3.356 \(\int \frac{(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=141 \[ \frac{2 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{4 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

(2*Sqrt[2]*a^3*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(7/2)*f) - (8*a^3)/
(5*d^2*f*(d*Tan[e + f*x])^(3/2)) - (4*a^3)/(d^3*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(5*d*f*
(d*Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.221937, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3565, 3628, 3529, 3532, 205} \[ \frac{2 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{4 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

(2*Sqrt[2]*a^3*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(7/2)*f) - (8*a^3)/
(5*d^2*f*(d*Tan[e + f*x])^(3/2)) - (4*a^3)/(d^3*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(5*d*f*
(d*Tan[e + f*x])^(5/2))

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 205

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

Rubi steps

\begin{align*} \int \frac{(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 \int \frac{6 a^3 d^2+5 a^3 d^2 \tan (e+f x)+a^3 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{5 d^3}\\ &=-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 \int \frac{5 a^3 d^3-5 a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{5 d^5}\\ &=-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{4 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 \int \frac{-5 a^3 d^4-5 a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{5 d^7}\\ &=-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{4 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac{\left (20 a^6 d\right ) \operatorname{Subst}\left (\int \frac{1}{50 a^6 d^8+d x^2} \, dx,x,\frac{-5 a^3 d^4+5 a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{2 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{4 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.24293, size = 271, normalized size = 1.92 \[ -\frac{a^3 (\cot (e+f x)+1)^3 \left (8 \sin (e+f x) \cos ^2(e+f x) \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\tan ^2(e+f x)\right )+5 \left (24 \sin ^3(e+f x) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(e+f x)\right )+8 \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\tan ^2(e+f x)\right )+\sqrt{2} \cos ^3(e+f x) \tan ^{\frac{7}{2}}(e+f x) \left (2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+\log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )\right )\right )}{20 d^3 f \sqrt{d \tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

-(a^3*(1 + Cot[e + f*x])^3*(8*Cos[e + f*x]^2*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[e + f*x]^2]*Sin[e + f*x] +
5*(8*Cos[e + f*x]*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 24*Hypergeometric2F1[-1/4,
1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^3 + Sqrt[2]*Cos[e + f*x]^3*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] -
2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - Log[1 + Sqrt[2
]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x]^(7/2))))/(20*d^3*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*T
an[e + f*x]])

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Maple [B]  time = 0.023, size = 409, normalized size = 2.9 \begin{align*} -{\frac{{a}^{3}\sqrt{2}}{2\,f{d}^{4}}\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}^{4}}\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}^{4}}\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}^{3}}\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}^{3}}\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}^{3}}\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}}{5\,fd} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{{a}^{3}}{{d}^{2}f \left ( d\tan \left ( fx+e \right ) \right ) ^{3/2}}}-4\,{\frac{{a}^{3}}{f{d}^{3}\sqrt{d\tan \left ( fx+e \right ) }}} \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))^(7/2),x)

[Out]

-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*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^4*(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^4*(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^3/(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^3/(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^3/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(
d*tan(f*x+e))^(1/2)+1)-2/5/f*a^3/d/(d*tan(f*x+e))^(5/2)-2*a^3/d^2/f/(d*tan(f*x+e))^(3/2)-4*a^3/d^3/f/(d*tan(f*
x+e))^(1/2)

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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))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76953, size = 682, normalized size = 4.84 \begin{align*} \left [\frac{5 \, \sqrt{2} a^{3} d \sqrt{-\frac{1}{d}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-\frac{1}{d}}{\left (\tan \left (f x + e\right ) - 1\right )} - \tan \left (f x + e\right )^{2} + 4 \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 2 \,{\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}, -\frac{2 \,{\left (5 \, \sqrt{2} a^{3} \sqrt{d} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} +{\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}\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))^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{7}{2}}}\, dx + \int \frac{3 \tan{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{7}{2}}}\, dx + \int \frac{3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{7}{2}}}\, dx + \int \frac{\tan ^{3}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{7}{2}}}\, dx\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))**(7/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-7/2), x) + Integral(3*tan(e + f*x)/(d*tan(e + f*x))**(7/2), x) + Integral(3
*tan(e + f*x)**2/(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(7/2), x))

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Giac [B]  time = 1.39365, size = 435, normalized size = 3.09 \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^{5} 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^{5} 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^{5} 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^{5} f} - \frac{2 \,{\left (10 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} + 5 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}\right )}}{5 \, \sqrt{d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \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))^(7/2),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^5*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^5*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^5*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^5*f) - 2/5*(10*a^3*d^2*tan(f*x + e)^2 + 5*a^3*d^2*tan(f*x + e) + a^3*d^2)/(sqrt(d*tan(f*x + e))*d^5*f
*tan(f*x + e)^2)