3.1.68 \(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [68]

3.1.68.1 Optimal result

Integrand size = 24, antiderivative size = 119 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {7 x}{8 a^3}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

output

-7/8*x/a^3-I*ln(cos(d*x+c))/a^3/d-1/6*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^3+ 
3/8*I*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^2+7/8*I/d/(a^3+I*a^3*tan(d*x+c))
 

3.1.68.2 Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) (51 \cos (c+d x)+\cos (3 (c+d x)) (-51+90 \log (i-\tan (c+d x))+6 \log (i+\tan (c+d x)))+i (81 \sin (c+d x)+(-55+90 \log (i-\tan (c+d x))+6 \log (i+\tan (c+d x))) \sin (3 (c+d x))))}{96 a^3 d (-i+\tan (c+d x))^3} \]

input

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]
 

output

-1/96*(Sec[c + d*x]^3*(51*Cos[c + d*x] + Cos[3*(c + d*x)]*(-51 + 90*Log[I 
- Tan[c + d*x]] + 6*Log[I + Tan[c + d*x]]) + I*(81*Sin[c + d*x] + (-55 + 9 
0*Log[I - Tan[c + d*x]] + 6*Log[I + Tan[c + d*x]])*Sin[3*(c + d*x)])))/(a^ 
3*d*(-I + Tan[c + d*x])^3)
 

3.1.68.3 Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4072, 27, 3042, 3956, 4009, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+i a \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {3 \tan ^2(c+d x) (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^2 (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {2 \tan (c+d x) \left (4 \tan (c+d x) a^2+3 i a^2\right )}{i \tan (c+d x) a+a}dx}{4 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x) \left (4 \tan (c+d x) a^2+3 i a^2\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x) \left (4 \tan (c+d x) a^2+3 i a^2\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4072

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {i \int -\frac {7 a^3 \tan (c+d x)}{i \tan (c+d x) a+a}dx}{a}-4 i a \int \tan (c+d x)dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {7 i a^2 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx-4 i a \int \tan (c+d x)dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {7 i a^2 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx-4 i a \int \tan (c+d x)dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {7 i a^2 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx+\frac {4 i a \log (\cos (c+d x))}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {7 i a^2 \left (-\frac {i \int 1dx}{2 a}-\frac {1}{2 d (a+i a \tan (c+d x))}\right )+\frac {4 i a \log (\cos (c+d x))}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {3 i a \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {7 i a^2 \left (-\frac {1}{2 d (a+i a \tan (c+d x))}-\frac {i x}{2 a}\right )+\frac {4 i a \log (\cos (c+d x))}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}\)

input

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]
 

output

-1/6*Tan[c + d*x]^3/(d*(a + I*a*Tan[c + d*x])^3) + ((((3*I)/4)*a*Tan[c + d 
*x]^2)/(d*(a + I*a*Tan[c + d*x])^2) - (((4*I)*a*Log[Cos[c + d*x]])/d + (7* 
I)*a^2*(((-1/2*I)*x)/a - 1/(2*d*(a + I*a*Tan[c + d*x]))))/(2*a^2))/(2*a^2)
 

3.1.68.3.1 Defintions of rubi rules used

rule 24

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4009

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] + Simp[(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 4041

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

rule 4072

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_ 
.)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ 
b)   Int[Tan[e + f*x], x], x] + Simp[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 4078

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), 
 x] + Simp[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]
 

3.1.68.4 Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {15 x}{8 a^{3}}+\frac {11 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}-\frac {2 c}{a^{3} d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\)	\(92\)
derivativedivides	\(\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\)	\(95\)
default	\(\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\)	\(95\)
norman	\(\frac {-\frac {7 x}{8 a}+\frac {17 \left (\tan ^{5}\left (d x +c \right )\right )}{8 a d}-\frac {21 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}-\frac {21 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {7 x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {17 i}{12 a d}+\frac {7 \tan \left (d x +c \right )}{8 a d}+\frac {7 \left (\tan ^{3}\left (d x +c \right )\right )}{3 a d}+\frac {3 i \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {15 i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}\)	\(176\)

input

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

output

-15/8*x/a^3+11/16*I/a^3/d*exp(-2*I*(d*x+c))-5/32*I/a^3/d*exp(-4*I*(d*x+c)) 
+1/48*I/a^3/d*exp(-6*I*(d*x+c))-2/a^3/d*c-I/a^3/d*ln(exp(2*I*(d*x+c))+1)
 

3.1.68.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (180 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 i \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 66 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

input

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

output

-1/96*(180*d*x*e^(6*I*d*x + 6*I*c) + 96*I*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d 
*x + 2*I*c) + 1) - 66*I*e^(4*I*d*x + 4*I*c) + 15*I*e^(2*I*d*x + 2*I*c) - 2 
*I)*e^(-6*I*d*x - 6*I*c)/(a^3*d)
 

3.1.68.6 Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (16896 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 3840 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 15 e^{6 i c} + 11 e^{4 i c} - 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {15}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {15 x}{8 a^{3}} - \frac {i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \]

input

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**3,x)
 

output

Piecewise(((16896*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) - 3840*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*((-15*exp(6* 
I*c) + 11*exp(4*I*c) - 5*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) + 15/(8*a**3 
)), True)) - 15*x/(8*a**3) - I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)
 

3.1.68.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

input

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

output

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

3.1.68.8 Giac [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {90 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {165 i \, \tan \left (d x + c\right )^{3} + 291 \, \tan \left (d x + c\right )^{2} - 171 i \, \tan \left (d x + c\right ) - 29}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

input

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

output

-1/96*(-6*I*log(tan(d*x + c) + I)/a^3 - 90*I*log(tan(d*x + c) - I)/a^3 + ( 
165*I*tan(d*x + c)^3 + 291*tan(d*x + c)^2 - 171*I*tan(d*x + c) - 29)/(a^3* 
(tan(d*x + c) - I)^3))/d
 

3.1.68.9 Mupad [B] (verification not implemented)

Time = 4.86 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {27\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^3}-\frac {17{}\mathrm {i}}{12\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,17{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,15{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d} \]

input

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^3,x)
 

output

(log(tan(c + d*x) - 1i)*15i)/(16*a^3*d) - ((27*tan(c + d*x))/(8*a^3) - 17i 
/(12*a^3) + (tan(c + d*x)^2*17i)/(8*a^3))/(d*(tan(c + d*x)*3i - 3*tan(c + 
d*x)^2 - tan(c + d*x)^3*1i + 1)) + (log(tan(c + d*x) + 1i)*1i)/(16*a^3*d)