3.1.69 \(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [69]

Optimal. Leaf size=92 \[ \frac {i x}{8 a^3}+\frac {3}{8 a^3 d (1+i \tan (c+d x))}+\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {1}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3627, 3621, 3607, 8} \begin {gather*} \frac {3}{8 a^3 d (1+i \tan (c+d x))}+\frac {i x}{8 a^3}+\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {1}{8 a d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*x)/a^3 + 3/(8*a^3*d*(1 + I*Tan[c + d*x])) + ((I/6)*Tan[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^3) - 1/(8*
a*d*(a + I*a*Tan[c + d*x])^2)

Rule 8

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

Rule 3607

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

Rule 3621

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^m/(2*a^3*f*m)), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*
Simp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a
*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rule 3627

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

Rubi steps

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

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Mathematica [A]
time = 0.28, size = 91, normalized size = 0.99 \begin {gather*} -\frac {\sec ^3(c+d x) (-9 i \cos (c+d x)+2 (-i+6 d x) \cos (3 (c+d x))+27 \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[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-1/96*(Sec[c + d*x]^3*((-9*I)*Cos[c + d*x] + 2*(-I + 6*d*x)*Cos[3*(c + d*x)] + 27*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 = 0.13, size = 74, normalized size = 0.80

method	result	size
risch	\(\frac {i x}{8 a^{3}}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\)	\(60\)
derivativedivides	\(\frac {\frac {i}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {7 i}{8 \left (\tan \left (d x +c \right )-i\right )}+\frac {5}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(74\)
default	\(\frac {\frac {i}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {7 i}{8 \left (\tan \left (d x +c \right )-i\right )}+\frac {5}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(74\)
norman	\(\frac {\frac {5}{12 d a}+\frac {i x}{8 a}+\frac {5 \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}+\frac {3 \left (\tan ^{4}\left (d x +c \right )\right )}{2 d a}+\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}+\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {i x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}-\frac {i \tan \left (d x +c \right )}{8 d a}-\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}-\frac {7 i \left (\tan ^{5}\left (d x +c \right )\right )}{8 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^3*(1/6*I/(tan(d*x+c)-I)^3-7/8*I/(tan(d*x+c)-I)+5/8/(tan(d*x+c)-I)^2+1/16*ln(tan(d*x+c)-I)-1/16*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(tan(d*x+c)^3/(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 = 0.36, size = 54, normalized size = 0.59 \begin {gather*} \frac {{\left (12 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\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(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.24, size = 156, normalized size = 1.70 \begin {gather*} \begin {cases} \frac {\left (4608 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 2304 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 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 (i e^{6 i c} - 3 i e^{4 i c} + 3 i e^{2 i c} - i\right ) e^{- 6 i c}}{8 a^{3}} - \frac {i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {i x}{8 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.07, size = 81, normalized size = 0.88 \begin {gather*} \frac {\frac {6 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac {11 \, \tan \left (d x + c\right )^{3} + 51 i \, \tan \left (d x + c\right )^{2} + 75 \, \tan \left (d x + c\right ) - 29 i}{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(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*log(tan(d*x + c) - I)/a^3 - 6*log(I*tan(d*x + c) - 1)/a^3 - (11*tan(d*x + c)^3 + 51*I*tan(d*x + c)^2 +
75*tan(d*x + c) - 29*I)/(a^3*(tan(d*x + c) - I)^3))/d

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Mupad [B]
time = 3.77, size = 49, normalized size = 0.53 \begin {gather*} \frac {x\,1{}\mathrm {i}}{8\,a^3}+\frac {-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8}+\frac {\mathrm {tan}\left (c+d\,x\right )\,9{}\mathrm {i}}{8}+\frac {5}{12}}{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(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(x*1i)/(8*a^3) + ((tan(c + d*x)*9i)/8 - (7*tan(c + d*x)^2)/8 + 5/12)/(a^3*d*(tan(c + d*x)*1i + 1)^3)

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