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

Optimal. Leaf size=109 \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac{\sqrt{c} (3 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a f} \]

[Out]

((I*A + 3*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(2*Sqrt[2]*a*f) + ((I*A - B)*Sqrt[
c - I*c*Tan[e + f*x]])/(2*a*f*(1 + I*Tan[e + f*x]))

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Rubi [A]  time = 0.185213, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.093, Rules used = {3588, 78, 63, 208} \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac{\sqrt{c} (3 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a f} \]

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]),x]

[Out]

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

Mathematica [A]  time = 2.46795, size = 168, normalized size = 1.54 \[ \frac{(\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (2 (A+i B) \cos (e+f x) (\sin (f x)+i \cos (f x)) \sqrt{c-i c \tan (e+f x)}+\sqrt{2} \sqrt{c} (3 B+i A) (\cos (e)+i \sin (e)) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{4 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]

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]),x]

[Out]

((Cos[f*x] + I*Sin[f*x])*(A + B*Tan[e + f*x])*(Sqrt[2]*(I*A + 3*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/
(Sqrt[2]*Sqrt[c])]*(Cos[e] + I*Sin[e]) + 2*(A + I*B)*Cos[e + f*x]*(I*Cos[f*x] + Sin[f*x])*Sqrt[c - I*c*Tan[e +
f*x]]))/(4*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x]))

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Maple [A]  time = 0.136, size = 88, normalized size = 0.8 \begin{align*}{\frac{2\,ic}{af} \left ({\frac{1}{-c-ic\tan \left ( fx+e \right ) } \left ( -{\frac{A}{4}}-{\frac{i}{4}}B \right ) \sqrt{c-ic\tan \left ( fx+e \right ) }}+{\frac{ \left ( A-3\,iB \right ) \sqrt{2}}{8}{\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 ) } \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)),x)

[Out]

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

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.09983, size = 841, normalized size = 7.72 \begin{align*} \frac{{\left (\sqrt{\frac{1}{2}} a f \sqrt{-\frac{{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} +{\left (i \, A + 3 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt{\frac{1}{2}} a f \sqrt{-\frac{{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c}{a^{2} f^{2}}} -{\left (i \, A + 3 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt{2}{\left ({\left (i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a 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)),x, algorithm="fricas")

[Out]

1/4*(sqrt(1/2)*a*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log((sqrt(2)*sqrt(1/2)*(a*f*
e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c/(a^2*f^2)) + (I*A
+ 3*B)*c)*e^(-I*f*x - I*e)/(a*f)) - sqrt(1/2)*a*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c/(a^2*f^2))*e^(2*I*f*x + 2*I
*e)*log(-(sqrt(2)*sqrt(1/2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 - 6*I
*A*B - 9*B^2)*c/(a^2*f^2)) - (I*A + 3*B)*c)*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*((I*A - B)*e^(2*I*f*x + 2*I*e) +
I*A - B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*f)

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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)),x)

[Out]

Exception raised: AttributeError

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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}}{i \, a \tan \left (f x + e\right ) + a}\,{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)),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), x)