3.10.84 \(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx\) [984]

Optimal. Leaf size=199 \[ \frac {35 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 i}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

35/128*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(3/2)/f*2^(1/2)-35/64*I/a^2/c/f/(c-I*c*ta
n(f*x+e))^(1/2)-35/96*I/a^2/f/(c-I*c*tan(f*x+e))^(3/2)+1/4*I/a^2/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2)
+7/16*I/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.16, antiderivative size = 199, 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 \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}-\frac {35 i}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(((35*I)/64)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*c^(3/2)*f) - ((35*I)/96)/(a^2
*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/4)/(a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((7*I)/
16)/(a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - ((35*I)/64)/(a^2*c*f*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))^2 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {\int \cos ^4(e+f x) \sqrt {c-i c \tan (e+f x)} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {\left (7 i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{8 a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f}\\ &=-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 f}\\ &=-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 i}{64 a^2 c f \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 )}{128 a^2 c f}\\ &=-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 i}{64 a^2 c f \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 )}{64 a^2 c f}\\ &=\frac {35 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 i}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.98, size = 145, normalized size = 0.73 \begin {gather*} -\frac {i e^{-4 i (e+f x)} \left (-6-45 e^{2 i (e+f x)}+41 e^{4 i (e+f x)}+88 e^{6 i (e+f x)}+8 e^{8 i (e+f x)}-105 e^{4 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )\right ) \sqrt {c-i c \tan (e+f x)}}{384 a^2 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

((-1/384*I)*(-6 - 45*E^((2*I)*(e + f*x)) + 41*E^((4*I)*(e + f*x)) + 88*E^((6*I)*(e + f*x)) + 8*E^((8*I)*(e + f
*x)) - 105*E^((4*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c -
I*c*Tan[e + f*x]])/(a^2*c^2*E^((4*I)*(e + f*x))*f)

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Maple [A]
time = 0.33, size = 139, normalized size = 0.70

method	result	size
derivativedivides	\(-\frac {2 i c^{3} \left (\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\frac {-\frac {11 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {13 c \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{2}}\)	\(139\)
default	\(-\frac {2 i c^{3} \left (\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\frac {-\frac {11 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {13 c \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{2}}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/f/a^2*c^3*(3/16/c^4/(c-I*c*tan(f*x+e))^(1/2)+1/24/c^3/(c-I*c*tan(f*x+e))^(3/2)-1/16/c^4*(4*(-11/32*(c-I*c
*tan(f*x+e))^(3/2)+13/16*c*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^2+35/16*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 = 192, normalized size = 0.96 \begin {gather*} -\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} - 350 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c + 224 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} + 64 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c^{2}} + \frac {105 \, \sqrt {2} \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^{2} \sqrt {c}}\right )}}{768 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*I*(4*(105*(-I*c*tan(f*x + e) + c)^3 - 350*(-I*c*tan(f*x + e) + c)^2*c + 224*(-I*c*tan(f*x + e) + c)*c^2
+ 64*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*a^2 - 4*(-I*c*tan(f*x + e) + c)^(5/2)*a^2*c + 4*(-I*c*tan(f*x + e) +
c)^(3/2)*a^2*c^2) + 105*sqrt(2)*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^2*sqrt(c)))/(c*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. 334 vs. \(2 (155) = 310\).
time = 1.43, size = 334, normalized size = 1.68 \begin {gather*} \frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 88 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 41 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 45 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(-105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(-35/32*(sqrt(2)*sqrt(1/2)*(I*a
^2*c*f*e^(2*I*f*x + 2*I*e) + I*a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^3*f^2)) - I)*e^(-I*f*x
- I*e)/(a^2*c*f)) + 105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(-35/32*(sqrt(2)*s
qrt(1/2)*(-I*a^2*c*f*e^(2*I*f*x + 2*I*e) - I*a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^3*f^2))
- I)*e^(-I*f*x - I*e)/(a^2*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-8*I*e^(8*I*f*x + 8*I*e) - 88*I*
e^(6*I*f*x + 6*I*e) - 41*I*e^(4*I*f*x + 4*I*e) + 45*I*e^(2*I*f*x + 2*I*e) + 6*I))*e^(-4*I*f*x - 4*I*e)/(a^2*c^
2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [B]
time = 5.19, size = 182, normalized size = 0.91 \begin {gather*} -\frac {-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,175{}\mathrm {i}}{96\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{3\,a^2\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a^2\,f}}{-4\,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}+4\,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}}{128\,a^2\,{\left (-c\right )}^{3/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*35i)/(128*a^2*(-c)^(3/2)*f) - ((c^2*1i)/
(3*a^2*f) - ((c - c*tan(e + f*x)*1i)^2*175i)/(96*a^2*f) + ((c - c*tan(e + f*x)*1i)^3*35i)/(64*a^2*c*f) + (c*(c
- c*tan(e + f*x)*1i)*7i)/(6*a^2*f))/((c - c*tan(e + f*x)*1i)^(7/2) - 4*c*(c - c*tan(e + f*x)*1i)^(5/2) + 4*c^
2*(c - c*tan(e + f*x)*1i)^(3/2))

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