\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx\) [778]

Optimal result

Integrand size = 43, antiderivative size = 273 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {7 (9 i A-B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

7/256*(9*I*A-B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(5/2)/f*2^(1/2)-7/128*(9*I*A-B)/a^
2/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-7/160*(9*I*A-B)/a^2/f/(c-I*c*tan(f*x+e))^(5/2)+1/4*(I*A-B)/a^2/f/(1+I*tan(f*x
+e))^2/(c-I*c*tan(f*x+e))^(5/2)+1/16*(9*I*A-B)/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-7/192*(9*I*A-B)
/a^2/c/f/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 44, 53, 65, 214} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {7 (-B+9 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (-B+9 i A)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac {7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[In]

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

[Out]

(7*((9*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(128*Sqrt[2]*a^2*c^(5/2)*f) - (7*((9*I
)*A - B))/(160*a^2*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e
+ f*x])^(5/2)) + ((9*I)*A - B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((9*I)*A - B)
)/(192*a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (7*((9*I)*A - B))/(128*a^2*c^2*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 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)^3 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {((9 A+i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = \frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 A+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 )}{32 a f} \\ & = -\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(7 (9 A+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 )}{64 a f} \\ & = -\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(7 (9 A+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 )}{128 a c f} \\ & = -\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 A+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 c^2 f} \\ & = -\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(7 (9 i A-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 c^3 f} \\ & = \frac {7 (9 i A-B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 (9 i A-B)}{128 a^2 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 = 5.84 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.52 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) \left (35 (-9 i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) (\cos (e+f x)+i \sin (e+f x))+6 \cos (e+f x) (65 i A-25 B+2 i (A+9 i B) \cos (2 (e+f x))+2 (9 A+i B) \sin (2 (e+f x)))\right )}{960 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(Cos[e + f*x]*(35*((-9*I)*A + B)*Hypergeometric2F1[-3/2, 1, -1/2, (-1/2*I)*(I + Tan[e + f*x])]*(Cos[e + f*x] +
I*Sin[e + f*x]) + 6*Cos[e + f*x]*((65*I)*A - 25*B + (2*I)*(A + (9*I)*B)*Cos[2*(e + f*x)] + 2*(9*A + I*B)*Sin[
2*(e + f*x)])))/(960*a^2*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 i c^{2} \left (\frac {\frac {4 \left (-\frac {7 i B}{64}-\frac {15 A}{64}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{32} i B c +\frac {17}{32} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {7 \left (\frac {i B}{4}+\frac {9 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {3 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +3 A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{40 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\)	\(196\)
default	\(\frac {2 i c^{2} \left (\frac {\frac {4 \left (-\frac {7 i B}{64}-\frac {15 A}{64}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (\frac {9}{32} i B c +\frac {17}{32} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {7 \left (\frac {i B}{4}+\frac {9 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {3 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +3 A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{40 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\)	\(196\)

[In]

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

[Out]

2*I/f/a^2*c^2*(1/16/c^4*(4*((-7/64*I*B-15/64*A)*(c-I*c*tan(f*x+e))^(3/2)+(9/32*I*B*c+17/32*c*A)*(c-I*c*tan(f*x
+e))^(1/2))/(c+I*c*tan(f*x+e))^2+7/8*(1/4*I*B+9/4*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)))-3/16/c^4*A/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^3*(3*A-I*B)/(c-I*c*tan(f*x+e))^(3/2)-1/40/c^2*(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. 446 vs. \(2 (202) = 404\).

Time = 0.28 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.63 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (105 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {81 \, A^{2} + 18 i \, A B - B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {81 \, A^{2} + 18 i \, A B - B^{2}}{a^{4} c^{5} f^{2}}} + 9 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) - 105 \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {-\frac {81 \, A^{2} + 18 i \, A B - B^{2}}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {81 \, A^{2} + 18 i \, A B - B^{2}}{a^{4} c^{5} f^{2}}} - 9 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) - \sqrt {2} {\left (24 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + 16 \, {\left (12 i \, A + 7 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 8 \, {\left (129 i \, A + 19 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (-609 i \, A - 199 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 15 \, {\left (-19 i \, A + 11 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 30 i \, A + 30 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{3840 \, a^{2} c^{3} f} \]

[In]

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

[Out]

1/3840*(105*sqrt(1/2)*a^2*c^3*f*sqrt(-(81*A^2 + 18*I*A*B - B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(7/64*(s
qrt(2)*sqrt(1/2)*(a^2*c^2*f*e^(2*I*f*x + 2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(81*A^2 +
18*I*A*B - B^2)/(a^4*c^5*f^2)) + 9*I*A - B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) - 105*sqrt(1/2)*a^2*c^3*f*sqrt(-(81
*A^2 + 18*I*A*B - B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(-7/64*(sqrt(2)*sqrt(1/2)*(a^2*c^2*f*e^(2*I*f*x +
2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(81*A^2 + 18*I*A*B - B^2)/(a^4*c^5*f^2)) - 9*I*A
+ B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) - sqrt(2)*(24*(I*A + B)*e^(10*I*f*x + 10*I*e) + 16*(12*I*A + 7*B)*e^(8*I*f*
x + 8*I*e) + 8*(129*I*A + 19*B)*e^(6*I*f*x + 6*I*e) - (-609*I*A - 199*B)*e^(4*I*f*x + 4*I*e) + 15*(-19*I*A + 1
1*B)*e^(2*I*f*x + 2*I*e) - 30*I*A + 30*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=- \frac {\int \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]

[In]

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

[Out]

-(Integral(A/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*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)), x) + Integral(B*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*ta
n(e + f*x)**4 - 2*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)), x))/a*
*2

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.90 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (9 \, A + i \, B\right )} - 350 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (9 \, A + i \, B\right )} c + 224 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (9 \, A + i \, B\right )} c^{2} + 64 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (9 \, A + i \, B\right )} c^{3} + 384 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{2} c - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} c^{2} + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c^{3}} + \frac {105 \, \sqrt {2} {\left (9 \, A + i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} c^{\frac {3}{2}}}\right )}}{7680 \, c f} \]

[In]

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

[Out]

-1/7680*I*(4*(105*(-I*c*tan(f*x + e) + c)^4*(9*A + I*B) - 350*(-I*c*tan(f*x + e) + c)^3*(9*A + I*B)*c + 224*(-
I*c*tan(f*x + e) + c)^2*(9*A + I*B)*c^2 + 64*(-I*c*tan(f*x + e) + c)*(9*A + I*B)*c^3 + 384*(A - I*B)*c^4)/((-I
*c*tan(f*x + e) + c)^(9/2)*a^2*c - 4*(-I*c*tan(f*x + e) + c)^(7/2)*a^2*c^2 + 4*(-I*c*tan(f*x + e) + c)^(5/2)*a
^2*c^3) + 105*sqrt(2)*(9*A + 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^2*c^(3/2)))/(c*f)

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 9.88 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {-\frac {B\,c^2}{5}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{60}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{192\,c}+\frac {7\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{128\,c^2}+\frac {B\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{30}}{a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-4\,a^2\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+4\,a^2\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,21{}\mathrm {i}}{20\,a^2\,f}+\frac {A\,c^2\,1{}\mathrm {i}}{5\,a^2\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,105{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,63{}\mathrm {i}}{128\,a^2\,c^2\,f}+\frac {A\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{10\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,63{}\mathrm {i}}{256\,a^2\,{\left (-c\right )}^{5/2}\,f}-\frac {7\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{256\,a^2\,c^{5/2}\,f} \]

[In]

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

[Out]

((7*B*(c - c*tan(e + f*x)*1i)^2)/60 - (B*c^2)/5 - (35*B*(c - c*tan(e + f*x)*1i)^3)/(192*c) + (7*B*(c - c*tan(e
+ f*x)*1i)^4)/(128*c^2) + (B*c*(c - c*tan(e + f*x)*1i))/30)/(a^2*f*(c - c*tan(e + f*x)*1i)^(9/2) - 4*a^2*c*f*
(c - c*tan(e + f*x)*1i)^(7/2) + 4*a^2*c^2*f*(c - c*tan(e + f*x)*1i)^(5/2)) - ((A*(c - c*tan(e + f*x)*1i)^2*21i
)/(20*a^2*f) + (A*c^2*1i)/(5*a^2*f) - (A*(c - c*tan(e + f*x)*1i)^3*105i)/(64*a^2*c*f) + (A*(c - c*tan(e + f*x)
*1i)^4*63i)/(128*a^2*c^2*f) + (A*c*(c - c*tan(e + f*x)*1i)*3i)/(10*a^2*f))/((c - c*tan(e + f*x)*1i)^(9/2) - 4*
c*(c - c*tan(e + f*x)*1i)^(7/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(5/2)) - (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e
+ f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*63i)/(256*a^2*(-c)^(5/2)*f) - (7*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*
x)*1i)^(1/2))/(2*c^(1/2))))/(256*a^2*c^(5/2)*f)