3.96 \(\int \frac{\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx\)

Optimal. Leaf size=194 \[ -\frac{1}{256 d \left (a^8 \sin (c+d x)+a^8\right )}-\frac{1}{256 d \left (a^4 \sin (c+d x)+a^4\right )^2}-\frac{1}{192 a^5 d (a \sin (c+d x)+a)^3}-\frac{1}{128 d \left (a^2 \sin (c+d x)+a^2\right )^4}-\frac{1}{80 a^3 d (a \sin (c+d x)+a)^5}-\frac{1}{48 a^2 d (a \sin (c+d x)+a)^6}+\frac{\tanh ^{-1}(\sin (c+d x))}{256 a^8 d}-\frac{1}{28 a d (a \sin (c+d x)+a)^7}-\frac{1}{16 d (a \sin (c+d x)+a)^8} \]

[Out]

ArcTanh[Sin[c + d*x]]/(256*a^8*d) - 1/(16*d*(a + a*Sin[c + d*x])^8) - 1/(28*a*d*(a + a*Sin[c + d*x])^7) - 1/(4
8*a^2*d*(a + a*Sin[c + d*x])^6) - 1/(80*a^3*d*(a + a*Sin[c + d*x])^5) - 1/(192*a^5*d*(a + a*Sin[c + d*x])^3) -
1/(128*d*(a^2 + a^2*Sin[c + d*x])^4) - 1/(256*d*(a^4 + a^4*Sin[c + d*x])^2) - 1/(256*d*(a^8 + a^8*Sin[c + d*x
]))

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Rubi [A]  time = 0.112255, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2667, 44, 206} \[ -\frac{1}{256 d \left (a^8 \sin (c+d x)+a^8\right )}-\frac{1}{256 d \left (a^4 \sin (c+d x)+a^4\right )^2}-\frac{1}{192 a^5 d (a \sin (c+d x)+a)^3}-\frac{1}{128 d \left (a^2 \sin (c+d x)+a^2\right )^4}-\frac{1}{80 a^3 d (a \sin (c+d x)+a)^5}-\frac{1}{48 a^2 d (a \sin (c+d x)+a)^6}+\frac{\tanh ^{-1}(\sin (c+d x))}{256 a^8 d}-\frac{1}{28 a d (a \sin (c+d x)+a)^7}-\frac{1}{16 d (a \sin (c+d x)+a)^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(256*a^8*d) - 1/(16*d*(a + a*Sin[c + d*x])^8) - 1/(28*a*d*(a + a*Sin[c + d*x])^7) - 1/(4
8*a^2*d*(a + a*Sin[c + d*x])^6) - 1/(80*a^3*d*(a + a*Sin[c + d*x])^5) - 1/(192*a^5*d*(a + a*Sin[c + d*x])^3) -
1/(128*d*(a^2 + a^2*Sin[c + d*x])^4) - 1/(256*d*(a^4 + a^4*Sin[c + d*x])^2) - 1/(256*d*(a^8 + a^8*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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^9} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (\frac{1}{2 a (a+x)^9}+\frac{1}{4 a^2 (a+x)^8}+\frac{1}{8 a^3 (a+x)^7}+\frac{1}{16 a^4 (a+x)^6}+\frac{1}{32 a^5 (a+x)^5}+\frac{1}{64 a^6 (a+x)^4}+\frac{1}{128 a^7 (a+x)^3}+\frac{1}{256 a^8 (a+x)^2}+\frac{1}{256 a^8 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{1}{16 d (a+a \sin (c+d x))^8}-\frac{1}{28 a d (a+a \sin (c+d x))^7}-\frac{1}{48 a^2 d (a+a \sin (c+d x))^6}-\frac{1}{80 a^3 d (a+a \sin (c+d x))^5}-\frac{1}{192 a^5 d (a+a \sin (c+d x))^3}-\frac{1}{128 d \left (a^2+a^2 \sin (c+d x)\right )^4}-\frac{1}{256 d \left (a^4+a^4 \sin (c+d x)\right )^2}-\frac{1}{256 d \left (a^8+a^8 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{256 a^7 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{256 a^8 d}-\frac{1}{16 d (a+a \sin (c+d x))^8}-\frac{1}{28 a d (a+a \sin (c+d x))^7}-\frac{1}{48 a^2 d (a+a \sin (c+d x))^6}-\frac{1}{80 a^3 d (a+a \sin (c+d x))^5}-\frac{1}{192 a^5 d (a+a \sin (c+d x))^3}-\frac{1}{128 d \left (a^2+a^2 \sin (c+d x)\right )^4}-\frac{1}{256 d \left (a^4+a^4 \sin (c+d x)\right )^2}-\frac{1}{256 d \left (a^8+a^8 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.773059, size = 122, normalized size = 0.63 \[ -\frac{105 \sin ^7(c+d x)+840 \sin ^6(c+d x)+2975 \sin ^5(c+d x)+6160 \sin ^4(c+d x)+8351 \sin ^3(c+d x)+8008 \sin ^2(c+d x)+5993 \sin (c+d x)-105 \tanh ^{-1}(\sin (c+d x)) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^{16}+4096}{26880 a^8 d (\sin (c+d x)+1)^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4096 - 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^16 + 5993*Sin[c + d*x] + 8008*Sin[c +
d*x]^2 + 8351*Sin[c + d*x]^3 + 6160*Sin[c + d*x]^4 + 2975*Sin[c + d*x]^5 + 840*Sin[c + d*x]^6 + 105*Sin[c + d
*x]^7)/(26880*a^8*d*(1 + Sin[c + d*x])^8)

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Maple [A]  time = 0.122, size = 180, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{512\,d{a}^{8}}}-{\frac{1}{16\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{8}}}-{\frac{1}{28\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{7}}}-{\frac{1}{48\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{6}}}-{\frac{1}{80\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{128\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{192\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{256\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{256\,d{a}^{8} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{512\,d{a}^{8}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/512/d/a^8*ln(sin(d*x+c)-1)-1/16/d/a^8/(1+sin(d*x+c))^8-1/28/d/a^8/(1+sin(d*x+c))^7-1/48/d/a^8/(1+sin(d*x+c)
)^6-1/80/d/a^8/(1+sin(d*x+c))^5-1/128/d/a^8/(1+sin(d*x+c))^4-1/192/d/a^8/(1+sin(d*x+c))^3-1/256/d/a^8/(1+sin(d
*x+c))^2-1/256/d/a^8/(1+sin(d*x+c))+1/512/d/a^8*ln(1+sin(d*x+c))

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Maxima [A]  time = 0.968938, size = 288, normalized size = 1.48 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{7} + 840 \, \sin \left (d x + c\right )^{6} + 2975 \, \sin \left (d x + c\right )^{5} + 6160 \, \sin \left (d x + c\right )^{4} + 8351 \, \sin \left (d x + c\right )^{3} + 8008 \, \sin \left (d x + c\right )^{2} + 5993 \, \sin \left (d x + c\right ) + 4096\right )}}{a^{8} \sin \left (d x + c\right )^{8} + 8 \, a^{8} \sin \left (d x + c\right )^{7} + 28 \, a^{8} \sin \left (d x + c\right )^{6} + 56 \, a^{8} \sin \left (d x + c\right )^{5} + 70 \, a^{8} \sin \left (d x + c\right )^{4} + 56 \, a^{8} \sin \left (d x + c\right )^{3} + 28 \, a^{8} \sin \left (d x + c\right )^{2} + 8 \, a^{8} \sin \left (d x + c\right ) + a^{8}} - \frac{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{8}} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{8}}}{53760 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/53760*(2*(105*sin(d*x + c)^7 + 840*sin(d*x + c)^6 + 2975*sin(d*x + c)^5 + 6160*sin(d*x + c)^4 + 8351*sin(d*
x + c)^3 + 8008*sin(d*x + c)^2 + 5993*sin(d*x + c) + 4096)/(a^8*sin(d*x + c)^8 + 8*a^8*sin(d*x + c)^7 + 28*a^8
*sin(d*x + c)^6 + 56*a^8*sin(d*x + c)^5 + 70*a^8*sin(d*x + c)^4 + 56*a^8*sin(d*x + c)^3 + 28*a^8*sin(d*x + c)^
2 + 8*a^8*sin(d*x + c) + a^8) - 105*log(sin(d*x + c) + 1)/a^8 + 105*log(sin(d*x + c) - 1)/a^8)/d

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Fricas [B]  time = 1.96336, size = 1052, normalized size = 5.42 \begin{align*} \frac{1680 \, \cos \left (d x + c\right )^{6} - 17360 \, \cos \left (d x + c\right )^{4} + 45696 \, \cos \left (d x + c\right )^{2} + 105 \,{\left (\cos \left (d x + c\right )^{8} - 32 \, \cos \left (d x + c\right )^{6} + 160 \, \cos \left (d x + c\right )^{4} - 256 \, \cos \left (d x + c\right )^{2} - 8 \,{\left (\cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} + 24 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 128\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (\cos \left (d x + c\right )^{8} - 32 \, \cos \left (d x + c\right )^{6} + 160 \, \cos \left (d x + c\right )^{4} - 256 \, \cos \left (d x + c\right )^{2} - 8 \,{\left (\cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} + 24 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 128\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (105 \, \cos \left (d x + c\right )^{6} - 3290 \, \cos \left (d x + c\right )^{4} + 14616 \, \cos \left (d x + c\right )^{2} - 17424\right )} \sin \left (d x + c\right ) - 38208}{53760 \,{\left (a^{8} d \cos \left (d x + c\right )^{8} - 32 \, a^{8} d \cos \left (d x + c\right )^{6} + 160 \, a^{8} d \cos \left (d x + c\right )^{4} - 256 \, a^{8} d \cos \left (d x + c\right )^{2} + 128 \, a^{8} d - 8 \,{\left (a^{8} d \cos \left (d x + c\right )^{6} - 10 \, a^{8} d \cos \left (d x + c\right )^{4} + 24 \, a^{8} d \cos \left (d x + c\right )^{2} - 16 \, a^{8} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/53760*(1680*cos(d*x + c)^6 - 17360*cos(d*x + c)^4 + 45696*cos(d*x + c)^2 + 105*(cos(d*x + c)^8 - 32*cos(d*x
+ c)^6 + 160*cos(d*x + c)^4 - 256*cos(d*x + c)^2 - 8*(cos(d*x + c)^6 - 10*cos(d*x + c)^4 + 24*cos(d*x + c)^2 -
16)*sin(d*x + c) + 128)*log(sin(d*x + c) + 1) - 105*(cos(d*x + c)^8 - 32*cos(d*x + c)^6 + 160*cos(d*x + c)^4
- 256*cos(d*x + c)^2 - 8*(cos(d*x + c)^6 - 10*cos(d*x + c)^4 + 24*cos(d*x + c)^2 - 16)*sin(d*x + c) + 128)*log
(-sin(d*x + c) + 1) + 2*(105*cos(d*x + c)^6 - 3290*cos(d*x + c)^4 + 14616*cos(d*x + c)^2 - 17424)*sin(d*x + c)
- 38208)/(a^8*d*cos(d*x + c)^8 - 32*a^8*d*cos(d*x + c)^6 + 160*a^8*d*cos(d*x + c)^4 - 256*a^8*d*cos(d*x + c)^
2 + 128*a^8*d - 8*(a^8*d*cos(d*x + c)^6 - 10*a^8*d*cos(d*x + c)^4 + 24*a^8*d*cos(d*x + c)^2 - 16*a^8*d)*sin(d*
x + c))

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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)/(a+a*sin(d*x+c))**8,x)

[Out]

Timed out

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Giac [A]  time = 1.18536, size = 177, normalized size = 0.91 \begin{align*} \frac{\frac{840 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{8}} - \frac{840 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{8}} - \frac{2283 \, \sin \left (d x + c\right )^{8} + 19944 \, \sin \left (d x + c\right )^{7} + 77364 \, \sin \left (d x + c\right )^{6} + 175448 \, \sin \left (d x + c\right )^{5} + 258370 \, \sin \left (d x + c\right )^{4} + 261464 \, \sin \left (d x + c\right )^{3} + 192052 \, \sin \left (d x + c\right )^{2} + 114152 \, \sin \left (d x + c\right ) + 67819}{a^{8}{\left (\sin \left (d x + c\right ) + 1\right )}^{8}}}{430080 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/430080*(840*log(abs(sin(d*x + c) + 1))/a^8 - 840*log(abs(sin(d*x + c) - 1))/a^8 - (2283*sin(d*x + c)^8 + 199
44*sin(d*x + c)^7 + 77364*sin(d*x + c)^6 + 175448*sin(d*x + c)^5 + 258370*sin(d*x + c)^4 + 261464*sin(d*x + c)
^3 + 192052*sin(d*x + c)^2 + 114152*sin(d*x + c) + 67819)/(a^8*(sin(d*x + c) + 1)^8))/d