3.11.84 \(\int \frac {(a+i a \tan (e+f x))^2}{c+d \tan (e+f x)} \, dx\) [1084]

Optimal. Leaf size=106 \[ -\frac {a^2 c (c+i d) x}{(c-i d) d^2}+\frac {a^2 (c+2 i d) x}{d^2}+\frac {a^2 \log (\cos (e+f x))}{d f}-\frac {a^2 (i c-d) \log (c \cos (e+f x)+d \sin (e+f x))}{d (i c+d) f} \]

[Out]

-a^2*c*(c+I*d)*x/(c-I*d)/d^2+a^2*(c+2*I*d)*x/d^2+a^2*ln(cos(f*x+e))/d/f-a^2*(I*c-d)*ln(c*cos(f*x+e)+d*sin(f*x+
e))/d/(I*c+d)/f

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Rubi [A]
time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3622, 3556, 3565, 3611} \begin {gather*} -\frac {a^2 c x (c+i d)}{d^2 (c-i d)}+\frac {a^2 x (c+2 i d)}{d^2}-\frac {a^2 (-d+i c) \log (c \cos (e+f x)+d \sin (e+f x))}{d f (d+i c)}+\frac {a^2 \log (\cos (e+f x))}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*c*(c + I*d)*x)/((c - I*d)*d^2)) + (a^2*(c + (2*I)*d)*x)/d^2 + (a^2*Log[Cos[e + f*x]])/(d*f) - (a^2*(I*c
- d)*Log[c*Cos[e + f*x] + d*Sin[e + f*x]])/(d*(I*c + d)*f)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3565

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

Rule 3611

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

Rule 3622

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

Rubi steps

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

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Mathematica [A]
time = 1.81, size = 176, normalized size = 1.66 \begin {gather*} \frac {a^2 \left (8 d f x+2 (-i c+d) \text {ArcTan}\left (\frac {d \cos (3 e+f x)-c \sin (3 e+f x)}{c \cos (3 e+f x)+d \sin (3 e+f x)}\right )+(-2 i c-2 d) \text {ArcTan}(\tan (3 e+f x))+c \log \left (\cos ^2(e+f x)\right )-i d \log \left (\cos ^2(e+f x)\right )-c \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )-i d \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )\right )}{2 (c-i d) d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*d*f*x + 2*((-I)*c + d)*ArcTan[(d*Cos[3*e + f*x] - c*Sin[3*e + f*x])/(c*Cos[3*e + f*x] + d*Sin[3*e + f*
x])] + ((-2*I)*c - 2*d)*ArcTan[Tan[3*e + f*x]] + c*Log[Cos[e + f*x]^2] - I*d*Log[Cos[e + f*x]^2] - c*Log[(c*Co
s[e + f*x] + d*Sin[e + f*x])^2] - I*d*Log[(c*Cos[e + f*x] + d*Sin[e + f*x])^2]))/(2*(c - I*d)*d*f)

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Maple [A]
time = 0.21, size = 95, normalized size = 0.90

method	result	size
norman	\(\frac {2 a^{2} x}{-i d +c}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f \left (-i d +c \right )}-\frac {\left (i a^{2} d +a^{2} c \right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d f \left (-i d +c \right )}\)	\(84\)
derivativedivides	\(\frac {a^{2} \left (\frac {\left (-2 i c d -c^{2}+d^{2}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right ) d}+\frac {\frac {\left (2 i c -2 d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (2 i d +2 c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}\right )}{f}\)	\(95\)
default	\(\frac {a^{2} \left (\frac {\left (-2 i c d -c^{2}+d^{2}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (c^{2}+d^{2}\right ) d}+\frac {\frac {\left (2 i c -2 d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (2 i d +2 c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}\right )}{f}\)	\(95\)
risch	\(-\frac {2 a^{2} x}{i d -c}-\frac {2 i a^{2} x}{d}-\frac {2 i a^{2} e}{d f}+\frac {2 a^{2} e}{f \left (i d -c \right )}-\frac {2 i a^{2} c x}{d \left (i d -c \right )}-\frac {2 i a^{2} c e}{d f \left (i d -c \right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{d f}+\frac {i a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{f \left (i d -c \right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) c}{d f \left (i d -c \right )}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 117, normalized size = 1.10 \begin {gather*} \frac {\frac {2 \, {\left (a^{2} c + i \, a^{2} d\right )} {\left (f x + e\right )}}{c^{2} + d^{2}} - \frac {{\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{2} d + d^{3}} - \frac {{\left (-i \, a^{2} c + a^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.04, size = 86, normalized size = 0.81 \begin {gather*} \frac {{\left (-i \, a^{2} c + a^{2} d\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right ) + {\left (i \, a^{2} c + a^{2} d\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (i \, c d + d^{2}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 4.22, size = 92, normalized size = 0.87 \begin {gather*} \frac {a^{2} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{d f} - \frac {a^{2} \left (c + i d\right ) \log {\left (e^{2 i f x} + \frac {\left (a^{2} c + i a^{2} d + \frac {i a^{2} d \left (c + i d\right )}{c - i d}\right ) e^{- 2 i e}}{a^{2} c} \right )}}{d f \left (c - i d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(exp(2*I*f*x) + exp(-2*I*e))/(d*f) - a**2*(c + I*d)*log(exp(2*I*f*x) + (a**2*c + I*a**2*d + I*a**2*d*(
c + I*d)/(c - I*d))*exp(-2*I*e)/(a**2*c))/(d*f*(c - I*d))

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Giac [A]
time = 0.54, size = 127, normalized size = 1.20 \begin {gather*} \frac {\frac {a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{d} + \frac {4 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c - i \, d} + \frac {a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{d} - \frac {{\left (a^{2} c + i \, a^{2} d\right )} \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{c d - i \, d^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.75, size = 64, normalized size = 0.60 \begin {gather*} \frac {a^2\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{f\,\left (c-d\,1{}\mathrm {i}\right )}-\frac {a^2\,\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f\,\left (c-d\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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