3.1.62 \(\int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [62]

Optimal. Leaf size=128 \[ -\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {9 \tan (c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-3*arctanh(sin(d*x+c))/a^3/d+9/5*tan(d*x+c)/a^3/d-1/5*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-3/5*sec(d*x
+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+3*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.18, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3901, 4104, 4093, 3872, 3855, 3852, 8} \begin {gather*} \frac {9 \tan (c+d x)}{5 a^3 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {3 \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {\tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {3 \tan (c+d x) \sec ^2(c+d x)}{5 a d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(a^3*d) + (9*Tan[c + d*x])/(5*a^3*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Se
c[c + d*x])^3) - (3*Sec[c + d*x]^2*Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^2) + (3*Tan[c + d*x])/(d*(a^3 + a
^3*Sec[c + d*x]))

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 3872

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

Rule 3901

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Dist[d^2/(a*b*(2*m + 1)),
Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[
m])

Rule 4093

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

Rule 4104

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

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x) (3 a-6 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) \left (18 a^2-27 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec (c+d x) \left (-45 a^3+27 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {9 \int \sec ^2(c+d x) \, dx}{5 a^3}-\frac {3 \int \sec (c+d x) \, dx}{a^3}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {9 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 a^3 d}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {9 \tan (c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(294\) vs. \(2(128)=256\).
time = 1.33, size = 294, normalized size = 2.30 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+76 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+20 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \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 ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+8 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{5 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(Sec[c/2]*Sin[(d*x)/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 76*Co
s[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3
*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*
x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + Cos[(c + d*x)/2]*Tan[c/2] + 8*Cos[(c + d*x
)/2]^3*Tan[c/2]))/(5*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.09, size = 105, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
risch	\(\frac {2 i \left (15 \,{\mathrm e}^{6 i \left (d x +c \right )}+75 \,{\mathrm e}^{5 i \left (d x +c \right )}+160 \,{\mathrm e}^{4 i \left (d x +c \right )}+200 \,{\mathrm e}^{3 i \left (d x +c \right )}+189 \,{\mathrm e}^{2 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}+24\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\)	\(147\)
norman	\(\frac {\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {45 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {591 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {81 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {51 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(1/5*tan(1/2*d*x+1/2*c)^5+2*tan(1/2*d*x+1/2*c)^3+17*tan(1/2*d*x+1/2*c)-4/(tan(1/2*d*x+1/2*c)-1)+12*l
n(tan(1/2*d*x+1/2*c)-1)-4/(tan(1/2*d*x+1/2*c)+1)-12*ln(tan(1/2*d*x+1/2*c)+1))

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Maxima [A]
time = 0.28, size = 165, normalized size = 1.29 \begin {gather*} \frac {\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/20*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/
(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*lo
g(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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Fricas [A]
time = 3.96, size = 190, normalized size = 1.48 \begin {gather*} -\frac {15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (24 \, \cos \left (d x + c\right )^{3} + 57 \, \cos \left (d x + c\right )^{2} + 39 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{10 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/10*(15*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*log(sin(d*x + c) + 1) - 15*(co
s(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(24*cos(d*x + c)
^3 + 57*cos(d*x + c)^2 + 39*cos(d*x + c) + 5)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3
*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{5}{\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]
time = 0.52, size = 122, normalized size = 0.95 \begin {gather*} -\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/20*(60*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 40*tan(1/2*d*x
+ 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (a^12*tan(1/2*d*x + 1/2*c)^5 + 10*a^12*tan(1/2*d*x + 1/2*c)^3 +
85*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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Mupad [B]
time = 0.68, size = 111, normalized size = 0.87 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a/cos(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(2*a^3*d) + tan(c/2 + (d*x)/2)^5/(20*a^3*d) - (6*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - (2*
tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/2)^2 - a^3)) + (17*tan(c/2 + (d*x)/2))/(4*a^3*d)

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