3.1.50 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx\) [50]

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

[Out]

-80*a^8*ln(1-sin(d*x+c))/d-31*a^8*sin(d*x+c)/d-4*a^8*sin(d*x+c)^2/d-1/3*a^8*sin(d*x+c)^3/d+16*a^10/d/(a-a*sin(
d*x+c))^2-80*a^9/d/(a-a*sin(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, 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 = {2746, 45} \begin {gather*} \frac {16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac {80 a^9}{d (a-a \sin (c+d x))}-\frac {a^8 \sin ^3(c+d x)}{3 d}-\frac {4 a^8 \sin ^2(c+d x)}{d}-\frac {31 a^8 \sin (c+d x)}{d}-\frac {80 a^8 \log (1-\sin (c+d x))}{d} \end {gather*}

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 45

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])

Rule 2746

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])

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx &=\frac {a^5 \text {Subst}\left (\int \frac {(a+x)^5}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \text {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*}

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Mathematica [A]
time = 0.29, size = 73, normalized size = 0.66 \begin {gather*} \frac {a^8 \left (-80 \log (1-\sin (c+d x))-31 \sin (c+d x)-4 \sin ^2(c+d x)-\frac {1}{3} \sin ^3(c+d x)+\frac {16 (-4+5 \sin (c+d x))}{(-1+\sin (c+d x))^2}\right )}{d} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(526\) vs. \(2(108)=216\).
time = 0.20, size = 527, normalized size = 4.79 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/d*(a^8*(1/4*sin(d*x+c)^9/cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5/8*sin(d*x+c)^7-7/8*sin(d*x+c)^5-35/24*
sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+c)))+8*a^8*(1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+
c)^8/cos(d*x+c)^2-1/2*sin(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+28*a^8*(1/4*sin(d*x+c)^
7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x
+c)+tan(d*x+c)))+56*a^8*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+70*a^8*(1/4*sin(d*x+c)^5/cos(d*x+c)
^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+14*a^8*sin(d*x
+c)^4/cos(d*x+c)^4+28*a^8*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(s
ec(d*x+c)+tan(d*x+c)))+2*a^8/cos(d*x+c)^4+a^8*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c
)+tan(d*x+c))))

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Maxima [A]
time = 0.31, size = 95, normalized size = 0.86 \begin {gather*} -\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 {gather*}

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 = 0.39, size = 139, normalized size = 1.26 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (110) = 220\).
time = 6.20, size = 243, normalized size = 2.21 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 4.60, size = 96, normalized size = 0.87 \begin {gather*} -\frac {80\,a^8\,\ln \left (\sin \left (c+d\,x\right )-1\right )+31\,a^8\,\sin \left (c+d\,x\right )-\frac {80\,a^8\,\sin \left (c+d\,x\right )-64\,a^8}{{\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1}+4\,a^8\,{\sin \left (c+d\,x\right )}^2+\frac {a^8\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(80*a^8*log(sin(c + d*x) - 1) + 31*a^8*sin(c + d*x) - (80*a^8*sin(c + d*x) - 64*a^8)/(sin(c + d*x)^2 - 2*sin(
c + d*x) + 1) + 4*a^8*sin(c + d*x)^2 + (a^8*sin(c + d*x)^3)/3)/d

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