3.9.92 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x)) \, dx\) [892]

Optimal. Leaf size=12 \[ \frac {a c \tan (e+f x)}{f} \]

[Out]

a*c*tan(f*x+e)/f

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Rubi [A]
time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3603, 3852, 8} \begin {gather*} \frac {a c \tan (e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*Tan[e + f*x])/f

Rule 8

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

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x)) (c-i c \tan (e+f x)) \, dx &=(a c) \int \sec ^2(e+f x) \, dx\\ &=-\frac {(a c) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{f}\\ &=\frac {a c \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} \frac {a c \tan (e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*Tan[e + f*x])/f

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Maple [A]
time = 0.02, size = 13, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {a c \tan \left (f x +e \right )}{f}\)	\(13\)
default	\(\frac {a c \tan \left (f x +e \right )}{f}\)	\(13\)
norman	\(\frac {a c \tan \left (f x +e \right )}{f}\)	\(13\)
risch	\(\frac {2 i a c}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

a*c*tan(f*x+e)/f

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Maxima [A]
time = 0.52, size = 13, normalized size = 1.08 \begin {gather*} \frac {a c \tan \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

a*c*tan(f*x + e)/f

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Fricas [C] Result contains complex when optimal does not.
time = 1.12, size = 20, normalized size = 1.67 \begin {gather*} \frac {2 i \, a c}{f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

2*I*a*c/(f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [C] Result contains complex when optimal does not.
time = 0.08, size = 24, normalized size = 2.00 \begin {gather*} \frac {2 i a c}{f e^{2 i e} e^{2 i f x} + f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e)),x)

[Out]

2*I*a*c/(f*exp(2*I*e)*exp(2*I*f*x) + f)

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Giac [A]
time = 0.43, size = 13, normalized size = 1.08 \begin {gather*} \frac {a c \tan \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

a*c*tan(f*x + e)/f

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Mupad [B]
time = 4.69, size = 12, normalized size = 1.00 \begin {gather*} \frac {a\,c\,\mathrm {tan}\left (e+f\,x\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i),x)

[Out]

(a*c*tan(e + f*x))/f

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