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

Optimal. Leaf size=117 \[ -\frac{8 a^5 \tan (c+d x)}{d}+\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac{16 i a^5 \log (\cos (c+d x))}{d}+16 a^5 x+\frac{i a (a+i a \tan (c+d x))^4}{4 d} \]

[Out]

16*a^5*x - ((16*I)*a^5*Log[Cos[c + d*x]])/d - (8*a^5*Tan[c + d*x])/d + (((2*I)/3)*a^2*(a + I*a*Tan[c + d*x])^3
)/d + ((I/4)*a*(a + I*a*Tan[c + d*x])^4)/d + ((2*I)*a*(a^2 + I*a^2*Tan[c + d*x])^2)/d

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Rubi [A]  time = 0.0658658, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3478, 3477, 3475} \[ -\frac{8 a^5 \tan (c+d x)}{d}+\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac{16 i a^5 \log (\cos (c+d x))}{d}+16 a^5 x+\frac{i a (a+i a \tan (c+d x))^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^5,x]

[Out]

16*a^5*x - ((16*I)*a^5*Log[Cos[c + d*x]])/d - (8*a^5*Tan[c + d*x])/d + (((2*I)/3)*a^2*(a + I*a*Tan[c + d*x])^3
)/d + ((I/4)*a*(a + I*a*Tan[c + d*x])^4)/d + ((2*I)*a*(a^2 + I*a^2*Tan[c + d*x])^2)/d

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

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 (a+i a \tan (c+d x))^5 \, dx &=\frac{i a (a+i a \tan (c+d x))^4}{4 d}+(2 a) \int (a+i a \tan (c+d x))^4 \, dx\\ &=\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{i a (a+i a \tan (c+d x))^4}{4 d}+\left (4 a^2\right ) \int (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{i a (a+i a \tan (c+d x))^4}{4 d}+\left (8 a^3\right ) \int (a+i a \tan (c+d x))^2 \, dx\\ &=16 a^5 x-\frac{8 a^5 \tan (c+d x)}{d}+\frac{2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{i a (a+i a \tan (c+d x))^4}{4 d}+\left (16 i a^5\right ) \int \tan (c+d x) \, dx\\ &=16 a^5 x-\frac{16 i a^5 \log (\cos (c+d x))}{d}-\frac{8 a^5 \tan (c+d x)}{d}+\frac{2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac{2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac{i a (a+i a \tan (c+d x))^4}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.5629, size = 228, normalized size = 1.95 \[ \frac{a^5 \sec (c) \sec ^4(c+d x) \left (-70 \sin (c+2 d x)+30 \sin (3 c+2 d x)-25 \sin (3 c+4 d x)+48 d x \cos (3 c+2 d x)-18 i \cos (3 c+2 d x)+12 d x \cos (3 c+4 d x)+12 d x \cos (5 c+4 d x)-24 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+6 \cos (c+2 d x) \left (-4 i \log \left (\cos ^2(c+d x)\right )+8 d x-3 i\right )+\cos (c) \left (-36 i \log \left (\cos ^2(c+d x)\right )+72 d x-33 i\right )-6 i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-6 i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+75 \sin (c)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Sec[c]*Sec[c + d*x]^4*((-18*I)*Cos[3*c + 2*d*x] + 48*d*x*Cos[3*c + 2*d*x] + 12*d*x*Cos[3*c + 4*d*x] + 12*
d*x*Cos[5*c + 4*d*x] + 6*Cos[c + 2*d*x]*(-3*I + 8*d*x - (4*I)*Log[Cos[c + d*x]^2]) + Cos[c]*(-33*I + 72*d*x -
(36*I)*Log[Cos[c + d*x]^2]) - (24*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (6*I)*Cos[3*c + 4*d*x]*Log[Cos[c +
d*x]^2] - (6*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 75*Sin[c] - 70*Sin[c + 2*d*x] + 30*Sin[3*c + 2*d*x] -
25*Sin[3*c + 4*d*x]))/(12*d)

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Maple [A]  time = 0.005, size = 101, normalized size = 0.9 \begin{align*} -15\,{\frac{{a}^{5}\tan \left ( dx+c \right ) }{d}}+{\frac{{\frac{i}{4}}{a}^{5} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{5\,{a}^{5} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{\frac{11\,i}{2}}{a}^{5} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{8\,i{a}^{5}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+16\,{\frac{{a}^{5}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*a^5*tan(d*x+c)/d+1/4*I/d*a^5*tan(d*x+c)^4+5/3/d*a^5*tan(d*x+c)^3-11/2*I/d*a^5*tan(d*x+c)^2+8*I/d*a^5*ln(1+
tan(d*x+c)^2)+16/d*a^5*arctan(tan(d*x+c))

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Maxima [A]  time = 1.85101, size = 223, normalized size = 1.91 \begin{align*} a^{5} x + \frac{5 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{5}}{3 \, d} + \frac{10 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{5}}{d} + \frac{i \, a^{5}{\left (\frac{4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{4 \, d} + \frac{5 i \, a^{5}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac{5 i \, a^{5} \log \left (\sec \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^5*x + 5/3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^5/d + 10*(d*x + c - tan(d*x + c))*a^5/d + 1/4*I*
a^5*((4*sin(d*x + c)^2 - 3)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 2*log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*(
1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*log(sec(d*x + c))/d

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Fricas [A]  time = 1.06098, size = 527, normalized size = 4.5 \begin{align*} \frac{-192 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 432 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 352 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 100 i \, a^{5} +{\left (-48 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} - 192 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 288 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 192 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 48 i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-192*I*a^5*e^(6*I*d*x + 6*I*c) - 432*I*a^5*e^(4*I*d*x + 4*I*c) - 352*I*a^5*e^(2*I*d*x + 2*I*c) - 100*I*a^
5 + (-48*I*a^5*e^(8*I*d*x + 8*I*c) - 192*I*a^5*e^(6*I*d*x + 6*I*c) - 288*I*a^5*e^(4*I*d*x + 4*I*c) - 192*I*a^5
*e^(2*I*d*x + 2*I*c) - 48*I*a^5)*log(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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Sympy [A]  time = 3.30864, size = 185, normalized size = 1.58 \begin{align*} - \frac{16 i a^{5} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{64 i a^{5} e^{- 2 i c} e^{6 i d x}}{d} - \frac{144 i a^{5} e^{- 4 i c} e^{4 i d x}}{d} - \frac{352 i a^{5} e^{- 6 i c} e^{2 i d x}}{3 d} - \frac{100 i a^{5} 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((a+I*a*tan(d*x+c))**5,x)

[Out]

-16*I*a**5*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-64*I*a**5*exp(-2*I*c)*exp(6*I*d*x)/d - 144*I*a**5*exp(-4*I*c)
*exp(4*I*d*x)/d - 352*I*a**5*exp(-6*I*c)*exp(2*I*d*x)/(3*d) - 100*I*a**5*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 = 1.15091, size = 300, normalized size = 2.56 \begin{align*} \frac{-48 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 192 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 288 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 192 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 192 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 432 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 352 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 48 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 100 i \, a^{5}}{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((a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

1/3*(-48*I*a^5*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 192*I*a^5*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) - 288*I*a^5*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 192*I*a^5*e^(2*I*d*x + 2*I*c)*lo
g(e^(2*I*d*x + 2*I*c) + 1) - 192*I*a^5*e^(6*I*d*x + 6*I*c) - 432*I*a^5*e^(4*I*d*x + 4*I*c) - 352*I*a^5*e^(2*I*
d*x + 2*I*c) - 48*I*a^5*log(e^(2*I*d*x + 2*I*c) + 1) - 100*I*a^5)/(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)