3.8.10 \(\int \frac {A+B \tan (e+f x)}{a+i a \tan (e+f x)} \, dx\) [710]

Optimal. Leaf size=47 \[ \frac {(A-i B) x}{2 a}+\frac {i A-B}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3607, 8} \begin {gather*} \frac {-B+i A}{2 f (a+i a \tan (e+f x))}+\frac {x (A-i B)}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A - I*B)*x)/(2*a) + (I*A - B)/(2*f*(a + I*a*Tan[e + f*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3607

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a*f*m)), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {A+B \tan (e+f x)}{a+i a \tan (e+f x)} \, dx &=\frac {i A-B}{2 f (a+i a \tan (e+f x))}+\frac {(A-i B) \int 1 \, dx}{2 a}\\ &=\frac {(A-i B) x}{2 a}+\frac {i A-B}{2 f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(47)=94\).
time = 0.26, size = 102, normalized size = 2.17 \begin {gather*} \frac {\cos (e+f x) (A+B \tan (e+f x)) (A-2 i A f x+B (i-2 f x)+(B-2 i B f x+A (-i+2 f x)) \tan (e+f x))}{4 a f (A \cos (e+f x)+B \sin (e+f x)) (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 68, normalized size = 1.45

method	result	size
risch	\(-\frac {i x B}{2 a}+\frac {x A}{2 a}-\frac {{\mathrm e}^{-2 i \left (f x +e \right )} B}{4 a f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} A}{4 a f}\)	\(54\)
derivativedivides	\(\frac {-\frac {-\frac {A}{2}-\frac {i B}{2}}{-i+\tan \left (f x +e \right )}+\left (-\frac {i A}{4}-\frac {B}{4}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{4}}{f a}\)	\(68\)
default	\(\frac {-\frac {-\frac {A}{2}-\frac {i B}{2}}{-i+\tan \left (f x +e \right )}+\left (-\frac {i A}{4}-\frac {B}{4}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{4}}{f a}\)	\(68\)
norman	\(\frac {\frac {\left (-i B +A \right ) x}{2 a}-\frac {-i A +B}{2 a f}+\frac {\left (i B +A \right ) \tan \left (f x +e \right )}{2 a f}+\frac {\left (-i B +A \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{2 a}}{1+\tan ^{2}\left (f x +e \right )}\)	\(81\)

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)),x,method=_RETURNVERBOSE)

[Out]

1/f/a*(-(-1/2*A-1/2*I*B)/(-I+tan(f*x+e))+(-1/4*I*A-1/4*B)*ln(-I+tan(f*x+e))+1/4*I*(A-I*B)*ln(I+tan(f*x+e)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 4.52, size = 44, normalized size = 0.94 \begin {gather*} \frac {{\left (2 \, {\left (A - i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \end {gather*}

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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.12, size = 87, normalized size = 1.85 \begin {gather*} \begin {cases} \frac {\left (i A - B\right ) e^{- 2 i e} e^{- 2 i f x}}{4 a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {A - i B}{2 a} + \frac {\left (A e^{2 i e} + A - i B e^{2 i e} + i B\right ) e^{- 2 i e}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A - i B\right )}{2 a} \end {gather*}

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)),x)

[Out]

Piecewise(((I*A - B)*exp(-2*I*e)*exp(-2*I*f*x)/(4*a*f), Ne(a*f*exp(2*I*e), 0)), (x*(-(A - I*B)/(2*a) + (A*exp(
2*I*e) + A - I*B*exp(2*I*e) + I*B)*exp(-2*I*e)/(2*a)), True)) + x*(A - I*B)/(2*a)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (36) = 72\).
time = 0.54, size = 90, normalized size = 1.91 \begin {gather*} -\frac {\frac {{\left (i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a} + \frac {{\left (-i \, A - B\right )} \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a} + \frac {-i \, A \tan \left (f x + e\right ) - B \tan \left (f x + e\right ) - 3 \, A - i \, B}{a {\left (\tan \left (f x + e\right ) - i\right )}}}{4 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

-1/4*((I*A + B)*log(tan(f*x + e) - I)/a + (-I*A - B)*log(-I*tan(f*x + e) + 1)/a + (-I*A*tan(f*x + e) - B*tan(f
*x + e) - 3*A - I*B)/(a*(tan(f*x + e) - I)))/f

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Mupad [B]
time = 8.59, size = 45, normalized size = 0.96 \begin {gather*} -\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a}+\frac {-\frac {B}{2\,a}+\frac {A\,1{}\mathrm {i}}{2\,a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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