3.913 \(\int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^5} \, dx\)

Optimal. Leaf size=90 \[ \frac {i c^3}{3 a^2 f (a+i a \tan (e+f x))^3}-\frac {i a^3 c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )^4}+\frac {4 i c^3}{5 f (a+i a \tan (e+f x))^5} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 90, 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} \[ \frac {i c^3}{3 a^2 f (a+i a \tan (e+f x))^3}-\frac {i a^3 c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )^4}+\frac {4 i c^3}{5 f (a+i a \tan (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^5,x]

[Out]

(((4*I)/5)*c^3)/(f*(a + I*a*Tan[e + f*x])^5) + ((I/3)*c^3)/(a^2*f*(a + I*a*Tan[e + f*x])^3) - (I*a^3*c^3)/(f*(
a^2 + I*a^2*Tan[e + f*x])^4)

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))^5} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(a+i a \tan (e+f x))^8} \, dx\\ &=-\frac {\left (i c^3\right ) \operatorname {Subst}\left (\int \frac {(a-x)^2}{(a+x)^6} \, 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)^6}-\frac {4 a}{(a+x)^5}+\frac {1}{(a+x)^4}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=\frac {4 i c^3}{5 f (a+i a \tan (e+f x))^5}-\frac {i c^3}{a f (a+i a \tan (e+f x))^4}+\frac {i c^3}{3 a^2 f (a+i a \tan (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 2.93, size = 58, normalized size = 0.64 \[ \frac {c^3 (4 i \sin (2 (e+f x))+16 \cos (2 (e+f x))+15) (\sin (8 (e+f x))+i \cos (8 (e+f x)))}{240 a^5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^5,x]

[Out]

(c^3*(15 + 16*Cos[2*(e + f*x)] + (4*I)*Sin[2*(e + f*x)])*(I*Cos[8*(e + f*x)] + Sin[8*(e + f*x)]))/(240*a^5*f)

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fricas [A]  time = 0.44, size = 51, normalized size = 0.57 \[ \frac {{\left (10 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, c^{3}\right )} e^{\left (-10 i \, f x - 10 i \, e\right )}}{240 \, a^{5} f} \]

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))^5,x, algorithm="fricas")

[Out]

1/240*(10*I*c^3*e^(4*I*f*x + 4*I*e) + 15*I*c^3*e^(2*I*f*x + 2*I*e) + 6*I*c^3)*e^(-10*I*f*x - 10*I*e)/(a^5*f)

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giac [B]  time = 2.73, size = 174, normalized size = 1.93 \[ -\frac {2 \, {\left (15 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 30 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 140 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 170 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 282 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 170 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 140 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{15 \, a^{5} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{10}} \]

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))^5,x, algorithm="giac")

[Out]

-2/15*(15*c^3*tan(1/2*f*x + 1/2*e)^9 - 30*I*c^3*tan(1/2*f*x + 1/2*e)^8 - 140*c^3*tan(1/2*f*x + 1/2*e)^7 + 170*
I*c^3*tan(1/2*f*x + 1/2*e)^6 + 282*c^3*tan(1/2*f*x + 1/2*e)^5 - 170*I*c^3*tan(1/2*f*x + 1/2*e)^4 - 140*c^3*tan
(1/2*f*x + 1/2*e)^3 + 30*I*c^3*tan(1/2*f*x + 1/2*e)^2 + 15*c^3*tan(1/2*f*x + 1/2*e))/(a^5*f*(tan(1/2*f*x + 1/2
*e) - I)^10)

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maple [A]  time = 0.24, size = 52, normalized size = 0.58 \[ \frac {c^{3} \left (\frac {4}{5 \left (\tan \left (f x +e \right )-i\right )^{5}}-\frac {i}{\left (\tan \left (f x +e \right )-i\right )^{4}}-\frac {1}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}\right )}{f \,a^{5}} \]

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))^5,x)

[Out]

1/f*c^3/a^5*(4/5/(tan(f*x+e)-I)^5-I/(tan(f*x+e)-I)^4-1/3/(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))^5,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 = 4.86, size = 88, normalized size = 0.98 \[ \frac {c^3\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^2\,5{}\mathrm {i}+5\,\mathrm {tan}\left (e+f\,x\right )+2{}\mathrm {i}\right )}{15\,a^5\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (e+f\,x\right )}^4-{\mathrm {tan}\left (e+f\,x\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,5{}\mathrm {i}+1\right )} \]

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)^5,x)

[Out]

(c^3*(5*tan(e + f*x) - tan(e + f*x)^2*5i + 2i))/(15*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))

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sympy [A]  time = 0.59, size = 156, normalized size = 1.73 \[ \begin {cases} - \frac {\left (- 640 i a^{10} c^{3} f^{2} e^{18 i e} e^{- 6 i f x} - 960 i a^{10} c^{3} f^{2} e^{16 i e} e^{- 8 i f x} - 384 i a^{10} c^{3} f^{2} e^{14 i e} e^{- 10 i f x}\right ) e^{- 24 i e}}{15360 a^{15} f^{3}} & \text {for}\: 15360 a^{15} f^{3} e^{24 i e} \neq 0 \\\frac {x \left (c^{3} e^{4 i e} + 2 c^{3} e^{2 i e} + c^{3}\right ) e^{- 10 i e}}{4 a^{5}} & \text {otherwise} \end {cases} \]

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))**5,x)

[Out]

Piecewise((-(-640*I*a**10*c**3*f**2*exp(18*I*e)*exp(-6*I*f*x) - 960*I*a**10*c**3*f**2*exp(16*I*e)*exp(-8*I*f*x
) - 384*I*a**10*c**3*f**2*exp(14*I*e)*exp(-10*I*f*x))*exp(-24*I*e)/(15360*a**15*f**3), Ne(15360*a**15*f**3*exp
(24*I*e), 0)), (x*(c**3*exp(4*I*e) + 2*c**3*exp(2*I*e) + c**3)*exp(-10*I*e)/(4*a**5), True))

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