3.954 \(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=193 \[ -\frac {5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {15 x}{64 a^2 c^4}-\frac {3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac {i}{32 a^2 f (c-i c \tan (e+f x))^4} \]

[Out]

15/64*x/a^2/c^4-1/32*I/a^2/f/(c-I*c*tan(f*x+e))^4-1/16*I/a^2/c/f/(c-I*c*tan(f*x+e))^3-3/32*I/a^2/f/(c^2-I*c^2*
tan(f*x+e))^2+1/64*I/a^2/f/(c^2+I*c^2*tan(f*x+e))^2-5/32*I/a^2/f/(c^4-I*c^4*tan(f*x+e))+5/64*I/a^2/f/(c^4+I*c^
4*tan(f*x+e))

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Rubi [A]  time = 0.19, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3522, 3487, 44, 206} \[ -\frac {5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac {3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}+\frac {15 x}{64 a^2 c^4}-\frac {i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac {i}{32 a^2 f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(15*x)/(64*a^2*c^4) - (I/32)/(a^2*f*(c - I*c*Tan[e + f*x])^4) - (I/16)/(a^2*c*f*(c - I*c*Tan[e + f*x])^3) - ((
3*I)/32)/(a^2*f*(c^2 - I*c^2*Tan[e + f*x])^2) + (I/64)/(a^2*f*(c^2 + I*c^2*Tan[e + f*x])^2) - ((5*I)/32)/(a^2*
f*(c^4 - I*c^4*Tan[e + f*x])) + ((5*I)/64)/(a^2*f*(c^4 + I*c^4*Tan[e + f*x]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx &=\frac {\int \frac {\cos ^4(e+f x)}{(c-i c \tan (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {\left (i c^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{32 c^5 (c-x)^3}+\frac {5}{64 c^6 (c-x)^2}+\frac {1}{8 c^3 (c+x)^5}+\frac {3}{16 c^4 (c+x)^4}+\frac {3}{16 c^5 (c+x)^3}+\frac {5}{32 c^6 (c+x)^2}+\frac {15}{64 c^6 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=-\frac {i}{32 a^2 f (c-i c \tan (e+f x))^4}-\frac {i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {(15 i) \operatorname {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 c^3 f}\\ &=\frac {15 x}{64 a^2 c^4}-\frac {i}{32 a^2 f (c-i c \tan (e+f x))^4}-\frac {i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 139, normalized size = 0.72 \[ \frac {\sec ^2(e+f x) (\sin (4 (e+f x))-i \cos (4 (e+f x))) (-120 f x \sin (2 (e+f x))-30 i \sin (2 (e+f x))-32 i \sin (4 (e+f x))-3 i \sin (6 (e+f x))+(-30-120 i f x) \cos (2 (e+f x))+16 \cos (4 (e+f x))+\cos (6 (e+f x))-80)}{512 a^2 c^4 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(Sec[e + f*x]^2*((-I)*Cos[4*(e + f*x)] + Sin[4*(e + f*x)])*(-80 + (-30 - (120*I)*f*x)*Cos[2*(e + f*x)] + 16*Co
s[4*(e + f*x)] + Cos[6*(e + f*x)] - (30*I)*Sin[2*(e + f*x)] - 120*f*x*Sin[2*(e + f*x)] - (32*I)*Sin[4*(e + f*x
)] - (3*I)*Sin[6*(e + f*x)]))/(512*a^2*c^4*f*(-I + Tan[e + f*x])^2)

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fricas [A]  time = 0.52, size = 90, normalized size = 0.47 \[ \frac {{\left (120 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 80 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 24 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{512 \, a^{2} c^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/512*(120*f*x*e^(4*I*f*x + 4*I*e) - I*e^(12*I*f*x + 12*I*e) - 8*I*e^(10*I*f*x + 10*I*e) - 30*I*e^(8*I*f*x + 8
*I*e) - 80*I*e^(6*I*f*x + 6*I*e) + 24*I*e^(2*I*f*x + 2*I*e) + 2*I)*e^(-4*I*f*x - 4*I*e)/(a^2*c^4*f)

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giac [A]  time = 1.20, size = 149, normalized size = 0.77 \[ -\frac {-\frac {60 i \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2} c^{4}} + \frac {60 i \, \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2} c^{4}} + \frac {2 \, {\left (-45 i \, \tan \left (f x + e\right )^{2} - 110 \, \tan \left (f x + e\right ) + 69 i\right )}}{a^{2} c^{4} {\left (\tan \left (f x + e\right ) - i\right )}^{2}} + \frac {125 i \, \tan \left (f x + e\right )^{4} - 580 \, \tan \left (f x + e\right )^{3} - 1038 i \, \tan \left (f x + e\right )^{2} + 868 \, \tan \left (f x + e\right ) + 301 i}{a^{2} c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{512 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/512*(-60*I*log(-I*tan(f*x + e) + 1)/(a^2*c^4) + 60*I*log(-I*tan(f*x + e) - 1)/(a^2*c^4) + 2*(-45*I*tan(f*x
+ e)^2 - 110*tan(f*x + e) + 69*I)/(a^2*c^4*(tan(f*x + e) - I)^2) + (125*I*tan(f*x + e)^4 - 580*tan(f*x + e)^3
- 1038*I*tan(f*x + e)^2 + 868*tan(f*x + e) + 301*I)/(a^2*c^4*(tan(f*x + e) + I)^4))/f

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maple [A]  time = 0.28, size = 181, normalized size = 0.94 \[ \frac {3 i}{32 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{32 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {15 i \ln \left (\tan \left (f x +e \right )+i\right )}{128 f \,a^{2} c^{4}}-\frac {1}{16 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {5}{32 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )+i\right )}-\frac {15 i \ln \left (\tan \left (f x +e \right )-i\right )}{128 f \,a^{2} c^{4}}-\frac {i}{64 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {5}{64 f \,a^{2} c^{4} \left (\tan \left (f x +e \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*I/f/a^2/c^4/(tan(f*x+e)+I)^2-1/32*I/f/a^2/c^4/(tan(f*x+e)+I)^4+15/128*I/f/a^2/c^4*ln(tan(f*x+e)+I)-1/16/f
/a^2/c^4/(tan(f*x+e)+I)^3+5/32/f/a^2/c^4/(tan(f*x+e)+I)-15/128*I/f/a^2/c^4*ln(tan(f*x+e)-I)-1/64*I/f/a^2/c^4/(
tan(f*x+e)-I)^2+5/64/f/a^2/c^4/(tan(f*x+e)-I)

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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(1/(a+I*a*tan(f*x+e))^2/(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 = 6.05, size = 98, normalized size = 0.51 \[ \frac {15\,x}{64\,a^2\,c^4}-\frac {\frac {15\,{\mathrm {tan}\left (e+f\,x\right )}^5}{64}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,15{}\mathrm {i}}{32}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{32}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,25{}\mathrm {i}}{32}-\frac {17\,\mathrm {tan}\left (e+f\,x\right )}{64}+\frac {1}{4}{}\mathrm {i}}{a^2\,c^4\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*x)/(64*a^2*c^4) - ((tan(e + f*x)^2*25i)/32 - (17*tan(e + f*x))/64 + (5*tan(e + f*x)^3)/32 + (tan(e + f*x)^
4*15i)/32 + (15*tan(e + f*x)^5)/64 + 1i/4)/(a^2*c^4*f*(tan(e + f*x)*1i + 1)^2*(tan(e + f*x) + 1i)^4)

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sympy [A]  time = 0.66, size = 298, normalized size = 1.54 \[ \begin {cases} \frac {\left (- 8589934592 i a^{10} c^{20} f^{5} e^{14 i e} e^{8 i f x} - 68719476736 i a^{10} c^{20} f^{5} e^{12 i e} e^{6 i f x} - 257698037760 i a^{10} c^{20} f^{5} e^{10 i e} e^{4 i f x} - 687194767360 i a^{10} c^{20} f^{5} e^{8 i e} e^{2 i f x} + 206158430208 i a^{10} c^{20} f^{5} e^{4 i e} e^{- 2 i f x} + 17179869184 i a^{10} c^{20} f^{5} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{4398046511104 a^{12} c^{24} f^{6}} & \text {for}\: 4398046511104 a^{12} c^{24} f^{6} e^{6 i e} \neq 0 \\x \left (\frac {\left (e^{12 i e} + 6 e^{10 i e} + 15 e^{8 i e} + 20 e^{6 i e} + 15 e^{4 i e} + 6 e^{2 i e} + 1\right ) e^{- 4 i e}}{64 a^{2} c^{4}} - \frac {15}{64 a^{2} c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {15 x}{64 a^{2} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-8589934592*I*a**10*c**20*f**5*exp(14*I*e)*exp(8*I*f*x) - 68719476736*I*a**10*c**20*f**5*exp(12*I*
e)*exp(6*I*f*x) - 257698037760*I*a**10*c**20*f**5*exp(10*I*e)*exp(4*I*f*x) - 687194767360*I*a**10*c**20*f**5*e
xp(8*I*e)*exp(2*I*f*x) + 206158430208*I*a**10*c**20*f**5*exp(4*I*e)*exp(-2*I*f*x) + 17179869184*I*a**10*c**20*
f**5*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(4398046511104*a**12*c**24*f**6), Ne(4398046511104*a**12*c**24*f**6
*exp(6*I*e), 0)), (x*((exp(12*I*e) + 6*exp(10*I*e) + 15*exp(8*I*e) + 20*exp(6*I*e) + 15*exp(4*I*e) + 6*exp(2*I
*e) + 1)*exp(-4*I*e)/(64*a**2*c**4) - 15/(64*a**2*c**4)), True)) + 15*x/(64*a**2*c**4)

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