3.714 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=181 \[ -\frac{A+i B}{32 a c^4 f (-\tan (e+f x)+i)}+\frac{2 A+i B}{16 a c^4 f (\tan (e+f x)+i)}+\frac{-B+3 i A}{32 a c^4 f (\tan (e+f x)+i)^2}-\frac{B+i A}{16 a c^4 f (\tan (e+f x)+i)^4}+\frac{x (5 A+3 i B)}{32 a c^4}-\frac{A}{12 a c^4 f (\tan (e+f x)+i)^3} \]

[Out]

((5*A + (3*I)*B)*x)/(32*a*c^4) - (A + I*B)/(32*a*c^4*f*(I - Tan[e + f*x])) - (I*A + B)/(16*a*c^4*f*(I + Tan[e
+ f*x])^4) - A/(12*a*c^4*f*(I + Tan[e + f*x])^3) + ((3*I)*A - B)/(32*a*c^4*f*(I + Tan[e + f*x])^2) + (2*A + I*
B)/(16*a*c^4*f*(I + Tan[e + f*x]))

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Rubi [A]  time = 0.240197, antiderivative size = 181, normalized size of antiderivative = 1., 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 = {3588, 77, 203} \[ -\frac{A+i B}{32 a c^4 f (-\tan (e+f x)+i)}+\frac{2 A+i B}{16 a c^4 f (\tan (e+f x)+i)}+\frac{-B+3 i A}{32 a c^4 f (\tan (e+f x)+i)^2}-\frac{B+i A}{16 a c^4 f (\tan (e+f x)+i)^4}+\frac{x (5 A+3 i B)}{32 a c^4}-\frac{A}{12 a c^4 f (\tan (e+f x)+i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A + (3*I)*B)*x)/(32*a*c^4) - (A + I*B)/(32*a*c^4*f*(I - Tan[e + f*x])) - (I*A + B)/(16*a*c^4*f*(I + Tan[e
+ f*x])^4) - A/(12*a*c^4*f*(I + Tan[e + f*x])^3) + ((3*I)*A - B)/(32*a*c^4*f*(I + Tan[e + f*x])^2) + (2*A + I*
B)/(16*a*c^4*f*(I + Tan[e + f*x]))

Rule 3588

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]

Rule 77

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 203

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

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^2 (c-i c x)^5} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{-A-i B}{32 a^2 c^5 (-i+x)^2}+\frac{i A+B}{4 a^2 c^5 (i+x)^5}+\frac{A}{4 a^2 c^5 (i+x)^4}+\frac{-3 i A+B}{16 a^2 c^5 (i+x)^3}+\frac{-2 A-i B}{16 a^2 c^5 (i+x)^2}+\frac{5 A+3 i B}{32 a^2 c^5 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{A+i B}{32 a c^4 f (i-\tan (e+f x))}-\frac{i A+B}{16 a c^4 f (i+\tan (e+f x))^4}-\frac{A}{12 a c^4 f (i+\tan (e+f x))^3}+\frac{3 i A-B}{32 a c^4 f (i+\tan (e+f x))^2}+\frac{2 A+i B}{16 a c^4 f (i+\tan (e+f x))}+\frac{(5 A+3 i B) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{32 a c^4 f}\\ &=\frac{(5 A+3 i B) x}{32 a c^4}-\frac{A+i B}{32 a c^4 f (i-\tan (e+f x))}-\frac{i A+B}{16 a c^4 f (i+\tan (e+f x))^4}-\frac{A}{12 a c^4 f (i+\tan (e+f x))^3}+\frac{3 i A-B}{32 a c^4 f (i+\tan (e+f x))^2}+\frac{2 A+i B}{16 a c^4 f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.60032, size = 221, normalized size = 1.22 \[ \frac{\sec (e+f x) (\cos (4 (e+f x))+i \sin (4 (e+f x))) (-12 (15 A+i B) \cos (e+f x)+4 (-30 i A f x-5 A+18 B f x+3 i B) \cos (3 (e+f x))+60 i A \sin (e+f x)-20 i A \sin (3 (e+f x))-120 A f x \sin (3 (e+f x))-15 i A \sin (5 (e+f x))+9 A \cos (5 (e+f x))-36 B \sin (e+f x)-12 B \sin (3 (e+f x))-72 i B f x \sin (3 (e+f x))+9 B \sin (5 (e+f x))+15 i B \cos (5 (e+f x)))}{768 a c^4 f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)])*(-12*(15*A + I*B)*Cos[e + f*x] + 4*(-5*A + (3*I)*B - (30
*I)*A*f*x + 18*B*f*x)*Cos[3*(e + f*x)] + 9*A*Cos[5*(e + f*x)] + (15*I)*B*Cos[5*(e + f*x)] + (60*I)*A*Sin[e + f
*x] - 36*B*Sin[e + f*x] - (20*I)*A*Sin[3*(e + f*x)] - 12*B*Sin[3*(e + f*x)] - 120*A*f*x*Sin[3*(e + f*x)] - (72
*I)*B*f*x*Sin[3*(e + f*x)] - (15*I)*A*Sin[5*(e + f*x)] + 9*B*Sin[5*(e + f*x)]))/(768*a*c^4*f*(-I + Tan[e + f*x
]))

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Maple [A]  time = 0.073, size = 303, normalized size = 1.7 \begin{align*}{\frac{A}{32\,af{c}^{4} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{32}}B}{af{c}^{4} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{5\,i}{64}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) A}{af{c}^{4}}}+{\frac{3\,\ln \left ( \tan \left ( fx+e \right ) -i \right ) B}{64\,af{c}^{4}}}+{\frac{{\frac{i}{16}}B}{af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{A}{8\,af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{{\frac{5\,i}{64}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) A}{af{c}^{4}}}-{\frac{3\,\ln \left ( \tan \left ( fx+e \right ) +i \right ) B}{64\,af{c}^{4}}}-{\frac{A}{12\,af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{{\frac{3\,i}{32}}A}{af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{B}{32\,af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{{\frac{i}{16}}A}{af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-{\frac{B}{16\,af{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}} \end{align*}

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))/(c-I*c*tan(f*x+e))^4,x)

[Out]

1/32/f/a/c^4/(tan(f*x+e)-I)*A+1/32*I/f/a/c^4/(tan(f*x+e)-I)*B-5/64*I/f/a/c^4*ln(tan(f*x+e)-I)*A+3/64/f/a/c^4*l
n(tan(f*x+e)-I)*B+1/16*I/f/a/c^4/(tan(f*x+e)+I)*B+1/8/f/a/c^4/(tan(f*x+e)+I)*A+5/64*I/f/a/c^4*ln(tan(f*x+e)+I)
*A-3/64/f/a/c^4*ln(tan(f*x+e)+I)*B-1/12*A/a/c^4/f/(tan(f*x+e)+I)^3+3/32*I/f/a/c^4/(tan(f*x+e)+I)^2*A-1/32/f/a/
c^4/(tan(f*x+e)+I)^2*B-1/16*I/f/a/c^4/(tan(f*x+e)+I)^4*A-1/16/f/a/c^4/(tan(f*x+e)+I)^4*B

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

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))/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.05957, size = 343, normalized size = 1.9 \begin{align*} \frac{{\left (24 \,{\left (5 \, A + 3 i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-3 i \, A - 3 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-20 i \, A - 12 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-60 i \, A - 12 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-120 i \, A + 24 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, A - 12 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \end{align*}

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

[Out]

1/768*(24*(5*A + 3*I*B)*f*x*e^(2*I*f*x + 2*I*e) + (-3*I*A - 3*B)*e^(10*I*f*x + 10*I*e) + (-20*I*A - 12*B)*e^(8
*I*f*x + 8*I*e) + (-60*I*A - 12*B)*e^(6*I*f*x + 6*I*e) + (-120*I*A + 24*B)*e^(4*I*f*x + 4*I*e) + 12*I*A - 12*B
)*e^(-2*I*f*x - 2*I*e)/(a*c^4*f)

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Sympy [A]  time = 3.73697, size = 440, normalized size = 2.43 \begin{align*} \begin{cases} \frac{\left (\left (100663296 i A a^{4} c^{16} f^{4} - 100663296 B a^{4} c^{16} f^{4}\right ) e^{- 2 i f x} + \left (- 1006632960 i A a^{4} c^{16} f^{4} e^{4 i e} + 201326592 B a^{4} c^{16} f^{4} e^{4 i e}\right ) e^{2 i f x} + \left (- 503316480 i A a^{4} c^{16} f^{4} e^{6 i e} - 100663296 B a^{4} c^{16} f^{4} e^{6 i e}\right ) e^{4 i f x} + \left (- 167772160 i A a^{4} c^{16} f^{4} e^{8 i e} - 100663296 B a^{4} c^{16} f^{4} e^{8 i e}\right ) e^{6 i f x} + \left (- 25165824 i A a^{4} c^{16} f^{4} e^{10 i e} - 25165824 B a^{4} c^{16} f^{4} e^{10 i e}\right ) e^{8 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text{for}\: 6442450944 a^{5} c^{20} f^{5} e^{2 i e} \neq 0 \\x \left (- \frac{5 A + 3 i B}{32 a c^{4}} + \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^{- 2 i e}}{32 a c^{4}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (5 A + 3 i B\right )}{32 a c^{4}} \end{align*}

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))/(c-I*c*tan(f*x+e))**4,x)

[Out]

Piecewise((((100663296*I*A*a**4*c**16*f**4 - 100663296*B*a**4*c**16*f**4)*exp(-2*I*f*x) + (-1006632960*I*A*a**
4*c**16*f**4*exp(4*I*e) + 201326592*B*a**4*c**16*f**4*exp(4*I*e))*exp(2*I*f*x) + (-503316480*I*A*a**4*c**16*f*
*4*exp(6*I*e) - 100663296*B*a**4*c**16*f**4*exp(6*I*e))*exp(4*I*f*x) + (-167772160*I*A*a**4*c**16*f**4*exp(8*I
*e) - 100663296*B*a**4*c**16*f**4*exp(8*I*e))*exp(6*I*f*x) + (-25165824*I*A*a**4*c**16*f**4*exp(10*I*e) - 2516
5824*B*a**4*c**16*f**4*exp(10*I*e))*exp(8*I*f*x))*exp(-2*I*e)/(6442450944*a**5*c**20*f**5), Ne(6442450944*a**5
*c**20*f**5*exp(2*I*e), 0)), (x*(-(5*A + 3*I*B)/(32*a*c**4) + (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(-2*I*e)/(32*a*c**4)), True)) + x*(5*A + 3*I*B)/(32*a*c**4)

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Giac [A]  time = 1.43888, size = 298, normalized size = 1.65 \begin{align*} \frac{\frac{12 \,{\left (5 i \, A - 3 \, B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{4}} + \frac{12 \,{\left (-5 i \, A + 3 \, B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{4}} + \frac{12 \,{\left (5 \, A \tan \left (f x + e\right ) + 3 i \, B \tan \left (f x + e\right ) - 7 i \, A + 5 \, B\right )}}{a c^{4}{\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac{-125 i \, A \tan \left (f x + e\right )^{4} + 75 \, B \tan \left (f x + e\right )^{4} + 596 \, A \tan \left (f x + e\right )^{3} + 348 i \, B \tan \left (f x + e\right )^{3} + 1110 i \, A \tan \left (f x + e\right )^{2} - 618 \, B \tan \left (f x + e\right )^{2} - 996 \, A \tan \left (f x + e\right ) - 492 i \, B \tan \left (f x + e\right ) - 405 i \, A + 99 \, B}{a c^{4}{\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \end{align*}

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

[Out]

1/768*(12*(5*I*A - 3*B)*log(tan(f*x + e) + I)/(a*c^4) + 12*(-5*I*A + 3*B)*log(tan(f*x + e) - I)/(a*c^4) + 12*(
5*A*tan(f*x + e) + 3*I*B*tan(f*x + e) - 7*I*A + 5*B)/(a*c^4*(-I*tan(f*x + e) - 1)) + (-125*I*A*tan(f*x + e)^4
+ 75*B*tan(f*x + e)^4 + 596*A*tan(f*x + e)^3 + 348*I*B*tan(f*x + e)^3 + 1110*I*A*tan(f*x + e)^2 - 618*B*tan(f*
x + e)^2 - 996*A*tan(f*x + e) - 492*I*B*tan(f*x + e) - 405*I*A + 99*B)/(a*c^4*(tan(f*x + e) + I)^4))/f