3.10.9 \(\int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx\) [909]

Optimal. Leaf size=71 \[ -\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \]

[Out]

-4*c^3*x/a-4*I*c^3*ln(cos(f*x+e))/a/f+c^3*tan(f*x+e)/a/f+4*I*c^3/f/(a+I*a*tan(f*x+e))

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Rubi [A]
time = 0.08, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \begin {gather*} \frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}-\frac {4 c^3 x}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(71)=142\).
time = 1.01, size = 234, normalized size = 3.30 \begin {gather*} \frac {i c^3 \sec ^2(e+f x) \left (-i \cos (3 e+2 f x)+i \cos (e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \cos (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \cos (e) \left (-3+2 \log \left (\cos ^2(e+f x)\right )\right )+\sin (e)+8 \text {ArcTan}(\tan (f x)) \cos (e) \cos (e+f x) (\cos (e+f x)+i \sin (e+f x))-2 \sin (e+2 f x)-\log \left (\cos ^2(e+f x)\right ) \sin (e+2 f x)-\sin (3 e+2 f x)-\log \left (\cos ^2(e+f x)\right ) \sin (3 e+2 f x)\right )}{2 a f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*c^3*Sec[e + f*x]^2*((-I)*Cos[3*e + 2*f*x] + I*Cos[e + 2*f*x]*Log[Cos[e + f*x]^2] + I*Cos[3*e + 2*f*x]*L
og[Cos[e + f*x]^2] + I*Cos[e]*(-3 + 2*Log[Cos[e + f*x]^2]) + Sin[e] + 8*ArcTan[Tan[f*x]]*Cos[e]*Cos[e + f*x]*(
Cos[e + f*x] + I*Sin[e + f*x]) - 2*Sin[e + 2*f*x] - Log[Cos[e + f*x]^2]*Sin[e + 2*f*x] - Sin[3*e + 2*f*x] - Lo
g[Cos[e + f*x]^2]*Sin[3*e + 2*f*x]))/(a*f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.17, size = 44, normalized size = 0.62

method	result	size
derivativedivides	\(\frac {c^{3} \left (\tan \left (f x +e \right )+4 i \ln \left (\tan \left (f x +e \right )-i\right )+\frac {4}{\tan \left (f x +e \right )-i}\right )}{f a}\)	\(44\)
default	\(\frac {c^{3} \left (\tan \left (f x +e \right )+4 i \ln \left (\tan \left (f x +e \right )-i\right )+\frac {4}{\tan \left (f x +e \right )-i}\right )}{f a}\)	\(44\)
risch	\(\frac {2 i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a f}-\frac {8 c^{3} x}{a}-\frac {8 c^{3} e}{a f}+\frac {2 i c^{3}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {4 i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a f}\)	\(93\)
norman	\(\frac {\frac {4 i c^{3}}{a f}+\frac {c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}-\frac {4 c^{3} x}{a}-\frac {4 c^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{a}+\frac {5 c^{3} \tan \left (f x +e \right )}{a f}}{1+\tan ^{2}\left (f x +e \right )}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{a f}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*c^3/a*(tan(f*x+e)+4*I*ln(tan(f*x+e)-I)+4/(tan(f*x+e)-I))

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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((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),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 [A]
time = 1.01, size = 125, normalized size = 1.76 \begin {gather*} -\frac {2 \, {\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{3} + 2 \, {\left (2 \, c^{3} f x - i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.22, size = 133, normalized size = 1.87 \begin {gather*} \frac {2 i c^{3}}{a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {2 i c^{3} e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (\frac {8 c^{3}}{a} + \frac {\left (- 8 c^{3} e^{2 i e} + 4 c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {8 c^{3} x}{a} - \frac {4 i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*c**3/(a*f*exp(2*I*e)*exp(2*I*f*x) + a*f) + Piecewise((2*I*c**3*exp(-2*I*e)*exp(-2*I*f*x)/(a*f), Ne(a*f*exp
(2*I*e), 0)), (x*(8*c**3/a + (-8*c**3*exp(2*I*e) + 4*c**3)*exp(-2*I*e)/a), True)) - 8*c**3*x/a - 4*I*c**3*log(
exp(2*I*f*x) + exp(-2*I*e))/(a*f)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (68) = 136\).
time = 0.57, size = 184, normalized size = 2.59 \begin {gather*} \frac {2 \, {\left (-\frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} + \frac {4 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} + \frac {2 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 i \, c^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a} - \frac {2 \, {\left (3 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 i \, c^{3}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-2*I*c^3*log(tan(1/2*f*x + 1/2*e) + 1)/a + 4*I*c^3*log(tan(1/2*f*x + 1/2*e) - I)/a - 2*I*c^3*log(tan(1/2*f*
x + 1/2*e) - 1)/a + (2*I*c^3*tan(1/2*f*x + 1/2*e)^2 - c^3*tan(1/2*f*x + 1/2*e) - 2*I*c^3)/((tan(1/2*f*x + 1/2*
e)^2 - 1)*a) - 2*(3*I*c^3*tan(1/2*f*x + 1/2*e)^2 + 8*c^3*tan(1/2*f*x + 1/2*e) - 3*I*c^3)/(a*(tan(1/2*f*x + 1/2
*e) - I)^2))/f

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Mupad [B]
time = 4.78, size = 64, normalized size = 0.90 \begin {gather*} \frac {c^3\,\mathrm {tan}\left (e+f\,x\right )}{a\,f}+\frac {c^3\,4{}\mathrm {i}}{a\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,4{}\mathrm {i}}{a\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*tan(e + f*x))/(a*f) + (c^3*4i)/(a*f*(tan(e + f*x)*1i + 1)) + (c^3*log(tan(e + f*x) - 1i)*4i)/(a*f)

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