3.48 \(\int \sec ^3(c+d x) (a+a \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=121 \[ \frac{a^8 \sin ^5(c+d x)}{5 d}+\frac{2 a^8 \sin ^4(c+d x)}{d}+\frac{10 a^8 \sin ^3(c+d x)}{d}+\frac{36 a^8 \sin ^2(c+d x)}{d}+\frac{64 a^9}{d (a-a \sin (c+d x))}+\frac{129 a^8 \sin (c+d x)}{d}+\frac{192 a^8 \log (1-\sin (c+d x))}{d} \]

[Out]

(192*a^8*Log[1 - Sin[c + d*x]])/d + (129*a^8*Sin[c + d*x])/d + (36*a^8*Sin[c + d*x]^2)/d + (10*a^8*Sin[c + d*x
]^3)/d + (2*a^8*Sin[c + d*x]^4)/d + (a^8*Sin[c + d*x]^5)/(5*d) + (64*a^9)/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.0940941, antiderivative size = 121, 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} \[ \frac{a^8 \sin ^5(c+d x)}{5 d}+\frac{2 a^8 \sin ^4(c+d x)}{d}+\frac{10 a^8 \sin ^3(c+d x)}{d}+\frac{36 a^8 \sin ^2(c+d x)}{d}+\frac{64 a^9}{d (a-a \sin (c+d x))}+\frac{129 a^8 \sin (c+d x)}{d}+\frac{192 a^8 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^8,x]

[Out]

(192*a^8*Log[1 - Sin[c + d*x]])/d + (129*a^8*Sin[c + d*x])/d + (36*a^8*Sin[c + d*x]^2)/d + (10*a^8*Sin[c + d*x
]^3)/d + (2*a^8*Sin[c + d*x]^4)/d + (a^8*Sin[c + d*x]^5)/(5*d) + (64*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 ^3(c+d x) (a+a \sin (c+d x))^8 \, dx &=\frac{a^3 \operatorname{Subst}\left (\int \frac{(a+x)^6}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (129 a^4+\frac{64 a^6}{(a-x)^2}-\frac{192 a^5}{a-x}+72 a^3 x+30 a^2 x^2+8 a x^3+x^4\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{192 a^8 \log (1-\sin (c+d x))}{d}+\frac{129 a^8 \sin (c+d x)}{d}+\frac{36 a^8 \sin ^2(c+d x)}{d}+\frac{10 a^8 \sin ^3(c+d x)}{d}+\frac{2 a^8 \sin ^4(c+d x)}{d}+\frac{a^8 \sin ^5(c+d x)}{5 d}+\frac{64 a^9}{d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.259012, size = 111, normalized size = 0.92 \[ \frac{a^8 (1-\sin (c+d x)) (\sin (c+d x)+1) \sec ^2(c+d x) \left (\frac{1}{5} \sin ^5(c+d x)+2 \sin ^4(c+d x)+10 \sin ^3(c+d x)+36 \sin ^2(c+d x)+129 \sin (c+d x)+\frac{64}{1-\sin (c+d x)}+192 \log (1-\sin (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^8,x]

[Out]

(a^8*Sec[c + d*x]^2*(1 - Sin[c + d*x])*(1 + Sin[c + d*x])*(192*Log[1 - Sin[c + d*x]] + 64/(1 - Sin[c + d*x]) +
129*Sin[c + d*x] + 36*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3 + 2*Sin[c + d*x]^4 + Sin[c + d*x]^5/5))/d

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Maple [B]  time = 0.112, size = 345, normalized size = 2.9 \begin{align*} 192\,{\frac{{a}^{8}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-192\,{\frac{{a}^{8}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+68\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{385\,{a}^{8}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d}}+4\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d}}+{\frac{147\,{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{10\,d}}+34\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{119\,{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+28\,{\frac{{a}^{8} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+4\,{\frac{{a}^{8}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+14\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+28\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+35\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+14\,{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{8}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))^8,x)

[Out]

192/d*a^8*ln(cos(d*x+c))-192/d*a^8*ln(sec(d*x+c)+tan(d*x+c))+68*a^8*sin(d*x+c)^2/d+385/2*a^8*sin(d*x+c)/d+1/2/
d*a^8*sin(d*x+c)^7+4/d*a^8*sin(d*x+c)^6+147/10*a^8*sin(d*x+c)^5/d+34*a^8*sin(d*x+c)^4/d+119/2*a^8*sin(d*x+c)^3
/d+28/d*a^8*tan(d*x+c)^2+4/d*a^8/cos(d*x+c)^2+4/d*a^8*sin(d*x+c)^8/cos(d*x+c)^2+14/d*a^8*sin(d*x+c)^7/cos(d*x+
c)^2+28/d*a^8*sin(d*x+c)^6/cos(d*x+c)^2+35/d*a^8*sin(d*x+c)^5/cos(d*x+c)^2+14/d*a^8*sin(d*x+c)^3/cos(d*x+c)^2+
1/2/d*a^8*sec(d*x+c)*tan(d*x+c)+1/2/d*a^8*sin(d*x+c)^9/cos(d*x+c)^2

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Maxima [A]  time = 0.959338, size = 131, normalized size = 1.08 \begin{align*} \frac{a^{8} \sin \left (d x + c\right )^{5} + 10 \, a^{8} \sin \left (d x + c\right )^{4} + 50 \, a^{8} \sin \left (d x + c\right )^{3} + 180 \, a^{8} \sin \left (d x + c\right )^{2} + 960 \, a^{8} \log \left (\sin \left (d x + c\right ) - 1\right ) + 645 \, a^{8} \sin \left (d x + c\right ) - \frac{320 \, a^{8}}{\sin \left (d x + c\right ) - 1}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/5*(a^8*sin(d*x + c)^5 + 10*a^8*sin(d*x + c)^4 + 50*a^8*sin(d*x + c)^3 + 180*a^8*sin(d*x + c)^2 + 960*a^8*log
(sin(d*x + c) - 1) + 645*a^8*sin(d*x + c) - 320*a^8/(sin(d*x + c) - 1))/d

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Fricas [A]  time = 1.87999, size = 328, normalized size = 2.71 \begin{align*} -\frac{4 \, a^{8} \cos \left (d x + c\right )^{6} - 172 \, a^{8} \cos \left (d x + c\right )^{4} + 2192 \, a^{8} \cos \left (d x + c\right )^{2} - 1119 \, a^{8} - 3840 \,{\left (a^{8} \sin \left (d x + c\right ) - a^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) -{\left (36 \, a^{8} \cos \left (d x + c\right )^{4} - 592 \, a^{8} \cos \left (d x + c\right )^{2} - 2399 \, a^{8}\right )} \sin \left (d x + c\right )}{20 \,{\left (d \sin \left (d x + c\right ) - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

-1/20*(4*a^8*cos(d*x + c)^6 - 172*a^8*cos(d*x + c)^4 + 2192*a^8*cos(d*x + c)^2 - 1119*a^8 - 3840*(a^8*sin(d*x
+ c) - a^8)*log(-sin(d*x + c) + 1) - (36*a^8*cos(d*x + c)^4 - 592*a^8*cos(d*x + c)^2 - 2399*a^8)*sin(d*x + c))
/(d*sin(d*x + c) - 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)**3*(a+a*sin(d*x+c))**8,x)

[Out]

Timed out

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Giac [B]  time = 1.26611, size = 371, normalized size = 3.07 \begin{align*} -\frac{2 \,{\left (480 \, a^{8} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 960 \, a^{8} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{160 \,{\left (9 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 20 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{8}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{2}} - \frac{1096 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 645 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 5840 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 2780 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12120 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 4286 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12120 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2780 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5840 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 645 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1096 \, a^{8}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}\right )}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

-2/5*(480*a^8*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 960*a^8*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 160*(9*a^8*tan(1/
2*d*x + 1/2*c)^2 - 20*a^8*tan(1/2*d*x + 1/2*c) + 9*a^8)/(tan(1/2*d*x + 1/2*c) - 1)^2 - (1096*a^8*tan(1/2*d*x +
1/2*c)^10 + 645*a^8*tan(1/2*d*x + 1/2*c)^9 + 5840*a^8*tan(1/2*d*x + 1/2*c)^8 + 2780*a^8*tan(1/2*d*x + 1/2*c)^
7 + 12120*a^8*tan(1/2*d*x + 1/2*c)^6 + 4286*a^8*tan(1/2*d*x + 1/2*c)^5 + 12120*a^8*tan(1/2*d*x + 1/2*c)^4 + 27
80*a^8*tan(1/2*d*x + 1/2*c)^3 + 5840*a^8*tan(1/2*d*x + 1/2*c)^2 + 645*a^8*tan(1/2*d*x + 1/2*c) + 1096*a^8)/(ta
n(1/2*d*x + 1/2*c)^2 + 1)^5)/d