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

Optimal. Leaf size=159 \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{(5 B+3 i A) \sqrt{c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac{\sqrt{c} (5 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^2 f} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.212286, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{(5 B+3 i A) \sqrt{c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac{\sqrt{c} (5 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^2 f} \]

Antiderivative was successfully verified.

[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 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)) \sqrt{c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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} \tanh ^{-1}\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]  time = 2.84846, size = 206, normalized size = 1.3 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \left (2 \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) \sqrt{c-i c \tan (e+f x)} ((-3 A+5 i B) \sin (e+f x)+(B+7 i A) \cos (e+f x))+\sqrt{2} \sqrt{c} (5 B+3 i A) (\cos (2 e)+i \sin (2 e)) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{32 f (a+i a \tan (e+f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[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]

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

________________________________________________________________________________________

Maple [A]  time = 0.144, size = 121, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{3\,A-5\,iB}{32\,c} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+ \left ( -{\frac{5\,A}{16}}+{\frac{3\,i}{16}}B \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 3\,A-5\,iB \right ) \sqrt{2}}{64}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

________________________________________________________________________________________

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((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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.43613, size = 961, normalized size = 6.04 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-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((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]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)