3.79 \(\int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=119 \[ \frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan (c+d x)}{a^2 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {\tan ^3(c+d x) \sec (c+d x)}{2 a^2 d}+\frac {3 \tan (c+d x) \sec (c+d x)}{4 a^2 d}+\frac {x}{a^2} \]

[Out]

x/a^2-3/4*arctanh(sin(d*x+c))/a^2/d-tan(d*x+c)/a^2/d+3/4*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*tan(d*x+c)^3/a^2/d-1/
2*sec(d*x+c)*tan(d*x+c)^3/a^2/d+1/5*tan(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3888, 3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan (c+d x)}{a^2 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {\tan ^3(c+d x) \sec (c+d x)}{2 a^2 d}+\frac {3 \tan (c+d x) \sec (c+d x)}{4 a^2 d}+\frac {x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]

[Out]

x/a^2 - (3*ArcTanh[Sin[c + d*x]])/(4*a^2*d) - Tan[c + d*x]/(a^2*d) + (3*Sec[c + d*x]*Tan[c + d*x])/(4*a^2*d) +
Tan[c + d*x]^3/(3*a^2*d) - (Sec[c + d*x]*Tan[c + d*x]^3)/(2*a^2*d) + Tan[c + d*x]^5/(5*a^2*d)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 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 3770

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

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 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]

Rubi steps

\begin {align*} \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\frac {\int (-a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \tan ^4(c+d x)-2 a^2 \sec (c+d x) \tan ^4(c+d x)+a^2 \sec ^2(c+d x) \tan ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \tan ^4(c+d x) \, dx}{a^2}+\frac {\int \sec ^2(c+d x) \tan ^4(c+d x) \, dx}{a^2}-\frac {2 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{a^2}\\ &=\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\sec (c+d x) \tan ^3(c+d x)}{2 a^2 d}-\frac {\int \tan ^2(c+d x) \, dx}{a^2}+\frac {3 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{2 a^2}+\frac {\operatorname {Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {\tan (c+d x)}{a^2 d}+\frac {3 \sec (c+d x) \tan (c+d x)}{4 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\sec (c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d}-\frac {3 \int \sec (c+d x) \, dx}{4 a^2}+\frac {\int 1 \, dx}{a^2}\\ &=\frac {x}{a^2}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {3 \sec (c+d x) \tan (c+d x)}{4 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\sec (c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 5.91, size = 495, normalized size = 4.16 \[ \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-\frac {151 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {151 \sin \left (\frac {c}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {36 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {36 \sin \left (\frac {c}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {180 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {180 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {\sec (c) \sin \left (\frac {d x}{2}\right ) \left (333 \cos \left (2 c+\frac {3 d x}{2}\right )+287 \cos \left (2 c+\frac {5 d x}{2}\right )+67 \cos \left (4 c+\frac {7 d x}{2}\right )+68 \cos \left (4 c+\frac {9 d x}{2}\right )+293 \cos \left (\frac {d x}{2}\right )\right ) \sec ^5(c+d x)}{2 d}+\frac {\cos \left (\frac {c}{2}\right ) \sec (c) \left (-43 \sin \left (\frac {c}{2}+d x\right )-43 \sin \left (\frac {3 c}{2}+d x\right )-346 \sin \left (\frac {3 c}{2}+2 d x\right )+346 \sin \left (\frac {5 c}{2}+2 d x\right )+149 \sin \left (\frac {5 c}{2}+3 d x\right )+149 \sin \left (\frac {7 c}{2}+3 d x\right )+308 \sin \left (\frac {c}{2}\right )\right ) \sec ^4(c+d x)}{4 d}+240 x\right )}{60 a^2 (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(240*x + (180*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d - (180*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]])/d - ((293*Cos[(d*x)/2] + 333*Cos[2*c + (3*d*x)/2] + 287*Cos[2*c + (5*d*x)/2] +
67*Cos[4*c + (7*d*x)/2] + 68*Cos[4*c + (9*d*x)/2])*Sec[c]*Sec[c + d*x]^5*Sin[(d*x)/2])/(2*d) + (36*Sin[c/2])/
(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) - (151*Sin[c/2])/(d*(Cos[c/2] - Sin[c/2])*(C
os[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (36*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d
*x)/2])^4) - (151*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Cos[c/2]*Sec[
c]*Sec[c + d*x]^4*(308*Sin[c/2] - 43*Sin[c/2 + d*x] - 43*Sin[(3*c)/2 + d*x] - 346*Sin[(3*c)/2 + 2*d*x] + 346*S
in[(5*c)/2 + 2*d*x] + 149*Sin[(5*c)/2 + 3*d*x] + 149*Sin[(7*c)/2 + 3*d*x]))/(4*d)))/(60*a^2*(1 + Sec[c + d*x])
^2)

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fricas [A]  time = 0.76, size = 117, normalized size = 0.98 \[ \frac {120 \, d x \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) + 45 \, \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (68 \, \cos \left (d x + c\right )^{4} - 75 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) - 12\right )} \sin \left (d x + c\right )}{120 \, a^{2} d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(120*d*x*cos(d*x + c)^5 - 45*cos(d*x + c)^5*log(sin(d*x + c) + 1) + 45*cos(d*x + c)^5*log(-sin(d*x + c)
+ 1) - 2*(68*cos(d*x + c)^4 - 75*cos(d*x + c)^3 + 4*cos(d*x + c)^2 + 30*cos(d*x + c) - 12)*sin(d*x + c))/(a^2*
d*cos(d*x + c)^5)

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giac [A]  time = 13.72, size = 136, normalized size = 1.14 \[ \frac {\frac {60 \, {\left (d x + c\right )}}{a^{2}} - \frac {45 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {45 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 530 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 328 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)/a^2 - 45*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + 45*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2
+ 2*(105*tan(1/2*d*x + 1/2*c)^9 - 530*tan(1/2*d*x + 1/2*c)^7 + 328*tan(1/2*d*x + 1/2*c)^5 - 110*tan(1/2*d*x +
1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^2))/d

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maple [B]  time = 0.66, size = 269, normalized size = 2.26 \[ -\frac {1}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {19}{12 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{2} d}-\frac {1}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {19}{12 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{2} d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x)

[Out]

-1/5/a^2/d/(tan(1/2*d*x+1/2*c)-1)^5-1/a^2/d/(tan(1/2*d*x+1/2*c)-1)^4-19/12/a^2/d/(tan(1/2*d*x+1/2*c)-1)^3-1/8/
a^2/d/(tan(1/2*d*x+1/2*c)-1)^2+7/4/a^2/d/(tan(1/2*d*x+1/2*c)-1)+3/4/a^2/d*ln(tan(1/2*d*x+1/2*c)-1)-1/5/a^2/d/(
tan(1/2*d*x+1/2*c)+1)^5+1/a^2/d/(tan(1/2*d*x+1/2*c)+1)^4-19/12/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3+1/8/a^2/d/(tan(1
/2*d*x+1/2*c)+1)^2+7/4/a^2/d/(tan(1/2*d*x+1/2*c)+1)-3/4/a^2/d*ln(tan(1/2*d*x+1/2*c)+1)+2/a^2/d*arctan(tan(1/2*
d*x+1/2*c))

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maxima [B]  time = 0.61, size = 301, normalized size = 2.53 \[ -\frac {\frac {2 \, {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {110 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {328 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {530 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {105 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )}}{a^{2} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} - \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*(2*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 110*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 328*sin(d*x + c)^5/(c
os(d*x + c) + 1)^5 - 530*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^2 -
5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10*a^2*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10
) - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 45*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 - 45*log
(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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mupad [B]  time = 2.26, size = 179, normalized size = 1.50 \[ \frac {x}{a^2}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-\frac {53\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {164\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^8/(a + a/cos(c + d*x))^2,x)

[Out]

x/a^2 - (3*atanh(tan(c/2 + (d*x)/2)))/(2*a^2*d) + (tan(c/2 + (d*x)/2)/2 - (11*tan(c/2 + (d*x)/2)^3)/3 + (164*t
an(c/2 + (d*x)/2)^5)/15 - (53*tan(c/2 + (d*x)/2)^7)/3 + (7*tan(c/2 + (d*x)/2)^9)/2)/(d*(5*a^2*tan(c/2 + (d*x)/
2)^2 - 10*a^2*tan(c/2 + (d*x)/2)^4 + 10*a^2*tan(c/2 + (d*x)/2)^6 - 5*a^2*tan(c/2 + (d*x)/2)^8 + a^2*tan(c/2 +
(d*x)/2)^10 - a^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{8}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**8/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**8/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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