3.96 \(\int \frac {\tan ^{10}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=169 \[ -\frac {3 \tan ^5(c+d x)}{5 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan (c+d x)}{a^3 d}+\frac {19 \tanh ^{-1}(\sin (c+d x))}{16 a^3 d}+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 a^3 d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{8 a^3 d}+\frac {3 \tan ^3(c+d x) \sec (c+d x)}{4 a^3 d}-\frac {17 \tan (c+d x) \sec (c+d x)}{16 a^3 d}-\frac {x}{a^3} \]

[Out]

-x/a^3+19/16*arctanh(sin(d*x+c))/a^3/d+tan(d*x+c)/a^3/d-17/16*sec(d*x+c)*tan(d*x+c)/a^3/d-1/8*sec(d*x+c)^3*tan
(d*x+c)/a^3/d-1/3*tan(d*x+c)^3/a^3/d+3/4*sec(d*x+c)*tan(d*x+c)^3/a^3/d+1/6*sec(d*x+c)^3*tan(d*x+c)^3/a^3/d-3/5
*tan(d*x+c)^5/a^3/d

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Rubi [A]  time = 0.27, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ -\frac {3 \tan ^5(c+d x)}{5 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan (c+d x)}{a^3 d}+\frac {19 \tanh ^{-1}(\sin (c+d x))}{16 a^3 d}+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 a^3 d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{8 a^3 d}+\frac {3 \tan ^3(c+d x) \sec (c+d x)}{4 a^3 d}-\frac {17 \tan (c+d x) \sec (c+d x)}{16 a^3 d}-\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^10/(a + a*Sec[c + d*x])^3,x]

[Out]

-(x/a^3) + (19*ArcTanh[Sin[c + d*x]])/(16*a^3*d) + Tan[c + d*x]/(a^3*d) - (17*Sec[c + d*x]*Tan[c + d*x])/(16*a
^3*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(8*a^3*d) - Tan[c + d*x]^3/(3*a^3*d) + (3*Sec[c + d*x]*Tan[c + d*x]^3)/(
4*a^3*d) + (Sec[c + d*x]^3*Tan[c + d*x]^3)/(6*a^3*d) - (3*Tan[c + d*x]^5)/(5*a^3*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 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 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 ^{10}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac {\int (-a+a \sec (c+d x))^3 \tan ^4(c+d x) \, dx}{a^6}\\ &=\frac {\int \left (-a^3 \tan ^4(c+d x)+3 a^3 \sec (c+d x) \tan ^4(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^4(c+d x)+a^3 \sec ^3(c+d x) \tan ^4(c+d x)\right ) \, dx}{a^6}\\ &=-\frac {\int \tan ^4(c+d x) \, dx}{a^3}+\frac {\int \sec ^3(c+d x) \tan ^4(c+d x) \, dx}{a^3}+\frac {3 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^2(c+d x) \tan ^4(c+d x) \, dx}{a^3}\\ &=-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {3 \sec (c+d x) \tan ^3(c+d x)}{4 a^3 d}+\frac {\sec ^3(c+d x) \tan ^3(c+d x)}{6 a^3 d}-\frac {\int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{2 a^3}+\frac {\int \tan ^2(c+d x) \, dx}{a^3}-\frac {9 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{4 a^3}-\frac {3 \operatorname {Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac {\tan (c+d x)}{a^3 d}-\frac {9 \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{8 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {3 \sec (c+d x) \tan ^3(c+d x)}{4 a^3 d}+\frac {\sec ^3(c+d x) \tan ^3(c+d x)}{6 a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}+\frac {\int \sec ^3(c+d x) \, dx}{8 a^3}-\frac {\int 1 \, dx}{a^3}+\frac {9 \int \sec (c+d x) \, dx}{8 a^3}\\ &=-\frac {x}{a^3}+\frac {9 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {17 \sec (c+d x) \tan (c+d x)}{16 a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{8 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {3 \sec (c+d x) \tan ^3(c+d x)}{4 a^3 d}+\frac {\sec ^3(c+d x) \tan ^3(c+d x)}{6 a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}+\frac {\int \sec (c+d x) \, dx}{16 a^3}\\ &=-\frac {x}{a^3}+\frac {19 \tanh ^{-1}(\sin (c+d x))}{16 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {17 \sec (c+d x) \tan (c+d x)}{16 a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{8 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {3 \sec (c+d x) \tan ^3(c+d x)}{4 a^3 d}+\frac {\sec ^3(c+d x) \tan ^3(c+d x)}{6 a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.92, size = 303, normalized size = 1.79 \[ -\frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (9120 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) \sec ^6(c+d x) (-210 \sin (2 c+d x)-1440 \sin (c+2 d x)+1200 \sin (3 c+2 d x)+865 \sin (2 c+3 d x)+865 \sin (4 c+3 d x)-1296 \sin (3 c+4 d x)-240 \sin (5 c+4 d x)+435 \sin (4 c+5 d x)+435 \sin (6 c+5 d x)-176 \sin (5 c+6 d x)+2400 d x \cos (c)+1800 d x \cos (c+2 d x)+1800 d x \cos (3 c+2 d x)+720 d x \cos (3 c+4 d x)+720 d x \cos (5 c+4 d x)+120 d x \cos (5 c+6 d x)+120 d x \cos (7 c+6 d x)+1760 \sin (c)-210 \sin (d x))\right )}{960 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^10/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/960*(Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(9120*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]) + Sec[c]*Sec[c + d*x]^6*(2400*d*x*Cos[c] + 1800*d*x*Cos[c + 2*d*x] + 1800*d*x*Cos[3*c
+ 2*d*x] + 720*d*x*Cos[3*c + 4*d*x] + 720*d*x*Cos[5*c + 4*d*x] + 120*d*x*Cos[5*c + 6*d*x] + 120*d*x*Cos[7*c +
6*d*x] + 1760*Sin[c] - 210*Sin[d*x] - 210*Sin[2*c + d*x] - 1440*Sin[c + 2*d*x] + 1200*Sin[3*c + 2*d*x] + 865*S
in[2*c + 3*d*x] + 865*Sin[4*c + 3*d*x] - 1296*Sin[3*c + 4*d*x] - 240*Sin[5*c + 4*d*x] + 435*Sin[4*c + 5*d*x] +
435*Sin[6*c + 5*d*x] - 176*Sin[5*c + 6*d*x])))/(a^3*d*(1 + Sec[c + d*x])^3)

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fricas [A]  time = 0.54, size = 127, normalized size = 0.75 \[ -\frac {480 \, d x \cos \left (d x + c\right )^{6} - 285 \, \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) + 285 \, \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (176 \, \cos \left (d x + c\right )^{5} - 435 \, \cos \left (d x + c\right )^{4} + 208 \, \cos \left (d x + c\right )^{3} + 110 \, \cos \left (d x + c\right )^{2} - 144 \, \cos \left (d x + c\right ) + 40\right )} \sin \left (d x + c\right )}{480 \, a^{3} d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^10/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/480*(480*d*x*cos(d*x + c)^6 - 285*cos(d*x + c)^6*log(sin(d*x + c) + 1) + 285*cos(d*x + c)^6*log(-sin(d*x +
c) + 1) - 2*(176*cos(d*x + c)^5 - 435*cos(d*x + c)^4 + 208*cos(d*x + c)^3 + 110*cos(d*x + c)^2 - 144*cos(d*x +
c) + 40)*sin(d*x + c))/(a^3*d*cos(d*x + c)^6)

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giac [A]  time = 64.84, size = 149, normalized size = 0.88 \[ -\frac {\frac {240 \, {\left (d x + c\right )}}{a^{3}} - \frac {285 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {285 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 3135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1746 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 366 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 95 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6} a^{3}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^10/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/240*(240*(d*x + c)/a^3 - 285*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 285*log(abs(tan(1/2*d*x + 1/2*c) - 1)
)/a^3 + 2*(525*tan(1/2*d*x + 1/2*c)^11 - 3135*tan(1/2*d*x + 1/2*c)^9 + 1746*tan(1/2*d*x + 1/2*c)^7 - 366*tan(1
/2*d*x + 1/2*c)^5 - 95*tan(1/2*d*x + 1/2*c)^3 + 45*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^6*a^3))
/d

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maple [B]  time = 0.82, size = 312, normalized size = 1.85 \[ \frac {1}{6 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}+\frac {11}{10 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {11}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {11}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {5}{16 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {35}{16 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {19 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a^{3} d}-\frac {1}{6 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {11}{10 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {11}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {11}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5}{16 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {35}{16 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {19 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a^{3} d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^10/(a+a*sec(d*x+c))^3,x)

[Out]

1/6/a^3/d/(tan(1/2*d*x+1/2*c)-1)^6+11/10/a^3/d/(tan(1/2*d*x+1/2*c)-1)^5+11/4/a^3/d/(tan(1/2*d*x+1/2*c)-1)^4+11
/4/a^3/d/(tan(1/2*d*x+1/2*c)-1)^3-5/16/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2-35/16/a^3/d/(tan(1/2*d*x+1/2*c)-1)-19/16
/a^3/d*ln(tan(1/2*d*x+1/2*c)-1)-1/6/a^3/d/(tan(1/2*d*x+1/2*c)+1)^6+11/10/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5-11/4/a
^3/d/(tan(1/2*d*x+1/2*c)+1)^4+11/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3+5/16/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2-35/16/a^
3/d/(tan(1/2*d*x+1/2*c)+1)+19/16/a^3/d*ln(tan(1/2*d*x+1/2*c)+1)-2/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.70, size = 343, normalized size = 2.03 \[ -\frac {\frac {2 \, {\left (\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {95 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {366 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {1746 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3135 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {525 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )}}{a^{3} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} + \frac {480 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {285 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {285 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^10/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/240*(2*(45*sin(d*x + c)/(cos(d*x + c) + 1) - 95*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 366*sin(d*x + c)^5/(c
os(d*x + c) + 1)^5 + 1746*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3135*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 525
*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)/(a^3 - 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^3*sin(d*x + c)
^4/(cos(d*x + c) + 1)^4 - 20*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 - 6*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 480*arctan(si
n(d*x + c)/(cos(d*x + c) + 1))/a^3 - 285*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 285*log(sin(d*x + c)/(
cos(d*x + c) + 1) - 1)/a^3)/d

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mupad [B]  time = 2.35, size = 208, normalized size = 1.23 \[ \frac {19\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^3\,d}-\frac {x}{a^3}-\frac {\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-\frac {209\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {291\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}-\frac {61\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^10/(a + a/cos(c + d*x))^3,x)

[Out]

(19*atanh(tan(c/2 + (d*x)/2)))/(8*a^3*d) - x/a^3 - ((3*tan(c/2 + (d*x)/2))/8 - (19*tan(c/2 + (d*x)/2)^3)/24 -
(61*tan(c/2 + (d*x)/2)^5)/20 + (291*tan(c/2 + (d*x)/2)^7)/20 - (209*tan(c/2 + (d*x)/2)^9)/8 + (35*tan(c/2 + (d
*x)/2)^11)/8)/(d*(15*a^3*tan(c/2 + (d*x)/2)^4 - 6*a^3*tan(c/2 + (d*x)/2)^2 - 20*a^3*tan(c/2 + (d*x)/2)^6 + 15*
a^3*tan(c/2 + (d*x)/2)^8 - 6*a^3*tan(c/2 + (d*x)/2)^10 + a^3*tan(c/2 + (d*x)/2)^12 + a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{10}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**10/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**10/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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