3.3 \(\int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=67 \[ \frac {i a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^2(c+d x)}{2 d}-\frac {i a \tan (c+d x)}{d}+\frac {a \log (\cos (c+d x))}{d}+i a x \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3528, 3525, 3475} \[ \frac {i a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^2(c+d x)}{2 d}-\frac {i a \tan (c+d x)}{d}+\frac {a \log (\cos (c+d x))}{d}+i a x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]

[Out]

I*a*x + (a*Log[Cos[c + d*x]])/d - (I*a*Tan[c + d*x])/d + (a*Tan[c + d*x]^2)/(2*d) + ((I/3)*a*Tan[c + d*x]^3)/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 ^3(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac {i a \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) (-i a+a \tan (c+d x)) \, dx\\ &=\frac {a \tan ^2(c+d x)}{2 d}+\frac {i a \tan ^3(c+d x)}{3 d}+\int \tan (c+d x) (-a-i a \tan (c+d x)) \, dx\\ &=i a x-\frac {i a \tan (c+d x)}{d}+\frac {a \tan ^2(c+d x)}{2 d}+\frac {i a \tan ^3(c+d x)}{3 d}-a \int \tan (c+d x) \, dx\\ &=i a x+\frac {a \log (\cos (c+d x))}{d}-\frac {i a \tan (c+d x)}{d}+\frac {a \tan ^2(c+d x)}{2 d}+\frac {i a \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 74, normalized size = 1.10 \[ \frac {i a \tan ^{-1}(\tan (c+d x))}{d}+\frac {i a \tan ^3(c+d x)}{3 d}-\frac {i a \tan (c+d x)}{d}+\frac {a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]

[Out]

(I*a*ArcTan[Tan[c + d*x]])/d - (I*a*Tan[c + d*x])/d + ((I/3)*a*Tan[c + d*x]^3)/d + (a*(2*Log[Cos[c + d*x]] + T
an[c + d*x]^2))/(2*d)

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fricas [B]  time = 0.43, size = 120, normalized size = 1.79 \[ \frac {18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c) + 3*(a*e^(6*I*d*x + 6*I*c) + 3*a*e^(4*I*d*x + 4*I*c)
+ 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4
*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 1.38, size = 156, normalized size = 2.33 \[ \frac {3 \, a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) +
1) + 9*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c
) + 3*a*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x
+ 2*I*c) + d)

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maple [A]  time = 0.02, size = 75, normalized size = 1.12 \[ -\frac {i a \tan \left (d x +c \right )}{d}+\frac {i a \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {i a \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.82, size = 59, normalized size = 0.88 \[ -\frac {-2 i \, a \tan \left (d x + c\right )^{3} - 3 \, a \tan \left (d x + c\right )^{2} - 6 i \, {\left (d x + c\right )} a + 3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 i \, a \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.69, size = 51, normalized size = 0.76 \[ -\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{3}+a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i),x)

[Out]

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

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sympy [B]  time = 0.41, size = 136, normalized size = 2.03 \[ \frac {a \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 18 i a e^{4 i c} e^{4 i d x} - 18 i a e^{2 i c} e^{2 i d x} - 8 i a}{- 3 i d e^{6 i c} e^{6 i d x} - 9 i d e^{4 i c} e^{4 i d x} - 9 i d e^{2 i c} e^{2 i d x} - 3 i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+I*a*tan(d*x+c)),x)

[Out]

a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-18*I*a*exp(4*I*c)*exp(4*I*d*x) - 18*I*a*exp(2*I*c)*exp(2*I*d*x) - 8*I*
a)/(-3*I*d*exp(6*I*c)*exp(6*I*d*x) - 9*I*d*exp(4*I*c)*exp(4*I*d*x) - 9*I*d*exp(2*I*c)*exp(2*I*d*x) - 3*I*d)

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