3.2.37 \(\int \frac {1}{(a+i a \tan (c+d x))^3} \, dx\) [137]

Optimal. Leaf size=88 \[ \frac {x}{8 a^3}+\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

1/8*x/a^3+1/6*I/d/(a+I*a*tan(d*x+c))^3+1/8*I/a/d/(a+I*a*tan(d*x+c))^2+1/8*I/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3560, 8} \begin {gather*} \frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x}{8 a^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{6 d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-3),x]

[Out]

x/(8*a^3) + (I/6)/(d*(a + I*a*Tan[c + d*x])^3) + (I/8)/(a*d*(a + I*a*Tan[c + d*x])^2) + (I/8)/(d*(a^3 + I*a^3*
Tan[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(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] && LtQ[n
, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{2 a}\\ &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3}\\ &=\frac {x}{8 a^3}+\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 93, normalized size = 1.06 \begin {gather*} \frac {i \sec ^3(c+d x) (27 i \cos (c+d x)+2 (i+6 d x) \cos (3 (c+d x))-9 \sin (c+d x)+2 \sin (3 (c+d x))+12 i d x \sin (3 (c+d x)))}{96 a^3 d (-i+\tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/96)*Sec[c + d*x]^3*((27*I)*Cos[c + d*x] + 2*(I + 6*d*x)*Cos[3*(c + d*x)] - 9*Sin[c + d*x] + 2*Sin[3*(c + d
*x)] + (12*I)*d*x*Sin[3*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]
time = 0.12, size = 75, normalized size = 0.85

method	result	size
risch	\(\frac {x}{8 a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\)	\(62\)
derivativedivides	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(75\)
default	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(75\)
norman	\(\frac {\frac {x}{8 a}+\frac {\tan ^{5}\left (d x +c \right )}{8 d a}+\frac {3 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}+\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {5 i}{12 d a}+\frac {7 \tan \left (d x +c \right )}{8 d a}+\frac {\tan ^{3}\left (d x +c \right )}{3 d a}-\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d/a^3*(-1/16*I*ln(tan(d*x+c)-I)-1/8*I/(tan(d*x+c)-I)^2-1/6/(tan(d*x+c)-I)^3+1/8/(tan(d*x+c)-I)+1/16*I*ln(tan
(d*x+c)+I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.37, size = 54, normalized size = 0.61 \begin {gather*} \frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(12*d*x*e^(6*I*d*x + 6*I*c) + 18*I*e^(4*I*d*x + 4*I*c) + 9*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I
*c)/(a^3*d)

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Sympy [A]
time = 0.16, size = 155, normalized size = 1.76 \begin {gather*} \begin {cases} \frac {\left (4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{8 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((4608*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 2304*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) + 512*I*a**
6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((exp(6*I*c
) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) - 1/(8*a**3)), True)) + x/(8*a**3)

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Giac [A]
time = 0.67, size = 80, normalized size = 0.91 \begin {gather*} -\frac {\frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {-11 i \, \tan \left (d x + c\right )^{3} - 45 \, \tan \left (d x + c\right )^{2} + 69 i \, \tan \left (d x + c\right ) + 51}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*I*log(tan(d*x + c) - I)/a^3 - 6*I*log(I*tan(d*x + c) - 1)/a^3 + (-11*I*tan(d*x + c)^3 - 45*tan(d*x +
c)^2 + 69*I*tan(d*x + c) + 51)/(a^3*(tan(d*x + c) - I)^3))/d

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Mupad [B]
time = 3.49, size = 50, normalized size = 0.57 \begin {gather*} \frac {x}{8\,a^3}-\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{8}+\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{8}-\frac {5}{12}{}\mathrm {i}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

x/(8*a^3) - ((3*tan(c + d*x))/8 + (tan(c + d*x)^2*1i)/8 - 5i/12)/(a^3*d*(tan(c + d*x)*1i + 1)^3)

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