3.2.96 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \, dx\) [196]

Optimal. Leaf size=103 \[ \frac {a^2 (3 c+2 d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac {2 a^2 (3 c+2 d) \tan (e+f x)}{3 f}+\frac {a^2 (3 c+2 d) \sec (e+f x) \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f} \]

[Out]

1/2*a^2*(3*c+2*d)*arctanh(sin(f*x+e))/f+2/3*a^2*(3*c+2*d)*tan(f*x+e)/f+1/6*a^2*(3*c+2*d)*sec(f*x+e)*tan(f*x+e)
/f+1/3*d*(a+a*sec(f*x+e))^2*tan(f*x+e)/f

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Rubi [A]
time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4086, 3873, 3852, 8, 4131, 3855} \begin {gather*} \frac {2 a^2 (3 c+2 d) \tan (e+f x)}{3 f}+\frac {a^2 (3 c+2 d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac {a^2 (3 c+2 d) \tan (e+f x) \sec (e+f x)}{6 f}+\frac {d \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x]),x]

[Out]

(a^2*(3*c + 2*d)*ArcTanh[Sin[e + f*x]])/(2*f) + (2*a^2*(3*c + 2*d)*Tan[e + f*x])/(3*f) + (a^2*(3*c + 2*d)*Sec[
e + f*x]*Tan[e + f*x])/(6*f) + (d*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(3*f)

Rule 8

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4086

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*B*m + A*b*(m + 1))/(b
*(m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B
, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(481\) vs. \(2(103)=206\).
time = 6.30, size = 481, normalized size = 4.67 \begin {gather*} \frac {a^2 \cos ^3(e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) (1+\sec (e+f x))^2 (c+d \sec (e+f x)) \left (-6 (3 c+2 d) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+6 (3 c+2 d) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {2 d \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {(3 c+7 d) \cos \left (\frac {e}{2}\right )-(3 c+5 d) \sin \left (\frac {e}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {4 (6 c+5 d) \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {2 d \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {(3 c+7 d) \cos \left (\frac {e}{2}\right )+(3 c+5 d) \sin \left (\frac {e}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {4 (6 c+5 d) \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}\right )}{48 f (d+c \cos (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x]),x]

[Out]

(a^2*Cos[e + f*x]^3*Sec[(e + f*x)/2]^4*(1 + Sec[e + f*x])^2*(c + d*Sec[e + f*x])*(-6*(3*c + 2*d)*Log[Cos[(e +
f*x)/2] - Sin[(e + f*x)/2]] + 6*(3*c + 2*d)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (2*d*Sin[(f*x)/2])/((Co
s[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3) + ((3*c + 7*d)*Cos[e/2] - (3*c + 5*d)*Sin[e/2])/((
Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2) + (4*(6*c + 5*d)*Sin[(f*x)/2])/((Cos[e/2] - Sin[
e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])) + (2*d*Sin[(f*x)/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^3) - ((3*c + 7*d)*Cos[e/2] + (3*c + 5*d)*Sin[e/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^2) + (4*(6*c + 5*d)*Sin[(f*x)/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]))))/(48*f*(d + c*Cos[e + f*x]))

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Maple [A]
time = 0.24, size = 145, normalized size = 1.41

method	result	size
derivativedivides	\(\frac {a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-a^{2} d \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+2 a^{2} c \tan \left (f x +e \right )+2 a^{2} d \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} d \tan \left (f x +e \right )}{f}\)	\(145\)
default	\(\frac {a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-a^{2} d \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+2 a^{2} c \tan \left (f x +e \right )+2 a^{2} d \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} d \tan \left (f x +e \right )}{f}\)	\(145\)
norman	\(\frac {\frac {8 a^{2} \left (3 c +2 d \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {a^{2} \left (3 c +2 d \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {a^{2} \left (5 c +6 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {a^{2} \left (3 c +2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {a^{2} \left (3 c +2 d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\)	\(149\)
risch	\(-\frac {i a^{2} \left (3 c \,{\mathrm e}^{5 i \left (f x +e \right )}+6 d \,{\mathrm e}^{5 i \left (f x +e \right )}-12 c \,{\mathrm e}^{4 i \left (f x +e \right )}-6 d \,{\mathrm e}^{4 i \left (f x +e \right )}-24 \,{\mathrm e}^{2 i \left (f x +e \right )} c -24 d \,{\mathrm e}^{2 i \left (f x +e \right )}-3 \,{\mathrm e}^{i \left (f x +e \right )} c -6 d \,{\mathrm e}^{i \left (f x +e \right )}-12 c -10 d \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {3 a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 f}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) d}{f}-\frac {3 a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) d}{f}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*(a^2*c*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))-a^2*d*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)+
2*a^2*c*tan(f*x+e)+2*a^2*d*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))+a^2*c*ln(sec(f*x+e)+tan(f
*x+e))+a^2*d*tan(f*x+e))

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Maxima [A]
time = 0.29, size = 181, normalized size = 1.76 \begin {gather*} \frac {4 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} d - 3 \, a^{2} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 6 \, a^{2} d {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 12 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 24 \, a^{2} c \tan \left (f x + e\right ) + 12 \, a^{2} d \tan \left (f x + e\right )}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*d - 3*a^2*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x +
e) + 1) + log(sin(f*x + e) - 1)) - 6*a^2*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log
(sin(f*x + e) - 1)) + 12*a^2*c*log(sec(f*x + e) + tan(f*x + e)) + 24*a^2*c*tan(f*x + e) + 12*a^2*d*tan(f*x + e
))/f

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Fricas [A]
time = 1.17, size = 146, normalized size = 1.42 \begin {gather*} \frac {3 \, {\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} d + 2 \, {\left (6 \, a^{2} c + 5 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(3*a^2*c + 2*a^2*d)*cos(f*x + e)^3*log(sin(f*x + e) + 1) - 3*(3*a^2*c + 2*a^2*d)*cos(f*x + e)^3*log(-s
in(f*x + e) + 1) + 2*(2*a^2*d + 2*(6*a^2*c + 5*a^2*d)*cos(f*x + e)^2 + 3*(a^2*c + 2*a^2*d)*cos(f*x + e))*sin(f
*x + e))/(f*cos(f*x + e)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int c \sec {\left (e + f x \right )}\, dx + \int 2 c \sec ^{2}{\left (e + f x \right )}\, dx + \int c \sec ^{3}{\left (e + f x \right )}\, dx + \int d \sec ^{2}{\left (e + f x \right )}\, dx + \int 2 d \sec ^{3}{\left (e + f x \right )}\, dx + \int d \sec ^{4}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c+d*sec(f*x+e)),x)

[Out]

a**2*(Integral(c*sec(e + f*x), x) + Integral(2*c*sec(e + f*x)**2, x) + Integral(c*sec(e + f*x)**3, x) + Integr
al(d*sec(e + f*x)**2, x) + Integral(2*d*sec(e + f*x)**3, x) + Integral(d*sec(e + f*x)**4, x))

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Giac [A]
time = 0.51, size = 178, normalized size = 1.73 \begin {gather*} \frac {3 \, {\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 6 \, a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 24 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 16 \, a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 18 \, a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(3*a^2*c + 2*a^2*d)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*(3*a^2*c + 2*a^2*d)*log(abs(tan(1/2*f*x + 1/
2*e) - 1)) - 2*(9*a^2*c*tan(1/2*f*x + 1/2*e)^5 + 6*a^2*d*tan(1/2*f*x + 1/2*e)^5 - 24*a^2*c*tan(1/2*f*x + 1/2*e
)^3 - 16*a^2*d*tan(1/2*f*x + 1/2*e)^3 + 15*a^2*c*tan(1/2*f*x + 1/2*e) + 18*a^2*d*tan(1/2*f*x + 1/2*e))/(tan(1/
2*f*x + 1/2*e)^2 - 1)^3)/f

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Mupad [B]
time = 4.52, size = 161, normalized size = 1.56 \begin {gather*} \frac {2\,a^2\,\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,c}{2}+d\right )}{6\,c+4\,d}\right )\,\left (\frac {3\,c}{2}+d\right )}{f}-\frac {\left (3\,a^2\,c+2\,a^2\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (-8\,a^2\,c-\frac {16\,a^2\,d}{3}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (5\,a^2\,c+6\,a^2\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))^2*(c + d/cos(e + f*x)))/cos(e + f*x),x)

[Out]

(2*a^2*atanh((4*tan(e/2 + (f*x)/2)*((3*c)/2 + d))/(6*c + 4*d))*((3*c)/2 + d))/f - (tan(e/2 + (f*x)/2)*(5*a^2*c
+ 6*a^2*d) + tan(e/2 + (f*x)/2)^5*(3*a^2*c + 2*a^2*d) - tan(e/2 + (f*x)/2)^3*(8*a^2*c + (16*a^2*d)/3))/(f*(3*
tan(e/2 + (f*x)/2)^2 - 3*tan(e/2 + (f*x)/2)^4 + tan(e/2 + (f*x)/2)^6 - 1))

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