\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx\) [784]

Optimal result

Integrand size = 43, antiderivative size = 245 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {5 (7 i A+5 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}-\frac {5 (7 i A+5 B)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[Out]

5/256*(7*I*A+5*B)*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)-5/128*(7*I*A+5*B
)/a^3/f/(c-I*c*tan(f*x+e))^(1/2)+1/6*(I*A-B)/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))^3+1/48*(7*I*A+5*B
)/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))^2+5/192*(7*I*A+5*B)/a^3/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(
f*x+e))

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 44, 53, 65, 214} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {5 (5 B+7 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}-\frac {5 (5 B+7 i A)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {5 (5 B+7 i A)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {5 B+7 i A}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(5*((7*I)*A + 5*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(128*Sqrt[2]*a^3*Sqrt[c]*f) - (5*((7
*I)*A + 5*B))/(128*a^3*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*Sqrt[c - I*c*
Tan[e + f*x]]) + ((7*I)*A + 5*B)/(48*a^3*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) + (5*((7*I)*A +
5*B))/(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 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*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {((7 A-5 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{12 f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 A-5 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 )}{96 a f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 A-5 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 )}{128 a^2 f} \\ & = -\frac {5 (7 i A+5 B)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 A-5 i B)) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f} \\ & = -\frac {5 (7 i A+5 B)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(5 (7 i A+5 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 )}{128 a^2 c f} \\ & = \frac {5 (7 i A+5 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}-\frac {5 (7 i A+5 B)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.07 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.86 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sec ^3(e+f x) \left (2 \sqrt {c} ((-125 A+7 i B) \cos (e+f x)+8 (5 A-7 i B) \cos (3 (e+f x))-i (7 A-5 i B) (7 \sin (e+f x)-8 \sin (3 (e+f x))))-15 \sqrt {2} (7 A-5 i B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{768 a^3 \sqrt {c} f (-i+\tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(Sec[e + f*x]^3*(2*Sqrt[c]*((-125*A + (7*I)*B)*Cos[e + f*x] + 8*(5*A - (7*I)*B)*Cos[3*(e + f*x)] - I*(7*A - (5
*I)*B)*(7*Sin[e + f*x] - 8*Sin[3*(e + f*x)])) - 15*Sqrt[2]*(7*A - (5*I)*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/
(Sqrt[2]*Sqrt[c])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]]))/(768*a^3*Sqrt[c]*f*(-I
+ Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (-\frac {-i B +A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {8 \left (-\frac {9 i B}{128}+\frac {19 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {7}{24} i B c -\frac {17}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {7}{32} i B \,c^{2}+\frac {29}{32} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {5 \left (-\frac {5 i B}{8}+\frac {7 A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{3}}\right )}{f \,a^{3}}\)	\(178\)
default	\(\frac {2 i c^{3} \left (-\frac {-i B +A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {8 \left (-\frac {9 i B}{128}+\frac {19 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {7}{24} i B c -\frac {17}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {7}{32} i B \,c^{2}+\frac {29}{32} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {5 \left (-\frac {5 i B}{8}+\frac {7 A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{3}}\right )}{f \,a^{3}}\)	\(178\)

[In]

int((A+B*tan(f*x+e))/(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^3*(-1/16/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(1/2)+1/16/c^3*(8*((-9/128*I*B+19/128*A)*(c-I*c*tan(f*x+e)
)^(5/2)+(7/24*I*B*c-17/24*c*A)*(c-I*c*tan(f*x+e))^(3/2)+(-7/32*I*B*c^2+29/32*c^2*A)*(c-I*c*tan(f*x+e))^(1/2))/
(c+I*c*tan(f*x+e))^3+5/4*(-5/8*I*B+7/8*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)
)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (190) = 380\).

Time = 0.27 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} + 7 i \, A + 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} - 7 i \, A - 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) - \sqrt {2} {\left (48 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, {\left (-13 i \, A + 9 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (125 i \, A + 7 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-23 i \, A + 11 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, A + 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{768 \, a^{3} c f} \]

[In]

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

[Out]

1/768*(15*sqrt(1/2)*a^3*c*f*sqrt(-(49*A^2 - 70*I*A*B - 25*B^2)/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(5/64*(sqrt
(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 - 70*I*A*B -
25*B^2)/(a^6*c*f^2)) + 7*I*A + 5*B)*e^(-I*f*x - I*e)/(a^3*f)) - 15*sqrt(1/2)*a^3*c*f*sqrt(-(49*A^2 - 70*I*A*B
- 25*B^2)/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(-5/64*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*s
qrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 - 70*I*A*B - 25*B^2)/(a^6*c*f^2)) - 7*I*A - 5*B)*e^(-I*f*x - I*
e)/(a^3*f)) - sqrt(2)*(48*(I*A + B)*e^(8*I*f*x + 8*I*e) + 3*(-13*I*A + 9*B)*e^(6*I*f*x + 6*I*e) - (125*I*A + 7
*B)*e^(4*I*f*x + 4*I*e) + 2*(-23*I*A + 11*B)*e^(2*I*f*x + 2*I*e) - 8*I*A + 8*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) +
1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {i \left (\int \frac {A}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a^{3}} \]

[In]

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

[Out]

I*(Integral(A/(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) + Integral(B*tan(e + f*x)/(sq
rt(-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

Maxima [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.98 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (7 \, A - 5 i \, B\right )} c - 80 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (7 \, A - 5 i \, B\right )} c^{2} + 132 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (7 \, A - 5 i \, B\right )} c^{3} - 384 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{2} - 8 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c^{3}} + \frac {15 \, \sqrt {2} {\left (7 \, A - 5 i \, B\right )} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}}\right )}}{1536 \, c f} \]

[In]

integrate((A+B*tan(f*x+e))/(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*(15*(-I*c*tan(f*x + e) + c)^3*(7*A - 5*I*B)*c - 80*(-I*c*tan(f*x + e) + c)^2*(7*A - 5*I*B)*c^2 +
132*(-I*c*tan(f*x + e) + c)*(7*A - 5*I*B)*c^3 - 384*(A - I*B)*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) + 15*sqrt(2)*(7*A - 5*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^3)/(c*f)

Giac [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.61 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {B\,c^3-\frac {25\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{128}+\frac {25\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{24}-\frac {55\,B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{32}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-8\,a^3\,c^3\,f\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{128\,a^3\,f}-\frac {A\,c^3\,1{}\mathrm {i}}{a^3\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,35{}\mathrm {i}}{24\,a^3\,f}+\frac {A\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,77{}\mathrm {i}}{32\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+8\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{256\,a^3\,\sqrt {-c}\,f}+\frac {25\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{256\,a^3\,\sqrt {c}\,f} \]

[In]

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

[Out]

(B*c^3 - (25*B*(c - c*tan(e + f*x)*1i)^3)/128 + (25*B*c*(c - c*tan(e + f*x)*1i)^2)/24 - (55*B*c^2*(c - c*tan(e
+ f*x)*1i))/32)/(a^3*f*(c - c*tan(e + f*x)*1i)^(7/2) - 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^(5/2) - 8*a^3*c^3*f*
(c - c*tan(e + f*x)*1i)^(1/2) + 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)^(3/2)) + ((A*(c - c*tan(e + f*x)*1i)^3*35
i)/(128*a^3*f) - (A*c^3*1i)/(a^3*f) - (A*c*(c - c*tan(e + f*x)*1i)^2*35i)/(24*a^3*f) + (A*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)*A*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) + (25*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/
2))/(2*c^(1/2))))/(256*a^3*c^(1/2)*f)