\(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [58]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {9 x}{4 a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

9/4*x/a^2+2*I*ln(cos(d*x+c))/a^2/d-9/4*tan(d*x+c)/a^2/d+I*tan(d*x+c)^2/a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^3
/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3639, 3676, 3606, 3556} \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {2 i \log (\cos (c+d x))}{a^2 d}+\frac {9 x}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(9*x)/(4*a^2) + ((2*I)*Log[Cos[c + d*x]])/(a^2*d) - (9*Tan[c + d*x])/(4*a^2*d) + (I*Tan[c + d*x]^2)/(a^2*d*(1
+ I*Tan[c + d*x])) - Tan[c + d*x]^3/(4*d*(a + I*a*Tan[c + d*x])^2)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^2(c+d x) (-3 a+5 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan (c+d x) \left (-16 i a^2-18 a^2 \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = \frac {9 x}{4 a^2}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \int \tan (c+d x) \, dx}{a^2} \\ & = \frac {9 x}{4 a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-i (18+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x)))+2 (9+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan (c+d x)+i (-10+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan ^2(c+d x)+8 \tan ^3(c+d x)}{8 a^2 d (-i+\tan (c+d x))^2} \]

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/8*((-I)*(18 + 17*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]]) + 2*(9 + 17*Log[I - Tan[c + d*x]] - Log[I +
Tan[c + d*x]])*Tan[c + d*x] + I*(-10 + 17*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + 8*T
an[c + d*x]^3)/(a^2*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(-\frac {\tan \left (d x +c \right )}{a^{2} d}-\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\)	\(90\)
default	\(-\frac {\tan \left (d x +c \right )}{a^{2} d}-\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\)	\(90\)
risch	\(\frac {17 x}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {4 c}{a^{2} d}-\frac {2 i}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(96\)
norman	\(\frac {\frac {9 x}{4 a}-\frac {\tan ^{5}\left (d x +c \right )}{a d}+\frac {9 x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}+\frac {9 x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}-\frac {3 i}{2 a d}-\frac {9 \tan \left (d x +c \right )}{4 a d}-\frac {15 \left (\tan ^{3}\left (d x +c \right )\right )}{4 a d}-\frac {2 i \left (\tan ^{2}\left (d x +c \right )\right )}{d a}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}\)	\(145\)

[In]

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-tan(d*x+c)/a^2/d-I/d/a^2*ln(1+tan(d*x+c)^2)+9/4/d/a^2*arctan(tan(d*x+c))-1/4*I/d/a^2/(tan(d*x+c)-I)^2-7/4/d/a
^2/(tan(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 4 \, {\left (17 \, d x - 11 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 32 \, {\left (-i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(68*d*x*e^(6*I*d*x + 6*I*c) + 4*(17*d*x - 11*I)*e^(4*I*d*x + 4*I*c) - 32*(-I*e^(6*I*d*x + 6*I*c) - I*e^(4
*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 11*I*e^(2*I*d*x + 2*I*c) + I)/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2
*d*e^(4*I*d*x + 4*I*c))

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 48 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (17 e^{4 i c} - 6 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {17}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \frac {17 x}{4 a^{2}} + \frac {2 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-48*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*
d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((17*exp(4*I*c) - 6*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) - 17/(4*a**2)
), True)) - 2*I/(a**2*d*exp(2*I*c)*exp(2*I*d*x) + a**2*d) + 17*x/(4*a**2) + 2*I*log(exp(2*I*d*x) + exp(-2*I*c)
)/(a**2*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 1.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {34 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {16 \, \tan \left (d x + c\right )}{a^{2}} + \frac {-51 i \, \tan \left (d x + c\right )^{2} - 74 \, \tan \left (d x + c\right ) + 27 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(-2*I*log(tan(d*x + c) + I)/a^2 + 34*I*log(tan(d*x + c) - I)/a^2 + 16*tan(d*x + c)/a^2 + (-51*I*tan(d*x
+ c)^2 - 74*tan(d*x + c) + 27*I)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,17{}\mathrm {i}}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}-\frac {\frac {3}{2\,a^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,7{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(log(tan(c + d*x) + 1i)*1i)/(8*a^2*d) - (log(tan(c + d*x) - 1i)*17i)/(8*a^2*d) - tan(c + d*x)/(a^2*d) - ((tan(
c + d*x)*7i)/(4*a^2) + 3/(2*a^2))/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))