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

Optimal. Leaf size=147 \[ \frac{7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{15}{16 a^4 d (1+i \tan (c+d x))}-\frac{\log (\cos (c+d x))}{a^4 d}+\frac{15 i x}{16 a^4}-\frac{\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.351491, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {3558, 3595, 3589, 3475, 12, 3526, 8} \[ \frac{7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{15}{16 a^4 d (1+i \tan (c+d x))}-\frac{\log (\cos (c+d x))}{a^4 d}+\frac{15 i x}{16 a^4}-\frac{\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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]

Rule 3589

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, 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 3526

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 8

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

Rubi steps

\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=-\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]  time = 0.419962, size = 126, normalized size = 0.86 \[ \frac{\sec ^4(c+d x) (112 \cos (2 (c+d x))+i (96 \sin (2 (c+d x))+120 i d x \sin (4 (c+d x))+\sin (4 (c+d x))+\cos (4 (c+d x)) (128 i \log (\cos (c+d x))+120 d x+i)-128 \sin (4 (c+d x)) \log (\cos (c+d x))+32 i))}{128 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(112*Cos[2*(c + d*x)] + I*(32*I + Cos[4*(c + d*x)]*(I + 120*d*x + (128*I)*Log[Cos[c + d*x]]) +
96*Sin[2*(c + d*x)] + Sin[4*(c + d*x)] + (120*I)*d*x*Sin[4*(c + d*x)] - 128*Log[Cos[c + d*x]]*Sin[4*(c + d*x)
])))/(128*a^4*d*(-I + Tan[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.027, size = 116, normalized size = 0.8 \begin{align*}{\frac{{\frac{3\,i}{4}}}{d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{49\,i}{16}}}{d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{1}{8\,d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}+{\frac{31}{16\,d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{31\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{32\,d{a}^{4}}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{32\,d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*I/d/a^4/(tan(d*x+c)-I)^3-49/16*I/d/a^4/(tan(d*x+c)-I)-1/8/d/a^4/(tan(d*x+c)-I)^4+31/16/d/a^4/(tan(d*x+c)-I
)^2+31/32/d/a^4*ln(tan(d*x+c)-I)+1/32/d/a^4*ln(tan(d*x+c)+I)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 2.49744, size = 273, normalized size = 1.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 5.25534, size = 221, normalized size = 1.5 \begin{align*} \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}\: 131072 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(131072*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)

________________________________________________________________________________________

Giac [A]  time = 3.5986, size = 120, normalized size = 0.82 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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