3.1075 \(\int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=91 \[ \frac {(c+i d) (3 d+i c)}{4 a^2 f (1+i \tan (e+f x))}+\frac {x (c-i d)^2}{4 a^2}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3540, 3526, 8} \[ \frac {(c+i d) (3 d+i c)}{4 a^2 f (1+i \tan (e+f x))}+\frac {x (c-i d)^2}{4 a^2}+\frac {i (c+i d)^2}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)^2*x)/(4*a^2) + ((c + I*d)*(I*c + 3*d))/(4*a^2*f*(1 + I*Tan[e + f*x])) + ((I/4)*(c + I*d)^2)/(f*(a +
I*a*Tan[e + f*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 1.31, size = 134, normalized size = 1.47 \[ -\frac {\sec ^2(e+f x) \left (\left (c^2 (1+4 i f x)+2 c d (4 f x+i)+d^2 (-1-4 i f x)\right ) \sin (2 (e+f x))+\left (c^2 (4 f x+i)+c d (-2-8 i f x)-d^2 (4 f x+i)\right ) \cos (2 (e+f x))+4 i \left (c^2+d^2\right )\right )}{16 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(Sec[e + f*x]^2*((4*I)*(c^2 + d^2) + (c*d*(-2 - (8*I)*f*x) + c^2*(I + 4*f*x) - d^2*(I + 4*f*x))*Cos[2*(e
+ f*x)] + (d^2*(-1 - (4*I)*f*x) + c^2*(1 + (4*I)*f*x) + 2*c*d*(I + 4*f*x))*Sin[2*(e + f*x)]))/(a^2*f*(-I + Ta
n[e + f*x])^2)

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fricas [A]  time = 0.42, size = 80, normalized size = 0.88 \[ \frac {{\left ({\left (4 \, c^{2} - 8 i \, c d - 4 \, d^{2}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{2} - 2 \, c d - i \, d^{2} + {\left (4 i \, c^{2} + 4 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*c^2 - 8*I*c*d - 4*d^2)*f*x*e^(4*I*f*x + 4*I*e) + I*c^2 - 2*c*d - I*d^2 + (4*I*c^2 + 4*I*d^2)*e^(2*I*f
*x + 2*I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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giac [B]  time = 0.69, size = 176, normalized size = 1.93 \[ -\frac {\frac {2 \, {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2}} + \frac {2 \, {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2}} + \frac {-3 i \, c^{2} \tan \left (f x + e\right )^{2} - 6 \, c d \tan \left (f x + e\right )^{2} + 3 i \, d^{2} \tan \left (f x + e\right )^{2} - 10 \, c^{2} \tan \left (f x + e\right ) + 20 i \, c d \tan \left (f x + e\right ) - 6 \, d^{2} \tan \left (f x + e\right ) + 11 i \, c^{2} + 6 \, c d + 5 i \, d^{2}}{a^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(-I*c^2 - 2*c*d + I*d^2)*log(-I*tan(f*x + e) + 1)/a^2 + 2*(I*c^2 + 2*c*d - I*d^2)*log(-I*tan(f*x + e)
- 1)/a^2 + (-3*I*c^2*tan(f*x + e)^2 - 6*c*d*tan(f*x + e)^2 + 3*I*d^2*tan(f*x + e)^2 - 10*c^2*tan(f*x + e) + 2
0*I*c*d*tan(f*x + e) - 6*d^2*tan(f*x + e) + 11*I*c^2 + 6*c*d + 5*I*d^2)/(a^2*(tan(f*x + e) - I)^2))/f

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maple [B]  time = 0.22, size = 263, normalized size = 2.89 \[ \frac {\ln \left (\tan \left (f x +e \right )+i\right ) c d}{4 f \,a^{2}}+\frac {i \ln \left (\tan \left (f x +e \right )+i\right ) c^{2}}{8 f \,a^{2}}-\frac {i \ln \left (\tan \left (f x +e \right )+i\right ) d^{2}}{8 f \,a^{2}}-\frac {i c d}{2 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {c^{2}}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {3 d^{2}}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {c d}{2 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i c^{2}}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {i d^{2}}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) d^{2}}{8 f \,a^{2}}-\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) c^{2}}{8 f \,a^{2}}-\frac {\ln \left (\tan \left (f x +e \right )-i\right ) c d}{4 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^2,x)

[Out]

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

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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((c+d*tan(f*x+e))^2/(a+I*a*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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mupad [B]  time = 5.24, size = 93, normalized size = 1.02 \[ -\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^2}{4\,a^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c\,d}{2\,a^2}+\frac {c^2\,1{}\mathrm {i}}{4\,a^2}+\frac {d^2\,3{}\mathrm {i}}{4\,a^2}\right )+\frac {c^2}{2\,a^2}+\frac {d^2}{2\,a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e + f*x)*((c^2*1i)/(4*a^2) + (d^2*3i)/(4*a^2) + (c*d)/(2*a^2)) + c^2/(2*a^2) + d^2/(2*a^2))/(f*(2*tan(e +
f*x) + tan(e + f*x)^2*1i - 1i)) - (x*(c*1i + d)^2)/(4*a^2)

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sympy [A]  time = 0.48, size = 260, normalized size = 2.86 \[ \begin {cases} \frac {\left (\left (16 i a^{2} c^{2} f e^{4 i e} + 16 i a^{2} d^{2} f e^{4 i e}\right ) e^{- 2 i f x} + \left (4 i a^{2} c^{2} f e^{2 i e} - 8 a^{2} c d f e^{2 i e} - 4 i a^{2} d^{2} f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{64 a^{4} f^{2}} & \text {for}\: 64 a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {c^{2} - 2 i c d - d^{2}}{4 a^{2}} + \frac {\left (c^{2} e^{4 i e} + 2 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{4 i e} + 2 i c d - d^{2} e^{4 i e} + 2 d^{2} e^{2 i e} - d^{2}\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- c^{2} + 2 i c d + d^{2}\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((16*I*a**2*c**2*f*exp(4*I*e) + 16*I*a**2*d**2*f*exp(4*I*e))*exp(-2*I*f*x) + (4*I*a**2*c**2*f*exp(2
*I*e) - 8*a**2*c*d*f*exp(2*I*e) - 4*I*a**2*d**2*f*exp(2*I*e))*exp(-4*I*f*x))*exp(-6*I*e)/(64*a**4*f**2), Ne(64
*a**4*f**2*exp(6*I*e), 0)), (x*(-(c**2 - 2*I*c*d - d**2)/(4*a**2) + (c**2*exp(4*I*e) + 2*c**2*exp(2*I*e) + c**
2 - 2*I*c*d*exp(4*I*e) + 2*I*c*d - d**2*exp(4*I*e) + 2*d**2*exp(2*I*e) - d**2)*exp(-4*I*e)/(4*a**2)), True)) -
x*(-c**2 + 2*I*c*d + d**2)/(4*a**2)

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