3.2.59 \(\int \frac {(a+i a \tan (e+f x))^3}{\sqrt {d \tan (e+f x)}} \, dx\) [159]

Optimal. Leaf size=107 \[ -\frac {8 \sqrt [4]{-1} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f} \]

[Out]

-8*(-1)^(1/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f/d^(1/2)-16/3*a^3*(d*tan(f*x+e))^(1/2)/d/f-
2/3*(d*tan(f*x+e))^(1/2)*(a^3+I*a^3*tan(f*x+e))/d/f

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Rubi [A]
time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3637, 3673, 3614, 211} \begin {gather*} -\frac {8 \sqrt [4]{-1} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {d \tan (e+f x)}}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f) - (16*a^3*Sqrt[d*Tan[e + f*x
]])/(3*d*f) - (2*Sqrt[d*Tan[e + f*x]]*(a^3 + I*a^3*Tan[e + f*x]))/(3*d*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 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]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3673

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {d \tan (e+f x)}} \, dx &=-\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac {(2 a) \int \frac {(a+i a \tan (e+f x)) (2 a d+4 i a d \tan (e+f x))}{\sqrt {d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac {(2 a) \int \frac {6 a^2 d+6 i a^2 d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac {\left (48 a^5 d\right ) \text {Subst}\left (\int \frac {1}{6 a^2 d^2-6 i a^2 d x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}\\ \end {align*}

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Mathematica [A]
time = 2.59, size = 154, normalized size = 1.44 \begin {gather*} -\frac {2 a^3 e^{-3 i (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) \left (-12 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+(9+i \tan (e+f x)) \sqrt {i \tan (e+f x)}\right ) \sqrt {d \tan (e+f x)}}{3 d \sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(-12*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e
+ f*x)))]] + (9 + I*Tan[e + f*x])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(3*d*E^((3*I)*(e + f*x))*Sqrt[(-
1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]*f)

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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. 310 vs. \(2 (88 ) = 176\).
time = 0.12, size = 311, normalized size = 2.91

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-3 d \sqrt {d \tan \left (f x +e \right )}+4 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(311\)
default	\(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-3 d \sqrt {d \tan \left (f x +e \right )}+4 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/f*a^3/d^2*(-1/3*I*(d*tan(f*x+e))^(3/2)-3*d*(d*tan(f*x+e))^(1/2)+4*d^2*(1/8/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*I/(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. 208 vs. \(2 (91) = 182\).
time = 0.56, size = 208, normalized size = 1.94 \begin {gather*} \frac {3 \, a^{3} d {\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 )} + \frac {2 \, {\left (-i \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} - 9 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d\right )}}{d}}{3 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/3*(3*a^3*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)
+ (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)) + 2*(-I*(d*tan(f*x + e))^(3/2)*a^3 - 9*sqrt(d*tan(
f*x + e))*a^3*d)/d)/(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. 340 vs. \(2 (91) = 182\).
time = 0.37, size = 340, normalized size = 3.18 \begin {gather*} \frac {3 \, \sqrt {-\frac {64 i \, a^{6}}{d f^{2}}} {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {-\frac {64 i \, a^{6}}{d f^{2}}} {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 )}}{4 \, a^{3}}\right ) - 3 \, \sqrt {-\frac {64 i \, a^{6}}{d f^{2}}} {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt {-\frac {64 i \, a^{6}}{d f^{2}}} {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 )}}{4 \, a^{3}}\right ) - 16 \, {\left (5 \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, a^{3}\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}}}{12 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + sqrt
(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*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^3) - 3*sqrt(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*log(1/4*(-8*I*a^
3*d*e^(2*I*f*x + 2*I*e) - sqrt(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*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^3) - 16*(5*a^3*e^(2*I*f*x + 2*I*e) + 4*a^3)*sqrt(
(-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d*f*e^(2*I*f*x + 2*I*e) + d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(1/2),x)

[Out]

-I*a**3*(Integral(I/sqrt(d*tan(e + f*x)), x) + Integral(-3*tan(e + f*x)/sqrt(d*tan(e + f*x)), x) + Integral(ta
n(e + f*x)**3/sqrt(d*tan(e + f*x)), x) + Integral(-3*I*tan(e + f*x)**2/sqrt(d*tan(e + f*x)), x))

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Giac [A]
time = 0.62, size = 130, normalized size = 1.21 \begin {gather*} \frac {8 i \, \sqrt {2} a^{3} \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 )}{\sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{5} f^{2} \tan \left (f x + e\right ) + 9 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{5} f^{2}\right )}}{3 \, d^{6} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

8*I*sqrt(2)*a^3*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)))/(
sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - 2/3*(I*sqrt(d*tan(f*x + e))*a^3*d^5*f^2*tan(f*x + e) + 9*sqrt(d*tan(f*x + e))
*a^3*d^5*f^2)/(d^6*f^3)

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Mupad [B]
time = 4.38, size = 81, normalized size = 0.76 \begin {gather*} -\frac {6\,a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{d\,f}-\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{\sqrt {-d}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^3/(d*tan(e + f*x))^(1/2),x)

[Out]

(16i^(1/2)*a^3*atan((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*(-d)^(1/2)))*2i)/((-d)^(1/2)*f) - (a^3*(d*tan(e + f*
x))^(3/2)*2i)/(3*d^2*f) - (6*a^3*(d*tan(e + f*x))^(1/2))/(d*f)

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