3.1 \(\int \tan ^5(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=102 \[ \frac{i a \tan ^5(c+d x)}{5 d}+\frac{a \tan ^4(c+d x)}{4 d}-\frac{i a \tan ^3(c+d x)}{3 d}-\frac{a \tan ^2(c+d x)}{2 d}+\frac{i a \tan (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d}-i a x \]

[Out]

(-I)*a*x - (a*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 + (a*Tan[c + d*x]^4)/(4*d) + ((I/5)*a*Tan[c + d*x]^5)/d

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Rubi [A]  time = 0.108247, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3528, 3525, 3475} \[ \frac{i a \tan ^5(c+d x)}{5 d}+\frac{a \tan ^4(c+d x)}{4 d}-\frac{i a \tan ^3(c+d x)}{3 d}-\frac{a \tan ^2(c+d x)}{2 d}+\frac{i a \tan (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d}-i a x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5*(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 + (a*Tan[c + d*x]^4)/(4*d) + ((I/5)*a*Tan[c + d*x]^5)/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 ^5(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac{i a \tan ^5(c+d x)}{5 d}+\int \tan ^4(c+d x) (-i a+a \tan (c+d x)) \, dx\\ &=\frac{a \tan ^4(c+d x)}{4 d}+\frac{i a \tan ^5(c+d x)}{5 d}+\int \tan ^3(c+d x) (-a-i a \tan (c+d x)) \, dx\\ &=-\frac{i a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^4(c+d x)}{4 d}+\frac{i a \tan ^5(c+d x)}{5 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}+\frac{a \tan ^4(c+d x)}{4 d}+\frac{i a \tan ^5(c+d x)}{5 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}+\frac{a \tan ^4(c+d x)}{4 d}+\frac{i a \tan ^5(c+d x)}{5 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}+\frac{a \tan ^4(c+d x)}{4 d}+\frac{i a \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.29265, size = 104, normalized size = 1.02 \[ \frac{i a \tan ^5(c+d x)}{5 d}-\frac{i a \tan ^3(c+d x)}{3 d}-\frac{i a \tan ^{-1}(\tan (c+d x))}{d}+\frac{i a \tan (c+d x)}{d}-\frac{a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5*(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 + ((I/5)*a*Tan[c + d*x]^5)
/d - (a*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d)

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Maple [A]  time = 0.007, size = 104, normalized size = 1. \begin{align*}{\frac{ia\tan \left ( dx+c \right ) }{d}}+{\frac{{\frac{i}{5}}a \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{d}}+{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{\frac{i}{3}}a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d}}-{\frac{ia\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)^5*(a+I*a*tan(d*x+c)),x)

[Out]

I*a*tan(d*x+c)/d+1/5*I*a*tan(d*x+c)^5/d+1/4*a*tan(d*x+c)^4/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 = 2.31828, size = 109, normalized size = 1.07 \begin{align*} -\frac{-12 i \, a \tan \left (d x + c\right )^{5} - 15 \, a \tan \left (d x + c\right )^{4} + 20 i \, a \tan \left (d x + c\right )^{3} + 30 \, a \tan \left (d x + c\right )^{2} + 60 i \,{\left (d x + c\right )} a - 30 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 i \, a \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(-12*I*a*tan(d*x + c)^5 - 15*a*tan(d*x + c)^4 + 20*I*a*tan(d*x + c)^3 + 30*a*tan(d*x + c)^2 + 60*I*(d*x
+ c)*a - 30*a*log(tan(d*x + c)^2 + 1) - 60*I*a*tan(d*x + c))/d

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Fricas [B]  time = 2.12487, size = 583, normalized size = 5.72 \begin{align*} -\frac{150 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 300 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 400 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 200 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \,{\left (a e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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 ) + 46 \, a}{15 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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)^5*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/15*(150*a*e^(8*I*d*x + 8*I*c) + 300*a*e^(6*I*d*x + 6*I*c) + 400*a*e^(4*I*d*x + 4*I*c) + 200*a*e^(2*I*d*x +
2*I*c) + 15*(a*e^(10*I*d*x + 10*I*c) + 5*a*e^(8*I*d*x + 8*I*c) + 10*a*e^(6*I*d*x + 6*I*c) + 10*a*e^(4*I*d*x +
4*I*c) + 5*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) + 1) + 46*a)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8
*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [B]  time = 8.98525, size = 206, normalized size = 2.02 \begin{align*} - \frac{a \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{10 a e^{- 2 i c} e^{8 i d x}}{d} - \frac{20 a e^{- 4 i c} e^{6 i d x}}{d} - \frac{80 a e^{- 6 i c} e^{4 i d x}}{3 d} - \frac{40 a e^{- 8 i c} e^{2 i d x}}{3 d} - \frac{46 a e^{- 10 i c}}{15 d}}{e^{10 i d x} + 5 e^{- 2 i c} e^{8 i d x} + 10 e^{- 4 i c} e^{6 i d x} + 10 e^{- 6 i c} e^{4 i d x} + 5 e^{- 8 i c} e^{2 i d x} + e^{- 10 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5*(a+I*a*tan(d*x+c)),x)

[Out]

-a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-10*a*exp(-2*I*c)*exp(8*I*d*x)/d - 20*a*exp(-4*I*c)*exp(6*I*d*x)/d - 8
0*a*exp(-6*I*c)*exp(4*I*d*x)/(3*d) - 40*a*exp(-8*I*c)*exp(2*I*d*x)/(3*d) - 46*a*exp(-10*I*c)/(15*d))/(exp(10*I
*d*x) + 5*exp(-2*I*c)*exp(8*I*d*x) + 10*exp(-4*I*c)*exp(6*I*d*x) + 10*exp(-6*I*c)*exp(4*I*d*x) + 5*exp(-8*I*c)
*exp(2*I*d*x) + exp(-10*I*c))

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Giac [B]  time = 3.59196, size = 340, normalized size = 3.33 \begin{align*} -\frac{15 \, a e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 75 \, 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 ) + 150 \, 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 ) + 150 \, 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 ) + 75 \, 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 ) + 150 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 300 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 400 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 200 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 46 \, a}{15 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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)^5*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/15*(15*a*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 75*a*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) + 150*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 150*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 75*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 150*a*e^(8*I*d*x + 8*I*c) + 300*a*e^(6*
I*d*x + 6*I*c) + 400*a*e^(4*I*d*x + 4*I*c) + 200*a*e^(2*I*d*x + 2*I*c) + 15*a*log(e^(2*I*d*x + 2*I*c) + 1) + 4
6*a)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c)
+ 5*d*e^(2*I*d*x + 2*I*c) + d)