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

Optimal. Leaf size=160 \[ -\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac{\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}-\frac{7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}+\frac{7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}-\frac{7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}+\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

(7*ArcTanh[Sin[c + d*x]])/(256*a*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]
)/(128*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x]^3)/(96*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]^5)/(80*a*d) + (Sec[c
+ d*x]^3*Tan[c + d*x]^7)/(10*a*d) - Tan[c + d*x]^10/(10*a*d)

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Rubi [A]  time = 0.270722, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2835, 2611, 3768, 3770, 2607, 30} \[ -\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac{\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}-\frac{7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}+\frac{7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}-\frac{7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}+\frac{7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTanh[Sin[c + d*x]])/(256*a*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]
)/(128*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x]^3)/(96*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]^5)/(80*a*d) + (Sec[c
+ d*x]^3*Tan[c + d*x]^7)/(10*a*d) - Tan[c + d*x]^10/(10*a*d)

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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

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

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^3(c+d x) \tan ^8(c+d x) \, dx}{a}-\frac{\int \sec ^2(c+d x) \tan ^9(c+d x) \, dx}{a}\\ &=\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{7 \int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{10 a}-\frac{\operatorname{Subst}\left (\int x^9 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{16 a}\\ &=\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}-\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{\tan ^{10}(c+d x)}{10 a d}-\frac{7 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{32 a}\\ &=-\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}-\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=\frac{7 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}-\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{7 \int \sec (c+d x) \, dx}{256 a}\\ &=\frac{7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac{7 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac{7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac{7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}-\frac{7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac{\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac{\tan ^{10}(c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 2.44905, size = 121, normalized size = 0.76 \[ \frac{\frac{-210 \sin ^8(c+d x)+3630 \sin ^7(c+d x)+2050 \sin ^6(c+d x)-5630 \sin ^5(c+d x)-3838 \sin ^4(c+d x)+3842 \sin ^3(c+d x)+2862 \sin ^2(c+d x)-978 \sin (c+d x)-768}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}+210 \tanh ^{-1}(\sin (c+d x))}{7680 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(210*ArcTanh[Sin[c + d*x]] + (-768 - 978*Sin[c + d*x] + 2862*Sin[c + d*x]^2 + 3842*Sin[c + d*x]^3 - 3838*Sin[c
+ d*x]^4 - 5630*Sin[c + d*x]^5 + 2050*Sin[c + d*x]^6 + 3630*Sin[c + d*x]^7 - 210*Sin[c + d*x]^8)/((-1 + Sin[c
+ d*x])^4*(1 + Sin[c + d*x])^5))/(7680*a*d)

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Maple [A]  time = 0.107, size = 198, normalized size = 1.2 \begin{align*}{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}+{\frac{5}{192\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{37}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{7}{64\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{7\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{512\,da}}-{\frac{1}{160\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{11}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{47}{384\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{93}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{35}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{7\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{512\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x)

[Out]

1/256/d/a/(sin(d*x+c)-1)^4+5/192/d/a/(sin(d*x+c)-1)^3+37/512/d/a/(sin(d*x+c)-1)^2+7/64/a/d/(sin(d*x+c)-1)-7/51
2/a/d*ln(sin(d*x+c)-1)-1/160/d/a/(1+sin(d*x+c))^5+11/256/d/a/(1+sin(d*x+c))^4-47/384/d/a/(1+sin(d*x+c))^3+93/5
12/a/d/(1+sin(d*x+c))^2-35/256/a/d/(1+sin(d*x+c))+7/512*ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.02855, size = 289, normalized size = 1.81 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{8} - 1815 \, \sin \left (d x + c\right )^{7} - 1025 \, \sin \left (d x + c\right )^{6} + 2815 \, \sin \left (d x + c\right )^{5} + 1919 \, \sin \left (d x + c\right )^{4} - 1921 \, \sin \left (d x + c\right )^{3} - 1431 \, \sin \left (d x + c\right )^{2} + 489 \, \sin \left (d x + c\right ) + 384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/7680*(2*(105*sin(d*x + c)^8 - 1815*sin(d*x + c)^7 - 1025*sin(d*x + c)^6 + 2815*sin(d*x + c)^5 + 1919*sin(d*
x + c)^4 - 1921*sin(d*x + c)^3 - 1431*sin(d*x + c)^2 + 489*sin(d*x + c) + 384)/(a*sin(d*x + c)^9 + a*sin(d*x +
c)^8 - 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^4 - 4*a*sin(d*x + c)^3
- 4*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) - 105*log(sin(d*x + c) + 1)/a + 105*log(sin(d*x + c) - 1)/a)/d

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Fricas [A]  time = 2.02497, size = 533, normalized size = 3.33 \begin{align*} -\frac{210 \, \cos \left (d x + c\right )^{8} + 1210 \, \cos \left (d x + c\right )^{6} - 1052 \, \cos \left (d x + c\right )^{4} + 496 \, \cos \left (d x + c\right )^{2} - 105 \,{\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \,{\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (1815 \, \cos \left (d x + c\right )^{6} - 2630 \, \cos \left (d x + c\right )^{4} + 1736 \, \cos \left (d x + c\right )^{2} - 432\right )} \sin \left (d x + c\right ) - 96}{7680 \,{\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/7680*(210*cos(d*x + c)^8 + 1210*cos(d*x + c)^6 - 1052*cos(d*x + c)^4 + 496*cos(d*x + c)^2 - 105*(cos(d*x +
c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)
*log(-sin(d*x + c) + 1) + 2*(1815*cos(d*x + c)^6 - 2630*cos(d*x + c)^4 + 1736*cos(d*x + c)^2 - 432)*sin(d*x +
c) - 96)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

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

[Out]

Timed out

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Giac [A]  time = 1.4009, size = 211, normalized size = 1.32 \begin{align*} \frac{\frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{5 \,{\left (175 \, \sin \left (d x + c\right )^{4} - 28 \, \sin \left (d x + c\right )^{3} - 522 \, \sin \left (d x + c\right )^{2} + 588 \, \sin \left (d x + c\right ) - 189\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{959 \, \sin \left (d x + c\right )^{5} + 8995 \, \sin \left (d x + c\right )^{4} + 20810 \, \sin \left (d x + c\right )^{3} + 21810 \, \sin \left (d x + c\right )^{2} + 11055 \, \sin \left (d x + c\right ) + 2211}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/30720*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1))/a + 5*(175*sin(d*x + c)^4 - 28*sin(
d*x + c)^3 - 522*sin(d*x + c)^2 + 588*sin(d*x + c) - 189)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 + 899
5*sin(d*x + c)^4 + 20810*sin(d*x + c)^3 + 21810*sin(d*x + c)^2 + 11055*sin(d*x + c) + 2211)/(a*(sin(d*x + c) +
1)^5))/d