3.736 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=99 \[ -\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac {5 A \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac {5 A \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac {5 A x}{16 a^3 c^3} \]

[Out]

5/16*A*x/a^3/c^3+5/16*A*cos(f*x+e)*sin(f*x+e)/a^3/c^3/f+5/24*A*cos(f*x+e)^3*sin(f*x+e)/a^3/c^3/f-1/6*cos(f*x+e
)^6*(B-A*tan(f*x+e))/a^3/c^3/f

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Rubi [A]  time = 0.15, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3588, 73, 639, 199, 205} \[ -\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac {5 A \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac {5 A \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac {5 A x}{16 a^3 c^3} \]

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])^3),x]

[Out]

(5*A*x)/(16*a^3*c^3) + (5*A*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*c^3*f) + (5*A*Cos[e + f*x]^3*Sin[e + f*x])/(24*
a^3*c^3*f) - (Cos[e + f*x]^6*(B - A*Tan[e + f*x]))/(6*a^3*c^3*f)

Rule 73

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]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(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 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))^3 (c-i c \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{\left (a c+a c x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac {(5 A) \operatorname {Subst}\left (\int \frac {1}{\left (a c+a c x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac {(5 A) \operatorname {Subst}\left (\int \frac {1}{\left (a c+a c x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 a c f}\\ &=\frac {5 A \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac {5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac {(5 A) \operatorname {Subst}\left (\int \frac {1}{a c+a c x^2} \, dx,x,\tan (e+f x)\right )}{16 a^2 c^2 f}\\ &=\frac {5 A x}{16 a^3 c^3}+\frac {5 A \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac {5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 63, normalized size = 0.64 \[ \frac {A (45 \sin (2 (e+f x))+9 \sin (4 (e+f x))+\sin (6 (e+f x))+60 e+60 f x)-32 B \cos ^6(e+f x)}{192 a^3 c^3 f} \]

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])^3),x]

[Out]

(-32*B*Cos[e + f*x]^6 + A*(60*e + 60*f*x + 45*Sin[2*(e + f*x)] + 9*Sin[4*(e + f*x)] + Sin[6*(e + f*x)]))/(192*
a^3*c^3*f)

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fricas [C]  time = 1.12, size = 126, normalized size = 1.27 \[ \frac {{\left (120 \, A f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + {\left (-9 i \, A - 6 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-45 i \, A - 15 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (45 i \, A - 15 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (9 i \, A - 6 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{3} f} \]

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))^3/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/384*(120*A*f*x*e^(6*I*f*x + 6*I*e) + (-I*A - B)*e^(12*I*f*x + 12*I*e) + (-9*I*A - 6*B)*e^(10*I*f*x + 10*I*e)
+ (-45*I*A - 15*B)*e^(8*I*f*x + 8*I*e) + (45*I*A - 15*B)*e^(4*I*f*x + 4*I*e) + (9*I*A - 6*B)*e^(2*I*f*x + 2*I
*e) + I*A - B)*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)

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giac [A]  time = 3.17, size = 79, normalized size = 0.80 \[ \frac {\frac {15 \, {\left (f x + e\right )} A}{a^{3} c^{3}} + \frac {15 \, A \tan \left (f x + e\right )^{5} + 40 \, A \tan \left (f x + e\right )^{3} + 33 \, A \tan \left (f x + e\right ) - 8 \, B}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3} c^{3}}}{48 \, f} \]

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))^3/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/48*(15*(f*x + e)*A/(a^3*c^3) + (15*A*tan(f*x + e)^5 + 40*A*tan(f*x + e)^3 + 33*A*tan(f*x + e) - 8*B)/((tan(f
*x + e)^2 + 1)^3*a^3*c^3))/f

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maple [C]  time = 0.33, size = 330, normalized size = 3.33 \[ \frac {5 i A \ln \left (\tan \left (f x +e \right )+i\right )}{32 f \,a^{3} c^{3}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )}-\frac {i B}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )}-\frac {A}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {i B}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {B}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {i A}{16 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {5 i A \ln \left (\tan \left (f x +e \right )-i\right )}{32 f \,a^{3} c^{3}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {i B}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {A}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {i B}{48 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {B}{32 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i A}{16 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )-i\right )^{2}} \]

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))^3,x)

[Out]

5/32*I/f/a^3/c^3*A*ln(tan(f*x+e)+I)+5/32/f/a^3/c^3/(tan(f*x+e)+I)*A-1/32*I/f/a^3/c^3/(tan(f*x+e)+I)*B-1/48/f/a
^3/c^3/(tan(f*x+e)+I)^3*A+1/48*I/f/a^3/c^3/(tan(f*x+e)+I)^3*B+1/32/f/a^3/c^3/(tan(f*x+e)+I)^2*B+1/16*I/f/a^3/c
^3/(tan(f*x+e)+I)^2*A-5/32*I/f/a^3/c^3*A*ln(tan(f*x+e)-I)+5/32/f/a^3/c^3/(tan(f*x+e)-I)*A+1/32*I/f/a^3/c^3/(ta
n(f*x+e)-I)*B-1/48/f/a^3/c^3/(tan(f*x+e)-I)^3*A-1/48*I/f/a^3/c^3/(tan(f*x+e)-I)^3*B+1/32/f/a^3/c^3/(tan(f*x+e)
-I)^2*B-1/16*I/f/a^3/c^3/(tan(f*x+e)-I)^2*A

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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))^3/(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.

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mupad [B]  time = 8.72, size = 64, normalized size = 0.65 \[ \frac {5\,A\,x}{16\,a^3\,c^3}+\frac {{\cos \left (e+f\,x\right )}^6\,\left (\frac {5\,A\,{\mathrm {tan}\left (e+f\,x\right )}^5}{16}+\frac {5\,A\,{\mathrm {tan}\left (e+f\,x\right )}^3}{6}+\frac {11\,A\,\mathrm {tan}\left (e+f\,x\right )}{16}-\frac {B}{6}\right )}{a^3\,c^3\,f} \]

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)^3),x)

[Out]

(5*A*x)/(16*a^3*c^3) + (cos(e + f*x)^6*((11*A*tan(e + f*x))/16 - B/6 + (5*A*tan(e + f*x)^3)/6 + (5*A*tan(e + f
*x)^5)/16))/(a^3*c^3*f)

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sympy [A]  time = 1.26, size = 510, normalized size = 5.15 \[ \frac {5 A x}{16 a^{3} c^{3}} + \begin {cases} \frac {\left (\left (103079215104 i A a^{15} c^{15} f^{5} e^{6 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{6 i e}\right ) e^{- 6 i f x} + \left (927712935936 i A a^{15} c^{15} f^{5} e^{8 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{8 i e}\right ) e^{- 4 i f x} + \left (4638564679680 i A a^{15} c^{15} f^{5} e^{10 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{10 i e}\right ) e^{- 2 i f x} + \left (- 4638564679680 i A a^{15} c^{15} f^{5} e^{14 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{14 i e}\right ) e^{2 i f x} + \left (- 927712935936 i A a^{15} c^{15} f^{5} e^{16 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{16 i e}\right ) e^{4 i f x} + \left (- 103079215104 i A a^{15} c^{15} f^{5} e^{18 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{18 i e}\right ) e^{6 i f x}\right ) e^{- 12 i e}}{39582418599936 a^{18} c^{18} f^{6}} & \text {for}\: 39582418599936 a^{18} c^{18} f^{6} e^{12 i e} \neq 0 \\x \left (- \frac {5 A}{16 a^{3} c^{3}} + \frac {\left (A e^{12 i e} + 6 A e^{10 i e} + 15 A e^{8 i e} + 20 A e^{6 i e} + 15 A e^{4 i e} + 6 A e^{2 i e} + A - i B e^{12 i e} - 4 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{4 i e} + 4 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{64 a^{3} c^{3}}\right ) & \text {otherwise} \end {cases} \]

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))**3,x)

[Out]

5*A*x/(16*a**3*c**3) + Piecewise((((103079215104*I*A*a**15*c**15*f**5*exp(6*I*e) - 103079215104*B*a**15*c**15*
f**5*exp(6*I*e))*exp(-6*I*f*x) + (927712935936*I*A*a**15*c**15*f**5*exp(8*I*e) - 618475290624*B*a**15*c**15*f*
*5*exp(8*I*e))*exp(-4*I*f*x) + (4638564679680*I*A*a**15*c**15*f**5*exp(10*I*e) - 1546188226560*B*a**15*c**15*f
**5*exp(10*I*e))*exp(-2*I*f*x) + (-4638564679680*I*A*a**15*c**15*f**5*exp(14*I*e) - 1546188226560*B*a**15*c**1
5*f**5*exp(14*I*e))*exp(2*I*f*x) + (-927712935936*I*A*a**15*c**15*f**5*exp(16*I*e) - 618475290624*B*a**15*c**1
5*f**5*exp(16*I*e))*exp(4*I*f*x) + (-103079215104*I*A*a**15*c**15*f**5*exp(18*I*e) - 103079215104*B*a**15*c**1
5*f**5*exp(18*I*e))*exp(6*I*f*x))*exp(-12*I*e)/(39582418599936*a**18*c**18*f**6), Ne(39582418599936*a**18*c**1
8*f**6*exp(12*I*e), 0)), (x*(-5*A/(16*a**3*c**3) + (A*exp(12*I*e) + 6*A*exp(10*I*e) + 15*A*exp(8*I*e) + 20*A*e
xp(6*I*e) + 15*A*exp(4*I*e) + 6*A*exp(2*I*e) + A - I*B*exp(12*I*e) - 4*I*B*exp(10*I*e) - 5*I*B*exp(8*I*e) + 5*
I*B*exp(4*I*e) + 4*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(64*a**3*c**3)), True))

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