\(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [775]

Optimal result

Integrand size = 43, antiderivative size = 159 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {(3 i A+5 B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \]

[Out]

1/32*(3*I*A+5*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a^2/f*2^(1/2)+1/4*(I*A-B)*(c-I*
c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))^2+1/16*(3*I*A+5*B)*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 44, 65, 214} \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\sqrt {c} (5 B+3 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(5 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.53 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\sqrt {2} (3 A-5 i B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) \sec ^2(e+f x) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+2 (-7 i A-B+(3 A-5 i B) \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{32 a^2 f (-i+\tan (e+f x))^2} \]

[In]

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

[Out]

(Sqrt[2]*(3*A - (5*I)*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*Sec[e + f*x]^2*((-I)*Co
s[2*(e + f*x)] + Sin[2*(e + f*x)]) + 2*((-7*I)*A - B + (3*A - (5*I)*B)*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]
])/(32*a^2*f*(-I + Tan[e + f*x])^2)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 i c^{2} \left (\frac {-\frac {\left (-5 i B +3 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{32 c}+4 \left (\frac {5 A}{64}-\frac {3 i B}{64}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\left (-5 i B +3 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\)	\(120\)
default	\(\frac {2 i c^{2} \left (\frac {-\frac {\left (-5 i B +3 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{32 c}+4 \left (\frac {5 A}{64}-\frac {3 i B}{64}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\left (-5 i B +3 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\)	\(120\)

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2*I/f/a^2*c^2*(4*(-1/128/c*(3*A-5*I*B)*(c-I*c*tan(f*x+e))^(3/2)+(5/64*A-3/64*I*B)*(c-I*c*tan(f*x+e))^(1/2))/(c
+I*c*tan(f*x+e))^2+1/64/c^(3/2)*(3*A-5*I*B)*2^(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. 362 vs. \(2 (122) = 244\).

Time = 0.28 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.28 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} + {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} - {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} {\left ({\left (5 i \, A + 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (7 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} f} \]

[In]

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

[Out]

1/32*(sqrt(1/2)*a^2*f*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(sqrt(2)*sqrt
(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*
c/(a^4*f^2)) + (3*I*A + 5*B)*c)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(1/2)*a^2*f*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*
c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*c/(a^4*f^2)) - (3*I*A + 5*B)*c)*e^(-I*f*x - I*e)/(a^2*f
)) + sqrt(2)*((5*I*A + 3*B)*e^(4*I*f*x + 4*I*e) + (7*I*A + B)*e^(2*I*f*x + 2*I*e) + 2*I*A - 2*B)*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

Sympy [F]

\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {A \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

[In]

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

[Out]

-(Integral(A*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(B*sqrt(-I*c*t
an(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} {\left (3 \, A - 5 i \, B\right )} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (3 \, A - 5 i \, B\right )} c^{2} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (5 \, A - 3 i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{64 \, c f} \]

[In]

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

[Out]

-1/64*I*(sqrt(2)*(3*A - 5*I*B)*c^(3/2)*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^2 + 4*((-I*c*tan(f*x + e) + c)^(3/2)*(3*A - 5*I*B)*c^2 - 2*sqrt(-I*c*tan(f*x
+ e) + c)*(5*A - 3*I*B)*c^3)/((-I*c*tan(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c
*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.83 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {5\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{16}-\frac {3\,B\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{8}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\frac {A\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,A\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{32\,a^2\,f}+\frac {5\,\sqrt {2}\,B\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{32\,a^2\,f} \]

[In]

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

[Out]

((A*c^2*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(8*a^2*f) - (A*c*(c - c*tan(e + f*x)*1i)^(3/2)*3i)/(16*a^2*f))/((c -
c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2) - ((5*B*c*(c - c*tan(e + f*x)*1i)^(3/2))/16 - (3*
B*c^2*(c - c*tan(e + f*x)*1i)^(1/2))/8)/(4*a^2*c^2*f + a^2*f*(c - c*tan(e + f*x)*1i)^2 - 4*a^2*c*f*(c - c*tan(
e + f*x)*1i)) + (2^(1/2)*A*(-c)^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(32*a^2
*f) + (5*2^(1/2)*B*c^(1/2)*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(32*a^2*f)