3.1.60 \(\int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [60]

Optimal. Leaf size=59 \[ -\frac {x}{4 a^2}+\frac {3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac {i}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-1/4*x/a^2+3/4*I/a^2/d/(1+I*tan(d*x+c))-1/4*I/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3621, 3607, 8} \begin {gather*} \frac {3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac {x}{4 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/4*x/a^2 + ((3*I)/4)/(a^2*d*(1 + I*Tan[c + d*x])) - (I/4)/(d*(a + I*a*Tan[c + d*x])^2)

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]

Rule 3621

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^m/(2*a^3*f*m)), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*
Simp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a
*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {i}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {a-2 i a \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{2 a^2}\\ &=\frac {3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac {i}{4 d (a+i a \tan (c+d x))^2}-\frac {\int 1 \, dx}{4 a^2}\\ &=-\frac {x}{4 a^2}+\frac {3 i}{4 a^2 d (1+i \tan (c+d x))}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 68, normalized size = 1.15 \begin {gather*} \frac {\sec ^2(c+d x) (-4 i+(i+4 d x) \cos (2 (c+d x))+(1+4 i d x) \sin (2 (c+d x)))}{16 a^2 d (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*(-4*I + (I + 4*d*x)*Cos[2*(c + d*x)] + (1 + (4*I)*d*x)*Sin[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[
c + d*x])^2)

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Maple [A]
time = 0.13, size = 62, normalized size = 1.05

method	result	size
risch	\(-\frac {x}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}\)	\(44\)
derivativedivides	\(\frac {\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}+\frac {3}{4 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(62\)
default	\(\frac {\frac {i}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}+\frac {3}{4 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(62\)
norman	\(\frac {\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{d a}-\frac {x}{4 a}-\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}+\frac {i}{2 d a}+\frac {\tan \left (d x +c \right )}{4 d a}+\frac {3 \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(1/4*I/(tan(d*x+c)-I)^2+1/8*I*ln(tan(d*x+c)-I)+3/4/(tan(d*x+c)-I)-1/8*I*ln(tan(d*x+c)+I))

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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(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,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 = 0.40, size = 43, normalized size = 0.73 \begin {gather*} -\frac {{\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/16*(4*d*x*e^(4*I*d*x + 4*I*c) - 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [A]
time = 0.16, size = 117, normalized size = 1.98 \begin {gather*} \begin {cases} \frac {\left (16 i a^{2} d e^{4 i c} e^{- 2 i d x} - 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- e^{4 i c} + 2 e^{2 i c} - 1\right ) e^{- 4 i c}}{4 a^{2}} + \frac {1}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x}{4 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((16*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((-exp(4*I*c) + 2*exp(2*I*c) - 1)*exp(-4*I*c)/(4*a**2) + 1/(4*a**2)), T
rue)) - x/(4*a**2)

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Giac [A]
time = 0.68, size = 72, normalized size = 1.22 \begin {gather*} -\frac {\frac {2 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {2 i \, \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a^{2}} + \frac {3 i \, \tan \left (d x + c\right )^{2} - 6 \, \tan \left (d x + c\right ) + 5 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*I*log(-I*tan(d*x + c) + 1)/a^2 - 2*I*log(-I*tan(d*x + c) - 1)/a^2 + (3*I*tan(d*x + c)^2 - 6*tan(d*x +
c) + 5*I)/(a^2*(tan(d*x + c) - I)^2))/d

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Mupad [B]
time = 3.97, size = 39, normalized size = 0.66 \begin {gather*} -\frac {x}{4\,a^2}-\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{4}-\frac {1}{2}{}\mathrm {i}}{a^2\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

- x/(4*a^2) - ((3*tan(c + d*x))/4 - 1i/2)/(a^2*d*(tan(c + d*x)*1i + 1)^2)

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