\(\int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [76]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {15 i x}{16 a^4}-\frac {\log (\cos (c+d x))}{a^4 d}+\frac {15}{16 a^4 d (1+i \tan (c+d x))}+\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3} \]

[Out]

15/16*I*x/a^4-ln(cos(d*x+c))/a^4/d+15/16/a^4/d/(1+I*tan(d*x+c))+7/16*tan(d*x+c)^2/a^4/d/(1+I*tan(d*x+c))^2-1/8
*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^4+1/4*I*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^3

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3639, 3676, 3670, 3556, 12, 3607, 8} \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {15}{16 a^4 d (1+i \tan (c+d x))}-\frac {\log (\cos (c+d x))}{a^4 d}+\frac {15 i x}{16 a^4}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3} \]

[In]

Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((15*I)/16)*x)/a^4 - Log[Cos[c + d*x]]/(a^4*d) + 15/(16*a^4*d*(1 + I*Tan[c + d*x])) + (7*Tan[c + d*x]^2)/(16*
a^4*d*(1 + I*Tan[c + d*x])^2) - Tan[c + d*x]^4/(8*d*(a + I*a*Tan[c + d*x])^4) + ((I/4)*Tan[c + d*x]^3)/(a*d*(a
+ I*a*Tan[c + d*x])^3)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, 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 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 3670

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

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 ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac {\int \frac {\tan ^3(c+d x) (-4 a+8 i a \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = -\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^2(c+d x) \left (-36 i a^2-48 a^2 \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = \frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x) \left (168 a^3-192 i a^3 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = \frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}+\frac {i \int \frac {360 i a^4 \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{192 a^7}+\frac {\int \tan (c+d x) \, dx}{a^4} \\ & = -\frac {\log (\cos (c+d x))}{a^4 d}+\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}-\frac {15 \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {\log (\cos (c+d x))}{a^4 d}+\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}+\frac {15}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {(15 i) \int 1 \, dx}{16 a^4} \\ & = \frac {15 i x}{16 a^4}-\frac {\log (\cos (c+d x))}{a^4 d}+\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3}+\frac {15}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (-16+56 \cos (2 (c+d x))+2 \cos (4 (c+d x)) (-20+31 \log (i-\tan (c+d x))+\log (i+\tan (c+d x)))+48 i \sin (2 (c+d x))+i (-39+62 \log (i-\tan (c+d x))+2 \log (i+\tan (c+d x))) \sin (4 (c+d x)))}{64 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*(-16 + 56*Cos[2*(c + d*x)] + 2*Cos[4*(c + d*x)]*(-20 + 31*Log[I - Tan[c + d*x]] + Log[I + Tan[
c + d*x]]) + (48*I)*Sin[2*(c + d*x)] + I*(-39 + 62*Log[I - Tan[c + d*x]] + 2*Log[I + Tan[c + d*x]])*Sin[4*(c +
d*x)]))/(64*a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {31 i x}{16 a^{4}}+\frac {13 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{4 a^{4} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}+\frac {2 i c}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{4} d}\)	\(107\)
derivativedivides	\(\frac {3 i}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\)	\(115\)
default	\(\frac {3 i}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\)	\(115\)
norman	\(\frac {\frac {5 \left (\tan ^{6}\left (d x +c \right )\right )}{a d}+\frac {15 i x}{16 a}+\frac {7}{4 a d}+\frac {35 \left (\tan ^{4}\left (d x +c \right )\right )}{4 a d}-\frac {15 i \tan \left (d x +c \right )}{16 d a}-\frac {55 i \left (\tan ^{3}\left (d x +c \right )\right )}{16 d a}-\frac {73 i \left (\tan ^{5}\left (d x +c \right )\right )}{16 a d}-\frac {49 i \left (\tan ^{7}\left (d x +c \right )\right )}{16 a d}+\frac {15 i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {45 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {15 i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {15 i x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {13 \left (\tan ^{2}\left (d x +c \right )\right )}{2 a d}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}\)	\(227\)

[In]

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

31/16*I*x/a^4+13/16/a^4/d*exp(-2*I*(d*x+c))-1/4/a^4/d*exp(-4*I*(d*x+c))+1/16/a^4/d*exp(-6*I*(d*x+c))-1/128/a^4
/d*exp(-8*I*(d*x+c))+2*I/a^4/d*c-1/a^4/d*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (248 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 128 \, e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 104 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 32 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{128 \, a^{4} d} \]

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/128*(248*I*d*x*e^(8*I*d*x + 8*I*c) - 128*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 104*e^(6*I*d*x +
6*I*c) - 32*e^(4*I*d*x + 4*I*c) + 8*e^(2*I*d*x + 2*I*c) - 1)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

Sympy [A] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (106496 a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 32768 a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 8192 a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1024 a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{131072 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (31 i e^{8 i c} - 26 i e^{6 i c} + 16 i e^{4 i c} - 6 i e^{2 i c} + i\right ) e^{- 8 i c}}{16 a^{4}} - \frac {31 i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{16 a^{4}} - \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} \]

[In]

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((106496*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 32768*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 8192*a*
*12*d**3*exp(14*I*c)*exp(-6*I*d*x) - 1024*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(131072*a**16*d**
4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((31*I*exp(8*I*c) - 26*I*exp(6*I*c) + 16*I*exp(4*I*c) - 6*I*exp(2*I*c)
+ I)*exp(-8*I*c)/(16*a**4) - 31*I/(16*a**4)), True)) + 31*I*x/(16*a**4) - log(exp(2*I*d*x) + exp(-2*I*c))/(a**
4*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 2.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {12 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {372 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {775 \, \tan \left (d x + c\right )^{4} - 1924 i \, \tan \left (d x + c\right )^{3} - 1866 \, \tan \left (d x + c\right )^{2} + 772 i \, \tan \left (d x + c\right ) + 103}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/384*(12*log(tan(d*x + c) + I)/a^4 + 372*log(tan(d*x + c) - I)/a^4 - (775*tan(d*x + c)^4 - 1924*I*tan(d*x + c
)^3 - 1866*tan(d*x + c)^2 + 772*I*tan(d*x + c) + 103)/(a^4*(tan(d*x + c) - I)^4))/d

Mupad [B] (verification not implemented)

Time = 4.58 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{32\,a^4\,d}+\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,97{}\mathrm {i}}{16\,a^4}+\frac {7}{4\,a^4}-\frac {29\,{\mathrm {tan}\left (c+d\,x\right )}^2}{4\,a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,49{}\mathrm {i}}{16\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]

[In]

int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(31*log(tan(c + d*x) - 1i))/(32*a^4*d) + log(tan(c + d*x) + 1i)/(32*a^4*d) + ((tan(c + d*x)*97i)/(16*a^4) + 7/
(4*a^4) - (29*tan(c + d*x)^2)/(4*a^4) - (tan(c + d*x)^3*49i)/(16*a^4))/(d*(tan(c + d*x)*4i - 6*tan(c + d*x)^2
- tan(c + d*x)^3*4i + tan(c + d*x)^4 + 1))