3.4 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=49 \[ \frac {i a \tan ^2(c+d x)}{2 d}+\frac {a \tan (c+d x)}{d}+\frac {i a \log (\cos (c+d x))}{d}-a x \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, 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 = {3528, 3525, 3475} \[ \frac {i a \tan ^2(c+d x)}{2 d}+\frac {a \tan (c+d x)}{d}+\frac {i a \log (\cos (c+d x))}{d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3475

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

Rule 3525

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

Rule 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 53, normalized size = 1.08 \[ -\frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \tan (c+d x)}{d}+\frac {i a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 85, normalized size = 1.73 \[ \frac {4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.05, size = 107, normalized size = 2.18 \[ \frac {i \, a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 59, normalized size = 1.20 \[ \frac {i a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \tan \left (d x +c \right )}{d}-\frac {i a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 48, normalized size = 0.98 \[ -\frac {-i \, a \tan \left (d x + c\right )^{2} + 2 \, {\left (d x + c\right )} a + i \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.75, size = 39, normalized size = 0.80 \[ \frac {a\,\left (2\,\mathrm {tan}\left (c+d\,x\right )-\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.32, size = 88, normalized size = 1.80 \[ \frac {i a \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 4 a e^{2 i c} e^{2 i d x} - 2 a}{i d e^{4 i c} e^{4 i d x} + 2 i d e^{2 i c} e^{2 i d x} + i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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