3.2 \(\int \tan ^4(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=83 \[ \frac{i a \tan ^4(c+d x)}{4 d}+\frac{a \tan ^3(c+d x)}{3 d}-\frac{i a \tan ^2(c+d x)}{2 d}-\frac{a \tan (c+d x)}{d}-\frac{i a \log (\cos (c+d x))}{d}+a x \]

[Out]

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

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Rubi [A]  time = 0.0825816, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3528, 3525, 3475} \[ \frac{i a \tan ^4(c+d x)}{4 d}+\frac{a \tan ^3(c+d x)}{3 d}-\frac{i a \tan ^2(c+d x)}{2 d}-\frac{a \tan (c+d x)}{d}-\frac{i a \log (\cos (c+d x))}{d}+a x \]

Antiderivative was successfully verified.

[In]

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

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 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 3475

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

Rubi steps

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

Mathematica [A]  time = 0.202331, size = 81, normalized size = 0.98 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^{-1}(\tan (c+d x))}{d}-\frac{a \tan (c+d x)}{d}-\frac{i a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4*(a + I*a*Tan[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.004, size = 88, normalized size = 1.1 \begin{align*} -{\frac{a\tan \left ( dx+c \right ) }{d}}+{\frac{{\frac{i}{4}}a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{\frac{i}{2}}a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{\frac{i}{2}}a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 2.18875, size = 95, normalized size = 1.14 \begin{align*} -\frac{-3 i \, a \tan \left (d x + c\right )^{4} - 4 \, a \tan \left (d x + c\right )^{3} + 6 i \, a \tan \left (d x + c\right )^{2} - 12 \,{\left (d x + c\right )} a - 6 i \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, a \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(-3*I*a*tan(d*x + c)^4 - 4*a*tan(d*x + c)^3 + 6*I*a*tan(d*x + c)^2 - 12*(d*x + c)*a - 6*I*a*log(tan(d*x
+ c)^2 + 1) + 12*a*tan(d*x + c))/d

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Fricas [B]  time = 2.1451, size = 489, normalized size = 5.89 \begin{align*} \frac{-24 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 36 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 32 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-3 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 12 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 18 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 i \, a}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(-24*I*a*e^(6*I*d*x + 6*I*c) - 36*I*a*e^(4*I*d*x + 4*I*c) - 32*I*a*e^(2*I*d*x + 2*I*c) + (-3*I*a*e^(8*I*d*
x + 8*I*c) - 12*I*a*e^(6*I*d*x + 6*I*c) - 18*I*a*e^(4*I*d*x + 4*I*c) - 12*I*a*e^(2*I*d*x + 2*I*c) - 3*I*a)*log
(e^(2*I*d*x + 2*I*c) + 1) - 8*I*a)/(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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Sympy [B]  time = 6.11669, size = 175, normalized size = 2.11 \begin{align*} - \frac{i a \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{8 i a e^{- 2 i c} e^{6 i d x}}{d} - \frac{12 i a e^{- 4 i c} e^{4 i d x}}{d} - \frac{32 i a e^{- 6 i c} e^{2 i d x}}{3 d} - \frac{8 i a e^{- 8 i c}}{3 d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4*(a+I*a*tan(d*x+c)),x)

[Out]

-I*a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-8*I*a*exp(-2*I*c)*exp(6*I*d*x)/d - 12*I*a*exp(-4*I*c)*exp(4*I*d*x)/
d - 32*I*a*exp(-6*I*c)*exp(2*I*d*x)/(3*d) - 8*I*a*exp(-8*I*c)/(3*d))/(exp(8*I*d*x) + 4*exp(-2*I*c)*exp(6*I*d*x
) + 6*exp(-4*I*c)*exp(4*I*d*x) + 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

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Giac [B]  time = 2.39208, size = 275, normalized size = 3.31 \begin{align*} \frac{-3 i \, a 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 i \, 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 ) - 18 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 ) - 12 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 ) - 24 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 36 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 32 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 i \, a}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(-3*I*a*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) - 18*I*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) - 24*I*a*e^(6*I*d*x + 6*I*c) - 36*I*a*e^(4*I*d*x + 4*I*c) - 32*I*a*e^(2*I*d*x + 2*I*c) - 3*I*a*
log(e^(2*I*d*x + 2*I*c) + 1) - 8*I*a)/(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)