3.10.45 \(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx\) [945]

Optimal. Leaf size=161 \[ \frac {5 x}{16 a^2 c^3}-\frac {i}{24 a^2 f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 c f (c-i c \tan (e+f x))^2}+\frac {i}{32 a^2 c f (c+i c \tan (e+f x))^2}-\frac {3 i}{16 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{8 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 161, 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 = {3603, 3568, 46, 212} \begin {gather*} -\frac {3 i}{16 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{8 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )}+\frac {5 x}{16 a^2 c^3}-\frac {3 i}{32 a^2 c f (c-i c \tan (e+f x))^2}+\frac {i}{32 a^2 c f (c+i c \tan (e+f x))^2}-\frac {i}{24 a^2 f (c-i c \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(16*a^2*c^3) - (I/24)/(a^2*f*(c - I*c*Tan[e + f*x])^3) - ((3*I)/32)/(a^2*c*f*(c - I*c*Tan[e + f*x])^2) +
(I/32)/(a^2*c*f*(c + I*c*Tan[e + f*x])^2) - ((3*I)/16)/(a^2*f*(c^3 - I*c^3*Tan[e + f*x])) + (I/8)/(a^2*f*(c^3
+ I*c^3*Tan[e + f*x]))

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

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 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))^2 (c-i c \tan (e+f x))^3} \, dx &=\frac {\int \frac {\cos ^4(e+f x)}{c-i c \tan (e+f x)} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \left (\frac {1}{16 c^4 (c-x)^3}+\frac {1}{8 c^5 (c-x)^2}+\frac {1}{8 c^3 (c+x)^4}+\frac {3}{16 c^4 (c+x)^3}+\frac {3}{16 c^5 (c+x)^2}+\frac {5}{16 c^5 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=-\frac {i}{24 a^2 f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 c f (c-i c \tan (e+f x))^2}+\frac {i}{32 a^2 c f (c+i c \tan (e+f x))^2}-\frac {3 i}{16 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{8 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )}+\frac {(5 i) \text {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{16 a^2 c^2 f}\\ &=\frac {5 x}{16 a^2 c^3}-\frac {i}{24 a^2 f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a^2 c f (c-i c \tan (e+f x))^2}+\frac {i}{32 a^2 c f (c+i c \tan (e+f x))^2}-\frac {3 i}{16 a^2 f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{8 a^2 f \left (c^3+i c^3 \tan (e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.97, size = 111, normalized size = 0.69 \begin {gather*} \frac {(\cos (e+f x)+i \sin (e+f x)) (60 (-i+2 f x) \cos (e+f x)+15 i \cos (3 (e+f x))+i \cos (5 (e+f x))+60 \sin (e+f x)-120 i f x \sin (e+f x)+45 \sin (3 (e+f x))+5 \sin (5 (e+f x)))}{384 a^2 c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.18, size = 105, normalized size = 0.65

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

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))^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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))^2/(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.

________________________________________________________________________________________

Fricas [A]
time = 0.85, size = 85, normalized size = 0.53 \begin {gather*} \frac {{\left (120 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 15 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 60 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 30 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{3} f} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.34, size = 258, normalized size = 1.60 \begin {gather*} \begin {cases} \frac {\left (- 33554432 i a^{8} c^{12} f^{4} e^{12 i e} e^{6 i f x} - 251658240 i a^{8} c^{12} f^{4} e^{10 i e} e^{4 i f x} - 1006632960 i a^{8} c^{12} f^{4} e^{8 i e} e^{2 i f x} + 503316480 i a^{8} c^{12} f^{4} e^{4 i e} e^{- 2 i f x} + 50331648 i a^{8} c^{12} f^{4} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{6442450944 a^{10} c^{15} f^{5}} & \text {for}\: a^{10} c^{15} f^{5} e^{6 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^{- 4 i e}}{32 a^{2} c^{3}} - \frac {5}{16 a^{2} c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{2} c^{3}} \end {gather*}

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))**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.67, size = 137, normalized size = 0.85 \begin {gather*} -\frac {-\frac {30 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{3}} + \frac {30 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{3}} + \frac {3 \, {\left (15 i \, \tan \left (f x + e\right )^{2} + 38 \, \tan \left (f x + e\right ) - 25 i\right )}}{a^{2} c^{3} {\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^{2} c^{3} {\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))^2/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 5.25, size = 87, normalized size = 0.54 \begin {gather*} \frac {5\,x}{16\,a^2\,c^3}-\frac {\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^4}{16}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,5{}\mathrm {i}}{16}+\frac {25\,{\mathrm {tan}\left (e+f\,x\right )}^2}{48}+\frac {\mathrm {tan}\left (e+f\,x\right )\,25{}\mathrm {i}}{48}+\frac {1}{6}}{a^2\,c^3\,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 )}^3} \end {gather*}

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)^3),x)

[Out]

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

________________________________________________________________________________________