3.10.31 \(\int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^2} \, dx\) [931]

Optimal. Leaf size=101 \[ \frac {6 a^4 x}{c^2}-\frac {6 i a^4 \log (\cos (e+f x))}{c^2 f}-\frac {a^4 \tan (e+f x)}{c^2 f}-\frac {4 i a^4}{f (c-i c \tan (e+f x))^2}+\frac {12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )} \]

[Out]

6*a^4*x/c^2-6*I*a^4*ln(cos(f*x+e))/c^2/f-a^4*tan(f*x+e)/c^2/f-4*I*a^4/f/(c-I*c*tan(f*x+e))^2+12*I*a^4/f/(c^2-I
*c^2*tan(f*x+e))

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Rubi [A]
time = 0.10, antiderivative size = 101, 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 {a^4 \tan (e+f x)}{c^2 f}+\frac {12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )}-\frac {6 i a^4 \log (\cos (e+f x))}{c^2 f}+\frac {6 a^4 x}{c^2}-\frac {4 i a^4}{f (c-i c \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^2,x]

[Out]

(6*a^4*x)/c^2 - ((6*I)*a^4*Log[Cos[e + f*x]])/(c^2*f) - (a^4*Tan[e + f*x])/(c^2*f) - ((4*I)*a^4)/(f*(c - I*c*T
an[e + f*x])^2) + ((12*I)*a^4)/(f*(c^2 - I*c^2*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 {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^2} \, dx &=\left (a^4 c^4\right ) \int \frac {\sec ^8(e+f x)}{(c-i c \tan (e+f x))^6} \, dx\\ &=\frac {\left (i a^4\right ) \text {Subst}\left (\int \frac {(c-x)^3}{(c+x)^3} \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=\frac {\left (i a^4\right ) \text {Subst}\left (\int \left (-1+\frac {8 c^3}{(c+x)^3}-\frac {12 c^2}{(c+x)^2}+\frac {6 c}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=\frac {6 a^4 x}{c^2}-\frac {6 i a^4 \log (\cos (e+f x))}{c^2 f}-\frac {a^4 \tan (e+f x)}{c^2 f}-\frac {4 i a^4}{f (c-i c \tan (e+f x))^2}+\frac {12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )}\\ \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. \(374\) vs. \(2(101)=202\).
time = 1.55, size = 374, normalized size = 3.70 \begin {gather*} \frac {a^4 \sec (e) \sec (e+f x) (\cos (2 (e+3 f x))+i \sin (2 (e+3 f x))) \left (-3 i \cos (2 e+3 f x)+6 f x \cos (2 e+3 f x)-i \cos (4 e+3 f x)+6 f x \cos (4 e+3 f x)+\cos (f x) \left (7 i+6 f x-3 i \log \left (\cos ^2(e+f x)\right )\right )+\cos (2 e+f x) \left (9 i+6 f x-3 i \log \left (\cos ^2(e+f x)\right )\right )-3 i \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 i \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )+\sin (f x)-6 i f x \sin (f x)-3 \log \left (\cos ^2(e+f x)\right ) \sin (f x)+3 \sin (2 e+f x)-6 i f x \sin (2 e+f x)-3 \log \left (\cos ^2(e+f x)\right ) \sin (2 e+f x)-\sin (2 e+3 f x)-6 i f x \sin (2 e+3 f x)-3 \log \left (\cos ^2(e+f x)\right ) \sin (2 e+3 f x)+\sin (4 e+3 f x)-6 i f x \sin (4 e+3 f x)-3 \log \left (\cos ^2(e+f x)\right ) \sin (4 e+3 f x)\right )}{4 c^2 f (\cos (f x)+i \sin (f x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^2,x]

[Out]

(a^4*Sec[e]*Sec[e + f*x]*(Cos[2*(e + 3*f*x)] + I*Sin[2*(e + 3*f*x)])*((-3*I)*Cos[2*e + 3*f*x] + 6*f*x*Cos[2*e
+ 3*f*x] - I*Cos[4*e + 3*f*x] + 6*f*x*Cos[4*e + 3*f*x] + Cos[f*x]*(7*I + 6*f*x - (3*I)*Log[Cos[e + f*x]^2]) +
Cos[2*e + f*x]*(9*I + 6*f*x - (3*I)*Log[Cos[e + f*x]^2]) - (3*I)*Cos[2*e + 3*f*x]*Log[Cos[e + f*x]^2] - (3*I)*
Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] + Sin[f*x] - (6*I)*f*x*Sin[f*x] - 3*Log[Cos[e + f*x]^2]*Sin[f*x] + 3*Sin[
2*e + f*x] - (6*I)*f*x*Sin[2*e + f*x] - 3*Log[Cos[e + f*x]^2]*Sin[2*e + f*x] - Sin[2*e + 3*f*x] - (6*I)*f*x*Si
n[2*e + 3*f*x] - 3*Log[Cos[e + f*x]^2]*Sin[2*e + 3*f*x] + Sin[4*e + 3*f*x] - (6*I)*f*x*Sin[4*e + 3*f*x] - 3*Lo
g[Cos[e + f*x]^2]*Sin[4*e + 3*f*x]))/(4*c^2*f*(Cos[f*x] + I*Sin[f*x])^4)

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Maple [A]
time = 0.21, size = 60, normalized size = 0.59

method	result	size
derivativedivides	\(\frac {a^{4} \left (-\tan \left (f x +e \right )+\frac {4 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}+6 i \ln \left (\tan \left (f x +e \right )+i\right )-\frac {12}{\tan \left (f x +e \right )+i}\right )}{f \,c^{2}}\)	\(60\)
default	\(\frac {a^{4} \left (-\tan \left (f x +e \right )+\frac {4 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}+6 i \ln \left (\tan \left (f x +e \right )+i\right )-\frac {12}{\tan \left (f x +e \right )+i}\right )}{f \,c^{2}}\)	\(60\)
risch	\(-\frac {i a^{4} {\mathrm e}^{4 i \left (f x +e \right )}}{c^{2} f}+\frac {4 i a^{4} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} f}-\frac {12 a^{4} e}{f \,c^{2}}-\frac {2 i a^{4}}{f \,c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {6 i a^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f \,c^{2}}\)	\(105\)
norman	\(\frac {\frac {8 i a^{4}}{c f}+\frac {6 a^{4} x}{c}+\frac {12 a^{4} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}+\frac {6 a^{4} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}-\frac {5 a^{4} \tan \left (f x +e \right )}{c f}-\frac {14 a^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{c f}-\frac {a^{4} \left (\tan ^{5}\left (f x +e \right )\right )}{c f}+\frac {16 i a^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}}{c \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {3 i a^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{c^{2} f}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/f*a^4/c^2*(-tan(f*x+e)+4*I/(tan(f*x+e)+I)^2+6*I*ln(tan(f*x+e)+I)-12/(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((a+I*a*tan(f*x+e))^4/(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 [A]
time = 1.03, size = 111, normalized size = 1.10 \begin {gather*} \frac {-i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{4} - 6 \, {\left (i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{4}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.33, size = 155, normalized size = 1.53 \begin {gather*} - \frac {2 i a^{4}}{c^{2} f e^{2 i e} e^{2 i f x} + c^{2} f} - \frac {6 i a^{4} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin {cases} \frac {- i a^{4} c^{2} f e^{4 i e} e^{4 i f x} + 4 i a^{4} c^{2} f e^{2 i e} e^{2 i f x}}{c^{4} f^{2}} & \text {for}\: c^{4} f^{2} \neq 0 \\\frac {x \left (4 a^{4} e^{4 i e} - 8 a^{4} e^{2 i e}\right )}{c^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**4/(c-I*c*tan(f*x+e))**2,x)

[Out]

-2*I*a**4/(c**2*f*exp(2*I*e)*exp(2*I*f*x) + c**2*f) - 6*I*a**4*log(exp(2*I*f*x) + exp(-2*I*e))/(c**2*f) + Piec
ewise(((-I*a**4*c**2*f*exp(4*I*e)*exp(4*I*f*x) + 4*I*a**4*c**2*f*exp(2*I*e)*exp(2*I*f*x))/(c**4*f**2), Ne(c**4
*f**2, 0)), (x*(4*a**4*exp(4*I*e) - 8*a**4*exp(2*I*e))/c**2, 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. 217 vs. \(2 (95) = 190\).
time = 0.71, size = 217, normalized size = 2.15 \begin {gather*} -\frac {\frac {6 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{2}} - \frac {12 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{2}} + \frac {6 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{2}} - \frac {2 \, {\left (3 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 i \, a^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2}} + \frac {25 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 108 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 182 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 108 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 25 i \, a^{4}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{4}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-(6*I*a^4*log(tan(1/2*f*x + 1/2*e) + 1)/c^2 - 12*I*a^4*log(tan(1/2*f*x + 1/2*e) + I)/c^2 + 6*I*a^4*log(tan(1/2
*f*x + 1/2*e) - 1)/c^2 - 2*(3*I*a^4*tan(1/2*f*x + 1/2*e)^2 + a^4*tan(1/2*f*x + 1/2*e) - 3*I*a^4)/((tan(1/2*f*x
+ 1/2*e)^2 - 1)*c^2) + (25*I*a^4*tan(1/2*f*x + 1/2*e)^4 - 108*a^4*tan(1/2*f*x + 1/2*e)^3 - 182*I*a^4*tan(1/2*
f*x + 1/2*e)^2 + 108*a^4*tan(1/2*f*x + 1/2*e) + 25*I*a^4)/(c^2*(tan(1/2*f*x + 1/2*e) + I)^4))/f

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Mupad [B]
time = 4.68, size = 90, normalized size = 0.89 \begin {gather*} -\frac {\frac {12\,a^4\,\mathrm {tan}\left (e+f\,x\right )}{c^2}+\frac {a^4\,8{}\mathrm {i}}{c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}-\frac {a^4\,\mathrm {tan}\left (e+f\,x\right )}{c^2\,f}+\frac {a^4\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,6{}\mathrm {i}}{c^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^4/(c - c*tan(e + f*x)*1i)^2,x)

[Out]

(a^4*log(tan(e + f*x) + 1i)*6i)/(c^2*f) - (a^4*tan(e + f*x))/(c^2*f) - ((a^4*8i)/c^2 + (12*a^4*tan(e + f*x))/c
^2)/(f*(tan(e + f*x)*2i + tan(e + f*x)^2 - 1))

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