3.10.79 \(\int \frac {1}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx\) [979]

Optimal. Leaf size=210 \[ \frac {35 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}-\frac {35 i}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[Out]

35/256*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)/c^(1/2)-35/128*I/a^3/f/(c-I*c*tan
(f*x+e))^(1/2)+1/6*I/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))^3+7/48*I/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(
1+I*tan(f*x+e))^2+35/192*I/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))

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Rubi [A]
time = 0.15, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3603, 3568, 44, 53, 65, 212} \begin {gather*} -\frac {35 i}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

(((35*I)/128)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*Sqrt[c]*f) - ((35*I)/128)/(a
^3*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]) + ((7*I)/48
)/(a^3*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) + ((35*I)/192)/(a^3*f*(1 + I*Tan[e + f*x])*Sqrt[c
- I*c*Tan[e + f*x]])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {\left (7 i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {\left (35 i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{96 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(35 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac {35 i}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(35 i) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac {35 i}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(35 i) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{128 a^3 f}\\ &=\frac {35 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}-\frac {35 i}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.07, size = 146, normalized size = 0.70 \begin {gather*} -\frac {i \sec ^2(e+f x) \left (125+105 e^{2 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )+85 \cos (2 (e+f x))-40 \cos (4 (e+f x))-7 i \sin (2 (e+f x))-56 i \sin (4 (e+f x))\right ) \sqrt {c-i c \tan (e+f x)}}{768 a^3 c f (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

((-1/768*I)*Sec[e + f*x]^2*(125 + 105*E^((2*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2
*I)*(e + f*x))]] + 85*Cos[2*(e + f*x)] - 40*Cos[4*(e + f*x)] - (7*I)*Sin[2*(e + f*x)] - (56*I)*Sin[4*(e + f*x)
])*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*c*f*(-I + Tan[e + f*x])^2)

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Maple [A]
time = 0.38, size = 139, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {2 i c^{4} \left (-\frac {1}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {\frac {19 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{16}-\frac {17 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {29 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{3}}\)	\(139\)
default	\(\frac {2 i c^{4} \left (-\frac {1}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {\frac {19 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{16}-\frac {17 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {29 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{3}}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^4*(-1/16/c^4/(c-I*c*tan(f*x+e))^(1/2)+1/16/c^4*(8*(19/128*(c-I*c*tan(f*x+e))^(5/2)-17/24*c*(c-I*c*
tan(f*x+e))^(3/2)+29/32*c^2*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3+35/32*2^(1/2)/c^(1/2)*arctanh(1/2*(
c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))

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Maxima [A]
time = 0.52, size = 217, normalized size = 1.03 \begin {gather*} -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c - 560 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} + 924 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} - 384 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{2} - 8 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c^{3}} + \frac {105 \, \sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}}\right )}}{1536 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1536*I*(4*(105*(-I*c*tan(f*x + e) + c)^3*c - 560*(-I*c*tan(f*x + e) + c)^2*c^2 + 924*(-I*c*tan(f*x + e) + c
)*c^3 - 384*c^4)/((-I*c*tan(f*x + e) + c)^(7/2)*a^3 - 6*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c + 12*(-I*c*tan(f*x
+ e) + c)^(3/2)*a^3*c^2 - 8*sqrt(-I*c*tan(f*x + e) + c)*a^3*c^3) + 105*sqrt(2)*sqrt(c)*log(-(sqrt(2)*sqrt(c)
- sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^3)/(c*f)

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Fricas [A]
time = 1.61, size = 320, normalized size = 1.52 \begin {gather*} \frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {\frac {1}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {\frac {1}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-48 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 39 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 125 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 46 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{768 \, a^{3} c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(-105*I*sqrt(1/2)*a^3*c*f*sqrt(1/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(-35/64*(sqrt(2)*sqrt(1/2)*(I*a^3*f
*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c*f^2)) - I)*e^(-I*f*x - I*e)/(a
^3*f)) + 105*I*sqrt(1/2)*a^3*c*f*sqrt(1/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(-35/64*(sqrt(2)*sqrt(1/2)*(-I*a^3
*f*e^(2*I*f*x + 2*I*e) - I*a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c*f^2)) - I)*e^(-I*f*x - I*e)/
(a^3*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-48*I*e^(8*I*f*x + 8*I*e) + 39*I*e^(6*I*f*x + 6*I*e) + 1
25*I*e^(4*I*f*x + 4*I*e) + 46*I*e^(2*I*f*x + 2*I*e) + 8*I))*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {1}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*Integral(1/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*I*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 -
3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*sqrt(-I*c*tan(e + f*x) + c)), x)/a**3

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^3*sqrt(-I*c*tan(f*x + e) + c)), x)

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Mupad [B]
time = 5.01, size = 202, normalized size = 0.96 \begin {gather*} \frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{128\,a^3\,f}-\frac {c^3\,1{}\mathrm {i}}{a^3\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,35{}\mathrm {i}}{24\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,77{}\mathrm {i}}{32\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+8\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{256\,a^3\,\sqrt {-c}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c - c*tan(e + f*x)*1i)^3*35i)/(128*a^3*f) - (c^3*1i)/(a^3*f) - (c*(c - c*tan(e + f*x)*1i)^2*35i)/(24*a^3*f)
+ (c^2*(c - c*tan(e + f*x)*1i)*77i)/(32*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(5/2) - (c - c*tan(e + f*x)*1i)^
(7/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(1/2) - 12*c^2*(c - c*tan(e + f*x)*1i)^(3/2)) - (2^(1/2)*atan((2^(1/2)*(
c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*35i)/(256*a^3*(-c)^(1/2)*f)

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