3.357 \(\int \frac{(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac{4 a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{9/2} f}-\frac{2 \left (a^3 \tan (e+f x)+a^3\right )}{7 d f (d \tan (e+f x))^{7/2}} \]

[Out]

(-2*Sqrt[2]*a^3*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(9/2)*f) - (32*a^
3)/(35*d^2*f*(d*Tan[e + f*x])^(5/2)) - (4*a^3)/(3*d^3*f*(d*Tan[e + f*x])^(3/2)) + (4*a^3)/(d^4*f*Sqrt[d*Tan[e
+ f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2))

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Rubi [A]  time = 0.270471, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3565, 3628, 3529, 3532, 208} \[ \frac{4 a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{9/2} f}-\frac{2 \left (a^3 \tan (e+f x)+a^3\right )}{7 d f (d \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

(-2*Sqrt[2]*a^3*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(9/2)*f) - (32*a^
3)/(35*d^2*f*(d*Tan[e + f*x])^(5/2)) - (4*a^3)/(3*d^3*f*(d*Tan[e + f*x])^(3/2)) + (4*a^3)/(d^4*f*Sqrt[d*Tan[e
+ f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2))

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx &=-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac{2 \int \frac{8 a^3 d^2+7 a^3 d^2 \tan (e+f x)+a^3 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{7/2}} \, dx}{7 d^3}\\ &=-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac{2 \int \frac{7 a^3 d^3-7 a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{7 d^5}\\ &=-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac{2 \int \frac{-7 a^3 d^4-7 a^3 d^4 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{7 d^7}\\ &=-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{4 a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac{2 \int \frac{-7 a^3 d^5+7 a^3 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{7 d^9}\\ &=-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{4 a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac{\left (28 a^6 d\right ) \operatorname{Subst}\left (\int \frac{1}{-98 a^6 d^{10}+d x^2} \, dx,x,\frac{-7 a^3 d^5-7 a^3 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=-\frac{2 \sqrt{2} a^3 \tanh ^{-1}\left (\frac{\sqrt{d}+\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{9/2} f}-\frac{32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac{4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac{4 a^3}{d^4 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.848465, size = 175, normalized size = 1.06 \[ -\frac{a^3 \cos (e+f x) (\cot (e+f x)+1)^3 \sqrt{d \tan (e+f x)} \left (70 \sin ^2(e+f x) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(e+f x)\right )+35 \sin (2 (e+f x)) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\tan ^2(e+f x)\right )+42 \cos ^2(e+f x) \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\tan ^2(e+f x)\right )+10 \cos ^2(e+f x) \cot (e+f x) \, _2F_1\left (-\frac{7}{4},1;-\frac{3}{4};-\tan ^2(e+f x)\right )\right )}{35 d^5 f (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

-(a^3*Cos[e + f*x]*(1 + Cot[e + f*x])^3*(10*Cos[e + f*x]^2*Cot[e + f*x]*Hypergeometric2F1[-7/4, 1, -3/4, -Tan[
e + f*x]^2] + 42*Cos[e + f*x]^2*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[e + f*x]^2] + 70*Hypergeometric2F1[-1/4,
1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 35*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[2*(e + f*x)
])*Sqrt[d*Tan[e + f*x]])/(35*d^5*f*(Cos[e + f*x] + Sin[e + f*x])^3)

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Maple [B]  time = 0.029, size = 430, normalized size = 2.6 \begin{align*} -{\frac{{a}^{3}\sqrt{2}}{2\,f{d}^{5}}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }-{\frac{{a}^{3}\sqrt{2}}{f{d}^{5}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{a}^{3}\sqrt{2}}{f{d}^{5}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{a}^{3}\sqrt{2}}{2\,f{d}^{4}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{a}^{3}\sqrt{2}}{f{d}^{4}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{a}^{3}\sqrt{2}}{f{d}^{4}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2\,{a}^{3}}{7\,fd} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}}-{\frac{6\,{a}^{3}}{5\,{d}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}+4\,{\frac{{a}^{3}}{f{d}^{4}\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{4\,{a}^{3}}{3\,{d}^{3}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x)

[Out]

-1/2/f*a^3/d^5*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*t
an(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/f*a^3/d^5*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2
)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/f*a^3/d^5*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e
))^(1/2)+1)+1/2/f*a^3/d^4/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^
(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/f*a^3/d^4/(d^2)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/f*a^3/d^4/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(
d*tan(f*x+e))^(1/2)+1)-2/7/f*a^3/d/(d*tan(f*x+e))^(7/2)-6/5*a^3/d^2/f/(d*tan(f*x+e))^(5/2)+4*a^3/d^4/f/(d*tan(
f*x+e))^(1/2)-4/3*a^3/d^3/f/(d*tan(f*x+e))^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80372, size = 765, normalized size = 4.64 \begin{align*} \left [\frac{105 \, \sqrt{2} a^{3} \sqrt{d} \log \left (\frac{\tan \left (f x + e\right )^{2} - \frac{2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt{d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + 2 \,{\left (210 \, a^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} \tan \left (f x + e\right ) - 15 \, a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{105 \, d^{5} f \tan \left (f x + e\right )^{4}}, \frac{2 \,{\left (105 \, \sqrt{2} a^{3} d \sqrt{-\frac{1}{d}} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-\frac{1}{d}}{\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{4} +{\left (210 \, a^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} \tan \left (f x + e\right ) - 15 \, a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}\right )}}{105 \, d^{5} f \tan \left (f x + e\right )^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*sqrt(2)*a^3*sqrt(d)*log((tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) + 1)/sqrt(d
) + 4*tan(f*x + e) + 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^4 + 2*(210*a^3*tan(f*x + e)^3 - 70*a^3*tan(f*x + e)
^2 - 63*a^3*tan(f*x + e) - 15*a^3)*sqrt(d*tan(f*x + e)))/(d^5*f*tan(f*x + e)^4), 2/105*(105*sqrt(2)*a^3*d*sqrt
(-1/d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-1/d)*(tan(f*x + e) + 1)/tan(f*x + e))*tan(f*x + e)^4 + (2
10*a^3*tan(f*x + e)^3 - 70*a^3*tan(f*x + e)^2 - 63*a^3*tan(f*x + e) - 15*a^3)*sqrt(d*tan(f*x + e)))/(d^5*f*tan
(f*x + e)^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))**3/(d*tan(f*x+e))**(9/2),x)

[Out]

Timed out

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Giac [B]  time = 1.49852, size = 459, normalized size = 2.78 \begin{align*} -\frac{\sqrt{2}{\left (a^{3} d \sqrt{{\left | d \right |}} + a^{3}{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{2 \, d^{6} f} + \frac{\sqrt{2}{\left (a^{3} d \sqrt{{\left | d \right |}} + a^{3}{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{2 \, d^{6} f} - \frac{{\left (\sqrt{2} a^{3} d \sqrt{{\left | d \right |}} - \sqrt{2} a^{3}{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} + 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{d^{6} f} - \frac{{\left (\sqrt{2} a^{3} d \sqrt{{\left | d \right |}} - \sqrt{2} a^{3}{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} - 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{d^{6} f} + \frac{2 \,{\left (210 \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} d^{3} \tan \left (f x + e\right ) - 15 \, a^{3} d^{3}\right )}}{105 \, \sqrt{d \tan \left (f x + e\right )} d^{7} f \tan \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(ab
s(d)) + abs(d))/(d^6*f) + 1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqr
t(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^6*f) - (sqrt(2)*a^3*d*sqrt(abs(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arc
tan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d^6*f) - (sqrt(2)*a^3*d*sqrt(ab
s(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs
(d)))/(d^6*f) + 2/105*(210*a^3*d^3*tan(f*x + e)^3 - 70*a^3*d^3*tan(f*x + e)^2 - 63*a^3*d^3*tan(f*x + e) - 15*a
^3*d^3)/(sqrt(d*tan(f*x + e))*d^7*f*tan(f*x + e)^3)