3.10.2 \(\int \frac {(c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\) [902]

Optimal. Leaf size=58 \[ \frac {2 i c^2}{3 f (a+i a \tan (e+f x))^3}-\frac {i c^2}{2 a f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 58, 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 {2 i c^2}{3 f (a+i a \tan (e+f x))^3}-\frac {i c^2}{2 a f (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^2/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(((2*I)/3)*c^2)/(f*(a + I*a*Tan[e + f*x])^3) - ((I/2)*c^2)/(a*f*(a + I*a*Tan[e + f*x])^2)

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))^2}{(a+i a \tan (e+f x))^3} \, dx &=\left (a^2 c^2\right ) \int \frac {\sec ^4(e+f x)}{(a+i a \tan (e+f x))^5} \, dx\\ &=-\frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {a-x}{(a+x)^4} \, dx,x,i a \tan (e+f x)\right )}{a f}\\ &=-\frac {\left (i c^2\right ) \text {Subst}\left (\int \left (\frac {2 a}{(a+x)^4}-\frac {1}{(a+x)^3}\right ) \, dx,x,i a \tan (e+f x)\right )}{a f}\\ &=\frac {2 i c^2}{3 f (a+i a \tan (e+f x))^3}-\frac {i c^2}{2 a f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.97, size = 53, normalized size = 0.91 \begin {gather*} \frac {c^2 (5 \cos (e+f x)+i \sin (e+f x)) (i \cos (5 (e+f x))+\sin (5 (e+f x)))}{24 a^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^2/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(c^2*(5*Cos[e + f*x] + I*Sin[e + f*x])*(I*Cos[5*(e + f*x)] + Sin[5*(e + f*x)]))/(24*a^3*f)

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Maple [A]
time = 0.19, size = 39, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {c^{2} \left (\frac {i}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {2}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}\right )}{f \,a^{3}}\)	\(39\)
default	\(\frac {c^{2} \left (\frac {i}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {2}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}\right )}{f \,a^{3}}\)	\(39\)
risch	\(\frac {i c^{2} {\mathrm e}^{-4 i \left (f x +e \right )}}{8 a^{3} f}+\frac {i c^{2} {\mathrm e}^{-6 i \left (f x +e \right )}}{12 a^{3} f}\)	\(44\)
norman	\(\frac {\frac {c^{2} \tan \left (f x +e \right )}{a f}-\frac {2 i c^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{a f}+\frac {i c^{2}}{6 a f}-\frac {5 c^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {i c^{2} \left (\tan ^{4}\left (f x +e \right )\right )}{2 a f}}{a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

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

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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))^2/(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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Fricas [A]
time = 0.89, size = 39, normalized size = 0.67 \begin {gather*} \frac {{\left (3 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{2}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/24*(3*I*c^2*e^(2*I*f*x + 2*I*e) + 2*I*c^2)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (46) = 92\).
time = 0.23, size = 107, normalized size = 1.84 \begin {gather*} \begin {cases} \frac {\left (12 i a^{3} c^{2} f e^{6 i e} e^{- 4 i f x} + 8 i a^{3} c^{2} f e^{4 i e} e^{- 6 i f x}\right ) e^{- 10 i e}}{96 a^{6} f^{2}} & \text {for}\: a^{6} f^{2} e^{10 i e} \neq 0 \\\frac {x \left (c^{2} e^{2 i e} + c^{2}\right ) e^{- 6 i e}}{2 a^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**2/(a+I*a*tan(f*x+e))**3,x)

[Out]

Piecewise(((12*I*a**3*c**2*f*exp(6*I*e)*exp(-4*I*f*x) + 8*I*a**3*c**2*f*exp(4*I*e)*exp(-6*I*f*x))*exp(-10*I*e)
/(96*a**6*f**2), Ne(a**6*f**2*exp(10*I*e), 0)), (x*(c**2*exp(2*I*e) + c**2)*exp(-6*I*e)/(2*a**3), True))

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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. 106 vs. \(2 (48) = 96\).
time = 0.69, size = 106, normalized size = 1.83 \begin {gather*} -\frac {2 \, {\left (3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 i \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 8 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 i \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-2/3*(3*c^2*tan(1/2*f*x + 1/2*e)^5 - 3*I*c^2*tan(1/2*f*x + 1/2*e)^4 - 8*c^2*tan(1/2*f*x + 1/2*e)^3 + 3*I*c^2*t
an(1/2*f*x + 1/2*e)^2 + 3*c^2*tan(1/2*f*x + 1/2*e))/(a^3*f*(tan(1/2*f*x + 1/2*e) - I)^6)

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Mupad [B]
time = 4.62, size = 56, normalized size = 0.97 \begin {gather*} \frac {c^2\,\left (3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}{6\,a^3\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^2/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

(c^2*(3*tan(e + f*x) + 1i))/(6*a^3*f*(tan(e + f*x)*3i - 3*tan(e + f*x)^2 - tan(e + f*x)^3*1i + 1))

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