3.1.61 \(\int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [61]

Optimal. Leaf size=189 \[ -\frac {7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a}{192 d (a-a \sin (c+d x))^3}-\frac {1}{32 d (a-a \sin (c+d x))^2}+\frac {a^3}{80 d (a+a \sin (c+d x))^5}-\frac {5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac {19 a}{96 d (a+a \sin (c+d x))^3}-\frac {1}{4 d (a+a \sin (c+d x))^2}+\frac {21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {35}{256 d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

-7/128*arctanh(sin(d*x+c))/a^2/d+1/192*a/d/(a-a*sin(d*x+c))^3-1/32/d/(a-a*sin(d*x+c))^2+1/80*a^3/d/(a+a*sin(d*
x+c))^5-5/64*a^2/d/(a+a*sin(d*x+c))^4+19/96*a/d/(a+a*sin(d*x+c))^3-1/4/d/(a+a*sin(d*x+c))^2+21/256/d/(a^2-a^2*
sin(d*x+c))+35/256/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]
time = 0.11, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2786, 90, 212} \begin {gather*} \frac {a^3}{80 d (a \sin (c+d x)+a)^5}-\frac {5 a^2}{64 d (a \sin (c+d x)+a)^4}+\frac {21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {35}{256 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a}{192 d (a-a \sin (c+d x))^3}+\frac {19 a}{96 d (a \sin (c+d x)+a)^3}-\frac {1}{32 d (a-a \sin (c+d x))^2}-\frac {1}{4 d (a \sin (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(128*a^2*d) + a/(192*d*(a - a*Sin[c + d*x])^3) - 1/(32*d*(a - a*Sin[c + d*x])^2) +
a^3/(80*d*(a + a*Sin[c + d*x])^5) - (5*a^2)/(64*d*(a + a*Sin[c + d*x])^4) + (19*a)/(96*d*(a + a*Sin[c + d*x])^
3) - 1/(4*d*(a + a*Sin[c + d*x])^2) + 21/(256*d*(a^2 - a^2*Sin[c + d*x])) + 35/(256*d*(a^2 + a^2*Sin[c + d*x])
)

Rule 90

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\text {Subst}\left (\int \frac {x^7}{(a-x)^4 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {a}{64 (a-x)^4}-\frac {1}{16 (a-x)^3}+\frac {21}{256 a (a-x)^2}-\frac {a^3}{16 (a+x)^6}+\frac {5 a^2}{16 (a+x)^5}-\frac {19 a}{32 (a+x)^4}+\frac {1}{2 (a+x)^3}-\frac {35}{256 a (a+x)^2}-\frac {7}{128 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a}{192 d (a-a \sin (c+d x))^3}-\frac {1}{32 d (a-a \sin (c+d x))^2}+\frac {a^3}{80 d (a+a \sin (c+d x))^5}-\frac {5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac {19 a}{96 d (a+a \sin (c+d x))^3}-\frac {1}{4 d (a+a \sin (c+d x))^2}+\frac {21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {35}{256 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {7 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a d}\\ &=-\frac {7 \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a}{192 d (a-a \sin (c+d x))^3}-\frac {1}{32 d (a-a \sin (c+d x))^2}+\frac {a^3}{80 d (a+a \sin (c+d x))^5}-\frac {5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac {19 a}{96 d (a+a \sin (c+d x))^3}-\frac {1}{4 d (a+a \sin (c+d x))^2}+\frac {21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {35}{256 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.06, size = 112, normalized size = 0.59 \begin {gather*} -\frac {210 \tanh ^{-1}(\sin (c+d x))-\frac {2 \left (-144-393 \sin (c+d x)+78 \sin ^2(c+d x)+1039 \sin ^3(c+d x)+560 \sin ^4(c+d x)-815 \sin ^5(c+d x)-750 \sin ^6(c+d x)+105 \sin ^7(c+d x)\right )}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^5}}{3840 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/3840*(210*ArcTanh[Sin[c + d*x]] - (2*(-144 - 393*Sin[c + d*x] + 78*Sin[c + d*x]^2 + 1039*Sin[c + d*x]^3 + 5
60*Sin[c + d*x]^4 - 815*Sin[c + d*x]^5 - 750*Sin[c + d*x]^6 + 105*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^3*(1 +
Sin[c + d*x])^5))/(a^2*d)

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Maple [A]
time = 0.27, size = 127, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {-\frac {1}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {21}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {19}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d \,a^{2}}\)	\(127\)
default	\(\frac {-\frac {1}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {21}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {19}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d \,a^{2}}\)	\(127\)
risch	\(\frac {i \left (-2525 \,{\mathrm e}^{3 i \left (d x +c \right )}-2529 \,{\mathrm e}^{5 i \left (d x +c \right )}+2529 \,{\mathrm e}^{11 i \left (d x +c \right )}+2525 \,{\mathrm e}^{13 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}-2084 i {\mathrm e}^{6 i \left (d x +c \right )}-4205 \,{\mathrm e}^{7 i \left (d x +c \right )}+4205 \,{\mathrm e}^{9 i \left (d x +c \right )}-1500 i {\mathrm e}^{14 i \left (d x +c \right )}-2084 i {\mathrm e}^{10 i \left (d x +c \right )}-1500 i {\mathrm e}^{2 i \left (d x +c \right )}+16560 i {\mathrm e}^{8 i \left (d x +c \right )}+4520 i {\mathrm e}^{4 i \left (d x +c \right )}+4520 i {\mathrm e}^{12 i \left (d x +c \right )}+105 \,{\mathrm e}^{15 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d \,a^{2}}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a^{2} d}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-1/192/(sin(d*x+c)-1)^3-1/32/(sin(d*x+c)-1)^2-21/256/(sin(d*x+c)-1)+7/256*ln(sin(d*x+c)-1)+1/80/(1+si
n(d*x+c))^5-5/64/(1+sin(d*x+c))^4+19/96/(1+sin(d*x+c))^3-1/4/(1+sin(d*x+c))^2+35/256/(1+sin(d*x+c))-7/256*ln(1
+sin(d*x+c)))

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Maxima [A]
time = 0.28, size = 202, normalized size = 1.07 \begin {gather*} \frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{7} - 750 \, \sin \left (d x + c\right )^{6} - 815 \, \sin \left (d x + c\right )^{5} + 560 \, \sin \left (d x + c\right )^{4} + 1039 \, \sin \left (d x + c\right )^{3} + 78 \, \sin \left (d x + c\right )^{2} - 393 \, \sin \left (d x + c\right ) - 144\right )}}{a^{2} \sin \left (d x + c\right )^{8} + 2 \, a^{2} \sin \left (d x + c\right )^{7} - 2 \, a^{2} \sin \left (d x + c\right )^{6} - 6 \, a^{2} \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{3840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3840*(2*(105*sin(d*x + c)^7 - 750*sin(d*x + c)^6 - 815*sin(d*x + c)^5 + 560*sin(d*x + c)^4 + 1039*sin(d*x +
c)^3 + 78*sin(d*x + c)^2 - 393*sin(d*x + c) - 144)/(a^2*sin(d*x + c)^8 + 2*a^2*sin(d*x + c)^7 - 2*a^2*sin(d*x
+ c)^6 - 6*a^2*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*a^2*sin(d*x + c) - a^2) - 105*
log(sin(d*x + c) + 1)/a^2 + 105*log(sin(d*x + c) - 1)/a^2)/d

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Fricas [A]
time = 0.40, size = 218, normalized size = 1.15 \begin {gather*} \frac {1500 \, \cos \left (d x + c\right )^{6} - 3380 \, \cos \left (d x + c\right )^{4} + 2104 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 500 \, \cos \left (d x + c\right )^{4} - 276 \, \cos \left (d x + c\right )^{2} + 64\right )} \sin \left (d x + c\right ) - 512}{3840 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 2 \, a^{2} d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3840*(1500*cos(d*x + c)^6 - 3380*cos(d*x + c)^4 + 2104*cos(d*x + c)^2 - 105*(cos(d*x + c)^8 - 2*cos(d*x + c)
^6*sin(d*x + c) - 2*cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8 - 2*cos(d*x + c)^6*sin(d*x + c
) - 2*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(105*cos(d*x + c)^6 + 500*cos(d*x + c)^4 - 276*cos(d*x + c)^2
+ 64)*sin(d*x + c) - 512)/(a^2*d*cos(d*x + c)^8 - 2*a^2*d*cos(d*x + c)^6*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^
6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**7/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]
time = 14.21, size = 146, normalized size = 0.77 \begin {gather*} -\frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {10 \, {\left (77 \, \sin \left (d x + c\right )^{3} - 105 \, \sin \left (d x + c\right )^{2} + 27 \, \sin \left (d x + c\right ) + 9\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {959 \, \sin \left (d x + c\right )^{5} + 6895 \, \sin \left (d x + c\right )^{4} + 14150 \, \sin \left (d x + c\right )^{3} + 13710 \, \sin \left (d x + c\right )^{2} + 6555 \, \sin \left (d x + c\right ) + 1251}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{15360 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/15360*(420*log(abs(sin(d*x + c) + 1))/a^2 - 420*log(abs(sin(d*x + c) - 1))/a^2 + 10*(77*sin(d*x + c)^3 - 10
5*sin(d*x + c)^2 + 27*sin(d*x + c) + 9)/(a^2*(sin(d*x + c) - 1)^3) - (959*sin(d*x + c)^5 + 6895*sin(d*x + c)^4
+ 14150*sin(d*x + c)^3 + 13710*sin(d*x + c)^2 + 6555*sin(d*x + c) + 1251)/(a^2*(sin(d*x + c) + 1)^5))/d

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Mupad [B]
time = 10.46, size = 444, normalized size = 2.35 \begin {gather*} \frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{16}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}-\frac {49\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{24}-\frac {693\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{320}+\frac {791\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{240}+\frac {1207\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {123\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+\frac {1207\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}+\frac {791\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{240}-\frac {693\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{320}-\frac {49\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{24}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+64\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-90\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+64\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )}-\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^7/(a + a*sin(c + d*x))^2,x)

[Out]

((7*tan(c/2 + (d*x)/2))/64 + (7*tan(c/2 + (d*x)/2)^2)/16 + (7*tan(c/2 + (d*x)/2)^3)/192 - (49*tan(c/2 + (d*x)/
2)^4)/24 - (693*tan(c/2 + (d*x)/2)^5)/320 + (791*tan(c/2 + (d*x)/2)^6)/240 + (1207*tan(c/2 + (d*x)/2)^7)/192 +
(123*tan(c/2 + (d*x)/2)^8)/4 + (1207*tan(c/2 + (d*x)/2)^9)/192 + (791*tan(c/2 + (d*x)/2)^10)/240 - (693*tan(c
/2 + (d*x)/2)^11)/320 - (49*tan(c/2 + (d*x)/2)^12)/24 + (7*tan(c/2 + (d*x)/2)^13)/192 + (7*tan(c/2 + (d*x)/2)^
14)/16 + (7*tan(c/2 + (d*x)/2)^15)/64)/(d*(36*a^2*tan(c/2 + (d*x)/2)^5 - 20*a^2*tan(c/2 + (d*x)/2)^4 - 20*a^2*
tan(c/2 + (d*x)/2)^3 + 64*a^2*tan(c/2 + (d*x)/2)^6 - 20*a^2*tan(c/2 + (d*x)/2)^7 - 90*a^2*tan(c/2 + (d*x)/2)^8
- 20*a^2*tan(c/2 + (d*x)/2)^9 + 64*a^2*tan(c/2 + (d*x)/2)^10 + 36*a^2*tan(c/2 + (d*x)/2)^11 - 20*a^2*tan(c/2
+ (d*x)/2)^12 - 20*a^2*tan(c/2 + (d*x)/2)^13 + 4*a^2*tan(c/2 + (d*x)/2)^15 + a^2*tan(c/2 + (d*x)/2)^16 + a^2 +
4*a^2*tan(c/2 + (d*x)/2))) - (7*atanh(tan(c/2 + (d*x)/2)))/(64*a^2*d)

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