\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\) [779]

Optimal result

Integrand size = 43, antiderivative size = 291 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {35 (i A-5 B) c^{9/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^3 f}+\frac {35 (i A-5 B) c^4 \sqrt {c-i c \tan (e+f x)}}{8 a^3 f}+\frac {35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

-35/8*(I*A-5*B)*c^(9/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)+35/8*(I*A-5*B)*c^4
*(c-I*c*tan(f*x+e))^(1/2)/a^3/f+35/48*(I*A-5*B)*c^3*(c-I*c*tan(f*x+e))^(3/2)/a^3/f+7/16*(I*A-5*B)*c^2*(c-I*c*t
an(f*x+e))^(5/2)/a^3/f/(1+I*tan(f*x+e))-1/8*(I*A-5*B)*c*(c-I*c*tan(f*x+e))^(7/2)/a^3/f/(1+I*tan(f*x+e))^2+1/6*
(I*A-B)*(c-I*c*tan(f*x+e))^(9/2)/a^3/f/(1+I*tan(f*x+e))^3

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 291, 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, 43, 52, 65, 214} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {35 c^{9/2} (-5 B+i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^3 f}+\frac {35 c^4 (-5 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^3 f}+\frac {35 c^3 (-5 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 c^2 (-5 B+i A) (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {c (-5 B+i A) (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(-35*(I*A - 5*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(4*Sqrt[2]*a^3*f) + (35*(I*A -
5*B)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(8*a^3*f) + (35*(I*A - 5*B)*c^3*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f)
+ (7*(I*A - 5*B)*c^2*(c - I*c*Tan[e + f*x])^(5/2))/(16*a^3*f*(1 + I*Tan[e + f*x])) - ((I*A - 5*B)*c*(c - I*c*
Tan[e + f*x])^(7/2))/(8*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(6*a^3*f*(1 +
I*Tan[e + f*x])^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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) (c-i c x)^{7/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {((A+5 i B) c) \text {Subst}\left (\int \frac {(c-i c x)^{7/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = -\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {\left (7 (A+5 i B) c^2\right ) \text {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{16 a f} \\ & = \frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (35 (A+5 i B) c^3\right ) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{32 a^2 f} \\ & = \frac {35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (35 (A+5 i B) c^4\right ) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a^2 f} \\ & = \frac {35 (i A-5 B) c^4 \sqrt {c-i c \tan (e+f x)}}{8 a^3 f}+\frac {35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (35 (A+5 i B) c^5\right ) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 a^2 f} \\ & = \frac {35 (i A-5 B) c^4 \sqrt {c-i c \tan (e+f x)}}{8 a^3 f}+\frac {35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left (35 (i A-5 B) c^4\right ) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{4 a^2 f} \\ & = -\frac {35 (i A-5 B) c^{9/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^3 f}+\frac {35 (i A-5 B) c^4 \sqrt {c-i c \tan (e+f x)}}{8 a^3 f}+\frac {35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac {7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac {(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.27 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {105 \sqrt {2} (A+5 i B) c^{9/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) \sec ^3(e+f x) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+2 c^4 \sqrt {c-i c \tan (e+f x)} \left (-67 A-327 i B+2 (-85 i A+441 B) \tan (e+f x)+3 (53 A+249 i B) \tan ^2(e+f x)+8 i (3 A+19 i B) \tan ^3(e+f x)+8 i B \tan ^4(e+f x)\right )}{24 a^3 f (-i+\tan (e+f x))^3} \]

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(105*Sqrt[2]*(A + (5*I)*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*Sec[e + f*x]^3*(Cos[3
*(e + f*x)] + I*Sin[3*(e + f*x)]) + 2*c^4*Sqrt[c - I*c*Tan[e + f*x]]*(-67*A - (327*I)*B + 2*((-85*I)*A + 441*B
)*Tan[e + f*x] + 3*(53*A + (249*I)*B)*Tan[e + f*x]^2 + (8*I)*(3*A + (19*I)*B)*Tan[e + f*x]^3 + (8*I)*B*Tan[e +
f*x]^4))/(24*a^3*f*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+7 i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A -8 c^{2} \left (\frac {8 \left (-\frac {81 i B}{512}-\frac {29 A}{512}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {53}{96} i B c +\frac {17}{96} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {63}{128} i B \,c^{2}-\frac {19}{128} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {A}{8}+\frac {5 i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )\right )}{f \,a^{3}}\)	\(207\)
default	\(\frac {2 i c^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+7 i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A -8 c^{2} \left (\frac {8 \left (-\frac {81 i B}{512}-\frac {29 A}{512}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {53}{96} i B c +\frac {17}{96} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {63}{128} i B \,c^{2}-\frac {19}{128} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {A}{8}+\frac {5 i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )\right )}{f \,a^{3}}\)	\(207\)

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2*I/f/a^3*c^3*(1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)+7*I*(c-I*c*tan(f*x+e))^(1/2)*B*c+(c-I*c*tan(f*x+e))^(1/2)*c*A-
8*c^2*(8*((-81/512*I*B-29/512*A)*(c-I*c*tan(f*x+e))^(5/2)+(53/96*I*B*c+17/96*c*A)*(c-I*c*tan(f*x+e))^(3/2)+(-6
3/128*I*B*c^2-19/128*c^2*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3+35/16*(1/8*A+5/8*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))))

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. 471 vs. \(2 (228) = 456\).

Time = 0.28 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {105 \, \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \log \left (-\frac {35 \, {\left ({\left (i \, A - 5 \, B\right )} c^{5} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{3} f}\right ) - 105 \, \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \log \left (-\frac {35 \, {\left ({\left (i \, A - 5 \, B\right )} c^{5} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (A^{2} + 10 i \, A B - 25 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{3} f}\right ) + \sqrt {2} {\left (105 \, {\left (-i \, A + 5 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + 140 \, {\left (-i \, A + 5 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 21 \, {\left (-i \, A + 5 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, {\left (i \, A - 5 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, A + B\right )} c^{4}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{24 \, {\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )}} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/24*(105*sqrt(1/2)*(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))*sqrt(-(A^2 + 10*I*A*B - 25*B^2)*c
^9/(a^6*f^2))*log(-35/2*((I*A - 5*B)*c^5 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(A^2 +
10*I*A*B - 25*B^2)*c^9/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) - 105*sqrt(1/2)
*(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))*sqrt(-(A^2 + 10*I*A*B - 25*B^2)*c^9/(a^6*f^2))*log(-3
5/2*((I*A - 5*B)*c^5 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(A^2 + 10*I*A*B - 25*B^2)*c
^9/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*(105*(-I*A + 5*B)*c^4*e^(
8*I*f*x + 8*I*e) + 140*(-I*A + 5*B)*c^4*e^(6*I*f*x + 6*I*e) + 21*(-I*A + 5*B)*c^4*e^(4*I*f*x + 4*I*e) + 6*(I*A
- 5*B)*c^4*e^(2*I*f*x + 2*I*e) + 8*(-I*A + B)*c^4)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(a^3*f*e^(8*I*f*x + 8*I
*e) + a^3*f*e^(6*I*f*x + 6*I*e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(9/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \, {\left (\frac {105 \, \sqrt {2} {\left (A + 5 i \, B\right )} c^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} - \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (29 \, A + 81 i \, B\right )} c^{6} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (17 \, A + 53 i \, B\right )} c^{7} + 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (19 \, A + 63 i \, B\right )} c^{8}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}} + \frac {32 \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B c^{4} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 7 i \, B\right )} c^{5}\right )}}{a^{3}}\right )}}{48 \, c f} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

1/48*I*(105*sqrt(2)*(A + 5*I*B)*c^(11/2)*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^3 - 4*(3*(-I*c*tan(f*x + e) + c)^(5/2)*(29*A + 81*I*B)*c^6 - 16*(-I*c*tan(f
*x + e) + c)^(3/2)*(17*A + 53*I*B)*c^7 + 12*sqrt(-I*c*tan(f*x + e) + c)*(19*A + 63*I*B)*c^8)/((-I*c*tan(f*x +
e) + c)^3*a^3 - 6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3) + 32*(I*(-
I*c*tan(f*x + e) + c)^(3/2)*B*c^4 + 3*sqrt(-I*c*tan(f*x + e) + c)*(A + 7*I*B)*c^5)/a^3)/(c*f)

Giac [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(9/2)/(I*a*tan(f*x + e) + a)^3, x)

Mupad [B] (verification not implemented)

Time = 9.17 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.52 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {A\,c^7\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,19{}\mathrm {i}}{a^3\,f}-\frac {A\,c^6\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,68{}\mathrm {i}}{3\,a^3\,f}+\frac {A\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,29{}\mathrm {i}}{4\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {63\,B\,c^7\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\frac {212\,B\,c^6\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3}+\frac {81\,B\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{4}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,f}-\frac {14\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^3\,f}-\frac {2\,B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^3\,f}-\frac {\sqrt {2}\,A\,{\left (-c\right )}^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{8\,a^3\,f}-\frac {\sqrt {2}\,B\,c^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,175{}\mathrm {i}}{8\,a^3\,f} \]

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(9/2))/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((A*c^7*(c - c*tan(e + f*x)*1i)^(1/2)*19i)/(a^3*f) - (A*c^6*(c - c*tan(e + f*x)*1i)^(3/2)*68i)/(3*a^3*f) + (A*
c^5*(c - c*tan(e + f*x)*1i)^(5/2)*29i)/(4*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*
1i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) - (63*B*c^7*(c - c*tan(e + f*x)*1i)^(1/2) - (212*B*c^6*(c - c*tan(e +
f*x)*1i)^(3/2))/3 + (81*B*c^5*(c - c*tan(e + f*x)*1i)^(5/2))/4)/(8*a^3*c^3*f - a^3*f*(c - c*tan(e + f*x)*1i)^
3 + 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^2 - 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)) + (A*c^4*(c - c*tan(e + f*x)*1
i)^(1/2)*2i)/(a^3*f) - (14*B*c^4*(c - c*tan(e + f*x)*1i)^(1/2))/(a^3*f) - (2*B*c^3*(c - c*tan(e + f*x)*1i)^(3/
2))/(3*a^3*f) - (2^(1/2)*A*(-c)^(9/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*35i)/(8*a^3
*f) - (2^(1/2)*B*c^(9/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(2*c^(1/2)))*175i)/(8*a^3*f)