3.8.35 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx\) [735]

Optimal. Leaf size=185 \[ \frac {(5 A-i B) x}{16 a^3 c^2}+\frac {A+i B}{24 a^3 c^2 f (i-\tan (e+f x))^3}-\frac {3 i A-B}{32 a^3 c^2 f (i-\tan (e+f x))^2}-\frac {3 A}{16 a^3 c^2 f (i-\tan (e+f x))}+\frac {i A+B}{32 a^3 c^2 f (i+\tan (e+f x))^2}+\frac {2 A-i B}{16 a^3 c^2 f (i+\tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3669, 78, 209} \begin {gather*} \frac {2 A-i B}{16 a^3 c^2 f (\tan (e+f x)+i)}-\frac {-B+3 i A}{32 a^3 c^2 f (-\tan (e+f x)+i)^2}+\frac {B+i A}{32 a^3 c^2 f (\tan (e+f x)+i)^2}+\frac {A+i B}{24 a^3 c^2 f (-\tan (e+f x)+i)^3}+\frac {x (5 A-i B)}{16 a^3 c^2}-\frac {3 A}{16 a^3 c^2 f (-\tan (e+f x)+i)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 209

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

Rule 3669

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 1.13, size = 217, normalized size = 1.17 \begin {gather*} \frac {\sec ^3(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (12 (A (-5+10 i f x)+B (-i+2 f x)) \cos (e+f x)+3 (5 A-9 i B) \cos (3 (e+f x))+A \cos (5 (e+f x))-5 i B \cos (5 (e+f x))+60 i A \sin (e+f x)-12 B \sin (e+f x)-120 A f x \sin (e+f x)+24 i B f x \sin (e+f x)+45 i A \sin (3 (e+f x))+9 B \sin (3 (e+f x))+5 i A \sin (5 (e+f x))+B \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[(A + B*Tan[e + f*x])/((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)])*(12*(A*(-5 + (10*I)*f*x) + B*(-I + 2*f*x))*Cos[e + f*x
] + 3*(5*A - (9*I)*B)*Cos[3*(e + f*x)] + A*Cos[5*(e + f*x)] - (5*I)*B*Cos[5*(e + f*x)] + (60*I)*A*Sin[e + f*x]
- 12*B*Sin[e + f*x] - 120*A*f*x*Sin[e + f*x] + (24*I)*B*f*x*Sin[e + f*x] + (45*I)*A*Sin[3*(e + f*x)] + 9*B*Si
n[3*(e + f*x)] + (5*I)*A*Sin[5*(e + f*x)] + B*Sin[5*(e + f*x)]))/(384*a^3*c^2*f*(-I + Tan[e + f*x])^3)

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Maple [A]
time = 0.30, size = 148, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\left (-\frac {5 i A}{32}-\frac {B}{32}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )+\frac {3 A}{16 \left (-i+\tan \left (f x +e \right )\right )}-\frac {\frac {3 i A}{16}-\frac {B}{16}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{8}+\frac {i B}{8}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {\frac {i B}{16}-\frac {A}{8}}{i+\tan \left (f x +e \right )}+\left (\frac {5 i A}{32}+\frac {B}{32}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {-\frac {i A}{16}-\frac {B}{16}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}}{f \,a^{3} c^{2}}\)	\(148\)
default	\(\frac {\left (-\frac {5 i A}{32}-\frac {B}{32}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )+\frac {3 A}{16 \left (-i+\tan \left (f x +e \right )\right )}-\frac {\frac {3 i A}{16}-\frac {B}{16}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{8}+\frac {i B}{8}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {\frac {i B}{16}-\frac {A}{8}}{i+\tan \left (f x +e \right )}+\left (\frac {5 i A}{32}+\frac {B}{32}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {-\frac {i A}{16}-\frac {B}{16}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}}{f \,a^{3} c^{2}}\)	\(148\)
norman	\(\frac {\frac {\left (-i B +5 A \right ) x}{16 a c}-\frac {-i A +B}{6 a c f}+\frac {\left (-i B +5 A \right ) \left (\tan ^{3}\left (f x +e \right )\right )}{6 a c f}+\frac {\left (-i B +5 A \right ) \left (\tan ^{5}\left (f x +e \right )\right )}{16 a c f}+\frac {3 \left (-i B +5 A \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{16 a c}+\frac {3 \left (-i B +5 A \right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{16 a c}+\frac {\left (-i B +5 A \right ) x \left (\tan ^{6}\left (f x +e \right )\right )}{16 a c}+\frac {\left (i B +11 A \right ) \tan \left (f x +e \right )}{16 a c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3} a^{2} c}\)	\(209\)
risch	\(-\frac {i x B}{16 a^{3} c^{2}}+\frac {5 x A}{16 a^{3} c^{2}}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{192 a^{3} c^{2} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{192 a^{3} c^{2} f}-\frac {\cos \left (4 f x +4 e \right ) B}{32 a^{3} c^{2} f}+\frac {i \cos \left (4 f x +4 e \right ) A}{32 a^{3} c^{2} f}+\frac {i \sin \left (4 f x +4 e \right ) B}{64 a^{3} c^{2} f}+\frac {3 \sin \left (4 f x +4 e \right ) A}{64 a^{3} c^{2} f}-\frac {5 \cos \left (2 f x +2 e \right ) B}{64 a^{3} c^{2} f}+\frac {5 i \cos \left (2 f x +2 e \right ) A}{64 a^{3} c^{2} f}-\frac {i \sin \left (2 f x +2 e \right ) B}{64 a^{3} c^{2} f}+\frac {15 \sin \left (2 f x +2 e \right ) A}{64 a^{3} c^{2} f}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))/(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*((-5/32*I*A-1/32*B)*ln(-I+tan(f*x+e))+3/16*A/(-I+tan(f*x+e))-1/2*(3/16*I*A-1/16*B)/(-I+tan(f*x+e))
^2-1/3*(1/8*A+1/8*I*B)/(-I+tan(f*x+e))^3-(1/16*I*B-1/8*A)/(I+tan(f*x+e))+(5/32*I*A+1/32*B)*ln(I+tan(f*x+e))-1/
2*(-1/16*I*A-1/16*B)/(I+tan(f*x+e))^2)

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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((A+B*tan(f*x+e))/(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 = 3.01, size = 121, normalized size = 0.65 \begin {gather*} \frac {{\left (24 \, {\left (5 \, A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 6 \, {\left (5 i \, A + 3 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 12 \, {\left (-5 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-5 i \, A + 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\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((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/384*(24*(5*A - I*B)*f*x*e^(6*I*f*x + 6*I*e) - 3*(I*A + B)*e^(10*I*f*x + 10*I*e) - 6*(5*I*A + 3*B)*e^(8*I*f*x
+ 8*I*e) - 12*(-5*I*A + B)*e^(4*I*f*x + 4*I*e) - 3*(-5*I*A + 3*B)*e^(2*I*f*x + 2*I*e) + 2*I*A - 2*B)*e^(-6*I*
f*x - 6*I*e)/(a^3*c^2*f)

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Sympy [A]
time = 0.46, size = 452, normalized size = 2.44 \begin {gather*} \begin {cases} \frac {\left (\left (33554432 i A a^{12} c^{8} f^{4} e^{6 i e} - 33554432 B a^{12} c^{8} f^{4} e^{6 i e}\right ) e^{- 6 i f x} + \left (251658240 i A a^{12} c^{8} f^{4} e^{8 i e} - 150994944 B a^{12} c^{8} f^{4} e^{8 i e}\right ) e^{- 4 i f x} + \left (1006632960 i A a^{12} c^{8} f^{4} e^{10 i e} - 201326592 B a^{12} c^{8} f^{4} e^{10 i e}\right ) e^{- 2 i f x} + \left (- 503316480 i A a^{12} c^{8} f^{4} e^{14 i e} - 301989888 B a^{12} c^{8} f^{4} e^{14 i e}\right ) e^{2 i f x} + \left (- 50331648 i A a^{12} c^{8} f^{4} e^{16 i e} - 50331648 B a^{12} c^{8} f^{4} e^{16 i e}\right ) e^{4 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 {5 A - i B}{16 a^{3} c^{2}} + \frac {\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{32 a^{3} c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (5 A - i B\right )}{16 a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)

[Out]

Piecewise((((33554432*I*A*a**12*c**8*f**4*exp(6*I*e) - 33554432*B*a**12*c**8*f**4*exp(6*I*e))*exp(-6*I*f*x) +
(251658240*I*A*a**12*c**8*f**4*exp(8*I*e) - 150994944*B*a**12*c**8*f**4*exp(8*I*e))*exp(-4*I*f*x) + (100663296
0*I*A*a**12*c**8*f**4*exp(10*I*e) - 201326592*B*a**12*c**8*f**4*exp(10*I*e))*exp(-2*I*f*x) + (-503316480*I*A*a
**12*c**8*f**4*exp(14*I*e) - 301989888*B*a**12*c**8*f**4*exp(14*I*e))*exp(2*I*f*x) + (-50331648*I*A*a**12*c**8
*f**4*exp(16*I*e) - 50331648*B*a**12*c**8*f**4*exp(16*I*e))*exp(4*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*(-(5*A - I*B)/(16*a**3*c**2) + (A*exp(10*I*e) + 5*A*exp(8*I*e
) + 10*A*exp(6*I*e) + 10*A*exp(4*I*e) + 5*A*exp(2*I*e) + A - I*B*exp(10*I*e) - 3*I*B*exp(8*I*e) - 2*I*B*exp(6*
I*e) + 2*I*B*exp(4*I*e) + 3*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(32*a**3*c**2)), True)) + x*(5*A - I*B)/(16*a**3
*c**2)

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Giac [A]
time = 0.94, size = 219, normalized size = 1.18 \begin {gather*} -\frac {\frac {6 \, {\left (-5 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{2}} + \frac {6 \, {\left (5 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{2}} + \frac {3 \, {\left (-15 i \, A \tan \left (f x + e\right )^{2} - 3 \, B \tan \left (f x + e\right )^{2} + 38 \, A \tan \left (f x + e\right ) - 10 i \, B \tan \left (f x + e\right ) + 25 i \, A + 9 \, B\right )}}{a^{3} c^{2} {\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{2}} + \frac {-55 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 201 \, A \tan \left (f x + e\right )^{2} + 33 i \, B \tan \left (f x + e\right )^{2} + 255 i \, A \tan \left (f x + e\right ) + 27 \, B \tan \left (f x + e\right ) + 117 \, A + 3 i \, B}{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((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/192*(6*(-5*I*A - B)*log(tan(f*x + e) + I)/(a^3*c^2) + 6*(5*I*A + B)*log(tan(f*x + e) - I)/(a^3*c^2) + 3*(-1
5*I*A*tan(f*x + e)^2 - 3*B*tan(f*x + e)^2 + 38*A*tan(f*x + e) - 10*I*B*tan(f*x + e) + 25*I*A + 9*B)/(a^3*c^2*(
-I*tan(f*x + e) + 1)^2) + (-55*I*A*tan(f*x + e)^3 - 11*B*tan(f*x + e)^3 - 201*A*tan(f*x + e)^2 + 33*I*B*tan(f*
x + e)^2 + 255*I*A*tan(f*x + e) + 27*B*tan(f*x + e) + 117*A + 3*I*B)/(a^3*c^2*(tan(f*x + e) - I)^3))/f

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Mupad [B]
time = 9.47, size = 208, normalized size = 1.12 \begin {gather*} \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {25\,A}{48\,a^3\,c^2}-\frac {B\,5{}\mathrm {i}}{48\,a^3\,c^2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {5\,A}{16\,a^3\,c^2}-\frac {B\,1{}\mathrm {i}}{16\,a^3\,c^2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {B}{16\,a^3\,c^2}+\frac {A\,5{}\mathrm {i}}{16\,a^3\,c^2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {5\,B}{48\,a^3\,c^2}+\frac {A\,25{}\mathrm {i}}{48\,a^3\,c^2}\right )-\frac {B}{6\,a^3\,c^2}+\frac {A\,1{}\mathrm {i}}{6\,a^3\,c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,2{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^2),x)

[Out]

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

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