3.955 \(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=223 \[ -\frac {5 i}{32 a^3 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {15 i}{128 a^3 f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {35 x}{128 a^3 c^4}-\frac {5 i}{64 a^3 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {5 i}{128 a^3 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {i}{24 a^3 c f (c-i c \tan (e+f x))^3}+\frac {i}{96 a^3 c f (c+i c \tan (e+f x))^3}-\frac {i}{64 a^3 f (c-i c \tan (e+f x))^4} \]

[Out]

35/128*x/a^3/c^4-1/64*I/a^3/f/(c-I*c*tan(f*x+e))^4-1/24*I/a^3/c/f/(c-I*c*tan(f*x+e))^3+1/96*I/a^3/c/f/(c+I*c*t
an(f*x+e))^3-5/64*I/a^3/f/(c^2-I*c^2*tan(f*x+e))^2+5/128*I/a^3/f/(c^2+I*c^2*tan(f*x+e))^2-5/32*I/a^3/f/(c^4-I*
c^4*tan(f*x+e))+15/128*I/a^3/f/(c^4+I*c^4*tan(f*x+e))

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Rubi [A]  time = 0.21, antiderivative size = 223, 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} \[ -\frac {5 i}{32 a^3 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {15 i}{128 a^3 f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac {5 i}{64 a^3 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {5 i}{128 a^3 f \left (c^2+i c^2 \tan (e+f x)\right )^2}+\frac {35 x}{128 a^3 c^4}-\frac {i}{24 a^3 c f (c-i c \tan (e+f x))^3}+\frac {i}{96 a^3 c f (c+i c \tan (e+f x))^3}-\frac {i}{64 a^3 f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(35*x)/(128*a^3*c^4) - (I/64)/(a^3*f*(c - I*c*Tan[e + f*x])^4) - (I/24)/(a^3*c*f*(c - I*c*Tan[e + f*x])^3) + (
I/96)/(a^3*c*f*(c + I*c*Tan[e + f*x])^3) - ((5*I)/64)/(a^3*f*(c^2 - I*c^2*Tan[e + f*x])^2) + ((5*I)/128)/(a^3*
f*(c^2 + I*c^2*Tan[e + f*x])^2) - ((5*I)/32)/(a^3*f*(c^4 - I*c^4*Tan[e + f*x])) + ((15*I)/128)/(a^3*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))^3 (c-i c \tan (e+f x))^4} \, dx &=\frac {\int \frac {\cos ^6(e+f x)}{c-i c \tan (e+f x)} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \left (\frac {1}{32 c^5 (c-x)^4}+\frac {5}{64 c^6 (c-x)^3}+\frac {15}{128 c^7 (c-x)^2}+\frac {1}{16 c^4 (c+x)^5}+\frac {1}{8 c^5 (c+x)^4}+\frac {5}{32 c^6 (c+x)^3}+\frac {5}{32 c^7 (c+x)^2}+\frac {35}{128 c^7 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=-\frac {i}{64 a^3 f (c-i c \tan (e+f x))^4}-\frac {i}{24 a^3 c f (c-i c \tan (e+f x))^3}+\frac {i}{96 a^3 c f (c+i c \tan (e+f x))^3}-\frac {5 i}{64 a^3 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {5 i}{128 a^3 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {5 i}{32 a^3 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {15 i}{128 a^3 f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {(35 i) \operatorname {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 c^3 f}\\ &=\frac {35 x}{128 a^3 c^4}-\frac {i}{64 a^3 f (c-i c \tan (e+f x))^4}-\frac {i}{24 a^3 c f (c-i c \tan (e+f x))^3}+\frac {i}{96 a^3 c f (c+i c \tan (e+f x))^3}-\frac {5 i}{64 a^3 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {5 i}{128 a^3 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac {5 i}{32 a^3 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {15 i}{128 a^3 f \left (c^4+i c^4 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.97, size = 133, normalized size = 0.60 \[ \frac {(\cos (e+f x)+i \sin (e+f x)) (-840 i f x \sin (e+f x)+420 \sin (e+f x)+378 \sin (3 (e+f x))+70 \sin (5 (e+f x))+7 \sin (7 (e+f x))+420 (2 f x-i) \cos (e+f x)+126 i \cos (3 (e+f x))+14 i \cos (5 (e+f x))+i \cos (7 (e+f x)))}{3072 a^3 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4),x]

[Out]

((Cos[e + f*x] + I*Sin[e + f*x])*(420*(-I + 2*f*x)*Cos[e + f*x] + (126*I)*Cos[3*(e + f*x)] + (14*I)*Cos[5*(e +
f*x)] + I*Cos[7*(e + f*x)] + 420*Sin[e + f*x] - (840*I)*f*x*Sin[e + f*x] + 378*Sin[3*(e + f*x)] + 70*Sin[5*(e
+ f*x)] + 7*Sin[7*(e + f*x)]))/(3072*a^3*c^4*f)

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fricas [A]  time = 0.41, size = 101, normalized size = 0.45 \[ \frac {{\left (840 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (14 i \, f x + 14 i \, e\right )} - 28 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 126 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 420 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 252 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 42 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{3072 \, a^{3} c^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/3072*(840*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(14*I*f*x + 14*I*e) - 28*I*e^(12*I*f*x + 12*I*e) - 126*I*e^(10*I*f
*x + 10*I*e) - 420*I*e^(8*I*f*x + 8*I*e) + 252*I*e^(4*I*f*x + 4*I*e) + 42*I*e^(2*I*f*x + 2*I*e) + 4*I)*e^(-6*I
*f*x - 6*I*e)/(a^3*c^4*f)

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giac [A]  time = 1.94, size = 160, normalized size = 0.72 \[ -\frac {\frac {420 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{4}} - \frac {420 i \, \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3} c^{4}} - \frac {2 \, {\left (385 \, \tan \left (f x + e\right )^{3} - 1335 i \, \tan \left (f x + e\right )^{2} - 1575 \, \tan \left (f x + e\right ) + 641 i\right )}}{a^{3} c^{4} {\left (i \, \tan \left (f x + e\right ) + 1\right )}^{3}} + \frac {875 i \, \tan \left (f x + e\right )^{4} - 3980 \, \tan \left (f x + e\right )^{3} - 6930 i \, \tan \left (f x + e\right )^{2} + 5548 \, \tan \left (f x + e\right ) + 1771 i}{a^{3} c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{3072 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/3072*(420*I*log(tan(f*x + e) - I)/(a^3*c^4) - 420*I*log(I*tan(f*x + e) - 1)/(a^3*c^4) - 2*(385*tan(f*x + e)
^3 - 1335*I*tan(f*x + e)^2 - 1575*tan(f*x + e) + 641*I)/(a^3*c^4*(I*tan(f*x + e) + 1)^3) + (875*I*tan(f*x + e)
^4 - 3980*tan(f*x + e)^3 - 6930*I*tan(f*x + e)^2 + 5548*tan(f*x + e) + 1771*I)/(a^3*c^4*(tan(f*x + e) + I)^4))
/f

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maple [A]  time = 0.30, size = 203, normalized size = 0.91 \[ \frac {5 i}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {35 i \ln \left (\tan \left (f x +e \right )+i\right )}{256 f \,a^{3} c^{4}}-\frac {1}{24 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {5}{32 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )}-\frac {35 i \ln \left (\tan \left (f x +e \right )-i\right )}{256 f \,a^{3} c^{4}}-\frac {5 i}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{96 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {15}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*I/f/a^3/c^4/(tan(f*x+e)+I)^2-1/64*I/f/a^3/c^4/(tan(f*x+e)+I)^4+35/256*I/f/a^3/c^4*ln(tan(f*x+e)+I)-1/24/f
/a^3/c^4/(tan(f*x+e)+I)^3+5/32/f/a^3/c^4/(tan(f*x+e)+I)-35/256*I/f/a^3/c^4*ln(tan(f*x+e)-I)-5/128*I/f/a^3/c^4/
(tan(f*x+e)-I)^2-1/96/f/a^3/c^4/(tan(f*x+e)-I)^3+15/128/f/a^3/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))^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 = 7.10, size = 109, normalized size = 0.49 \[ \frac {35\,x}{128\,a^3\,c^4}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,35{}\mathrm {i}}{128}-\frac {35\,{\mathrm {tan}\left (e+f\,x\right )}^5}{128}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,35{}\mathrm {i}}{48}-\frac {35\,{\mathrm {tan}\left (e+f\,x\right )}^3}{48}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,77{}\mathrm {i}}{128}-\frac {77\,\mathrm {tan}\left (e+f\,x\right )}{128}+\frac {1}{8}{}\mathrm {i}}{a^3\,c^4\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(35*x)/(128*a^3*c^4) - ((tan(e + f*x)^2*77i)/128 - (77*tan(e + f*x))/128 - (35*tan(e + f*x)^3)/48 + (tan(e + f
*x)^4*35i)/48 - (35*tan(e + f*x)^5)/128 + (tan(e + f*x)^6*35i)/128 + 1i/8)/(a^3*c^4*f*(tan(e + f*x)*1i + 1)^3*
(tan(e + f*x) + 1i)^4)

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sympy [A]  time = 0.77, size = 337, normalized size = 1.51 \[ \begin {cases} - \frac {\left (10133099161583616 i a^{18} c^{24} f^{6} e^{20 i e} e^{8 i f x} + 94575592174780416 i a^{18} c^{24} f^{6} e^{18 i e} e^{6 i f x} + 425590164786511872 i a^{18} c^{24} f^{6} e^{16 i e} e^{4 i f x} + 1418633882621706240 i a^{18} c^{24} f^{6} e^{14 i e} e^{2 i f x} - 851180329573023744 i a^{18} c^{24} f^{6} e^{10 i e} e^{- 2 i f x} - 141863388262170624 i a^{18} c^{24} f^{6} e^{8 i e} e^{- 4 i f x} - 13510798882111488 i a^{18} c^{24} f^{6} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{10376293541461622784 a^{21} c^{28} f^{7}} & \text {for}\: 10376293541461622784 a^{21} c^{28} f^{7} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{14 i e} + 7 e^{12 i e} + 21 e^{10 i e} + 35 e^{8 i e} + 35 e^{6 i e} + 21 e^{4 i e} + 7 e^{2 i e} + 1\right ) e^{- 6 i e}}{128 a^{3} c^{4}} - \frac {35}{128 a^{3} c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {35 x}{128 a^{3} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(10133099161583616*I*a**18*c**24*f**6*exp(20*I*e)*exp(8*I*f*x) + 94575592174780416*I*a**18*c**24*f
**6*exp(18*I*e)*exp(6*I*f*x) + 425590164786511872*I*a**18*c**24*f**6*exp(16*I*e)*exp(4*I*f*x) + 14186338826217
06240*I*a**18*c**24*f**6*exp(14*I*e)*exp(2*I*f*x) - 851180329573023744*I*a**18*c**24*f**6*exp(10*I*e)*exp(-2*I
*f*x) - 141863388262170624*I*a**18*c**24*f**6*exp(8*I*e)*exp(-4*I*f*x) - 13510798882111488*I*a**18*c**24*f**6*
exp(6*I*e)*exp(-6*I*f*x))*exp(-12*I*e)/(10376293541461622784*a**21*c**28*f**7), Ne(10376293541461622784*a**21*
c**28*f**7*exp(12*I*e), 0)), (x*((exp(14*I*e) + 7*exp(12*I*e) + 21*exp(10*I*e) + 35*exp(8*I*e) + 35*exp(6*I*e)
+ 21*exp(4*I*e) + 7*exp(2*I*e) + 1)*exp(-6*I*e)/(128*a**3*c**4) - 35/(128*a**3*c**4)), True)) + 35*x/(128*a**
3*c**4)

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