3.76 \(\int \frac{\sin ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=73 \[ \frac{\cos ^7(c+d x)}{7 a^2 d}-\frac{\cos ^6(c+d x)}{3 a^2 d}+\frac{\cos ^4(c+d x)}{2 a^2 d}-\frac{\cos ^3(c+d x)}{3 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.159336, antiderivative size = 73, 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} \[ \frac{\cos ^7(c+d x)}{7 a^2 d}-\frac{\cos ^6(c+d x)}{3 a^2 d}+\frac{\cos ^4(c+d x)}{2 a^2 d}-\frac{\cos ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 1.76868, size = 53, normalized size = 0.73 \[ \frac{4 \sin ^8\left (\frac{1}{2} (c+d x)\right ) (17 \cos (c+d x)+10 \cos (2 (c+d x))+3 (\cos (3 (c+d x))+4))}{21 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^7/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*(17*Cos[c + d*x] + 10*Cos[2*(c + d*x)] + 3*(4 + Cos[3*(c + d*x)]))*Sin[(c + d*x)/2]^8)/(21*a^2*d)

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Maple [A]  time = 0.086, size = 50, normalized size = 0.7 \begin{align*} -{\frac{1}{d{a}^{2}} \left ({\frac{1}{3\, \left ( \sec \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{2\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{3\, \left ( \sec \left ( dx+c \right ) \right ) ^{6}}}-{\frac{1}{7\, \left ( \sec \left ( dx+c \right ) \right ) ^{7}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^7/(a+a*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.990104, size = 66, normalized size = 0.9 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{7} - 14 \, \cos \left (d x + c\right )^{6} + 21 \, \cos \left (d x + c\right )^{4} - 14 \, \cos \left (d x + c\right )^{3}}{42 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/42*(6*cos(d*x + c)^7 - 14*cos(d*x + c)^6 + 21*cos(d*x + c)^4 - 14*cos(d*x + c)^3)/(a^2*d)

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Fricas [A]  time = 1.77204, size = 126, normalized size = 1.73 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{7} - 14 \, \cos \left (d x + c\right )^{6} + 21 \, \cos \left (d x + c\right )^{4} - 14 \, \cos \left (d x + c\right )^{3}}{42 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/42*(6*cos(d*x + c)^7 - 14*cos(d*x + c)^6 + 21*cos(d*x + c)^4 - 14*cos(d*x + c)^3)/(a^2*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)**7/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.34276, size = 190, normalized size = 2.6 \begin{align*} -\frac{8 \,{\left (\frac{7 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{21 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{35 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{14 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{42 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1\right )}}{21 \, a^{2} d{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-8/21*(7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 21*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 35*(cos(d*x +
c) - 1)^3/(cos(d*x + c) + 1)^3 - 14*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 42*(cos(d*x + c) - 1)^5/(cos(d
*x + c) + 1)^5 - 1)/(a^2*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^7)