3.10.37 \(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx\) [937]

Optimal. Leaf size=114 \[ \frac {5 x}{16 a^3 c^2}+\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f} \]

[Out]

5/16*x/a^3/c^2+1/6*I*cos(f*x+e)^6/a^3/c^2/f+5/16*cos(f*x+e)*sin(f*x+e)/a^3/c^2/f+5/24*cos(f*x+e)^3*sin(f*x+e)/
a^3/c^2/f+1/6*cos(f*x+e)^5*sin(f*x+e)/a^3/c^2/f

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Rubi [A]
time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3603, 3567, 2715, 8} \begin {gather*} \frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^2 f}+\frac {5 x}{16 a^3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]

[Out]

(5*x)/(16*a^3*c^2) + ((I/6)*Cos[e + f*x]^6)/(a^3*c^2*f) + (5*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*c^2*f) + (5*Co
s[e + f*x]^3*Sin[e + f*x])/(24*a^3*c^2*f) + (Cos[e + f*x]^5*Sin[e + f*x])/(6*a^3*c^2*f)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3603

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

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Mathematica [A]
time = 0.97, size = 135, normalized size = 1.18 \begin {gather*} \frac {\sec ^3(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (60 i (i+2 f x) \cos (e+f x)+15 \cos (3 (e+f x))+\cos (5 (e+f x))+60 i \sin (e+f x)-120 f x \sin (e+f x)+45 i \sin (3 (e+f x))+5 i \sin (5 (e+f x)))}{384 a^3 c^2 f (-i+\tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]

[Out]

(Sec[e + f*x]^3*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((60*I)*(I + 2*f*x)*Cos[e + f*x] + 15*Cos[3*(e + f*x)]
+ Cos[5*(e + f*x)] + (60*I)*Sin[e + f*x] - 120*f*x*Sin[e + f*x] + (45*I)*Sin[3*(e + f*x)] + (5*I)*Sin[5*(e +
f*x)]))/(384*a^3*c^2*f*(-I + Tan[e + f*x])^3)

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Maple [A]
time = 0.21, size = 105, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {i}{32 \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{32}+\frac {1}{8 \tan \left (f x +e \right )+8 i}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{32}-\frac {3 i}{32 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{24 \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (f x +e \right )-i\right )}}{f \,a^{3} c^{2}}\)	\(105\)
default	\(\frac {\frac {i}{32 \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{32}+\frac {1}{8 \tan \left (f x +e \right )+8 i}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{32}-\frac {3 i}{32 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{24 \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (f x +e \right )-i\right )}}{f \,a^{3} c^{2}}\)	\(105\)
risch	\(\frac {5 x}{16 a^{3} c^{2}}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )}}{192 a^{3} c^{2} f}+\frac {i \cos \left (4 f x +4 e \right )}{32 a^{3} c^{2} f}+\frac {3 \sin \left (4 f x +4 e \right )}{64 a^{3} c^{2} f}+\frac {5 i \cos \left (2 f x +2 e \right )}{64 a^{3} c^{2} f}+\frac {15 \sin \left (2 f x +2 e \right )}{64 a^{3} c^{2} f}\)	\(114\)
norman	\(\frac {\frac {5 x}{16 a c}+\frac {i}{6 a c f}+\frac {11 \tan \left (f x +e \right )}{16 a c f}+\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{6 a c f}+\frac {5 \left (\tan ^{5}\left (f x +e \right )\right )}{16 a c f}+\frac {15 x \left (\tan ^{2}\left (f x +e \right )\right )}{16 a c}+\frac {15 x \left (\tan ^{4}\left (f x +e \right )\right )}{16 a c}+\frac {5 x \left (\tan ^{6}\left (f x +e \right )\right )}{16 a c}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3} a^{2} c}\)	\(148\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

1/f/a^3/c^2*(1/32*I/(tan(f*x+e)+I)^2+5/32*I*ln(tan(f*x+e)+I)+1/8/(tan(f*x+e)+I)-5/32*I*ln(tan(f*x+e)-I)-3/32*I
/(tan(f*x+e)-I)^2-1/24/(tan(f*x+e)-I)^3+3/16/(tan(f*x+e)-I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^2,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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Fricas [A]
time = 1.21, size = 85, normalized size = 0.75 \begin {gather*} \frac {{\left (120 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 60 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{2} f} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

1/384*(120*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(10*I*f*x + 10*I*e) - 30*I*e^(8*I*f*x + 8*I*e) + 60*I*e^(4*I*f*x +
4*I*e) + 15*I*e^(2*I*f*x + 2*I*e) + 2*I)*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)

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Sympy [A]
time = 0.35, size = 258, normalized size = 2.26 \begin {gather*} \begin {cases} \frac {\left (- 50331648 i a^{12} c^{8} f^{4} e^{16 i e} e^{4 i f x} - 503316480 i a^{12} c^{8} f^{4} e^{14 i e} e^{2 i f x} + 1006632960 i a^{12} c^{8} f^{4} e^{10 i e} e^{- 2 i f x} + 251658240 i a^{12} c^{8} f^{4} e^{8 i e} e^{- 4 i f x} + 33554432 i a^{12} c^{8} f^{4} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{6442450944 a^{15} c^{10} f^{5}} & \text {for}\: a^{15} c^{10} f^{5} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 6 i e}}{32 a^{3} c^{2}} - \frac {5}{16 a^{3} c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{3} c^{2}} \end {gather*}

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))**2,x)

[Out]

Piecewise(((-50331648*I*a**12*c**8*f**4*exp(16*I*e)*exp(4*I*f*x) - 503316480*I*a**12*c**8*f**4*exp(14*I*e)*exp
(2*I*f*x) + 1006632960*I*a**12*c**8*f**4*exp(10*I*e)*exp(-2*I*f*x) + 251658240*I*a**12*c**8*f**4*exp(8*I*e)*ex
p(-4*I*f*x) + 33554432*I*a**12*c**8*f**4*exp(6*I*e)*exp(-6*I*f*x))*exp(-12*I*e)/(6442450944*a**15*c**10*f**5),
Ne(a**15*c**10*f**5*exp(12*I*e), 0)), (x*((exp(10*I*e) + 5*exp(8*I*e) + 10*exp(6*I*e) + 10*exp(4*I*e) + 5*exp
(2*I*e) + 1)*exp(-6*I*e)/(32*a**3*c**2) - 5/(16*a**3*c**2)), True)) + 5*x/(16*a**3*c**2)

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Giac [A]
time = 0.67, size = 137, normalized size = 1.20 \begin {gather*} -\frac {-\frac {30 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{2}} + \frac {30 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{2}} + \frac {3 \, {\left (-15 i \, \tan \left (f x + e\right )^{2} + 38 \, \tan \left (f x + e\right ) + 25 i\right )}}{a^{3} c^{2} {\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{2}} - \frac {55 i \, \tan \left (f x + e\right )^{3} + 201 \, \tan \left (f x + e\right )^{2} - 255 i \, \tan \left (f x + e\right ) - 117}{a^{3} c^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{192 \, f} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

-1/192*(-30*I*log(tan(f*x + e) + I)/(a^3*c^2) + 30*I*log(tan(f*x + e) - I)/(a^3*c^2) + 3*(-15*I*tan(f*x + e)^2
+ 38*tan(f*x + e) + 25*I)/(a^3*c^2*(-I*tan(f*x + e) + 1)^2) - (55*I*tan(f*x + e)^3 + 201*tan(f*x + e)^2 - 255
*I*tan(f*x + e) - 117)/(a^3*c^2*(tan(f*x + e) - I)^3))/f

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Mupad [B]
time = 5.22, size = 88, normalized size = 0.77 \begin {gather*} \frac {5\,x}{16\,a^3\,c^2}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}}{16}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,25{}\mathrm {i}}{48}+\frac {25\,\mathrm {tan}\left (e+f\,x\right )}{48}+\frac {1}{6}{}\mathrm {i}}{a^3\,c^2\,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 )}^2} \end {gather*}

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)^2),x)

[Out]

(5*x)/(16*a^3*c^2) - ((25*tan(e + f*x))/48 + (tan(e + f*x)^2*25i)/48 + (5*tan(e + f*x)^3)/16 + (tan(e + f*x)^4
*5i)/16 + 1i/6)/(a^3*c^2*f*(tan(e + f*x)*1i + 1)^3*(tan(e + f*x) + 1i)^2)

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