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]

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

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Rubi [A]  time = 0.24, antiderivative size = 181, 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 = {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 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])

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]

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*}

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Mathematica [A]  time = 2.77, 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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fricas [A]  time = 1.09, size = 115, normalized size = 0.64 \[ \frac {{\left (24 \, {\left (5 \, A + 3 i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A - 3 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-20 i \, A - 12 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-60 i \, A - 12 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-120 i \, A + 24 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, A - 12 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \]

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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giac [A]  time = 2.81, size = 221, normalized size = 1.22 \[ \frac {\frac {12 \, {\left (5 i \, A - 3 \, B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{4}} + \frac {12 \, {\left (-5 i \, A + 3 \, B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{4}} + \frac {12 \, {\left (5 \, A \tan \left (f x + e\right ) + 3 i \, B \tan \left (f x + e\right ) - 7 i \, A + 5 \, B\right )}}{a c^{4} {\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac {-125 i \, A \tan \left (f x + e\right )^{4} + 75 \, B \tan \left (f x + e\right )^{4} + 596 \, A \tan \left (f x + e\right )^{3} + 348 i \, B \tan \left (f x + e\right )^{3} + 1110 i \, A \tan \left (f x + e\right )^{2} - 618 \, B \tan \left (f x + e\right )^{2} - 996 \, A \tan \left (f x + e\right ) - 492 i \, B \tan \left (f x + e\right ) - 405 i \, A + 99 \, B}{a c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \]

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

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maple [A]  time = 0.41, size = 303, normalized size = 1.67 \[ \frac {A}{8 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )}+\frac {i B}{16 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right ) A}{64 f a \,c^{4}}-\frac {3 \ln \left (\tan \left (f x +e \right )+i\right ) B}{64 f a \,c^{4}}+\frac {3 i A}{32 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {B}{32 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {B}{16 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {i A}{16 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {A}{12 a \,c^{4} f \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {A}{32 f a \,c^{4} \left (\tan \left (f x +e \right )-i\right )}+\frac {i B}{32 f a \,c^{4} \left (\tan \left (f x +e \right )-i\right )}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right ) A}{64 f a \,c^{4}}+\frac {3 \ln \left (\tan \left (f x +e \right )-i\right ) B}{64 f a \,c^{4}} \]

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]

1/8/f/a/c^4/(tan(f*x+e)+I)*A+1/16*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*ln
(tan(f*x+e)+I)*B+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/f/a/c^4/(tan(f*x+e)+I)
^4*B-1/16*I/f/a/c^4/(tan(f*x+e)+I)^4*A-1/12*A/a/c^4/f/(tan(f*x+e)+I)^3+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*ln(tan(f*x+e)-I)*B

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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((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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 9.26, size = 204, normalized size = 1.13 \[ -\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {3\,B}{32\,a\,c^4}+\frac {A\,5{}\mathrm {i}}{32\,a\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {5\,A}{32\,a\,c^4}+\frac {B\,3{}\mathrm {i}}{32\,a\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {9\,B}{32\,a\,c^4}+\frac {A\,15{}\mathrm {i}}{32\,a\,c^4}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {35\,A}{96\,a\,c^4}+\frac {B\,7{}\mathrm {i}}{32\,a\,c^4}\right )-\frac {A}{3\,a\,c^4}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^5-{\mathrm {tan}\left (e+f\,x\right )}^4\,3{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-3\,B+A\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a\,c^4} \]

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)*(c - c*tan(e + f*x)*1i)^4),x)

[Out]

- (tan(e + f*x)*((A*5i)/(32*a*c^4) - (3*B)/(32*a*c^4)) + tan(e + f*x)^4*((5*A)/(32*a*c^4) + (B*3i)/(32*a*c^4))
+ tan(e + f*x)^3*((A*15i)/(32*a*c^4) - (9*B)/(32*a*c^4)) - tan(e + f*x)^2*((35*A)/(96*a*c^4) + (B*7i)/(32*a*c
^4)) - A/(3*a*c^4))/(f*(3*tan(e + f*x) - tan(e + f*x)^2*2i + 2*tan(e + f*x)^3 - tan(e + f*x)^4*3i - tan(e + f*
x)^5 + 1i)) - (x*(A*5i - 3*B)*1i)/(32*a*c^4)

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sympy [A]  time = 0.79, size = 439, normalized size = 2.43 \[ \begin {cases} - \frac {\left (\left (- 100663296 i A a^{4} c^{16} f^{4} + 100663296 B a^{4} c^{16} f^{4}\right ) e^{- 2 i f x} + \left (1006632960 i A a^{4} c^{16} f^{4} e^{4 i e} - 201326592 B a^{4} c^{16} f^{4} e^{4 i e}\right ) e^{2 i f x} + \left (503316480 i A a^{4} c^{16} f^{4} e^{6 i e} + 100663296 B a^{4} c^{16} f^{4} e^{6 i e}\right ) e^{4 i f x} + \left (167772160 i A a^{4} c^{16} f^{4} e^{8 i e} + 100663296 B a^{4} c^{16} f^{4} e^{8 i e}\right ) e^{6 i f x} + \left (25165824 i A a^{4} c^{16} f^{4} e^{10 i e} + 25165824 B a^{4} c^{16} f^{4} e^{10 i e}\right ) e^{8 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text {for}\: 6442450944 a^{5} c^{20} f^{5} e^{2 i e} \neq 0 \\x \left (- \frac {5 A + 3 i B}{32 a c^{4}} + \frac {\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 2 i e}}{32 a c^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 5 A - 3 i B\right )}{32 a c^{4}} \]

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) + 251658
24*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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