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

Optimal. Leaf size=73 \[ \frac {\sin ^6(c+d x)}{6 a d}+\frac {\cos ^7(c+d x)}{7 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d} \]

[Out]

1/3*cos(d*x+c)^3/a/d-2/5*cos(d*x+c)^5/a/d+1/7*cos(d*x+c)^7/a/d+1/6*sin(d*x+c)^6/a/d

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Rubi [A]  time = 0.15, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3872, 2835, 2564, 30, 2565, 270} \[ \frac {\sin ^6(c+d x)}{6 a d}+\frac {\cos ^7(c+d x)}{7 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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 ^7(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^7(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=\frac {\int \cos (c+d x) \sin ^5(c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^5(c+d x) \, dx}{a}\\ &=\frac {\operatorname {Subst}\left (\int x^5 \, dx,x,\sin (c+d x)\right )}{a d}+\frac {\operatorname {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\sin ^6(c+d x)}{6 a d}+\frac {\operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\cos ^3(c+d x)}{3 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^7(c+d x)}{7 a d}+\frac {\sin ^6(c+d x)}{6 a d}\\ \end {align*}

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Mathematica [A]  time = 1.72, size = 52, normalized size = 0.71 \[ \frac {4 \sin ^8\left (\frac {1}{2} (c+d x)\right ) (197 \cos (c+d x)+85 \cos (2 (c+d x))+15 \cos (3 (c+d x))+123)}{105 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(123 + 197*Cos[c + d*x] + 85*Cos[2*(c + d*x)] + 15*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]^8)/(105*a*d)

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fricas [A]  time = 0.48, size = 69, normalized size = 0.95 \[ \frac {30 \, \cos \left (d x + c\right )^{7} - 35 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{4} + 70 \, \cos \left (d x + c\right )^{3} - 105 \, \cos \left (d x + c\right )^{2}}{210 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(30*cos(d*x + c)^7 - 35*cos(d*x + c)^6 - 84*cos(d*x + c)^5 + 105*cos(d*x + c)^4 + 70*cos(d*x + c)^3 - 10
5*cos(d*x + c)^2)/(a*d)

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giac [A]  time = 0.84, size = 119, normalized size = 1.63 \[ \frac {16 \, {\left (\frac {7 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {21 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {35 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {140 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1\right )}}{105 \, a d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/105*(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 - 140*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 1)/(a*d*((cos(d*x + c) - 1)/
(cos(d*x + c) + 1) - 1)^7)

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maple [A]  time = 0.46, size = 70, normalized size = 0.96 \[ -\frac {\frac {1}{6 \sec \left (d x +c \right )^{6}}+\frac {1}{2 \sec \left (d x +c \right )^{2}}-\frac {1}{7 \sec \left (d x +c \right )^{7}}-\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{4}}+\frac {2}{5 \sec \left (d x +c \right )^{5}}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 69, normalized size = 0.95 \[ \frac {30 \, \cos \left (d x + c\right )^{7} - 35 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{4} + 70 \, \cos \left (d x + c\right )^{3} - 105 \, \cos \left (d x + c\right )^{2}}{210 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(30*cos(d*x + c)^7 - 35*cos(d*x + c)^6 - 84*cos(d*x + c)^5 + 105*cos(d*x + c)^4 + 70*cos(d*x + c)^3 - 10
5*cos(d*x + c)^2)/(a*d)

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mupad [B]  time = 0.06, size = 84, normalized size = 1.15 \[ -\frac {\frac {{\cos \left (c+d\,x\right )}^2}{2\,a}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}-\frac {{\cos \left (c+d\,x\right )}^4}{2\,a}+\frac {2\,{\cos \left (c+d\,x\right )}^5}{5\,a}+\frac {{\cos \left (c+d\,x\right )}^6}{6\,a}-\frac {{\cos \left (c+d\,x\right )}^7}{7\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(c + d*x)^2/(2*a) - cos(c + d*x)^3/(3*a) - cos(c + d*x)^4/(2*a) + (2*cos(c + d*x)^5)/(5*a) + cos(c + d*x)
^6/(6*a) - cos(c + d*x)^7/(7*a))/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)**7/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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