3.74 \(\int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=137 \[ -\frac {(a-a \cos (c+d x))^{11}}{11 a^{13} d}+\frac {4 (a-a \cos (c+d x))^{10}}{5 a^{12} d}-\frac {25 (a-a \cos (c+d x))^9}{9 a^{11} d}+\frac {19 (a-a \cos (c+d x))^8}{4 a^{10} d}-\frac {4 (a-a \cos (c+d x))^7}{a^9 d}+\frac {4 (a-a \cos (c+d x))^6}{3 a^8 d} \]

[Out]

4/3*(a-a*cos(d*x+c))^6/a^8/d-4*(a-a*cos(d*x+c))^7/a^9/d+19/4*(a-a*cos(d*x+c))^8/a^10/d-25/9*(a-a*cos(d*x+c))^9
/a^11/d+4/5*(a-a*cos(d*x+c))^10/a^12/d-1/11*(a-a*cos(d*x+c))^11/a^13/d

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Rubi [A]  time = 0.19, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2836, 12, 88} \[ -\frac {(a-a \cos (c+d x))^{11}}{11 a^{13} d}+\frac {4 (a-a \cos (c+d x))^{10}}{5 a^{12} d}-\frac {25 (a-a \cos (c+d x))^9}{9 a^{11} d}+\frac {19 (a-a \cos (c+d x))^8}{4 a^{10} d}-\frac {4 (a-a \cos (c+d x))^7}{a^9 d}+\frac {4 (a-a \cos (c+d x))^6}{3 a^8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^11/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*(a - a*Cos[c + d*x])^6)/(3*a^8*d) - (4*(a - a*Cos[c + d*x])^7)/(a^9*d) + (19*(a - a*Cos[c + d*x])^8)/(4*a^1
0*d) - (25*(a - a*Cos[c + d*x])^9)/(9*a^11*d) + (4*(a - a*Cos[c + d*x])^10)/(5*a^12*d) - (a - a*Cos[c + d*x])^
11/(11*a^13*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^{11}(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {(-a-x)^5 x^2 (-a+x)^3}{a^2} \, dx,x,-a \cos (c+d x)\right )}{a^{11} d}\\ &=\frac {\operatorname {Subst}\left (\int (-a-x)^5 x^2 (-a+x)^3 \, dx,x,-a \cos (c+d x)\right )}{a^{13} d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-8 a^5 (-a-x)^5-28 a^4 (-a-x)^6-38 a^3 (-a-x)^7-25 a^2 (-a-x)^8-8 a (-a-x)^9-(-a-x)^{10}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^{13} d}\\ &=\frac {4 (a-a \cos (c+d x))^6}{3 a^8 d}-\frac {4 (a-a \cos (c+d x))^7}{a^9 d}+\frac {19 (a-a \cos (c+d x))^8}{4 a^{10} d}-\frac {25 (a-a \cos (c+d x))^9}{9 a^{11} d}+\frac {4 (a-a \cos (c+d x))^{10}}{5 a^{12} d}-\frac {(a-a \cos (c+d x))^{11}}{11 a^{13} d}\\ \end {align*}

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Mathematica [A]  time = 5.63, size = 72, normalized size = 0.53 \[ \frac {4 \sin ^{12}\left (\frac {1}{2} (c+d x)\right ) (4038 \cos (c+d x)+2586 \cos (2 (c+d x))+1189 \cos (3 (c+d x))+342 \cos (4 (c+d x))+45 \cos (5 (c+d x))+2360)}{495 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^11/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*(2360 + 4038*Cos[c + d*x] + 2586*Cos[2*(c + d*x)] + 1189*Cos[3*(c + d*x)] + 342*Cos[4*(c + d*x)] + 45*Cos[5
*(c + d*x)])*Sin[(c + d*x)/2]^12)/(495*a^2*d)

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fricas [A]  time = 1.51, size = 89, normalized size = 0.65 \[ \frac {180 \, \cos \left (d x + c\right )^{11} - 396 \, \cos \left (d x + c\right )^{10} - 440 \, \cos \left (d x + c\right )^{9} + 1485 \, \cos \left (d x + c\right )^{8} - 1980 \, \cos \left (d x + c\right )^{6} + 792 \, \cos \left (d x + c\right )^{5} + 990 \, \cos \left (d x + c\right )^{4} - 660 \, \cos \left (d x + c\right )^{3}}{1980 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^11/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/1980*(180*cos(d*x + c)^11 - 396*cos(d*x + c)^10 - 440*cos(d*x + c)^9 + 1485*cos(d*x + c)^8 - 1980*cos(d*x +
c)^6 + 792*cos(d*x + c)^5 + 990*cos(d*x + c)^4 - 660*cos(d*x + c)^3)/(a^2*d)

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giac [A]  time = 0.89, size = 185, normalized size = 1.35 \[ -\frac {64 \, {\left (\frac {11 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {55 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {165 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {330 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {462 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {198 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {990 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 1\right )}}{495 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^11/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-64/495*(11*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 55*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 165*(cos(d*
x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 330*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 462*(cos(d*x + c) - 1)^5/
(cos(d*x + c) + 1)^5 + 198*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 990*(cos(d*x + c) - 1)^7/(cos(d*x + c)
+ 1)^7 - 1)/(a^2*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^11)

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maple [A]  time = 0.69, size = 88, normalized size = 0.64 \[ -\frac {\frac {1}{\sec \left (d x +c \right )^{6}}-\frac {1}{11 \sec \left (d x +c \right )^{11}}+\frac {2}{9 \sec \left (d x +c \right )^{9}}+\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{4}}-\frac {3}{4 \sec \left (d x +c \right )^{8}}+\frac {1}{5 \sec \left (d x +c \right )^{10}}-\frac {2}{5 \sec \left (d x +c \right )^{5}}}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^11/(a+a*sec(d*x+c))^2,x)

[Out]

-1/d/a^2*(1/sec(d*x+c)^6-1/11/sec(d*x+c)^11+2/9/sec(d*x+c)^9+1/3/sec(d*x+c)^3-1/2/sec(d*x+c)^4-3/4/sec(d*x+c)^
8+1/5/sec(d*x+c)^10-2/5/sec(d*x+c)^5)

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maxima [A]  time = 0.33, size = 89, normalized size = 0.65 \[ \frac {180 \, \cos \left (d x + c\right )^{11} - 396 \, \cos \left (d x + c\right )^{10} - 440 \, \cos \left (d x + c\right )^{9} + 1485 \, \cos \left (d x + c\right )^{8} - 1980 \, \cos \left (d x + c\right )^{6} + 792 \, \cos \left (d x + c\right )^{5} + 990 \, \cos \left (d x + c\right )^{4} - 660 \, \cos \left (d x + c\right )^{3}}{1980 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^11/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/1980*(180*cos(d*x + c)^11 - 396*cos(d*x + c)^10 - 440*cos(d*x + c)^9 + 1485*cos(d*x + c)^8 - 1980*cos(d*x +
c)^6 + 792*cos(d*x + c)^5 + 990*cos(d*x + c)^4 - 660*cos(d*x + c)^3)/(a^2*d)

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mupad [B]  time = 0.09, size = 109, normalized size = 0.80 \[ -\frac {\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}-\frac {{\cos \left (c+d\,x\right )}^4}{2\,a^2}-\frac {2\,{\cos \left (c+d\,x\right )}^5}{5\,a^2}+\frac {{\cos \left (c+d\,x\right )}^6}{a^2}-\frac {3\,{\cos \left (c+d\,x\right )}^8}{4\,a^2}+\frac {2\,{\cos \left (c+d\,x\right )}^9}{9\,a^2}+\frac {{\cos \left (c+d\,x\right )}^{10}}{5\,a^2}-\frac {{\cos \left (c+d\,x\right )}^{11}}{11\,a^2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^11/(a + a/cos(c + d*x))^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**11/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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