3.2.35 \(\int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx\) [135]

Optimal. Leaf size=107 \[ -\frac {2 (-1)^{3/4} a d^{5/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3609, 3614, 211} \begin {gather*} -\frac {2 (-1)^{3/4} a d^{5/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(-2*(-1)^(3/4)*a*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((2*I)*a*d^2*Sqrt[d*Tan[e + f*
x]])/f + (2*a*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((2*I)/5)*a*(d*Tan[e + f*x])^(5/2))/f

Rule 211

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

Rule 3609

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

Rule 3614

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

Rubi steps

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

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Mathematica [A]
time = 2.40, size = 125, normalized size = 1.17 \begin {gather*} \frac {a d^2 \left (30 i \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+\sec ^2(e+f x) (-12 i-18 i \cos (2 (e+f x))+5 \sin (2 (e+f x))) \sqrt {i \tan (e+f x)}\right ) \sqrt {d \tan (e+f x)}}{15 f \sqrt {i \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(a*d^2*((30*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + Sec[e + f*x]^2*(-12*I - (
18*I)*Cos[2*(e + f*x)] + 5*Sin[2*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(15*f*Sqrt[I*Tan[e +
f*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (85 ) = 170\).
time = 0.24, size = 321, normalized size = 3.00

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\)	\(321\)
default	\(\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*a*(2/5*I*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-2*I*d^2*(d*tan(f*x+e))^(1/2)+2*d^3*(1/8*I/d*(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)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2
^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(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)))+2*arctan(2^
(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (87) = 174\).
time = 0.51, size = 218, normalized size = 2.04 \begin {gather*} \frac {15 \, a d^{4} {\left (\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 24 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} - 120 i \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{60 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

1/60*(15*a*d^4*((2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(
d) + (2*I - 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + (I +
1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - (I + 1)*sqrt(2)*log(d*tan(
f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) + 24*I*(d*tan(f*x + e))^(5/2)*a*d + 40*(d*tan(f*
x + e))^(3/2)*a*d^2 - 120*I*sqrt(d*tan(f*x + e))*a*d^3)/(d*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (87) = 174\).
time = 0.39, size = 391, normalized size = 3.65 \begin {gather*} -\frac {15 \, \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2}}\right ) - 15 \, \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2}}\right ) + 8 \, {\left (23 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 24 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 13 i \, a d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt(4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3*e^(2*I*f
*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(I*f*e^(2*I*f*x + 2*I*e) + I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(
2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*d^2)) - 15*sqrt(4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*
e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3*e^(2*I*f*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(-I*f*e^(2*I*f*x + 2*I*e)
- I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*d^2)) + 8*(2
3*I*a*d^2*e^(4*I*f*x + 4*I*e) + 24*I*a*d^2*e^(2*I*f*x + 2*I*e) + 13*I*a*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) +
I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2)*(a+I*a*tan(f*x+e)),x)

[Out]

I*a*(Integral(-I*(d*tan(e + f*x))**(5/2), x) + Integral((d*tan(e + f*x))**(5/2)*tan(e + f*x), x))

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Giac [A]
time = 0.65, size = 154, normalized size = 1.44 \begin {gather*} -\frac {2}{15} \, a d^{2} {\left (\frac {15 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {-3 i \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4} \tan \left (f x + e\right )^{2} - 5 \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4} \tan \left (f x + e\right ) + 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4}}{d^{10} f^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-2/15*a*d^2*(15*sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(
d^2)*sqrt(d)))/(f*(I*d/sqrt(d^2) + 1)) + (-3*I*sqrt(d*tan(f*x + e))*d^10*f^4*tan(f*x + e)^2 - 5*sqrt(d*tan(f*x
+ e))*d^10*f^4*tan(f*x + e) + 15*I*sqrt(d*tan(f*x + e))*d^10*f^4)/(d^10*f^5))

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Mupad [B]
time = 4.93, size = 114, normalized size = 1.07 \begin {gather*} \frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,f}-\frac {a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x)*1i),x)

[Out]

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

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