3.50 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=110 \[ -\frac{a^8 \sin ^3(c+d x)}{3 d}-\frac{4 a^8 \sin ^2(c+d x)}{d}+\frac{16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac{80 a^9}{d (a-a \sin (c+d x))}-\frac{31 a^8 \sin (c+d x)}{d}-\frac{80 a^8 \log (1-\sin (c+d x))}{d} \]

[Out]

(-80*a^8*Log[1 - Sin[c + d*x]])/d - (31*a^8*Sin[c + d*x])/d - (4*a^8*Sin[c + d*x]^2)/d - (a^8*Sin[c + d*x]^3)/
(3*d) + (16*a^10)/(d*(a - a*Sin[c + d*x])^2) - (80*a^9)/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.091071, antiderivative size = 110, normalized size of antiderivative = 1., 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} \[ -\frac{a^8 \sin ^3(c+d x)}{3 d}-\frac{4 a^8 \sin ^2(c+d x)}{d}+\frac{16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac{80 a^9}{d (a-a \sin (c+d x))}-\frac{31 a^8 \sin (c+d x)}{d}-\frac{80 a^8 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^8,x]

[Out]

(-80*a^8*Log[1 - Sin[c + d*x]])/d - (31*a^8*Sin[c + d*x])/d - (4*a^8*Sin[c + d*x]^2)/d - (a^8*Sin[c + d*x]^3)/
(3*d) + (16*a^10)/(d*(a - a*Sin[c + d*x])^2) - (80*a^9)/(d*(a - a*Sin[c + d*x]))

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])

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])

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{(a+x)^5}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^5 \operatorname{Subst}\left (\int \left (-31 a^2+\frac{32 a^5}{(a-x)^3}-\frac{80 a^4}{(a-x)^2}+\frac{80 a^3}{a-x}-8 a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{80 a^8 \log (1-\sin (c+d x))}{d}-\frac{31 a^8 \sin (c+d x)}{d}-\frac{4 a^8 \sin ^2(c+d x)}{d}-\frac{a^8 \sin ^3(c+d x)}{3 d}+\frac{16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac{80 a^9}{d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.445311, size = 73, normalized size = 0.66 \[ \frac{a^8 \left (-\frac{1}{3} \sin ^3(c+d x)-4 \sin ^2(c+d x)-31 \sin (c+d x)+\frac{16 (5 \sin (c+d x)-4)}{(\sin (c+d x)-1)^2}-80 \log (1-\sin (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^8,x]

[Out]

(a^8*(-80*Log[1 - Sin[c + d*x]] - 31*Sin[c + d*x] - 4*Sin[c + d*x]^2 - Sin[c + d*x]^3/3 + (16*(-4 + 5*Sin[c +
d*x]))/(-1 + Sin[c + d*x])^2))/d

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Maple [B]  time = 0.116, size = 503, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x)

[Out]

7/d*a^8*sin(d*x+c)^7/cos(d*x+c)^4+35/2/d*a^8*sin(d*x+c)^5/cos(d*x+c)^4+7/d*a^8*sin(d*x+c)^3/cos(d*x+c)^4+1/4/d
*a^8*tan(d*x+c)*sec(d*x+c)^3+14/d*a^8*sin(d*x+c)^4/cos(d*x+c)^4+1/4/d*a^8*sin(d*x+c)^9/cos(d*x+c)^4+2/d*a^8*si
n(d*x+c)^8/cos(d*x+c)^4-80/d*a^8*ln(cos(d*x+c))+80/d*a^8*ln(sec(d*x+c)+tan(d*x+c))-5/8/d*a^8*sin(d*x+c)^7-4/d*
a^8*sin(d*x+c)^6-12*a^8*sin(d*x+c)^2/d-665/24*a^8*sin(d*x+c)^3/d-6*a^8*sin(d*x+c)^4/d-91/8*a^8*sin(d*x+c)^5/d-
637/8*a^8*sin(d*x+c)/d+2/d*a^8/cos(d*x+c)^4+14/d*a^8*tan(d*x+c)^4-28/d*a^8*tan(d*x+c)^2-4/d*a^8*sin(d*x+c)^8/c
os(d*x+c)^2-21/2/d*a^8*sin(d*x+c)^7/cos(d*x+c)^2-35/4/d*a^8*sin(d*x+c)^5/cos(d*x+c)^2+7/2/d*a^8*sin(d*x+c)^3/c
os(d*x+c)^2+3/8/d*a^8*sec(d*x+c)*tan(d*x+c)-5/8/d*a^8*sin(d*x+c)^9/cos(d*x+c)^2

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Maxima [A]  time = 0.959797, size = 128, normalized size = 1.16 \begin{align*} -\frac{a^{8} \sin \left (d x + c\right )^{3} + 12 \, a^{8} \sin \left (d x + c\right )^{2} + 240 \, a^{8} \log \left (\sin \left (d x + c\right ) - 1\right ) + 93 \, a^{8} \sin \left (d x + c\right ) - \frac{48 \,{\left (5 \, a^{8} \sin \left (d x + c\right ) - 4 \, a^{8}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/3*(a^8*sin(d*x + c)^3 + 12*a^8*sin(d*x + c)^2 + 240*a^8*log(sin(d*x + c) - 1) + 93*a^8*sin(d*x + c) - 48*(5
*a^8*sin(d*x + c) - 4*a^8)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.8313, size = 344, normalized size = 3.13 \begin{align*} \frac{10 \, a^{8} \cos \left (d x + c\right )^{4} + 160 \, a^{8} \cos \left (d x + c\right )^{2} + 16 \, a^{8} - 240 \,{\left (a^{8} \cos \left (d x + c\right )^{2} + 2 \, a^{8} \sin \left (d x + c\right ) - 2 \, a^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (a^{8} \cos \left (d x + c\right )^{4} - 72 \, a^{8} \cos \left (d x + c\right )^{2} - 64 \, a^{8}\right )} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/3*(10*a^8*cos(d*x + c)^4 + 160*a^8*cos(d*x + c)^2 + 16*a^8 - 240*(a^8*cos(d*x + c)^2 + 2*a^8*sin(d*x + c) -
2*a^8)*log(-sin(d*x + c) + 1) + (a^8*cos(d*x + c)^4 - 72*a^8*cos(d*x + c)^2 - 64*a^8)*sin(d*x + c))/(d*cos(d*x
+ c)^2 + 2*d*sin(d*x + c) - 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(sec(d*x+c)**5*(a+a*sin(d*x+c))**8,x)

[Out]

Timed out

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Giac [B]  time = 1.23866, size = 328, normalized size = 2.98 \begin{align*} \frac{2 \,{\left (120 \, a^{8} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 240 \, a^{8} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{220 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 93 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 684 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 190 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 684 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 93 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 220 \, a^{8}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac{4 \,{\left (125 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 536 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 846 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 536 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 125 \, a^{8}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{4}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

2/3*(120*a^8*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 240*a^8*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (220*a^8*tan(1/2*d
*x + 1/2*c)^6 + 93*a^8*tan(1/2*d*x + 1/2*c)^5 + 684*a^8*tan(1/2*d*x + 1/2*c)^4 + 190*a^8*tan(1/2*d*x + 1/2*c)^
3 + 684*a^8*tan(1/2*d*x + 1/2*c)^2 + 93*a^8*tan(1/2*d*x + 1/2*c) + 220*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 + 4
*(125*a^8*tan(1/2*d*x + 1/2*c)^4 - 536*a^8*tan(1/2*d*x + 1/2*c)^3 + 846*a^8*tan(1/2*d*x + 1/2*c)^2 - 536*a^8*t
an(1/2*d*x + 1/2*c) + 125*a^8)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d