\(\int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx\) [923]

Optimal result

Integrand size = 31, antiderivative size = 87 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\frac {i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5}+\frac {i c^4 (a-i a \tan (e+f x))^4}{80 a^5 f (a+i a \tan (e+f x))^4} \]

[Out]

1/10*I*c^4*(1-I*tan(f*x+e))^4/f/(a+I*a*tan(f*x+e))^5+1/80*I*c^4*(a-I*a*tan(f*x+e))^4/a^5/f/(a+I*a*tan(f*x+e))^
4

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3603, 3568, 47, 37} \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\frac {i c^4 (a-i a \tan (e+f x))^4}{80 a^5 f (a+i a \tan (e+f x))^4}+\frac {i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5} \]

[In]

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

[Out]

((I/10)*c^4*(1 - I*Tan[e + f*x])^4)/(f*(a + I*a*Tan[e + f*x])^5) + ((I/80)*c^4*(a - I*a*Tan[e + f*x])^4)/(a^5*
f*(a + I*a*Tan[e + f*x])^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = \left (a^4 c^4\right ) \int \frac {\sec ^8(e+f x)}{(a+i a \tan (e+f x))^9} \, dx \\ & = -\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {(a-x)^3}{(a+x)^6} \, dx,x,i a \tan (e+f x)\right )}{a^3 f} \\ & = \frac {i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5}-\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {(a-x)^3}{(a+x)^5} \, dx,x,i a \tan (e+f x)\right )}{10 a^4 f} \\ & = \frac {i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5}+\frac {i c^4 (a-i a \tan (e+f x))^4}{80 a^5 f (a+i a \tan (e+f x))^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.56 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\frac {c^4 (9+i \tan (e+f x)) (i+\tan (e+f x))^4}{80 a^5 f (-i+\tan (e+f x))^5} \]

[In]

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

[Out]

(c^4*(9 + I*Tan[e + f*x])*(I + Tan[e + f*x])^4)/(80*a^5*f*(-I + Tan[e + f*x])^5)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51

method	result	size
risch	\(\frac {i c^{4} {\mathrm e}^{-8 i \left (f x +e \right )}}{16 a^{5} f}+\frac {i c^{4} {\mathrm e}^{-10 i \left (f x +e \right )}}{20 a^{5} f}\)	\(44\)
derivativedivides	\(\frac {c^{4} \left (\frac {8}{5 \left (\tan \left (f x +e \right )-i\right )^{5}}+\frac {i}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {2}{\left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {3 i}{\left (\tan \left (f x +e \right )-i\right )^{4}}\right )}{f \,a^{5}}\)	\(66\)
default	\(\frac {c^{4} \left (\frac {8}{5 \left (\tan \left (f x +e \right )-i\right )^{5}}+\frac {i}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {2}{\left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {3 i}{\left (\tan \left (f x +e \right )-i\right )^{4}}\right )}{f \,a^{5}}\)	\(66\)
norman	\(\frac {\frac {c^{4} \tan \left (f x +e \right )}{a f}-\frac {4 i c^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{a f}+\frac {13 i c^{4} \left (\tan ^{4}\left (f x +e \right )\right )}{a f}+\frac {i c^{4}}{10 a f}-\frac {9 c^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}+\frac {63 c^{4} \left (\tan ^{5}\left (f x +e \right )\right )}{5 a f}-\frac {3 c^{4} \left (\tan ^{7}\left (f x +e \right )\right )}{a f}+\frac {i c^{4} \left (\tan ^{8}\left (f x +e \right )\right )}{2 a f}-\frac {8 i c^{4} \left (\tan ^{6}\left (f x +e \right )\right )}{a f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{5} a^{4}}\)	\(183\)

[In]

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

[Out]

1/16*I*c^4/a^5/f*exp(-8*I*(f*x+e))+1/20*I*c^4/a^5/f*exp(-10*I*(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.43 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\frac {{\left (5 i \, c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{4}\right )} e^{\left (-10 i \, f x - 10 i \, e\right )}}{80 \, a^{5} f} \]

[In]

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

[Out]

1/80*(5*I*c^4*e^(2*I*f*x + 2*I*e) + 4*I*c^4)*e^(-10*I*f*x - 10*I*e)/(a^5*f)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\begin {cases} \frac {\left (20 i a^{5} c^{4} f e^{10 i e} e^{- 8 i f x} + 16 i a^{5} c^{4} f e^{8 i e} e^{- 10 i f x}\right ) e^{- 18 i e}}{320 a^{10} f^{2}} & \text {for}\: a^{10} f^{2} e^{18 i e} \neq 0 \\\frac {x \left (c^{4} e^{2 i e} + c^{4}\right ) e^{- 10 i e}}{2 a^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate((c-I*c*tan(f*x+e))**4/(a+I*a*tan(f*x+e))**5,x)

[Out]

Piecewise(((20*I*a**5*c**4*f*exp(10*I*e)*exp(-8*I*f*x) + 16*I*a**5*c**4*f*exp(8*I*e)*exp(-10*I*f*x))*exp(-18*I
*e)/(320*a**10*f**2), Ne(a**10*f**2*exp(18*I*e), 0)), (x*(c**4*exp(2*I*e) + c**4)*exp(-10*I*e)/(2*a**5), True)
)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (71) = 142\).

Time = 1.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.89 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=-\frac {2 \, {\left (5 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 5 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 50 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 35 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 98 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 50 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{5 \, a^{5} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{10}} \]

[In]

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

[Out]

-2/5*(5*c^4*tan(1/2*f*x + 1/2*e)^9 - 5*I*c^4*tan(1/2*f*x + 1/2*e)^8 - 50*c^4*tan(1/2*f*x + 1/2*e)^7 + 35*I*c^4
*tan(1/2*f*x + 1/2*e)^6 + 98*c^4*tan(1/2*f*x + 1/2*e)^5 - 35*I*c^4*tan(1/2*f*x + 1/2*e)^4 - 50*c^4*tan(1/2*f*x
+ 1/2*e)^3 + 5*I*c^4*tan(1/2*f*x + 1/2*e)^2 + 5*c^4*tan(1/2*f*x + 1/2*e))/(a^5*f*(tan(1/2*f*x + 1/2*e) - I)^1
0)

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx=\frac {c^4\,\left (-5\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,5{}\mathrm {i}+5\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}{10\,a^5\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (e+f\,x\right )}^4-{\mathrm {tan}\left (e+f\,x\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,5{}\mathrm {i}+1\right )} \]

[In]

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

[Out]

(c^4*(5*tan(e + f*x) - tan(e + f*x)^2*5i - 5*tan(e + f*x)^3 + 1i))/(10*a^5*f*(tan(e + f*x)*5i - 10*tan(e + f*x
)^2 - tan(e + f*x)^3*10i + 5*tan(e + f*x)^4 + tan(e + f*x)^5*1i + 1))