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

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

[Out]

-I*a^3/f/(c-I*c*tan(f*x+e))^4+4/3*I*a^3/c/f/(c-I*c*tan(f*x+e))^3-1/2*I*a^3/f/(c^2-I*c^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} \[ -\frac {i a^3}{2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {4 i a^3}{3 c f (c-i c \tan (e+f x))^3}-\frac {i a^3}{f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 2.28, size = 53, normalized size = 0.61 \[ \frac {a^3 (7 \cos (e+f x)-i \sin (e+f x)) (\sin (7 (e+f x))-i \cos (7 (e+f x)))}{48 c^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 37, normalized size = 0.43 \[ \frac {-3 i \, a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} - 4 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{48 \, c^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-3*I*a^3*e^(8*I*f*x + 8*I*e) - 4*I*a^3*e^(6*I*f*x + 6*I*e))/(c^4*f)

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giac [A]  time = 2.17, size = 140, normalized size = 1.61 \[ -\frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 17 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 17 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*a^3*tan(1/2*f*x + 1/2*e)^7 + 3*I*a^3*tan(1/2*f*x + 1/2*e)^6 - 17*a^3*tan(1/2*f*x + 1/2*e)^5 - 10*I*a^3
*tan(1/2*f*x + 1/2*e)^4 + 17*a^3*tan(1/2*f*x + 1/2*e)^3 + 3*I*a^3*tan(1/2*f*x + 1/2*e)^2 - 3*a^3*tan(1/2*f*x +
1/2*e))/(c^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 \[ \frac {a^{3} \left (\frac {4}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {i}{\left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {i}{2 \left (\tan \left (f x +e \right )+i\right )^{2}}\right )}{f \,c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+I*a*tan(f*x+e))^3/(c-I*c*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 = 4.77, size = 75, normalized size = 0.86 \[ \frac {a^3\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}{6\,c^4\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (e+f\,x\right )}^2-\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 97, normalized size = 1.11 \[ \begin {cases} \frac {- 12 i a^{3} c^{4} f e^{8 i e} e^{8 i f x} - 16 i a^{3} c^{4} f e^{6 i e} e^{6 i f x}}{192 c^{8} f^{2}} & \text {for}\: 192 c^{8} f^{2} \neq 0 \\\frac {x \left (a^{3} e^{8 i e} + a^{3} e^{6 i e}\right )}{2 c^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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