3.947 \(\int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=160 \[ -\frac {a^6 \tan (e+f x)}{c^4 f}+\frac {40 i a^6}{f \left (c^4-i c^4 \tan (e+f x)\right )}-\frac {10 i a^6 \log (\cos (e+f x))}{c^4 f}+\frac {10 a^6 x}{c^4}-\frac {40 i a^6}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {80 i a^6}{3 c f (c-i c \tan (e+f x))^3}-\frac {8 i a^6}{f (c-i c \tan (e+f x))^4} \]

[Out]

10*a^6*x/c^4-10*I*a^6*ln(cos(f*x+e))/c^4/f-a^6*tan(f*x+e)/c^4/f-8*I*a^6/f/(c-I*c*tan(f*x+e))^4+80/3*I*a^6/c/f/
(c-I*c*tan(f*x+e))^3-40*I*a^6/f/(c^2-I*c^2*tan(f*x+e))^2+40*I*a^6/f/(c^4-I*c^4*tan(f*x+e))

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Rubi [A]  time = 0.16, antiderivative size = 160, 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} \[ -\frac {a^6 \tan (e+f x)}{c^4 f}+\frac {40 i a^6}{f \left (c^4-i c^4 \tan (e+f x)\right )}-\frac {40 i a^6}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {10 i a^6 \log (\cos (e+f x))}{c^4 f}+\frac {10 a^6 x}{c^4}+\frac {80 i a^6}{3 c f (c-i c \tan (e+f x))^3}-\frac {8 i a^6}{f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(10*a^6*x)/c^4 - ((10*I)*a^6*Log[Cos[e + f*x]])/(c^4*f) - (a^6*Tan[e + f*x])/(c^4*f) - ((8*I)*a^6)/(f*(c - I*c
*Tan[e + f*x])^4) + (((80*I)/3)*a^6)/(c*f*(c - I*c*Tan[e + f*x])^3) - ((40*I)*a^6)/(f*(c^2 - I*c^2*Tan[e + f*x
])^2) + ((40*I)*a^6)/(f*(c^4 - I*c^4*Tan[e + f*x]))

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))^6}{(c-i c \tan (e+f x))^4} \, dx &=\left (a^6 c^6\right ) \int \frac {\sec ^{12}(e+f x)}{(c-i c \tan (e+f x))^{10}} \, dx\\ &=\frac {\left (i a^6\right ) \operatorname {Subst}\left (\int \frac {(c-x)^5}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{c^5 f}\\ &=\frac {\left (i a^6\right ) \operatorname {Subst}\left (\int \left (-1+\frac {32 c^5}{(c+x)^5}-\frac {80 c^4}{(c+x)^4}+\frac {80 c^3}{(c+x)^3}-\frac {40 c^2}{(c+x)^2}+\frac {10 c}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^5 f}\\ &=\frac {10 a^6 x}{c^4}-\frac {10 i a^6 \log (\cos (e+f x))}{c^4 f}-\frac {a^6 \tan (e+f x)}{c^4 f}-\frac {8 i a^6}{f (c-i c \tan (e+f x))^4}+\frac {80 i a^6}{3 c f (c-i c \tan (e+f x))^3}-\frac {40 i a^6}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {40 i a^6}{f \left (c^4-i c^4 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.45, size = 455, normalized size = 2.84 \[ \frac {a^6 \sec (e) \sec (e+f x) (\cos (4 (e+f x))+i \sin (4 (e+f x))) \left (40 \sin (2 e+f x)-60 i f x \sin (2 e+3 f x)+43 \sin (2 e+3 f x)-60 i f x \sin (4 e+3 f x)+55 \sin (4 e+3 f x)-60 i f x \sin (4 e+5 f x)-9 \sin (4 e+5 f x)-60 i f x \sin (6 e+5 f x)+3 \sin (6 e+5 f x)+20 i \cos (2 e+f x)+60 f x \cos (2 e+3 f x)+53 i \cos (2 e+3 f x)+60 f x \cos (4 e+3 f x)+65 i \cos (4 e+3 f x)+60 f x \cos (4 e+5 f x)-15 i \cos (4 e+5 f x)+60 f x \cos (6 e+5 f x)-3 i \cos (6 e+5 f x)-30 i \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-30 i \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-30 i \cos (4 e+5 f x) \log \left (\cos ^2(e+f x)\right )-30 i \cos (6 e+5 f x) \log \left (\cos ^2(e+f x)\right )-30 \sin (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-30 \sin (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-30 \sin (4 e+5 f x) \log \left (\cos ^2(e+f x)\right )-30 \sin (6 e+5 f x) \log \left (\cos ^2(e+f x)\right )+40 \sin (f x)+20 i \cos (f x)\right )}{24 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^6*Sec[e]*Sec[e + f*x]*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)])*((20*I)*Cos[f*x] + (20*I)*Cos[2*e + f*x] + (5
3*I)*Cos[2*e + 3*f*x] + 60*f*x*Cos[2*e + 3*f*x] + (65*I)*Cos[4*e + 3*f*x] + 60*f*x*Cos[4*e + 3*f*x] - (15*I)*C
os[4*e + 5*f*x] + 60*f*x*Cos[4*e + 5*f*x] - (3*I)*Cos[6*e + 5*f*x] + 60*f*x*Cos[6*e + 5*f*x] - (30*I)*Cos[2*e
+ 3*f*x]*Log[Cos[e + f*x]^2] - (30*I)*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] - (30*I)*Cos[4*e + 5*f*x]*Log[Cos[e
+ f*x]^2] - (30*I)*Cos[6*e + 5*f*x]*Log[Cos[e + f*x]^2] + 40*Sin[f*x] + 40*Sin[2*e + f*x] + 43*Sin[2*e + 3*f*
x] - (60*I)*f*x*Sin[2*e + 3*f*x] - 30*Log[Cos[e + f*x]^2]*Sin[2*e + 3*f*x] + 55*Sin[4*e + 3*f*x] - (60*I)*f*x*
Sin[4*e + 3*f*x] - 30*Log[Cos[e + f*x]^2]*Sin[4*e + 3*f*x] - 9*Sin[4*e + 5*f*x] - (60*I)*f*x*Sin[4*e + 5*f*x]
- 30*Log[Cos[e + f*x]^2]*Sin[4*e + 5*f*x] + 3*Sin[6*e + 5*f*x] - (60*I)*f*x*Sin[6*e + 5*f*x] - 30*Log[Cos[e +
f*x]^2]*Sin[6*e + 5*f*x]))/(24*c^4*f)

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fricas [A]  time = 0.46, size = 133, normalized size = 0.83 \[ \frac {-3 i \, a^{6} e^{\left (10 i \, f x + 10 i \, e\right )} + 5 i \, a^{6} e^{\left (8 i \, f x + 8 i \, e\right )} - 10 i \, a^{6} e^{\left (6 i \, f x + 6 i \, e\right )} + 30 i \, a^{6} e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{6} + {\left (-60 i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} - 60 i \, a^{6}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{4} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/6*(-3*I*a^6*e^(10*I*f*x + 10*I*e) + 5*I*a^6*e^(8*I*f*x + 8*I*e) - 10*I*a^6*e^(6*I*f*x + 6*I*e) + 30*I*a^6*e^
(4*I*f*x + 4*I*e) + 48*I*a^6*e^(2*I*f*x + 2*I*e) - 12*I*a^6 + (-60*I*a^6*e^(2*I*f*x + 2*I*e) - 60*I*a^6)*log(e
^(2*I*f*x + 2*I*e) + 1))/(c^4*f*e^(2*I*f*x + 2*I*e) + c^4*f)

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giac [B]  time = 3.64, size = 285, normalized size = 1.78 \[ -\frac {\frac {420 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{4}} - \frac {840 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{4}} + \frac {420 i \, a^{6} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{4}} - \frac {84 \, {\left (5 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 i \, a^{6}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{4}} + \frac {2283 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18936 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 69300 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 141512 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 183106 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 141512 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 69300 i \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 18936 \, a^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2283 i \, a^{6}}{c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{8}}}{42 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/42*(420*I*a^6*log(tan(1/2*f*x + 1/2*e) + 1)/c^4 - 840*I*a^6*log(tan(1/2*f*x + 1/2*e) + I)/c^4 + 420*I*a^6*l
og(tan(1/2*f*x + 1/2*e) - 1)/c^4 - 84*(5*I*a^6*tan(1/2*f*x + 1/2*e)^2 + a^6*tan(1/2*f*x + 1/2*e) - 5*I*a^6)/((
tan(1/2*f*x + 1/2*e)^2 - 1)*c^4) + (2283*I*a^6*tan(1/2*f*x + 1/2*e)^8 - 18936*a^6*tan(1/2*f*x + 1/2*e)^7 - 693
00*I*a^6*tan(1/2*f*x + 1/2*e)^6 + 141512*a^6*tan(1/2*f*x + 1/2*e)^5 + 183106*I*a^6*tan(1/2*f*x + 1/2*e)^4 - 14
1512*a^6*tan(1/2*f*x + 1/2*e)^3 - 69300*I*a^6*tan(1/2*f*x + 1/2*e)^2 + 18936*a^6*tan(1/2*f*x + 1/2*e) + 2283*I
*a^6)/(c^4*(tan(1/2*f*x + 1/2*e) + I)^8))/f

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maple [A]  time = 0.18, size = 131, normalized size = 0.82 \[ -\frac {a^{6} \tan \left (f x +e \right )}{c^{4} f}-\frac {40 a^{6}}{f \,c^{4} \left (\tan \left (f x +e \right )+i\right )}+\frac {10 i a^{6} \ln \left (\tan \left (f x +e \right )+i\right )}{f \,c^{4}}+\frac {80 a^{6}}{3 f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {40 i a^{6}}{f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {8 i a^{6}}{f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x)

[Out]

-a^6*tan(f*x+e)/c^4/f-40/f*a^6/c^4/(tan(f*x+e)+I)+10*I/f*a^6/c^4*ln(tan(f*x+e)+I)+80/3/f*a^6/c^4/(tan(f*x+e)+I
)^3+40*I/f*a^6/c^4/(tan(f*x+e)+I)^2-8*I/f*a^6/c^4/(tan(f*x+e)+I)^4

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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))^6/(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 = 6.95, size = 170, normalized size = 1.06 \[ \frac {a^6\,\left (10\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )-60\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+10\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4-76\,{\mathrm {tan}\left (e+f\,x\right )}^2-4\,{\mathrm {tan}\left (e+f\,x\right )}^4+\frac {56}{3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,197{}\mathrm {i}}{3}-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,40{}\mathrm {i}+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3\,40{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^3\,34{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^4\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^6*(10*log(tan(e + f*x) + 1i) - (tan(e + f*x)*197i)/3 - log(tan(e + f*x) + 1i)*tan(e + f*x)*40i - 60*log(tan
(e + f*x) + 1i)*tan(e + f*x)^2 + log(tan(e + f*x) + 1i)*tan(e + f*x)^3*40i + 10*log(tan(e + f*x) + 1i)*tan(e +
f*x)^4 - 76*tan(e + f*x)^2 + tan(e + f*x)^3*34i - 4*tan(e + f*x)^4 + tan(e + f*x)^5*1i + 56/3)*1i)/(c^4*f*(ta
n(e + f*x)*1i - 1)^4)

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sympy [A]  time = 0.89, size = 246, normalized size = 1.54 \[ \frac {2 i a^{6}}{- c^{4} f e^{2 i e} e^{2 i f x} - c^{4} f} - \frac {10 i a^{6} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{4} f} + \begin {cases} \frac {- 3 i a^{6} c^{12} f^{3} e^{8 i e} e^{8 i f x} + 8 i a^{6} c^{12} f^{3} e^{6 i e} e^{6 i f x} - 18 i a^{6} c^{12} f^{3} e^{4 i e} e^{4 i f x} + 48 i a^{6} c^{12} f^{3} e^{2 i e} e^{2 i f x}}{6 c^{16} f^{4}} & \text {for}\: 6 c^{16} f^{4} \neq 0 \\\frac {x \left (4 a^{6} e^{8 i e} - 8 a^{6} e^{6 i e} + 12 a^{6} e^{4 i e} - 16 a^{6} e^{2 i e}\right )}{c^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**6/(c-I*c*tan(f*x+e))**4,x)

[Out]

2*I*a**6/(-c**4*f*exp(2*I*e)*exp(2*I*f*x) - c**4*f) - 10*I*a**6*log(exp(2*I*f*x) + exp(-2*I*e))/(c**4*f) + Pie
cewise(((-3*I*a**6*c**12*f**3*exp(8*I*e)*exp(8*I*f*x) + 8*I*a**6*c**12*f**3*exp(6*I*e)*exp(6*I*f*x) - 18*I*a**
6*c**12*f**3*exp(4*I*e)*exp(4*I*f*x) + 48*I*a**6*c**12*f**3*exp(2*I*e)*exp(2*I*f*x))/(6*c**16*f**4), Ne(6*c**1
6*f**4, 0)), (x*(4*a**6*exp(8*I*e) - 8*a**6*exp(6*I*e) + 12*a**6*exp(4*I*e) - 16*a**6*exp(2*I*e))/c**4, True))

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