\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) [781]

Optimal result

Integrand size = 43, antiderivative size = 213 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {(i A+11 B) c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^3 f}-\frac {(i A+11 B) c^2 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac {(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

1/32*(I*A+11*B)*c^(5/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)-1/16*(I*A+11*B)*c^
2*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))+1/24*(I*A+11*B)*c*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(1+I*tan(f*
x+e))^2+1/6*(I*A-B)*(c-I*c*tan(f*x+e))^(5/2)/a^3/f/(1+I*tan(f*x+e))^3

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 43, 65, 214} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {c^{5/2} (11 B+i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^3 f}-\frac {c^2 (11 B+i A) \sqrt {c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac {c (11 B+i A) (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((I*A + 11*B)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a^3*f) - ((I*A + 11*B
)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f*(1 + I*Tan[e + f*x])) + ((I*A + 11*B)*c*(c - I*c*Tan[e + f*x])^(3/
2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(5/2))/(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 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)^{3/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {((A-11 i B) c) \text {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{12 f} \\ & = \frac {(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left ((A-11 i B) c^2\right ) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{16 a f} \\ & = -\frac {(i A+11 B) c^2 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac {(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {\left ((A-11 i B) c^3\right ) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{32 a^2 f} \\ & = -\frac {(i A+11 B) c^2 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac {(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {\left ((i A+11 B) c^2\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 )}{16 a^2 f} \\ & = \frac {(i A+11 B) c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^3 f}-\frac {(i A+11 B) c^2 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac {(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.50 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {c^2 \sec ^3(e+f x) \left (3 \sqrt {2} (A-11 i B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+2 \cos (e+f x) (2 (A-11 i B)+(5 A+41 i B) \cos (2 (e+f x))+(-11 i A-25 B) \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{96 a^3 f (-i+\tan (e+f x))^3} \]

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

-1/96*(c^2*Sec[e + f*x]^3*(3*Sqrt[2]*(A - (11*I)*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c
])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + 2*Cos[e + f*x]*(2*(A - (11*I)*B) + (5*A + (41*I)*B)*Cos[2*(e + f
*x)] + ((-11*I)*A - 25*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]]))/(a^3*f*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {8 \left (\frac {21 i B}{256}+\frac {A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (-\frac {11}{48} i B c +\frac {1}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {11}{64} i B \,c^{2}-\frac {1}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {\left (-\frac {11 i B}{8}+\frac {A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )}{f \,a^{3}}\)	\(147\)
default	\(\frac {2 i c^{3} \left (\frac {8 \left (\frac {21 i B}{256}+\frac {A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (-\frac {11}{48} i B c +\frac {1}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {11}{64} i B \,c^{2}-\frac {1}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {\left (-\frac {11 i B}{8}+\frac {A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )}{f \,a^{3}}\)	\(147\)

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2*I/f/a^3*c^3*(8*((21/256*I*B+1/256*A)*(c-I*c*tan(f*x+e))^(5/2)+(-11/48*I*B*c+1/48*c*A)*(c-I*c*tan(f*x+e))^(3/
2)+(11/64*I*B*c^2-1/64*c^2*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3+1/8*(-11/8*I*B+1/8*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. 403 vs. \(2 (166) = 332\).

Time = 0.27 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.89 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (i \, A + 11 \, B\right )} c^{3} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (i \, A + 11 \, B\right )} c^{3} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) - \sqrt {2} {\left (3 \, {\left (i \, A + 11 \, B\right )} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (-i \, A - 11 \, B\right )} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-5 i \, A - 7 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, A + B\right )} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/96*(3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 22*I*A*B - 121*B^2)*c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((I*A + 11
*B)*c^3 + 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(-(A^2 -
22*I*A*B - 121*B^2)*c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 22*I*A*B - 121
*B^2)*c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((I*A + 11*B)*c^3 - 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(-(A^2 - 22*I*A*B - 121*B^2)*c^5/(a^6*f^2)))*e^(-I*f*x - I*
e)/(a^3*f)) - sqrt(2)*(3*(I*A + 11*B)*c^2*e^(6*I*f*x + 6*I*e) - (-I*A - 11*B)*c^2*e^(4*I*f*x + 4*I*e) + 2*(-5*
I*A - 7*B)*c^2*e^(2*I*f*x + 2*I*e) + 8*(-I*A + B)*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)
/(a^3*f)

Sympy [F]

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

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - 11 i \, B\right )} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A + 21 i \, B\right )} c^{4} + 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 11 i \, B\right )} c^{5} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 11 i \, B\right )} c^{6}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{192 \, c f} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/192*I*(3*sqrt(2)*(A - 11*I*B)*c^(7/2)*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)*(A + 21*I*B)*c^4 + 16*(-I*c*tan(f*x
+ e) + c)^(3/2)*(A - 11*I*B)*c^5 - 12*sqrt(-I*c*tan(f*x + e) + c)*(A - 11*I*B)*c^6)/((-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))/(c*f)

Giac [F]

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

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(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)^(5/2)/(I*a*tan(f*x + e) + a)^3, x)

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {-\frac {A\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,a^3\,f}+\frac {A\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{3\,a^3\,f}+\frac {A\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{16\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {\frac {11\,B\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{4}-\frac {11\,B\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3}+\frac {21\,B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{16}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{32\,a^3\,f}+\frac {11\,\sqrt {2}\,B\,c^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{32\,a^3\,f} \]

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(5/2))/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((A*c^4*(c - c*tan(e + f*x)*1i)^(3/2)*1i)/(3*a^3*f) - (A*c^5*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(4*a^3*f) + (A*
c^3*(c - c*tan(e + f*x)*1i)^(5/2)*1i)/(16*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) - ((11*B*c^5*(c - c*tan(e + f*x)*1i)^(1/2))/4 - (11*B*c^4*(c - c*tan(
e + f*x)*1i)^(3/2))/3 + (21*B*c^3*(c - c*tan(e + f*x)*1i)^(5/2))/16)/(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)) + (2^(1/2)*A*(-c)^(5/2)*at
an((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*1i)/(32*a^3*f) + (11*2^(1/2)*B*c^(5/2)*atanh((2^(1/
2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(32*a^3*f)