3.898 \(\int \frac{\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ \frac{\tan ^{10}(c+d x)}{10 a d}+\frac{63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^9(c+d x) \sec (c+d x)}{10 a d}+\frac{9 \tan ^7(c+d x) \sec (c+d x)}{80 a d}-\frac{21 \tan ^5(c+d x) \sec (c+d x)}{160 a d}+\frac{21 \tan ^3(c+d x) \sec (c+d x)}{128 a d}-\frac{63 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

(63*ArcTanh[Sin[c + d*x]])/(256*a*d) - (63*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (21*Sec[c + d*x]*Tan[c + d*x
]^3)/(128*a*d) - (21*Sec[c + d*x]*Tan[c + d*x]^5)/(160*a*d) + (9*Sec[c + d*x]*Tan[c + d*x]^7)/(80*a*d) - (Sec[
c + d*x]*Tan[c + d*x]^9)/(10*a*d) + Tan[c + d*x]^10/(10*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.192562, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{\tan ^{10}(c+d x)}{10 a d}+\frac{63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{\tan ^9(c+d x) \sec (c+d x)}{10 a d}+\frac{9 \tan ^7(c+d x) \sec (c+d x)}{80 a d}-\frac{21 \tan ^5(c+d x) \sec (c+d x)}{160 a d}+\frac{21 \tan ^3(c+d x) \sec (c+d x)}{128 a d}-\frac{63 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^9/(a + a*Sin[c + d*x]),x]

[Out]

(63*ArcTanh[Sin[c + d*x]])/(256*a*d) - (63*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (21*Sec[c + d*x]*Tan[c + d*x
]^3)/(128*a*d) - (21*Sec[c + d*x]*Tan[c + d*x]^5)/(160*a*d) + (9*Sec[c + d*x]*Tan[c + d*x]^7)/(80*a*d) - (Sec[
c + d*x]*Tan[c + d*x]^9)/(10*a*d) + Tan[c + d*x]^10/(10*a*d)

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

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]

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 3770

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

Rubi steps

\begin{align*} \int \frac{\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^2(c+d x) \tan ^9(c+d x) \, dx}{a}-\frac{\int \sec (c+d x) \tan ^{10}(c+d x) \, dx}{a}\\ &=-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{9 \int \sec (c+d x) \tan ^8(c+d x) \, dx}{10 a}+\frac{\operatorname{Subst}\left (\int x^9 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}-\frac{63 \int \sec (c+d x) \tan ^6(c+d x) \, dx}{80 a}\\ &=-\frac{21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac{9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{21 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{32 a}\\ &=\frac{21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac{21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac{9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}-\frac{63 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{128 a}\\ &=-\frac{63 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac{21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac{21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac{9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}+\frac{63 \int \sec (c+d x) \, dx}{256 a}\\ &=\frac{63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac{63 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac{21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac{21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac{9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac{\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac{\tan ^{10}(c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 2.52125, size = 122, normalized size = 0.79 \[ \frac{\frac{2 \left (965 \sin ^8(c+d x)+325 \sin ^7(c+d x)-2045 \sin ^6(c+d x)-765 \sin ^5(c+d x)+1923 \sin ^4(c+d x)+643 \sin ^3(c+d x)-827 \sin ^2(c+d x)-187 \sin (c+d x)+128\right )}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}+630 \tanh ^{-1}(\sin (c+d x))}{2560 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^9/(a + a*Sin[c + d*x]),x]

[Out]

(630*ArcTanh[Sin[c + d*x]] + (2*(128 - 187*Sin[c + d*x] - 827*Sin[c + d*x]^2 + 643*Sin[c + d*x]^3 + 1923*Sin[c
+ d*x]^4 - 765*Sin[c + d*x]^5 - 2045*Sin[c + d*x]^6 + 325*Sin[c + d*x]^7 + 965*Sin[c + d*x]^8))/((-1 + Sin[c
+ d*x])^4*(1 + Sin[c + d*x])^5))/(2560*a*d)

________________________________________________________________________________________

Maple [A]  time = 0.105, size = 198, normalized size = 1.3 \begin{align*}{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}+{\frac{1}{32\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{57}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{65}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{63\,\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{13}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{23}{128\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{187}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{63\,\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)^9/(a+a*sin(d*x+c)),x)

[Out]

1/256/d/a/(sin(d*x+c)-1)^4+1/32/d/a/(sin(d*x+c)-1)^3+57/512/d/a/(sin(d*x+c)-1)^2+65/256/a/d/(sin(d*x+c)-1)-63/
512/a/d*ln(sin(d*x+c)-1)+1/160/d/a/(1+sin(d*x+c))^5-13/256/d/a/(1+sin(d*x+c))^4+23/128/d/a/(1+sin(d*x+c))^3-18
7/512/a/d/(1+sin(d*x+c))^2+1/2/a/d/(1+sin(d*x+c))+63/512*ln(1+sin(d*x+c))/a/d

________________________________________________________________________________________

Maxima [A]  time = 1.05284, size = 289, normalized size = 1.88 \begin{align*} \frac{\frac{2 \,{\left (965 \, \sin \left (d x + c\right )^{8} + 325 \, \sin \left (d x + c\right )^{7} - 2045 \, \sin \left (d x + c\right )^{6} - 765 \, \sin \left (d x + c\right )^{5} + 1923 \, \sin \left (d x + c\right )^{4} + 643 \, \sin \left (d x + c\right )^{3} - 827 \, \sin \left (d x + c\right )^{2} - 187 \, \sin \left (d x + c\right ) + 128\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{315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2560*(2*(965*sin(d*x + c)^8 + 325*sin(d*x + c)^7 - 2045*sin(d*x + c)^6 - 765*sin(d*x + c)^5 + 1923*sin(d*x +
c)^4 + 643*sin(d*x + c)^3 - 827*sin(d*x + c)^2 - 187*sin(d*x + c) + 128)/(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) + 315*log(sin(d*x + c) + 1)/a - 315*log(sin(d*x + c) - 1)/a)/d

________________________________________________________________________________________

Fricas [A]  time = 2.04701, size = 529, normalized size = 3.44 \begin{align*} \frac{1930 \, \cos \left (d x + c\right )^{8} - 3630 \, \cos \left (d x + c\right )^{6} + 3156 \, \cos \left (d x + c\right )^{4} - 1488 \, \cos \left (d x + c\right )^{2} + 315 \,{\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 ) - 315 \,{\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 (325 \, \cos \left (d x + c\right )^{6} - 210 \, \cos \left (d x + c\right )^{4} + 88 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 288}{2560 \,{\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)^9/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2560*(1930*cos(d*x + c)^8 - 3630*cos(d*x + c)^6 + 3156*cos(d*x + c)^4 - 1488*cos(d*x + c)^2 + 315*(cos(d*x +
c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) - 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8
)*log(-sin(d*x + c) + 1) - 2*(325*cos(d*x + c)^6 - 210*cos(d*x + c)^4 + 88*cos(d*x + c)^2 - 16)*sin(d*x + c) +
288)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.33374, size = 211, normalized size = 1.37 \begin{align*} \frac{\frac{1260 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{1260 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{5 \,{\left (525 \, \sin \left (d x + c\right )^{4} - 1580 \, \sin \left (d x + c\right )^{3} + 1818 \, \sin \left (d x + c\right )^{2} - 932 \, \sin \left (d x + c\right ) + 177\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{2877 \, \sin \left (d x + c\right )^{5} + 9265 \, \sin \left (d x + c\right )^{4} + 12030 \, \sin \left (d x + c\right )^{3} + 7430 \, \sin \left (d x + c\right )^{2} + 1965 \, \sin \left (d x + c\right ) + 113}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10240*(1260*log(abs(sin(d*x + c) + 1))/a - 1260*log(abs(sin(d*x + c) - 1))/a + 5*(525*sin(d*x + c)^4 - 1580*
sin(d*x + c)^3 + 1818*sin(d*x + c)^2 - 932*sin(d*x + c) + 177)/(a*(sin(d*x + c) - 1)^4) - (2877*sin(d*x + c)^5
+ 9265*sin(d*x + c)^4 + 12030*sin(d*x + c)^3 + 7430*sin(d*x + c)^2 + 1965*sin(d*x + c) + 113)/(a*(sin(d*x + c
) + 1)^5))/d