\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx\) [770]

Optimal result

Integrand size = 43, antiderivative size = 223 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {(7 i A-3 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i A-3 B}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

1/32*(7*I*A-3*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a/c^(5/2)/f*2^(1/2)+1/16*(-7*I*A+3*B)/a
/c^2/f/(c-I*c*tan(f*x+e))^(1/2)+1/20*(-7*I*A+3*B)/a/f/(c-I*c*tan(f*x+e))^(5/2)+1/2*(I*A-B)/a/f/(1+I*tan(f*x+e)
)/(c-I*c*tan(f*x+e))^(5/2)+1/24*(-7*I*A+3*B)/a/c/f/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 53, 65, 214} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {(-3 B+7 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {-3 B+7 i A}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {-3 B+7 i A}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {-3 B+7 i A}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((7*I)*A - 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a*c^(5/2)*f) - ((7*I)*A -
3*B)/(20*a*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2
)) - ((7*I)*A - 3*B)/(24*a*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((7*I)*A - 3*B)/(16*a*c^2*f*Sqrt[c - I*c*Tan[e
+ f*x]])

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)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {((7 A+3 i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = -\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 A+3 i B) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = -\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}+\frac {(7 A+3 i B) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{16 c f} \\ & = -\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i A-3 B}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 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 )}{32 c^2 f} \\ & = -\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i A-3 B}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 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 )}{16 c^3 f} \\ & = \frac {(7 i A-3 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i A-3 B}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.85 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {5 (7 A+3 i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) \sec ^2(e+f x)+6 i (3 i A-7 B+(7 A+3 i B) \tan (e+f x))}{120 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(5*(7*A + (3*I)*B)*Hypergeometric2F1[-3/2, 1, -1/2, (-1/2*I)*(I + Tan[e + f*x])]*Sec[e + f*x]^2 + (6*I)*((3*I)
*A - 7*B + (7*A + (3*I)*B)*Tan[e + f*x]))/(120*a*c^2*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^2*Sqrt[c - I*c*T
an[e + f*x]])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {7 A}{2}+\frac {3 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{16 c^{3}}-\frac {i B +3 A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {A}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{20 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f a}\)	\(168\)
default	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {7 A}{2}+\frac {3 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{16 c^{3}}-\frac {i B +3 A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {A}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{20 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f a}\)	\(168\)

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/f/a*c*(1/16/c^3*((1/4*I*B+1/4*A)*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+1/2*(7/2*A+3/2*I*B)*2
^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/16/c^3*(3*A+I*B)/(c-I*c*tan(f*x+e))^(1
/2)-1/12/c^2*A/(c-I*c*tan(f*x+e))^(3/2)-1/20/c*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2))

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. 418 vs. \(2 (172) = 344\).

Time = 0.29 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} + 7 i \, A - 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} - 7 i \, A + 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) - \sqrt {2} {\left (6 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, {\left (19 i \, A + 9 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (37 i \, A - 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-101 i \, A + 9 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 15 i \, A + 15 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \]

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/480*(15*sqrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(1/8*(sqrt
(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2*I*e) + a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A
*B - 9*B^2)/(a^2*c^5*f^2)) + 7*I*A - 3*B)*e^(-I*f*x - I*e)/(a*c^2*f)) - 15*sqrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 4
2*I*A*B - 9*B^2)/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/8*(sqrt(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2*I*e) +
a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2)) - 7*I*A + 3*B)*e^
(-I*f*x - I*e)/(a*c^2*f)) - sqrt(2)*(6*(I*A + B)*e^(8*I*f*x + 8*I*e) + 2*(19*I*A + 9*B)*e^(6*I*f*x + 6*I*e) +
4*(37*I*A - 3*B)*e^(4*I*f*x + 4*I*e) - (-101*I*A + 9*B)*e^(2*I*f*x + 2*I*e) - 15*I*A + 15*B)*sqrt(c/(e^(2*I*f*
x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c^3*f)

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=- \frac {i \left (\int \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a} \]

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (7 \, A + 3 i \, B\right )} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (7 \, A + 3 i \, B\right )} c - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (7 \, A + 3 i \, B\right )} c^{2} - 48 \, {\left (A - i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a c - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a c^{2}} + \frac {15 \, \sqrt {2} {\left (7 \, A + 3 i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a c^{\frac {3}{2}}}\right )}}{960 \, c f} \]

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/960*I*(4*(15*(-I*c*tan(f*x + e) + c)^3*(7*A + 3*I*B) - 20*(-I*c*tan(f*x + e) + c)^2*(7*A + 3*I*B)*c - 8*(-I
*c*tan(f*x + e) + c)*(7*A + 3*I*B)*c^2 - 48*(A - I*B)*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*a*c - 2*(-I*c*tan(f*
x + e) + c)^(5/2)*a*c^2) + 15*sqrt(2)*(7*A + 3*I*B)*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*c^(3/2)))/(c*f)

Giac [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)*(-I*c*tan(f*x + e) + c)^(5/2)), x)

Mupad [B] (verification not implemented)

Time = 10.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\frac {B\,c}{5}-\frac {B\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{10}-\frac {B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,c}+\frac {3\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{16\,c^2}}{a\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-2\,a\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\frac {A\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{30\,a\,f}+\frac {A\,c\,1{}\mathrm {i}}{5\,a\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{12\,a\,c\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,7{}\mathrm {i}}{16\,a\,c^2\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,7{}\mathrm {i}}{32\,a\,{\left (-c\right )}^{5/2}\,f}-\frac {3\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{32\,a\,c^{5/2}\,f} \]

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

((B*c)/5 - (B*(c - c*tan(e + f*x)*1i))/10 - (B*(c - c*tan(e + f*x)*1i)^2)/(4*c) + (3*B*(c - c*tan(e + f*x)*1i)
^3)/(16*c^2))/(a*f*(c - c*tan(e + f*x)*1i)^(7/2) - 2*a*c*f*(c - c*tan(e + f*x)*1i)^(5/2)) - ((A*(c - c*tan(e +
f*x)*1i)*7i)/(30*a*f) + (A*c*1i)/(5*a*f) + (A*(c - c*tan(e + f*x)*1i)^2*7i)/(12*a*c*f) - (A*(c - c*tan(e + f*
x)*1i)^3*7i)/(16*a*c^2*f))/(2*c*(c - c*tan(e + f*x)*1i)^(5/2) - (c - c*tan(e + f*x)*1i)^(7/2)) - (2^(1/2)*A*at
an((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*7i)/(32*a*(-c)^(5/2)*f) - (3*2^(1/2)*B*atanh((2^(1/
2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(32*a*c^(5/2)*f)