3.97 \(\int \frac{\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=99 \[ -\frac{\tan ^3(c+d x)}{a^3 d}-\frac{\tan (c+d x)}{a^3 d}-\frac{13 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a^3 d}+\frac{11 \tan (c+d x) \sec (c+d x)}{8 a^3 d}+\frac{x}{a^3} \]

[Out]

x/a^3 - (13*ArcTanh[Sin[c + d*x]])/(8*a^3*d) - Tan[c + d*x]/(a^3*d) + (11*Sec[c + d*x]*Tan[c + d*x])/(8*a^3*d)
+ (Sec[c + d*x]^3*Tan[c + d*x])/(4*a^3*d) - Tan[c + d*x]^3/(a^3*d)

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Rubi [A]  time = 0.204299, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3888, 3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ -\frac{\tan ^3(c+d x)}{a^3 d}-\frac{\tan (c+d x)}{a^3 d}-\frac{13 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a^3 d}+\frac{11 \tan (c+d x) \sec (c+d x)}{8 a^3 d}+\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^3 - (13*ArcTanh[Sin[c + d*x]])/(8*a^3*d) - Tan[c + d*x]/(a^3*d) + (11*Sec[c + d*x]*Tan[c + d*x])/(8*a^3*d)
+ (Sec[c + d*x]^3*Tan[c + d*x])/(4*a^3*d) - Tan[c + d*x]^3/(a^3*d)

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

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 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]

Rubi steps

\begin{align*} \int \frac{\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac{\int (-a+a \sec (c+d x))^3 \tan ^2(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (-a^3 \tan ^2(c+d x)+3 a^3 \sec (c+d x) \tan ^2(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^2(c+d x)+a^3 \sec ^3(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \tan ^2(c+d x) \, dx}{a^3}+\frac{\int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{a^3}+\frac{3 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx}{a^3}\\ &=-\frac{\tan (c+d x)}{a^3 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a^3 d}-\frac{\int \sec ^3(c+d x) \, dx}{4 a^3}+\frac{\int 1 \, dx}{a^3}-\frac{3 \int \sec (c+d x) \, dx}{2 a^3}-\frac{3 \operatorname{Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac{x}{a^3}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{11 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a^3 d}-\frac{\tan ^3(c+d x)}{a^3 d}-\frac{\int \sec (c+d x) \, dx}{8 a^3}\\ &=\frac{x}{a^3}-\frac{13 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{11 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a^3 d}-\frac{\tan ^3(c+d x)}{a^3 d}\\ \end{align*}

Mathematica [B]  time = 0.729249, size = 230, normalized size = 2.32 \[ \frac{\sec ^4(c+d x) \left (38 \sin (c+d x)-32 \sin (2 (c+d x))+22 \sin (3 (c+d x))+39 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \cos (2 (c+d x)) \left (13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 d x\right )+\cos (4 (c+d x)) \left (13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 d x\right )-39 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 d x\right )}{64 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(24*d*x + 39*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*Cos[2*(c + d*x)]*(8*d*x + 13*Log[Cos
[(c + d*x)/2] - Sin[(c + d*x)/2]] - 13*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[4*(c + d*x)]*(8*d*x + 1
3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 13*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 39*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] + 38*Sin[c + d*x] - 32*Sin[2*(c + d*x)] + 22*Sin[3*(c + d*x)]))/(64*a^3*d)

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Maple [B]  time = 0.088, size = 228, normalized size = 2.3 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{1}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{27}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{21}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{13}{8\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}+{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{27}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{21}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{13}{8\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3*arctan(tan(1/2*d*x+1/2*c))-1/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^4+3/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-27/8/
d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+21/8/d/a^3/(tan(1/2*d*x+1/2*c)+1)-13/8/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)+1/4/d/a^3
/(tan(1/2*d*x+1/2*c)-1)^4+3/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)^3+27/8/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2+21/8/d/a^3/(t
an(1/2*d*x+1/2*c)-1)+13/8/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [B]  time = 1.64603, size = 347, normalized size = 3.51 \begin{align*} \frac{\frac{2 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{13 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{21 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{3} - \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{16 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{13 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{13 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 13*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 + 21*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^3 - 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^
3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^8/(cos(d*
x + c) + 1)^8) + 16*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 13*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/
a^3 + 13*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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Fricas [A]  time = 1.19605, size = 265, normalized size = 2.68 \begin{align*} \frac{16 \, d x \cos \left (d x + c\right )^{4} - 13 \, \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + 13 \, \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (11 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{16 \, a^{3} d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*d*x*cos(d*x + c)^4 - 13*cos(d*x + c)^4*log(sin(d*x + c) + 1) + 13*cos(d*x + c)^4*log(-sin(d*x + c) +
1) + 2*(11*cos(d*x + c)^2 - 8*cos(d*x + c) + 2)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4)

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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(tan(d*x+c)**8/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 13.369, size = 166, normalized size = 1.68 \begin{align*} \frac{\frac{8 \,{\left (d x + c\right )}}{a^{3}} - \frac{13 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac{13 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{2 \,{\left (21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 13 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4} a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*(d*x + c)/a^3 - 13*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 13*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 +
2*(21*tan(1/2*d*x + 1/2*c)^7 + 3*tan(1/2*d*x + 1/2*c)^5 - 13*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1/2*c))
/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*a^3))/d