\right )^{10} - 190 \, a^{8} \cos \left (d x + c\right )^{8} + 1040 \, a^{8} \cos \left (d x + c\right )^{6} - 1568 \, a^{8} \cos \left (d x + c\right )^{4} + 256 \, a^{8} \cos \left (d x + c\right )^{2} + 512 \, a^{8}\right )} \sin \left (d x + c\right )}{55 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/55*(44*a^8*cos(d*x + c)^10 - 550*a^8*cos(d*x + c)^8 + 1760*a^8*cos(d*x + c)^6 - 1760*a^8*cos(d*x + c)^4 + (5
*a^8*cos(d*x + c)^10 - 190*a^8*cos(d*x + c)^8 + 1040*a^8*cos(d*x + c)^6 - 1568*a^8*cos(d*x + c)^4 + 256*a^8*co
s(d*x + c)^2 + 512*a^8)*sin(d*x + c))/d

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giac [B]  time = 1.57, size = 134, normalized size = 2.98 \[ -\frac {5 \, a^{8} \sin \left (d x + c\right )^{11} + 44 \, a^{8} \sin \left (d x + c\right )^{10} + 165 \, a^{8} \sin \left (d x + c\right )^{9} + 330 \, a^{8} \sin \left (d x + c\right )^{8} + 330 \, a^{8} \sin \left (d x + c\right )^{7} - 462 \, a^{8} \sin \left (d x + c\right )^{5} - 660 \, a^{8} \sin \left (d x + c\right )^{4} - 495 \, a^{8} \sin \left (d x + c\right )^{3} - 220 \, a^{8} \sin \left (d x + c\right )^{2} - 55 \, a^{8} \sin \left (d x + c\right )}{55 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

-1/55*(5*a^8*sin(d*x + c)^11 + 44*a^8*sin(d*x + c)^10 + 165*a^8*sin(d*x + c)^9 + 330*a^8*sin(d*x + c)^8 + 330*
a^8*sin(d*x + c)^7 - 462*a^8*sin(d*x + c)^5 - 660*a^8*sin(d*x + c)^4 - 495*a^8*sin(d*x + c)^3 - 220*a^8*sin(d*
x + c)^2 - 55*a^8*sin(d*x + c))/d

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maple [B]  time = 0.19, size = 463, normalized size = 10.29 \[ \frac {a^{8} \left (-\frac {\left (\sin ^{7}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{11}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{99}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{99}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{33}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{99}\right )+8 a^{8} \left (-\frac {\left (\sin ^{6}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{10}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{40}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{20}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{40}\right )+28 a^{8} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{9}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{63}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{21}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{63}\right )+56 a^{8} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{8}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{12}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{24}\right )+70 a^{8} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{7}-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}\right )+56 a^{8} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+28 a^{8} \left (-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-2 \left (\cos ^{4}\left (d x +c \right )\right ) a^{8}+\frac {a^{8} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))^8,x)

[Out]

1/d*(a^8*(-1/11*sin(d*x+c)^7*cos(d*x+c)^4-7/99*sin(d*x+c)^5*cos(d*x+c)^4-5/99*sin(d*x+c)^3*cos(d*x+c)^4-1/33*c
os(d*x+c)^4*sin(d*x+c)+1/99*(2+cos(d*x+c)^2)*sin(d*x+c))+8*a^8*(-1/10*sin(d*x+c)^6*cos(d*x+c)^4-3/40*sin(d*x+c
)^4*cos(d*x+c)^4-1/20*sin(d*x+c)^2*cos(d*x+c)^4-1/40*cos(d*x+c)^4)+28*a^8*(-1/9*sin(d*x+c)^5*cos(d*x+c)^4-5/63
*sin(d*x+c)^3*cos(d*x+c)^4-1/21*cos(d*x+c)^4*sin(d*x+c)+1/63*(2+cos(d*x+c)^2)*sin(d*x+c))+56*a^8*(-1/8*sin(d*x
+c)^4*cos(d*x+c)^4-1/12*sin(d*x+c)^2*cos(d*x+c)^4-1/24*cos(d*x+c)^4)+70*a^8*(-1/7*sin(d*x+c)^3*cos(d*x+c)^4-3/
35*cos(d*x+c)^4*sin(d*x+c)+1/35*(2+cos(d*x+c)^2)*sin(d*x+c))+56*a^8*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d
*x+c)^4)+28*a^8*(-1/5*cos(d*x+c)^4*sin(d*x+c)+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-2*cos(d*x+c)^4*a^8+1/3*a^8*(2+
cos(d*x+c)^2)*sin(d*x+c))

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maxima [B]  time = 0.33, size = 134, normalized size = 2.98 \[ -\frac {5 \, a^{8} \sin \left (d x + c\right )^{11} + 44 \, a^{8} \sin \left (d x + c\right )^{10} + 165 \, a^{8} \sin \left (d x + c\right )^{9} + 330 \, a^{8} \sin \left (d x + c\right )^{8} + 330 \, a^{8} \sin \left (d x + c\right )^{7} - 462 \, a^{8} \sin \left (d x + c\right )^{5} - 660 \, a^{8} \sin \left (d x + c\right )^{4} - 495 \, a^{8} \sin \left (d x + c\right )^{3} - 220 \, a^{8} \sin \left (d x + c\right )^{2} - 55 \, a^{8} \sin \left (d x + c\right )}{55 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/55*(5*a^8*sin(d*x + c)^11 + 44*a^8*sin(d*x + c)^10 + 165*a^8*sin(d*x + c)^9 + 330*a^8*sin(d*x + c)^8 + 330*
a^8*sin(d*x + c)^7 - 462*a^8*sin(d*x + c)^5 - 660*a^8*sin(d*x + c)^4 - 495*a^8*sin(d*x + c)^3 - 220*a^8*sin(d*
x + c)^2 - 55*a^8*sin(d*x + c))/d

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mupad [B]  time = 0.12, size = 132, normalized size = 2.93 \[ \frac {-\frac {a^8\,{\sin \left (c+d\,x\right )}^{11}}{11}-\frac {4\,a^8\,{\sin \left (c+d\,x\right )}^{10}}{5}-3\,a^8\,{\sin \left (c+d\,x\right )}^9-6\,a^8\,{\sin \left (c+d\,x\right )}^8-6\,a^8\,{\sin \left (c+d\,x\right )}^7+\frac {42\,a^8\,{\sin \left (c+d\,x\right )}^5}{5}+12\,a^8\,{\sin \left (c+d\,x\right )}^4+9\,a^8\,{\sin \left (c+d\,x\right )}^3+4\,a^8\,{\sin \left (c+d\,x\right )}^2+a^8\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + a*sin(c + d*x))^8,x)

[Out]

(a^8*sin(c + d*x) + 4*a^8*sin(c + d*x)^2 + 9*a^8*sin(c + d*x)^3 + 12*a^8*sin(c + d*x)^4 + (42*a^8*sin(c + d*x)
^5)/5 - 6*a^8*sin(c + d*x)^7 - 6*a^8*sin(c + d*x)^8 - 3*a^8*sin(c + d*x)^9 - (4*a^8*sin(c + d*x)^10)/5 - (a^8*
sin(c + d*x)^11)/11)/d

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sympy [A]  time = 50.56, size = 422, normalized size = 9.38 \[ \begin {cases} \frac {2 a^{8} \sin ^{11}{\left (c + d x \right )}}{99 d} + \frac {a^{8} \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{9 d} + \frac {8 a^{8} \sin ^{9}{\left (c + d x \right )}}{9 d} + \frac {4 a^{8} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {4 a^{8} \sin ^{7}{\left (c + d x \right )}}{d} - \frac {2 a^{8} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {14 a^{8} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {56 a^{8} \sin ^{5}{\left (c + d x \right )}}{15 d} - \frac {2 a^{8} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {14 a^{8} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {28 a^{8} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 a^{8} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{8} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{d} - \frac {28 a^{8} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {14 a^{8} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {a^{8} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a^{8} \cos ^{10}{\left (c + d x \right )}}{5 d} - \frac {7 a^{8} \cos ^{8}{\left (c + d x \right )}}{3 d} - \frac {14 a^{8} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {2 a^{8} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{8} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))**8,x)

[Out]

Piecewise((2*a**8*sin(c + d*x)**11/(99*d) + a**8*sin(c + d*x)**9*cos(c + d*x)**2/(9*d) + 8*a**8*sin(c + d*x)**
9/(9*d) + 4*a**8*sin(c + d*x)**7*cos(c + d*x)**2/d + 4*a**8*sin(c + d*x)**7/d - 2*a**8*sin(c + d*x)**6*cos(c +
d*x)**4/d + 14*a**8*sin(c + d*x)**5*cos(c + d*x)**2/d + 56*a**8*sin(c + d*x)**5/(15*d) - 2*a**8*sin(c + d*x)*
*4*cos(c + d*x)**6/d - 14*a**8*sin(c + d*x)**4*cos(c + d*x)**4/d + 28*a**8*sin(c + d*x)**3*cos(c + d*x)**2/(3*
d) + 2*a**8*sin(c + d*x)**3/(3*d) - a**8*sin(c + d*x)**2*cos(c + d*x)**8/d - 28*a**8*sin(c + d*x)**2*cos(c + d
*x)**6/(3*d) - 14*a**8*sin(c + d*x)**2*cos(c + d*x)**4/d + a**8*sin(c + d*x)*cos(c + d*x)**2/d - a**8*cos(c +
d*x)**10/(5*d) - 7*a**8*cos(c + d*x)**8/(3*d) - 14*a**8*cos(c + d*x)**6/(3*d) - 2*a**8*cos(c + d*x)**4/d, Ne(d
, 0)), (x*(a*sin(c) + a)**8*cos(c)**3, True))

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