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

Optimal. Leaf size=92 \[ \frac{i A \sqrt{c-i c \tan (e+f x)}}{c f \sqrt{a+i a \tan (e+f x)}}-\frac{B+i A}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.199729, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3588, 78, 37} \[ \frac{i A \sqrt{c-i c \tan (e+f x)}}{c f \sqrt{a+i a \tan (e+f x)}}-\frac{B+i A}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((I*A + B)/(f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])) + (I*A*Sqrt[c - I*c*Tan[e + f*x]])/(c*f
*Sqrt[a + I*a*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 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 3.77169, size = 77, normalized size = 0.84 \[ -\frac{\sqrt{c-i c \tan (e+f x)} (\cos (e+f x)+i \sin (e+f x)) (B \cos (e+f x)-A \sin (e+f x))}{c f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.193, size = 99, normalized size = 1.1 \begin{align*}{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+A\tan \left ( fx+e \right ) -B}{afc \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\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))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

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

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Maxima [A]  time = 2.52534, size = 170, normalized size = 1.85 \begin{align*} -\frac{{\left ({\left (2 \, A - 2 i \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) - 4 i \, B \cos \left (2 \, f x + 2 \, e\right ) - 2 \,{\left (-i \, A - B\right )} \sin \left (4 \, f x + 4 \, e\right ) + 4 \, B \sin \left (2 \, f x + 2 \, e\right ) - 2 \, A - 2 i \, B\right )} \sqrt{a} \sqrt{c}}{{\left (-4 i \, a c \cos \left (3 \, f x + 3 \, e\right ) - 4 i \, a c \cos \left (f x + e\right ) + 4 \, a c \sin \left (3 \, f x + 3 \, e\right ) + 4 \, a c \sin \left (f x + e\right )\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*A - 2*I*B)*cos(4*f*x + 4*e) - 4*I*B*cos(2*f*x + 2*e) - 2*(-I*A - B)*sin(4*f*x + 4*e) + 4*B*sin(2*f*x + 2*
e) - 2*A - 2*I*B)*sqrt(a)*sqrt(c)/((-4*I*a*c*cos(3*f*x + 3*e) - 4*I*a*c*cos(f*x + e) + 4*a*c*sin(3*f*x + 3*e)
+ 4*a*c*sin(f*x + e))*f)

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Fricas [A]  time = 1.31045, size = 290, normalized size = 3.15 \begin{align*} \frac{{\left ({\left (-i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, B e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, B e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, B e^{\left (i \, f x + i \, e\right )} + i \, A - B\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-i \, f x - i \, e\right )}}{2 \, a c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (e + f x \right )}}{\sqrt{a \left (i \tan{\left (e + f x \right )} + 1\right )} \sqrt{- c \left (i \tan{\left (e + f x \right )} - 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (f x + e\right ) + A}{\sqrt{i \, a \tan \left (f x + e\right ) + a} \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))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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