3.8.76 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx\) [776]

Optimal. Leaf size=195 \[ \frac {3 (5 i A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[Out]

3/64*(5*I*A+3*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/f*2^(1/2)/c^(1/2)-3/32*(5*I*A+3*B)/
a^2/f/(c-I*c*tan(f*x+e))^(1/2)+1/4*(I*A-B)/a^2/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))^2+1/16*(5*I*A+3*B)/
a^2/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))

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Rubi [A]
time = 0.17, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 44, 53, 65, 214} \begin {gather*} \frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}-\frac {3 (3 B+5 i A)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {3 (3 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

(3*((5*I)*A + 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(32*Sqrt[2]*a^2*Sqrt[c]*f) - (3*((5*
I)*A + 3*B))/(32*a^2*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Ta
n[e + f*x]]) + ((5*I)*A + 3*B)/(16*a^2*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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 3669

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

Rubi steps

\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {((5 A-3 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 A-3 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 A-3 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 (5 i A+3 B)) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a c f}\\ &=\frac {3 (5 i A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {3 (5 i A+3 B)}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.14, size = 160, normalized size = 0.82 \begin {gather*} \frac {(i \cos (e+f x)+\sin (e+f x)) \left (3 (5 A-3 i B) e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (-9 A-i B+2 (3 A-5 i B) \cos (2 (e+f x))+2 (5 i A+3 B) \sin (2 (e+f x)))\right ) \sqrt {c-i c \tan (e+f x)}}{64 a^2 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

((I*Cos[e + f*x] + Sin[e + f*x])*(3*(5*A - (3*I)*B)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt
[1 + E^((2*I)*(e + f*x))]] - 2*Cos[e + f*x]*(-9*A - I*B + 2*(3*A - (5*I)*B)*Cos[2*(e + f*x)] + 2*((5*I)*A + 3*
B)*Sin[2*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(64*a^2*c*f)

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Maple [A]
time = 0.35, size = 151, normalized size = 0.77

method	result	size
derivativedivides	\(-\frac {2 i c^{2} \left (-\frac {i B -A}{8 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\frac {4 \left (\frac {i B}{32}-\frac {7 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{16} A c +\frac {1}{16} i B c \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \left (-\frac {3 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{8 c^{2}}\right )}{f \,a^{2}}\)	\(151\)
default	\(-\frac {2 i c^{2} \left (-\frac {i B -A}{8 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\frac {4 \left (\frac {i B}{32}-\frac {7 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{16} A c +\frac {1}{16} i B c \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \left (-\frac {3 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{8 c^{2}}\right )}{f \,a^{2}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/f/a^2*c^2*(-1/8/c^2*(-A+I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/8/c^2*(4*((1/32*I*B-7/32*A)*(c-I*c*tan(f*x+e))^(3
/2)+(9/16*A*c+1/16*I*B*c)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^2+3/4*(-3/4*I*B+5/4*A)*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 = 0.54, size = 202, normalized size = 1.04 \begin {gather*} -\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (5 \, A - 3 i \, B\right )} \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^{2}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (5 \, A - 3 i \, B\right )} c - 10 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (5 \, A - 3 i \, B\right )} c^{2} + 32 \, {\left (A - i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c + 4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} c^{2}}\right )}}{128 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*I*(3*sqrt(2)*(5*A - 3*I*B)*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^2 + 4*(3*(-I*c*tan(f*x + e) + c)^2*(5*A - 3*I*B)*c - 10*(-I*c*tan(f*x + e)
+ c)*(5*A - 3*I*B)*c^2 + 32*(A - I*B)*c^3)/((-I*c*tan(f*x + e) + c)^(5/2)*a^2 - 4*(-I*c*tan(f*x + e) + c)^(3/
2)*a^2*c + 4*sqrt(-I*c*tan(f*x + e) + c)*a^2*c^2))/(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. 403 vs. \(2 (156) = 312\).
time = 3.92, size = 403, normalized size = 2.07 \begin {gather*} \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} + 5 i \, A + 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} - 30 i \, A B - 9 \, B^{2}}{a^{4} c f^{2}}} - 5 i \, A - 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - \sqrt {2} {\left (8 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (i \, A - 9 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (11 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, A + 2 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{64 \, a^{2} c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(3*sqrt(1/2)*a^2*c*f*sqrt(-(25*A^2 - 30*I*A*B - 9*B^2)/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(3/16*(sqrt(2)
*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 - 30*I*A*B - 9*
B^2)/(a^4*c*f^2)) + 5*I*A + 3*B)*e^(-I*f*x - I*e)/(a^2*f)) - 3*sqrt(1/2)*a^2*c*f*sqrt(-(25*A^2 - 30*I*A*B - 9*
B^2)/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/16*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/
(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 - 30*I*A*B - 9*B^2)/(a^4*c*f^2)) - 5*I*A - 3*B)*e^(-I*f*x - I*e)/(a^2
*f)) - sqrt(2)*(8*(I*A + B)*e^(6*I*f*x + 6*I*e) - (I*A - 9*B)*e^(4*I*f*x + 4*I*e) - (11*I*A - 3*B)*e^(2*I*f*x
+ 2*I*e) - 2*I*A + 2*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*c*f)

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

[Out]

-(Integral(A/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) - sqr
t(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*sqr
t(-I*c*tan(e + f*x) + c)*tan(e + f*x) - 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((A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^2*sqrt(-I*c*tan(f*x + e) + c)), x)

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Mupad [B]
time = 9.44, size = 305, normalized size = 1.56 \begin {gather*} -\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,15{}\mathrm {i}}{32\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{a^2\,f}-\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{16\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+4\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {B\,c^2+\frac {9\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{32}-\frac {15\,B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{16}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+4\,a^2\,c^2\,f\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{64\,a^2\,\sqrt {-c}\,f}+\frac {9\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{64\,a^2\,\sqrt {c}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(64*a^2*c^(1/2)*f) - (B*c^2 + (9*B*(c
- c*tan(e + f*x)*1i)^2)/32 - (15*B*c*(c - c*tan(e + f*x)*1i))/16)/(a^2*f*(c - c*tan(e + f*x)*1i)^(5/2) - 4*a^
2*c*f*(c - c*tan(e + f*x)*1i)^(3/2) + 4*a^2*c^2*f*(c - c*tan(e + f*x)*1i)^(1/2)) - (2^(1/2)*A*atan((2^(1/2)*(c
- c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*15i)/(64*a^2*(-c)^(1/2)*f) - ((A*(c - c*tan(e + f*x)*1i)^2*15i)/(
32*a^2*f) + (A*c^2*1i)/(a^2*f) - (A*c*(c - c*tan(e + f*x)*1i)*25i)/(16*a^2*f))/((c - c*tan(e + f*x)*1i)^(5/2)
- 4*c*(c - c*tan(e + f*x)*1i)^(3/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(1/2))

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