3.92 \(\int \frac{\sin ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=109 \[ -\frac{\cos ^9(c+d x)}{9 a^3 d}+\frac{3 \cos ^8(c+d x)}{8 a^3 d}-\frac{2 \cos ^7(c+d x)}{7 a^3 d}-\frac{\cos ^6(c+d x)}{3 a^3 d}+\frac{3 \cos ^5(c+d x)}{5 a^3 d}-\frac{\cos ^4(c+d x)}{4 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.178612, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2836, 12, 75} \[ -\frac{\cos ^9(c+d x)}{9 a^3 d}+\frac{3 \cos ^8(c+d x)}{8 a^3 d}-\frac{2 \cos ^7(c+d x)}{7 a^3 d}-\frac{\cos ^6(c+d x)}{3 a^3 d}+\frac{3 \cos ^5(c+d x)}{5 a^3 d}-\frac{\cos ^4(c+d x)}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]

[Out]

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

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]

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 12

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 2.86371, size = 100, normalized size = 0.92 \[ -\frac{-52920 \cos (c+d x)+37800 \cos (2 (c+d x))-18480 \cos (3 (c+d x))+3780 \cos (4 (c+d x))+3024 \cos (5 (c+d x))-4200 \cos (6 (c+d x))+2700 \cos (7 (c+d x))-945 \cos (8 (c+d x))+140 \cos (9 (c+d x))+34771}{322560 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]

[Out]

-(34771 - 52920*Cos[c + d*x] + 37800*Cos[2*(c + d*x)] - 18480*Cos[3*(c + d*x)] + 3780*Cos[4*(c + d*x)] + 3024*
Cos[5*(c + d*x)] - 4200*Cos[6*(c + d*x)] + 2700*Cos[7*(c + d*x)] - 945*Cos[8*(c + d*x)] + 140*Cos[9*(c + d*x)]
)/(322560*a^3*d)

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Maple [A]  time = 0.104, size = 69, normalized size = 0.6 \begin{align*}{\frac{1}{d{a}^{3}} \left ( -{\frac{1}{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3}{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{8}}}-{\frac{1}{3\, \left ( \sec \left ( dx+c \right ) \right ) ^{6}}}+{\frac{3}{5\, \left ( \sec \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2}{7\, \left ( \sec \left ( dx+c \right ) \right ) ^{7}}}-{\frac{1}{9\, \left ( \sec \left ( dx+c \right ) \right ) ^{9}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^9/(a+a*sec(d*x+c))^3,x)

[Out]

1/d/a^3*(-1/4/sec(d*x+c)^4+3/8/sec(d*x+c)^8-1/3/sec(d*x+c)^6+3/5/sec(d*x+c)^5-2/7/sec(d*x+c)^7-1/9/sec(d*x+c)^
9)

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Maxima [A]  time = 0.985532, size = 93, normalized size = 0.85 \begin{align*} -\frac{280 \, \cos \left (d x + c\right )^{9} - 945 \, \cos \left (d x + c\right )^{8} + 720 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 1512 \, \cos \left (d x + c\right )^{5} + 630 \, \cos \left (d x + c\right )^{4}}{2520 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2520*(280*cos(d*x + c)^9 - 945*cos(d*x + c)^8 + 720*cos(d*x + c)^7 + 840*cos(d*x + c)^6 - 1512*cos(d*x + c)
^5 + 630*cos(d*x + c)^4)/(a^3*d)

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Fricas [A]  time = 1.75617, size = 194, normalized size = 1.78 \begin{align*} -\frac{280 \, \cos \left (d x + c\right )^{9} - 945 \, \cos \left (d x + c\right )^{8} + 720 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 1512 \, \cos \left (d x + c\right )^{5} + 630 \, \cos \left (d x + c\right )^{4}}{2520 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2520*(280*cos(d*x + c)^9 - 945*cos(d*x + c)^8 + 720*cos(d*x + c)^7 + 840*cos(d*x + c)^6 - 1512*cos(d*x + c)
^5 + 630*cos(d*x + c)^4)/(a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**9/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.34183, size = 250, normalized size = 2.29 \begin{align*} \frac{32 \,{\left (\frac{36 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{144 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{336 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{504 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{630 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{105 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{315 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 4\right )}}{315 \, a^{3} d{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

32/315*(36*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 144*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 336*(cos(d*
x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 504*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 630*(cos(d*x + c) - 1)^5/
(cos(d*x + c) + 1)^5 - 105*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 315*(cos(d*x + c) - 1)^7/(cos(d*x + c)
+ 1)^7 - 4)/(a^3*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^9)