3.1.70 \(\int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [70]

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

[Out]

-1/8*x/a^3-1/6*I/d/(a+I*a*tan(d*x+c))^3+3/8*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 = 0.07, antiderivative size = 88, 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 = {3621, 3607, 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 {3 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[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {a-2 i a \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{2 a^2}\\ &=-\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {3 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 {3 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 {3 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 = 0.45, size = 91, normalized size = 1.03 \begin {gather*} \frac {\sec ^3(c+d x) (-9 \cos (c+d x)+2 (1-6 i d x) \cos (3 (c+d x))-3 i \sin (c+d x)-2 i \sin (3 (c+d x))+12 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]^2/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(-9*Cos[c + d*x] + 2*(1 - (6*I)*d*x)*Cos[3*(c + d*x)] - (3*I)*Sin[c + d*x] - (2*I)*Sin[3*(c +
d*x)] + 12*d*x*Sin[3*(c + d*x)]))/(96*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	\(-\frac {x}{8 a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\)	\(62\)
derivativedivides	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {3 i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{8 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(75\)
default	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {3 i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{8 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(75\)
norman	\(\frac {-\frac {x}{8 a}+\frac {i}{12 d a}-\frac {\tan ^{5}\left (d x +c \right )}{8 d a}-\frac {3 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}-\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {\tan \left (d x +c \right )}{8 d a}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2 d a}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(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)-3/8*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(tan(d*x+c)^2/(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 = 0.38, size = 54, normalized size = 0.61 \begin {gather*} -\frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 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(tan(d*x+c)^2/(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) - 6*I*e^(4*I*d*x + 4*I*c) - 3*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 = 0.21, size = 151, normalized size = 1.72 \begin {gather*} \begin {cases} \frac {\left (1536 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 768 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} + e^{4 i c} + 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(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((1536*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 768*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
) + exp(4*I*c) + 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 = 0.96, 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} - 21 i \, \tan \left (d x + c\right ) - 3}{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)^2/(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 - 21*I*tan(d*x + c) - 3)/(a^3*(tan(d*x + c) - I)^3))/d

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Mupad [B]
time = 3.76, size = 49, normalized size = 0.56 \begin {gather*} -\frac {x}{8\,a^3}+\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{8}-\frac {\mathrm {tan}\left (c+d\,x\right )}{8}+\frac {1}{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(tan(c + d*x)^2/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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