3.8.23 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^2} \, dx\) [723]

Optimal. Leaf size=71 \[ \frac {3 A x}{8 a^2 c^2}+\frac {3 A \cos (e+f x) \sin (e+f x)}{8 a^2 c^2 f}-\frac {\cos ^4(e+f x) (B-A \tan (e+f x))}{4 a^2 c^2 f} \]

[Out]

3/8*A*x/a^2/c^2+3/8*A*cos(f*x+e)*sin(f*x+e)/a^2/c^2/f-1/4*cos(f*x+e)^4*(B-A*tan(f*x+e))/a^2/c^2/f

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Rubi [A]
time = 0.10, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3669, 74, 653, 205, 211} \begin {gather*} -\frac {\cos ^4(e+f x) (B-A \tan (e+f x))}{4 a^2 c^2 f}+\frac {3 A \sin (e+f x) \cos (e+f x)}{8 a^2 c^2 f}+\frac {3 A x}{8 a^2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*x)/(8*a^2*c^2) + (3*A*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*c^2*f) - (Cos[e + f*x]^4*(B - A*Tan[e + f*x]))/(4
*a^2*c^2*f)

Rule 74

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

Rule 653

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)/(2*a*c*(p + 1)))*(a + c*x
^2)^(p + 1), x] + Dist[d*((2*p + 3)/(2*a*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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

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Mathematica [A]
time = 0.09, size = 53, normalized size = 0.75 \begin {gather*} \frac {-8 B \cos ^4(e+f x)+A (12 (e+f x)+8 \sin (2 (e+f x))+\sin (4 (e+f x)))}{32 a^2 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*B*Cos[e + f*x]^4 + A*(12*(e + f*x) + 8*Sin[2*(e + f*x)] + Sin[4*(e + f*x)]))/(32*a^2*c^2*f)

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Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 124, normalized size = 1.75

method	result	size
risch	\(\frac {3 A x}{8 a^{2} c^{2}}-\frac {B \cos \left (4 f x +4 e \right )}{32 a^{2} c^{2} f}+\frac {A \sin \left (4 f x +4 e \right )}{32 a^{2} c^{2} f}-\frac {B \cos \left (2 f x +2 e \right )}{8 a^{2} c^{2} f}+\frac {A \sin \left (2 f x +2 e \right )}{4 a^{2} c^{2} f}\)	\(96\)
norman	\(\frac {\frac {3 A x}{8 a c}-\frac {B}{4 a c f}+\frac {5 A \tan \left (f x +e \right )}{8 a c f}+\frac {3 A \left (\tan ^{3}\left (f x +e \right )\right )}{8 a c f}+\frac {3 A x \left (\tan ^{2}\left (f x +e \right )\right )}{4 a c}+\frac {3 A x \left (\tan ^{4}\left (f x +e \right )\right )}{8 a c}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{2} a c}\)	\(117\)
derivativedivides	\(\frac {-\frac {3 i A \ln \left (-i+\tan \left (f x +e \right )\right )}{16}-\frac {-\frac {3 A}{16}-\frac {i B}{16}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{8}-\frac {B}{8}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {i A}{8}-\frac {B}{8}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {3 A}{16}+\frac {i B}{16}}{i+\tan \left (f x +e \right )}+\frac {3 i A \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{2} c^{2}}\)	\(124\)
default	\(\frac {-\frac {3 i A \ln \left (-i+\tan \left (f x +e \right )\right )}{16}-\frac {-\frac {3 A}{16}-\frac {i B}{16}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{8}-\frac {B}{8}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {i A}{8}-\frac {B}{8}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {3 A}{16}+\frac {i B}{16}}{i+\tan \left (f x +e \right )}+\frac {3 i A \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{2} c^{2}}\)	\(124\)

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))^2/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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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))^2/(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 [C] Result contains complex when optimal does not.
time = 1.14, size = 95, normalized size = 1.34 \begin {gather*} \frac {{\left (24 \, A f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, {\left (2 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 4 \, {\left (-2 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{64 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/64*(24*A*f*x*e^(4*I*f*x + 4*I*e) + (-I*A - B)*e^(8*I*f*x + 8*I*e) - 4*(2*I*A + B)*e^(6*I*f*x + 6*I*e) - 4*(-
2*I*A + B)*e^(2*I*f*x + 2*I*e) + I*A - B)*e^(-4*I*f*x - 4*I*e)/(a^2*c^2*f)

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Sympy [A]
time = 0.37, size = 360, normalized size = 5.07 \begin {gather*} \frac {3 A x}{8 a^{2} c^{2}} + \begin {cases} \frac {\left (\left (16384 i A a^{6} c^{6} f^{3} e^{2 i e} - 16384 B a^{6} c^{6} f^{3} e^{2 i e}\right ) e^{- 4 i f x} + \left (131072 i A a^{6} c^{6} f^{3} e^{4 i e} - 65536 B a^{6} c^{6} f^{3} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 131072 i A a^{6} c^{6} f^{3} e^{8 i e} - 65536 B a^{6} c^{6} f^{3} e^{8 i e}\right ) e^{2 i f x} + \left (- 16384 i A a^{6} c^{6} f^{3} e^{10 i e} - 16384 B a^{6} c^{6} f^{3} e^{10 i e}\right ) e^{4 i f x}\right ) e^{- 6 i e}}{1048576 a^{8} c^{8} f^{4}} & \text {for}\: a^{8} c^{8} f^{4} e^{6 i e} \neq 0 \\x \left (- \frac {3 A}{8 a^{2} c^{2}} + \frac {\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{16 a^{2} c^{2}}\right ) & \text {otherwise} \end {cases} \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))**2/(c-I*c*tan(f*x+e))**2,x)

[Out]

3*A*x/(8*a**2*c**2) + Piecewise((((16384*I*A*a**6*c**6*f**3*exp(2*I*e) - 16384*B*a**6*c**6*f**3*exp(2*I*e))*ex
p(-4*I*f*x) + (131072*I*A*a**6*c**6*f**3*exp(4*I*e) - 65536*B*a**6*c**6*f**3*exp(4*I*e))*exp(-2*I*f*x) + (-131
072*I*A*a**6*c**6*f**3*exp(8*I*e) - 65536*B*a**6*c**6*f**3*exp(8*I*e))*exp(2*I*f*x) + (-16384*I*A*a**6*c**6*f*
*3*exp(10*I*e) - 16384*B*a**6*c**6*f**3*exp(10*I*e))*exp(4*I*f*x))*exp(-6*I*e)/(1048576*a**8*c**8*f**4), Ne(a*
*8*c**8*f**4*exp(6*I*e), 0)), (x*(-3*A/(8*a**2*c**2) + (A*exp(8*I*e) + 4*A*exp(6*I*e) + 6*A*exp(4*I*e) + 4*A*e
xp(2*I*e) + A - I*B*exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(16*a**2*c**2)), True)
)

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Giac [A]
time = 0.64, size = 67, normalized size = 0.94 \begin {gather*} \frac {\frac {3 \, {\left (f x + e\right )} A}{a^{2} c^{2}} + \frac {3 \, A \tan \left (f x + e\right )^{3} + 5 \, A \tan \left (f x + e\right ) - 2 \, B}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2} c^{2}}}{8 \, 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))^2/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/8*(3*(f*x + e)*A/(a^2*c^2) + (3*A*tan(f*x + e)^3 + 5*A*tan(f*x + e) - 2*B)/((tan(f*x + e)^2 + 1)^2*a^2*c^2))
/f

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Mupad [B]
time = 8.50, size = 53, normalized size = 0.75 \begin {gather*} \frac {3\,A\,x}{8\,a^2\,c^2}+\frac {{\cos \left (e+f\,x\right )}^4\,\left (\frac {3\,A\,{\mathrm {tan}\left (e+f\,x\right )}^3}{8}+\frac {5\,A\,\mathrm {tan}\left (e+f\,x\right )}{8}-\frac {B}{4}\right )}{a^2\,c^2\,f} \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)^2*(c - c*tan(e + f*x)*1i)^2),x)

[Out]

(3*A*x)/(8*a^2*c^2) + (cos(e + f*x)^4*((5*A*tan(e + f*x))/8 - B/4 + (3*A*tan(e + f*x)^3)/8))/(a^2*c^2*f)

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