3.14 \(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=93 \[ -\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+\frac {2 a^2 \log (\cos (c+d x))}{d}+2 i a^2 x \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3543, 3528, 3525, 3475} \[ -\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+\frac {2 a^2 \log (\cos (c+d x))}{d}+2 i a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3543

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 73, normalized size = 0.78 \[ \frac {a^2 \left (24 i \tan ^{-1}(\tan (c+d x))-3 \tan ^4(c+d x)+8 i \tan ^3(c+d x)+12 \tan ^2(c+d x)-24 i \tan (c+d x)+24 \log (\cos (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*((24*I)*ArcTan[Tan[c + d*x]] + 24*Log[Cos[c + d*x]] - (24*I)*Tan[c + d*x] + 12*Tan[c + d*x]^2 + (8*I)*Tan
[c + d*x]^3 - 3*Tan[c + d*x]^4))/(12*d)

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fricas [B]  time = 0.42, size = 174, normalized size = 1.87 \[ \frac {2 \, {\left (21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{2} + 3 \, {\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

2/3*(21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) + 8*a^2 + 3*(a^2*e^(
8*I*d*x + 8*I*c) + 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) + 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*lo
g(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e
^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 1.75, size = 222, normalized size = 2.39 \[ \frac {2 \, {\left (3 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a^{2}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

2/3*(3*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*
c) + 1) + 18*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) + 3*a^2*l
og(e^(2*I*d*x + 2*I*c) + 1) + 8*a^2)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c
) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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maple [A]  time = 0.02, size = 100, normalized size = 1.08 \[ -\frac {2 i a^{2} \tan \left (d x +c \right )}{d}-\frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {2 i a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {2 i a^{2} \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.58, size = 82, normalized size = 0.88 \[ -\frac {3 \, a^{2} \tan \left (d x + c\right )^{4} - 8 i \, a^{2} \tan \left (d x + c\right )^{3} - 12 \, a^{2} \tan \left (d x + c\right )^{2} - 24 i \, {\left (d x + c\right )} a^{2} + 12 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 24 i \, a^{2} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*(3*a^2*tan(d*x + c)^4 - 8*I*a^2*tan(d*x + c)^3 - 12*a^2*tan(d*x + c)^2 - 24*I*(d*x + c)*a^2 + 12*a^2*log
(tan(d*x + c)^2 + 1) + 24*I*a^2*tan(d*x + c))/d

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mupad [B]  time = 3.80, size = 73, normalized size = 0.78 \[ -\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3\,2{}\mathrm {i}}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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

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sympy [B]  time = 1.41, size = 172, normalized size = 1.85 \[ \frac {2 a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 42 a^{2} e^{6 i c} e^{6 i d x} - 72 a^{2} e^{4 i c} e^{4 i d x} - 58 a^{2} e^{2 i c} e^{2 i d x} - 16 a^{2}}{- 3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+I*a*tan(d*x+c))**2,x)

[Out]

2*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-42*a**2*exp(6*I*c)*exp(6*I*d*x) - 72*a**2*exp(4*I*c)*exp(4*I*d*x)
- 58*a**2*exp(2*I*c)*exp(2*I*d*x) - 16*a**2)/(-3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*d*x) - 1
8*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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