3.912 \(\int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=87 \[ \frac {i c^3}{2 f \left (a^2+i a^2 \tan (e+f x)\right )^2}-\frac {4 i c^3}{3 a f (a+i a \tan (e+f x))^3}+\frac {i c^3}{f (a+i a \tan (e+f x))^4} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ \frac {i c^3}{2 f \left (a^2+i a^2 \tan (e+f x)\right )^2}-\frac {4 i c^3}{3 a f (a+i a \tan (e+f x))^3}+\frac {i c^3}{f (a+i a \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*c^3)/(f*(a + I*a*Tan[e + f*x])^4) - (((4*I)/3)*c^3)/(a*f*(a + I*a*Tan[e + f*x])^3) + ((I/2)*c^3)/(f*(a^2 +
I*a^2*Tan[e + f*x])^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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]))

Rubi steps

\begin {align*} \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^4} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(a+i a \tan (e+f x))^7} \, dx\\ &=-\frac {\left (i c^3\right ) \operatorname {Subst}\left (\int \frac {(a-x)^2}{(a+x)^5} \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac {\left (i c^3\right ) \operatorname {Subst}\left (\int \left (\frac {4 a^2}{(a+x)^5}-\frac {4 a}{(a+x)^4}+\frac {1}{(a+x)^3}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i c^3}{f (a+i a \tan (e+f x))^4}-\frac {4 i c^3}{3 a f (a+i a \tan (e+f x))^3}+\frac {i c^3}{2 f \left (a^2+i a^2 \tan (e+f x)\right )^2}\\ \end {align*}

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Mathematica [A]  time = 2.32, size = 64, normalized size = 0.74 \[ -\frac {c^3 (\tan (e+f x)-7 i) \sec ^3(e+f x) (\cos (3 (e+f x))-i \sin (3 (e+f x)))}{48 a^4 f (\tan (e+f x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(c^3*Sec[e + f*x]^3*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)])*(-7*I + Tan[e + f*x]))/(a^4*f*(-I + Tan[e +
f*x])^4)

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fricas [A]  time = 0.42, size = 37, normalized size = 0.43 \[ \frac {{\left (4 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, c^{3}\right )} e^{\left (-8 i \, f x - 8 i \, e\right )}}{48 \, a^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(4*I*c^3*e^(2*I*f*x + 2*I*e) + 3*I*c^3)*e^(-8*I*f*x - 8*I*e)/(a^4*f)

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giac [A]  time = 1.95, size = 140, normalized size = 1.61 \[ -\frac {2 \, {\left (3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 3 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 17 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 10 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 17 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*c^3*tan(1/2*f*x + 1/2*e)^7 - 3*I*c^3*tan(1/2*f*x + 1/2*e)^6 - 17*c^3*tan(1/2*f*x + 1/2*e)^5 + 10*I*c^3
*tan(1/2*f*x + 1/2*e)^4 + 17*c^3*tan(1/2*f*x + 1/2*e)^3 - 3*I*c^3*tan(1/2*f*x + 1/2*e)^2 - 3*c^3*tan(1/2*f*x +
1/2*e))/(a^4*f*(tan(1/2*f*x + 1/2*e) - I)^8)

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maple [A]  time = 0.22, size = 53, normalized size = 0.61 \[ \frac {c^{3} \left (-\frac {i}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {i}{\left (\tan \left (f x +e \right )-i\right )^{4}}+\frac {4}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}\right )}{f \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^4,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 4.85, size = 77, normalized size = 0.89 \[ \frac {c^3\,\left (3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}{6\,a^4\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+4\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,6{}\mathrm {i}-4\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*(tan(e + f*x)*2i + 3*tan(e + f*x)^2 - 1))/(6*a^4*f*(4*tan(e + f*x)^3 - tan(e + f*x)^2*6i - 4*tan(e + f*x)
+ tan(e + f*x)^4*1i + 1i))

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sympy [A]  time = 0.51, size = 109, normalized size = 1.25 \[ \begin {cases} \frac {\left (16 i a^{4} c^{3} f e^{8 i e} e^{- 6 i f x} + 12 i a^{4} c^{3} f e^{6 i e} e^{- 8 i f x}\right ) e^{- 14 i e}}{192 a^{8} f^{2}} & \text {for}\: 192 a^{8} f^{2} e^{14 i e} \neq 0 \\\frac {x \left (c^{3} e^{2 i e} + c^{3}\right ) e^{- 8 i e}}{2 a^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**3/(a+I*a*tan(f*x+e))**4,x)

[Out]

Piecewise(((16*I*a**4*c**3*f*exp(8*I*e)*exp(-6*I*f*x) + 12*I*a**4*c**3*f*exp(6*I*e)*exp(-8*I*f*x))*exp(-14*I*e
)/(192*a**8*f**2), Ne(192*a**8*f**2*exp(14*I*e), 0)), (x*(c**3*exp(2*I*e) + c**3)*exp(-8*I*e)/(2*a**4), True))

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