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

Optimal. Leaf size=87 \[ \frac{i c^4 (a-i a \tan (e+f x))^4}{80 a^5 f (a+i a \tan (e+f x))^4}+\frac{i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5} \]

[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)

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Rubi [A]  time = 0.118551, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3522, 3487, 45, 37} \[ \frac{i c^4 (a-i a \tan (e+f x))^4}{80 a^5 f (a+i a \tan (e+f x))^4}+\frac{i c^4 (1-i \tan (e+f x))^4}{10 f (a+i a \tan (e+f x))^5} \]

Antiderivative was successfully verified.

[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 3522

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]))

Rule 3487

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 45

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 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]

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^5} \, dx &=\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 ) \operatorname{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 ) \operatorname{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]  time = 1.77434, size = 53, normalized size = 0.61 \[ \frac{c^4 (9 \cos (e+f x)+i \sin (e+f x)) (\sin (9 (e+f x))+i \cos (9 (e+f x)))}{80 a^5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 66, normalized size = 0.8 \begin{align*}{\frac{{c}^{4}}{f{a}^{5}} \left ({\frac{8}{5\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{5}}}+{\frac{{\frac{i}{2}}}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{3\,i}{ \left ( \tan \left ( fx+e \right ) -i \right ) ^{4}}}-2\, \left ( \tan \left ( fx+e \right ) -i \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*c^4/a^5*(8/5/(tan(f*x+e)-I)^5+1/2*I/(tan(f*x+e)-I)^2-3*I/(tan(f*x+e)-I)^4-2/(tan(f*x+e)-I)^3)

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.37009, size = 104, normalized size = 1.2 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 1.5263, size = 109, normalized size = 1.25 \begin{align*} \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}\: 320 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(320*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), T
rue))

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Giac [B]  time = 1.66166, size = 235, normalized size = 2.7 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)