3.19 \(\int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx\)

Optimal. Leaf size=192 \[ \frac {a^2 \sec ^{10}(c+d x)}{10 d}+\frac {2 a^2 \sec ^9(c+d x)}{9 d}-\frac {3 a^2 \sec ^8(c+d x)}{8 d}-\frac {8 a^2 \sec ^7(c+d x)}{7 d}+\frac {a^2 \sec ^6(c+d x)}{3 d}+\frac {12 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^4(c+d x)}{2 d}-\frac {8 a^2 \sec ^3(c+d x)}{3 d}-\frac {3 a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

[Out]

-a^2*ln(cos(d*x+c))/d+2*a^2*sec(d*x+c)/d-3/2*a^2*sec(d*x+c)^2/d-8/3*a^2*sec(d*x+c)^3/d+1/2*a^2*sec(d*x+c)^4/d+
12/5*a^2*sec(d*x+c)^5/d+1/3*a^2*sec(d*x+c)^6/d-8/7*a^2*sec(d*x+c)^7/d-3/8*a^2*sec(d*x+c)^8/d+2/9*a^2*sec(d*x+c
)^9/d+1/10*a^2*sec(d*x+c)^10/d

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Rubi [A]  time = 0.10, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 88} \[ \frac {a^2 \sec ^{10}(c+d x)}{10 d}+\frac {2 a^2 \sec ^9(c+d x)}{9 d}-\frac {3 a^2 \sec ^8(c+d x)}{8 d}-\frac {8 a^2 \sec ^7(c+d x)}{7 d}+\frac {a^2 \sec ^6(c+d x)}{3 d}+\frac {12 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^4(c+d x)}{2 d}-\frac {8 a^2 \sec ^3(c+d x)}{3 d}-\frac {3 a^2 \sec ^2(c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^9,x]

[Out]

-((a^2*Log[Cos[c + d*x]])/d) + (2*a^2*Sec[c + d*x])/d - (3*a^2*Sec[c + d*x]^2)/(2*d) - (8*a^2*Sec[c + d*x]^3)/
(3*d) + (a^2*Sec[c + d*x]^4)/(2*d) + (12*a^2*Sec[c + d*x]^5)/(5*d) + (a^2*Sec[c + d*x]^6)/(3*d) - (8*a^2*Sec[c
+ d*x]^7)/(7*d) - (3*a^2*Sec[c + d*x]^8)/(8*d) + (2*a^2*Sec[c + d*x]^9)/(9*d) + (a^2*Sec[c + d*x]^10)/(10*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(a-a x)^4 (a+a x)^6}{x^{11}} \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a^{10}}{x^{11}}+\frac {2 a^{10}}{x^{10}}-\frac {3 a^{10}}{x^9}-\frac {8 a^{10}}{x^8}+\frac {2 a^{10}}{x^7}+\frac {12 a^{10}}{x^6}+\frac {2 a^{10}}{x^5}-\frac {8 a^{10}}{x^4}-\frac {3 a^{10}}{x^3}+\frac {2 a^{10}}{x^2}+\frac {a^{10}}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^8 d}\\ &=-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {3 a^2 \sec ^2(c+d x)}{2 d}-\frac {8 a^2 \sec ^3(c+d x)}{3 d}+\frac {a^2 \sec ^4(c+d x)}{2 d}+\frac {12 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^6(c+d x)}{3 d}-\frac {8 a^2 \sec ^7(c+d x)}{7 d}-\frac {3 a^2 \sec ^8(c+d x)}{8 d}+\frac {2 a^2 \sec ^9(c+d x)}{9 d}+\frac {a^2 \sec ^{10}(c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 140, normalized size = 0.73 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-252 \sec ^{10}(c+d x)-560 \sec ^9(c+d x)+945 \sec ^8(c+d x)+2880 \sec ^7(c+d x)-840 \sec ^6(c+d x)-6048 \sec ^5(c+d x)-1260 \sec ^4(c+d x)+6720 \sec ^3(c+d x)+3780 \sec ^2(c+d x)-5040 \sec (c+d x)+2520 \log (\cos (c+d x))\right )}{10080 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^9,x]

[Out]

-1/10080*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(2520*Log[Cos[c + d*x]] - 5040*Sec[c + d*x] + 3780*Sec[c
+ d*x]^2 + 6720*Sec[c + d*x]^3 - 1260*Sec[c + d*x]^4 - 6048*Sec[c + d*x]^5 - 840*Sec[c + d*x]^6 + 2880*Sec[c
+ d*x]^7 + 945*Sec[c + d*x]^8 - 560*Sec[c + d*x]^9 - 252*Sec[c + d*x]^10))/d

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fricas [A]  time = 0.87, size = 156, normalized size = 0.81 \[ -\frac {2520 \, a^{2} \cos \left (d x + c\right )^{10} \log \left (-\cos \left (d x + c\right )\right ) - 5040 \, a^{2} \cos \left (d x + c\right )^{9} + 3780 \, a^{2} \cos \left (d x + c\right )^{8} + 6720 \, a^{2} \cos \left (d x + c\right )^{7} - 1260 \, a^{2} \cos \left (d x + c\right )^{6} - 6048 \, a^{2} \cos \left (d x + c\right )^{5} - 840 \, a^{2} \cos \left (d x + c\right )^{4} + 2880 \, a^{2} \cos \left (d x + c\right )^{3} + 945 \, a^{2} \cos \left (d x + c\right )^{2} - 560 \, a^{2} \cos \left (d x + c\right ) - 252 \, a^{2}}{2520 \, d \cos \left (d x + c\right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^9,x, algorithm="fricas")

[Out]

-1/2520*(2520*a^2*cos(d*x + c)^10*log(-cos(d*x + c)) - 5040*a^2*cos(d*x + c)^9 + 3780*a^2*cos(d*x + c)^8 + 672
0*a^2*cos(d*x + c)^7 - 1260*a^2*cos(d*x + c)^6 - 6048*a^2*cos(d*x + c)^5 - 840*a^2*cos(d*x + c)^4 + 2880*a^2*c
os(d*x + c)^3 + 945*a^2*cos(d*x + c)^2 - 560*a^2*cos(d*x + c) - 252*a^2)/(d*cos(d*x + c)^10)

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giac [A]  time = 25.57, size = 342, normalized size = 1.78 \[ \frac {2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {11477 \, a^{2} + \frac {119810 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {566865 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605720 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3031770 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2995020 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2171610 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {1114200 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {382545 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {78850 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {7381 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{10}}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^9,x, algorithm="giac")

[Out]

1/2520*(2520*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 2520*a^2*log(abs(-(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - 1)) + (11477*a^2 + 119810*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 566865*a^2*(cos(d*x
+ c) - 1)^2/(cos(d*x + c) + 1)^2 + 1605720*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3031770*a^2*(cos(d*
x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 2995020*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 2171610*a^2*(cos(
d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 1114200*a^2*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 382545*a^2*(cos
(d*x + c) - 1)^8/(cos(d*x + c) + 1)^8 + 78850*a^2*(cos(d*x + c) - 1)^9/(cos(d*x + c) + 1)^9 + 7381*a^2*(cos(d*
x + c) - 1)^10/(cos(d*x + c) + 1)^10)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^10)/d

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maple [A]  time = 0.86, size = 327, normalized size = 1.70 \[ \frac {a^{2} \left (\tan ^{8}\left (d x +c \right )\right )}{8 d}-\frac {a^{2} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}+\frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{9}}-\frac {2 a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{7}}+\frac {2 a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )^{5}}-\frac {2 a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{3}}+\frac {2 a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )}+\frac {256 a^{2} \cos \left (d x +c \right )}{315 d}+\frac {2 a^{2} \cos \left (d x +c \right ) \left (\sin ^{8}\left (d x +c \right )\right )}{9 d}+\frac {16 a^{2} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{63 d}+\frac {32 a^{2} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{105 d}+\frac {128 a^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{315 d}+\frac {a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{10 d \cos \left (d x +c \right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*tan(d*x+c)^9,x)

[Out]

1/8/d*a^2*tan(d*x+c)^8-1/6/d*a^2*tan(d*x+c)^6+1/4*a^2*tan(d*x+c)^4/d-1/2*a^2*tan(d*x+c)^2/d-a^2*ln(cos(d*x+c))
/d+2/9/d*a^2*sin(d*x+c)^10/cos(d*x+c)^9-2/63/d*a^2*sin(d*x+c)^10/cos(d*x+c)^7+2/105/d*a^2*sin(d*x+c)^10/cos(d*
x+c)^5-2/63/d*a^2*sin(d*x+c)^10/cos(d*x+c)^3+2/9/d*a^2*sin(d*x+c)^10/cos(d*x+c)+256/315*a^2*cos(d*x+c)/d+2/9/d
*a^2*cos(d*x+c)*sin(d*x+c)^8+16/63/d*a^2*cos(d*x+c)*sin(d*x+c)^6+32/105/d*a^2*cos(d*x+c)*sin(d*x+c)^4+128/315/
d*a^2*cos(d*x+c)*sin(d*x+c)^2+1/10/d*a^2*sin(d*x+c)^10/cos(d*x+c)^10

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maxima [A]  time = 0.32, size = 149, normalized size = 0.78 \[ -\frac {2520 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {5040 \, a^{2} \cos \left (d x + c\right )^{9} - 3780 \, a^{2} \cos \left (d x + c\right )^{8} - 6720 \, a^{2} \cos \left (d x + c\right )^{7} + 1260 \, a^{2} \cos \left (d x + c\right )^{6} + 6048 \, a^{2} \cos \left (d x + c\right )^{5} + 840 \, a^{2} \cos \left (d x + c\right )^{4} - 2880 \, a^{2} \cos \left (d x + c\right )^{3} - 945 \, a^{2} \cos \left (d x + c\right )^{2} + 560 \, a^{2} \cos \left (d x + c\right ) + 252 \, a^{2}}{\cos \left (d x + c\right )^{10}}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^9,x, algorithm="maxima")

[Out]

-1/2520*(2520*a^2*log(cos(d*x + c)) - (5040*a^2*cos(d*x + c)^9 - 3780*a^2*cos(d*x + c)^8 - 6720*a^2*cos(d*x +
c)^7 + 1260*a^2*cos(d*x + c)^6 + 6048*a^2*cos(d*x + c)^5 + 840*a^2*cos(d*x + c)^4 - 2880*a^2*cos(d*x + c)^3 -
945*a^2*cos(d*x + c)^2 + 560*a^2*cos(d*x + c) + 252*a^2)/cos(d*x + c)^10)/d

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mupad [B]  time = 5.14, size = 308, normalized size = 1.60 \[ \frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+\frac {272\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{3}-\frac {740\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3}+\frac {2252\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{5}-588\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {2000\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{7}-\frac {652\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {1150\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{63}-\frac {512\,a^2}{315}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{20}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+45\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-120\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-252\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+45\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^9*(a + a/cos(c + d*x))^2,x)

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d - ((1150*a^2*tan(c/2 + (d*x)/2)^2)/63 - (652*a^2*tan(c/2 + (d*x)/2)^4)/7
+ (2000*a^2*tan(c/2 + (d*x)/2)^6)/7 - 588*a^2*tan(c/2 + (d*x)/2)^8 + (2252*a^2*tan(c/2 + (d*x)/2)^10)/5 - (74
0*a^2*tan(c/2 + (d*x)/2)^12)/3 + (272*a^2*tan(c/2 + (d*x)/2)^14)/3 - 20*a^2*tan(c/2 + (d*x)/2)^16 + 2*a^2*tan(
c/2 + (d*x)/2)^18 - (512*a^2)/315)/(d*(45*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 - 120*tan(c/2 + (d*x)
/2)^6 + 210*tan(c/2 + (d*x)/2)^8 - 252*tan(c/2 + (d*x)/2)^10 + 210*tan(c/2 + (d*x)/2)^12 - 120*tan(c/2 + (d*x)
/2)^14 + 45*tan(c/2 + (d*x)/2)^16 - 10*tan(c/2 + (d*x)/2)^18 + tan(c/2 + (d*x)/2)^20 + 1))

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sympy [A]  time = 33.75, size = 314, normalized size = 1.64 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{8}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {2 a^{2} \tan ^{8}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{9 d} + \frac {a^{2} \tan ^{8}{\left (c + d x \right )}}{8 d} - \frac {a^{2} \tan ^{6}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} - \frac {16 a^{2} \tan ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{63 d} - \frac {a^{2} \tan ^{6}{\left (c + d x \right )}}{6 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {32 a^{2} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{105 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} - \frac {128 a^{2} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{315 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {256 a^{2} \sec {\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\relax (c )} + a\right )^{2} \tan ^{9}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*tan(d*x+c)**9,x)

[Out]

Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**8*sec(c + d*x)**2/(10*d) + 2*a**2*tan(c +
d*x)**8*sec(c + d*x)/(9*d) + a**2*tan(c + d*x)**8/(8*d) - a**2*tan(c + d*x)**6*sec(c + d*x)**2/(10*d) - 16*a**
2*tan(c + d*x)**6*sec(c + d*x)/(63*d) - a**2*tan(c + d*x)**6/(6*d) + a**2*tan(c + d*x)**4*sec(c + d*x)**2/(10*
d) + 32*a**2*tan(c + d*x)**4*sec(c + d*x)/(105*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2*sec(c +
d*x)**2/(10*d) - 128*a**2*tan(c + d*x)**2*sec(c + d*x)/(315*d) - a**2*tan(c + d*x)**2/(2*d) + a**2*sec(c + d*x
)**2/(10*d) + 256*a**2*sec(c + d*x)/(315*d), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**9, True))

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