3.1.16 \(\int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx\) [16]

Optimal. Leaf size=62 \[ -2 i a^2 x-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {i a^2 \tan (c+d x)}{d}+\frac {(a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3608, 3558, 3556} \begin {gather*} \frac {i a^2 \tan (c+d x)}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-2 i a^2 x+\frac {(a+i a \tan (c+d x))^2}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*I)*a^2*x - (2*a^2*Log[Cos[c + d*x]])/d + (I*a^2*Tan[c + d*x])/d + (a + I*a*Tan[c + d*x])^2/(2*d)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3558

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[b^2*(Tan[c + d*x]/d), x]) /; FreeQ[{a, b, c, d}, x]

Rule 3608

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

Rubi steps

\begin {align*} \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx &=\frac {(a+i a \tan (c+d x))^2}{2 d}-i \int (a+i a \tan (c+d x))^2 \, dx\\ &=-2 i a^2 x+\frac {i a^2 \tan (c+d x)}{d}+\frac {(a+i a \tan (c+d x))^2}{2 d}+\left (2 a^2\right ) \int \tan (c+d x) \, dx\\ &=-2 i a^2 x-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {i a^2 \tan (c+d x)}{d}+\frac {(a+i a \tan (c+d x))^2}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 51, normalized size = 0.82 \begin {gather*} \frac {a^2 \left (-4 i \text {ArcTan}(\tan (c+d x))-4 \log (\cos (c+d x))+4 i \tan (c+d x)-\tan ^2(c+d x)\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((-4*I)*ArcTan[Tan[c + d*x]] - 4*Log[Cos[c + d*x]] + (4*I)*Tan[c + d*x] - Tan[c + d*x]^2))/(2*d)

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Maple [A]
time = 0.04, size = 49, normalized size = 0.79

method	result	size
derivativedivides	\(\frac {a^{2} \left (2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (1+\tan ^{2}\left (d x +c \right )\right )-2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(49\)
default	\(\frac {a^{2} \left (2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (1+\tan ^{2}\left (d x +c \right )\right )-2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(49\)
norman	\(-\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-2 i a^{2} x +\frac {2 i a^{2} \tan \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(58\)
risch	\(\frac {4 i a^{2} c}{d}-\frac {2 a^{2} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 55, normalized size = 0.89 \begin {gather*} -\frac {a^{2} \tan \left (d x + c\right )^{2} + 4 i \, {\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 4 i \, a^{2} \tan \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.45, size = 93, normalized size = 1.50 \begin {gather*} -\frac {2 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a^{2} + {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.17, size = 88, normalized size = 1.42 \begin {gather*} - \frac {2 a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 6 a^{2} e^{2 i c} e^{2 i d x} - 4 a^{2}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-6*a**2*exp(2*I*c)*exp(2*I*d*x) - 4*a**2)/(d*exp(4*I*c)*exp(4*I*d
*x) + 2*d*exp(2*I*c)*exp(2*I*d*x) + d)

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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. 116 vs. \(2 (54) = 108\).
time = 0.61, size = 116, normalized size = 1.87 \begin {gather*} -\frac {2 \, {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 \, a^{2}\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.74, size = 40, normalized size = 0.65 \begin {gather*} \frac {a^2\,\left (4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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