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

Optimal. Leaf size=195 \[ \frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}-\frac{3 (3 B+5 i A)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{3 (3 B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f} \]

[Out]

(3*((5*I)*A + 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(32*Sqrt[2]*a^2*Sqrt[c]*f) - (3*((5*
I)*A + 3*B))/(32*a^2*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Ta
n[e + f*x]]) + ((5*I)*A + 3*B)/(16*a^2*f*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.247676, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}-\frac{3 (3 B+5 i A)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{3 (3 B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*((5*I)*A + 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(32*Sqrt[2]*a^2*Sqrt[c]*f) - (3*((5*
I)*A + 3*B))/(32*a^2*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Ta
n[e + f*x]]) + ((5*I)*A + 3*B)/(16*a^2*f*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])

Rule 3588

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]

Rule 78

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] || LtQ
[p, n]))))

Rule 51

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{((5 A-3 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 A-3 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 A-3 i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 i A+3 B)) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{32 a c f}\\ &=\frac{3 (5 i A+3 B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f}-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.74789, size = 160, normalized size = 0.82 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\sin (e+f x)+i \cos (e+f x)) \left (3 (5 A-3 i B) e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (2 (3 B+5 i A) \sin (2 (e+f x))+2 (3 A-5 i B) \cos (2 (e+f x))-9 A-i B)\right )}{64 a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[e + f*x] + Sin[e + f*x])*(3*(5*A - (3*I)*B)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt
[1 + E^((2*I)*(e + f*x))]] - 2*Cos[e + f*x]*(-9*A - I*B + 2*(3*A - (5*I)*B)*Cos[2*(e + f*x)] + 2*((5*I)*A + 3*
B)*Sin[2*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(64*a^2*c*f)

________________________________________________________________________________________

Maple [A]  time = 0.161, size = 152, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{8\,{c}^{2}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( -{\frac{i}{8}}B+{\frac{7\,A}{8}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{9\,Ac}{4}}-{\frac{i}{4}}Bc \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( -9\,iB+15\,A \right ) \sqrt{2}}{16}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) }-{\frac{-A+iB}{8\,{c}^{2}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.54952, size = 1031, normalized size = 5.29 \begin{align*} \frac{{\left (\sqrt{\frac{1}{2}} a^{2} c f \sqrt{-\frac{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c 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{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} + 15 i \, A + 9 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - \sqrt{\frac{1}{2}} a^{2} c f \sqrt{-\frac{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c 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{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} - 15 i \, A - 9 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) + \sqrt{2}{\left ({\left (-8 i \, A - 8 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A - 9 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (11 i \, A - 3 \, 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 )}}{64 \, a^{2} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(1/2)*a^2*c*f*sqrt(-(225*A^2 - 270*I*A*B - 81*B^2)/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(1/16*(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(-(225*A^2 - 270*I*A*B -
81*B^2)/(a^4*c*f^2)) + 15*I*A + 9*B)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(1/2)*a^2*c*f*sqrt(-(225*A^2 - 270*I*A*B
- 81*B^2)/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/16*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*s
qrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(225*A^2 - 270*I*A*B - 81*B^2)/(a^4*c*f^2)) - 15*I*A - 9*B)*e^(-I*f*x -
I*e)/(a^2*f)) + sqrt(2)*((-8*I*A - 8*B)*e^(6*I*f*x + 6*I*e) + (I*A - 9*B)*e^(4*I*f*x + 4*I*e) + (11*I*A - 3*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*c*f)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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