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

Optimal. Leaf size=135 \[ \frac {c^3 (-B+i A) (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {4 i B c^3}{a^3 f (-\tan (e+f x)+i)}-\frac {2 B c^3}{a^3 f (-\tan (e+f x)+i)^2}+\frac {B c^3 \log (\cos (e+f x))}{a^3 f}-\frac {i B c^3 x}{a^3} \]

[Out]

-I*B*c^3*x/a^3+B*c^3*ln(cos(f*x+e))/a^3/f-2*B*c^3/a^3/f/(-tan(f*x+e)+I)^2-4*I*B*c^3/a^3/f/(-tan(f*x+e)+I)+1/6*
(I*A-B)*c^3*(1-I*tan(f*x+e))^3/a^3/f/(1+I*tan(f*x+e))^3

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Rubi [A]  time = 0.16, antiderivative size = 135, 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, 78, 43} \[ \frac {c^3 (-B+i A) (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {4 i B c^3}{a^3 f (-\tan (e+f x)+i)}-\frac {2 B c^3}{a^3 f (-\tan (e+f x)+i)^2}+\frac {B c^3 \log (\cos (e+f x))}{a^3 f}-\frac {i B c^3 x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*B*c^3*x)/a^3 + (B*c^3*Log[Cos[e + f*x]])/(a^3*f) - (2*B*c^3)/(a^3*f*(I - Tan[e + f*x])^2) - ((4*I)*B*c^3
)/(a^3*f*(I - Tan[e + f*x])) + ((I*A - B)*c^3*(1 - I*Tan[e + f*x])^3)/(6*a^3*f*(1 + I*Tan[e + f*x])^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

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

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)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^2}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) c^3 (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {(i B c) \operatorname {Subst}\left (\int \frac {(c-i c x)^2}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) c^3 (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {(i B c) \operatorname {Subst}\left (\int \left (\frac {4 i c^2}{a^3 (-i+x)^3}+\frac {4 c^2}{a^3 (-i+x)^2}-\frac {i c^2}{a^3 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i B c^3 x}{a^3}+\frac {B c^3 \log (\cos (e+f x))}{a^3 f}-\frac {2 B c^3}{a^3 f (i-\tan (e+f x))^2}-\frac {4 i B c^3}{a^3 f (i-\tan (e+f x))}+\frac {(i A-B) c^3 (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 3.92, size = 145, normalized size = 1.07 \[ \frac {c^3 \sec ^3(e+f x) (-\cos (3 (e+f x)) (A-6 i B \log (\cos (e+f x))-6 B f x+i B)+i A \sin (3 (e+f x))+9 B \sin (e+f x)-B \sin (3 (e+f x))+6 i B f x \sin (3 (e+f x))-3 i B \cos (e+f x)-6 B \sin (3 (e+f x)) \log (\cos (e+f x)))}{6 a^3 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Sec[e + f*x]^3*((-3*I)*B*Cos[e + f*x] - Cos[3*(e + f*x)]*(A + I*B - 6*B*f*x - (6*I)*B*Log[Cos[e + f*x]])
+ 9*B*Sin[e + f*x] + I*A*Sin[3*(e + f*x)] - B*Sin[3*(e + f*x)] + (6*I)*B*f*x*Sin[3*(e + f*x)] - 6*B*Log[Cos[e
+ f*x]]*Sin[3*(e + f*x)]))/(6*a^3*f*(-I + Tan[e + f*x])^3)

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fricas [A]  time = 0.57, size = 103, normalized size = 0.76 \[ \frac {{\left (-12 i \, B c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, B c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 6 \, B c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, B c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A - B\right )} c^{3}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{6 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-12*I*B*c^3*f*x*e^(6*I*f*x + 6*I*e) + 6*B*c^3*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 6*B*c^3*
e^(4*I*f*x + 4*I*e) + 3*B*c^3*e^(2*I*f*x + 2*I*e) + (I*A - B)*c^3)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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giac [B]  time = 3.26, size = 255, normalized size = 1.89 \[ \frac {\frac {30 \, B c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3}} - \frac {60 \, B c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{3}} + \frac {30 \, B c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{3}} + \frac {147 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 60 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 942 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2445 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 200 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3620 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2445 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 60 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 942 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 147 \, B c^{3}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(30*B*c^3*log(tan(1/2*f*x + 1/2*e) + 1)/a^3 - 60*B*c^3*log(tan(1/2*f*x + 1/2*e) - I)/a^3 + 30*B*c^3*log(t
an(1/2*f*x + 1/2*e) - 1)/a^3 + (147*B*c^3*tan(1/2*f*x + 1/2*e)^6 - 60*A*c^3*tan(1/2*f*x + 1/2*e)^5 - 942*I*B*c
^3*tan(1/2*f*x + 1/2*e)^5 - 2445*B*c^3*tan(1/2*f*x + 1/2*e)^4 + 200*A*c^3*tan(1/2*f*x + 1/2*e)^3 + 3620*I*B*c^
3*tan(1/2*f*x + 1/2*e)^3 + 2445*B*c^3*tan(1/2*f*x + 1/2*e)^2 - 60*A*c^3*tan(1/2*f*x + 1/2*e) - 942*I*B*c^3*tan
(1/2*f*x + 1/2*e) - 147*B*c^3)/(a^3*(tan(1/2*f*x + 1/2*e) - I)^6))/f

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maple [A]  time = 0.24, size = 164, normalized size = 1.21 \[ \frac {2 i c^{3} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4 c^{3} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {5 i c^{3} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {c^{3} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {4 i c^{3} B}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {4 c^{3} A}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {c^{3} B \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f*c^3/a^3/(tan(f*x+e)-I)^2*A-4/f*c^3/a^3/(tan(f*x+e)-I)^2*B+5*I/f*c^3/a^3/(tan(f*x+e)-I)*B+1/f*c^3/a^3/(ta
n(f*x+e)-I)*A-4/3*I/f*c^3/a^3/(tan(f*x+e)-I)^3*B-4/3/f*c^3/a^3/(tan(f*x+e)-I)^3*A-1/f*c^3/a^3*B*ln(tan(f*x+e)-
I)

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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))*(c-I*c*tan(f*x+e))^3/(a+I*a*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 = 8.96, size = 149, normalized size = 1.10 \[ -\frac {c^3\,\left (18\,B\,\mathrm {tan}\left (e+f\,x\right )-B\,7{}\mathrm {i}-A-B\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}+3\,A\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,15{}\mathrm {i}+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,9{}\mathrm {i}-3\,B\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+9\,B\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{3\,a^3\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^3*(18*B*tan(e + f*x) - B*7i - A - B*log(- tan(e + f*x)*1i - 1)*3i + 3*A*tan(e + f*x)^2 + B*tan(e + f*x)^2*
15i + B*tan(e + f*x)^2*log(- tan(e + f*x)*1i - 1)*9i - 3*B*tan(e + f*x)^3*log(- tan(e + f*x)*1i - 1) + 9*B*tan
(e + f*x)*log(- tan(e + f*x)*1i - 1))*1i)/(3*a^3*f*(tan(e + f*x)*1i + 1)^3)

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sympy [A]  time = 0.91, size = 258, normalized size = 1.91 \[ - \frac {2 i B c^{3} x}{a^{3}} + \frac {B c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{3} f} + \begin {cases} - \frac {\left (12 B a^{6} c^{3} f^{2} e^{10 i e} e^{- 2 i f x} - 6 B a^{6} c^{3} f^{2} e^{8 i e} e^{- 4 i f x} + \left (- 2 i A a^{6} c^{3} f^{2} e^{6 i e} + 2 B a^{6} c^{3} f^{2} e^{6 i e}\right ) e^{- 6 i f x}\right ) e^{- 12 i e}}{12 a^{9} f^{3}} & \text {for}\: 12 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (\frac {2 i B c^{3}}{a^{3}} + \frac {i \left (- i A c^{3} - 2 B c^{3} e^{6 i e} + 2 B c^{3} e^{4 i e} - 2 B c^{3} e^{2 i e} + B c^{3}\right ) e^{- 6 i e}}{a^{3}}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*B*c**3*x/a**3 + B*c**3*log(exp(2*I*f*x) + exp(-2*I*e))/(a**3*f) + Piecewise((-(12*B*a**6*c**3*f**2*exp(10
*I*e)*exp(-2*I*f*x) - 6*B*a**6*c**3*f**2*exp(8*I*e)*exp(-4*I*f*x) + (-2*I*A*a**6*c**3*f**2*exp(6*I*e) + 2*B*a*
*6*c**3*f**2*exp(6*I*e))*exp(-6*I*f*x))*exp(-12*I*e)/(12*a**9*f**3), Ne(12*a**9*f**3*exp(12*I*e), 0)), (x*(2*I
*B*c**3/a**3 + I*(-I*A*c**3 - 2*B*c**3*exp(6*I*e) + 2*B*c**3*exp(4*I*e) - 2*B*c**3*exp(2*I*e) + B*c**3)*exp(-6
*I*e)/a**3), True))

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