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

Optimal. Leaf size=91 \[ \frac {\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^3 f}+\frac {5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac {5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac {5 x}{16 a^3 c^3} \]

[Out]

5/16*x/a^3/c^3+5/16*cos(f*x+e)*sin(f*x+e)/a^3/c^3/f+5/24*cos(f*x+e)^3*sin(f*x+e)/a^3/c^3/f+1/6*cos(f*x+e)^5*si
n(f*x+e)/a^3/c^3/f

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Rubi [A]  time = 0.09, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3522, 2635, 8} \[ \frac {\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^3 f}+\frac {5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac {5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac {5 x}{16 a^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(16*a^3*c^3) + (5*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*c^3*f) + (5*Cos[e + f*x]^3*Sin[e + f*x])/(24*a^3*c^
3*f) + (Cos[e + f*x]^5*Sin[e + f*x])/(6*a^3*c^3*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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))^3} \, dx &=\frac {\int \cos ^6(e+f x) \, dx}{a^3 c^3}\\ &=\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^3 f}+\frac {5 \int \cos ^4(e+f x) \, dx}{6 a^3 c^3}\\ &=\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^3 f}+\frac {5 \int \cos ^2(e+f x) \, dx}{8 a^3 c^3}\\ &=\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^3 f}+\frac {5 \int 1 \, dx}{16 a^3 c^3}\\ &=\frac {5 x}{16 a^3 c^3}+\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^3 f}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 0.54 \[ \frac {45 \sin (2 (e+f x))+9 \sin (4 (e+f x))+\sin (6 (e+f x))+60 e+60 f x}{192 a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*e + 60*f*x + 45*Sin[2*(e + f*x)] + 9*Sin[4*(e + f*x)] + Sin[6*(e + f*x)])/(192*a^3*c^3*f)

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fricas [C]  time = 0.43, size = 90, normalized size = 0.99 \[ \frac {{\left (120 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 9 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 45 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 45 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{3} f} \]

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))^3,x, algorithm="fricas")

[Out]

1/384*(120*f*x*e^(6*I*f*x + 6*I*e) - I*e^(12*I*f*x + 12*I*e) - 9*I*e^(10*I*f*x + 10*I*e) - 45*I*e^(8*I*f*x + 8
*I*e) + 45*I*e^(4*I*f*x + 4*I*e) + 9*I*e^(2*I*f*x + 2*I*e) + I)*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)

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giac [A]  time = 1.30, size = 72, normalized size = 0.79 \[ \frac {\frac {15 \, {\left (f x + e\right )}}{a^{3} c^{3}} + \frac {15 \, \tan \left (f x + e\right )^{5} + 40 \, \tan \left (f x + e\right )^{3} + 33 \, \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3} c^{3}}}{48 \, f} \]

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))^3,x, algorithm="giac")

[Out]

1/48*(15*(f*x + e)/(a^3*c^3) + (15*tan(f*x + e)^5 + 40*tan(f*x + e)^3 + 33*tan(f*x + e))/((tan(f*x + e)^2 + 1)
^3*a^3*c^3))/f

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maple [C]  time = 0.36, size = 180, normalized size = 1.98 \[ \frac {i}{16 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{32 f \,a^{3} c^{3}}-\frac {1}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {5}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{32 f \,a^{3} c^{3}}-\frac {i}{16 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {5}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )} \]

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))^3,x)

[Out]

1/16*I/f/a^3/c^3/(tan(f*x+e)+I)^2+5/32*I/f/a^3/c^3*ln(tan(f*x+e)+I)-1/48/f/a^3/c^3/(tan(f*x+e)+I)^3+5/32/f/a^3
/c^3/(tan(f*x+e)+I)-5/32*I/f/a^3/c^3*ln(tan(f*x+e)-I)-1/16*I/f/a^3/c^3/(tan(f*x+e)-I)^2-1/48/f/a^3/c^3/(tan(f*
x+e)-I)^3+5/32/f/a^3/c^3/(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))^3,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 = 4.72, size = 57, normalized size = 0.63 \[ \frac {5\,x}{16\,a^3\,c^3}+\frac {{\cos \left (e+f\,x\right )}^6\,\left (\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^5}{16}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{6}+\frac {11\,\mathrm {tan}\left (e+f\,x\right )}{16}\right )}{a^3\,c^3\,f} \]

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)^3),x)

[Out]

(5*x)/(16*a^3*c^3) + (cos(e + f*x)^6*((11*tan(e + f*x))/16 + (5*tan(e + f*x)^3)/6 + (5*tan(e + f*x)^5)/16))/(a
^3*c^3*f)

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sympy [A]  time = 0.60, size = 298, normalized size = 3.27 \[ \begin {cases} \frac {\left (- 103079215104 i a^{15} c^{15} f^{5} e^{18 i e} e^{6 i f x} - 927712935936 i a^{15} c^{15} f^{5} e^{16 i e} e^{4 i f x} - 4638564679680 i a^{15} c^{15} f^{5} e^{14 i e} e^{2 i f x} + 4638564679680 i a^{15} c^{15} f^{5} e^{10 i e} e^{- 2 i f x} + 927712935936 i a^{15} c^{15} f^{5} e^{8 i e} e^{- 4 i f x} + 103079215104 i a^{15} c^{15} f^{5} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{39582418599936 a^{18} c^{18} f^{6}} & \text {for}\: 39582418599936 a^{18} c^{18} f^{6} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{12 i e} + 6 e^{10 i e} + 15 e^{8 i e} + 20 e^{6 i e} + 15 e^{4 i e} + 6 e^{2 i e} + 1\right ) e^{- 6 i e}}{64 a^{3} c^{3}} - \frac {5}{16 a^{3} c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{3} c^{3}} \]

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))**3,x)

[Out]

Piecewise(((-103079215104*I*a**15*c**15*f**5*exp(18*I*e)*exp(6*I*f*x) - 927712935936*I*a**15*c**15*f**5*exp(16
*I*e)*exp(4*I*f*x) - 4638564679680*I*a**15*c**15*f**5*exp(14*I*e)*exp(2*I*f*x) + 4638564679680*I*a**15*c**15*f
**5*exp(10*I*e)*exp(-2*I*f*x) + 927712935936*I*a**15*c**15*f**5*exp(8*I*e)*exp(-4*I*f*x) + 103079215104*I*a**1
5*c**15*f**5*exp(6*I*e)*exp(-6*I*f*x))*exp(-12*I*e)/(39582418599936*a**18*c**18*f**6), Ne(39582418599936*a**18
*c**18*f**6*exp(12*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(-6*I*e)/(64*a**3*c**3) - 5/(16*a**3*c**3)), True)) + 5*x/(16*a**3*c**3)

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