3.13.13 \(\int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))} \, dx\) [1213]

Optimal. Leaf size=118 \[ \frac {(a c-b d) x}{\left (a^2+b^2\right ) \left (c^2+d^2\right )}+\frac {b^2 \log (a \cos (e+f x)+b \sin (e+f x))}{\left (a^2+b^2\right ) (b c-a d) f}-\frac {d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{(b c-a d) \left (c^2+d^2\right ) f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3652, 3611} \begin {gather*} \frac {x (a c-b d)}{\left (a^2+b^2\right ) \left (c^2+d^2\right )}+\frac {b^2 \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right ) (b c-a d)}-\frac {d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])),x]

[Out]

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

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 3652

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.37, size = 143, normalized size = 1.21 \begin {gather*} \frac {\frac {\log (i-\tan (e+f x))}{(a+i b) (i c-d)}-\frac {\log (i+\tan (e+f x))}{(i a+b) (c-i d)}+\frac {2 b^2 \log (a+b \tan (e+f x))}{\left (a^2+b^2\right ) (b c-a d)}+\frac {2 d^2 \log (c+d \tan (e+f x))}{(-b c+a d) \left (c^2+d^2\right )}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])),x]

[Out]

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

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Maple [A]
time = 0.26, size = 133, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-a d -b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a c -b d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{\left (a^{2}+b^{2}\right ) \left (c^{2}+d^{2}\right )}+\frac {d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (a d -b c \right ) \left (c^{2}+d^{2}\right )}-\frac {b^{2} \ln \left (a +b \tan \left (f x +e \right )\right )}{\left (a^{2}+b^{2}\right ) \left (a d -b c \right )}}{f}\)	\(133\)
default	\(\frac {\frac {\frac {\left (-a d -b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a c -b d \right ) \arctan \left (\tan \left (f x +e \right )\right )}{\left (a^{2}+b^{2}\right ) \left (c^{2}+d^{2}\right )}+\frac {d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{\left (a d -b c \right ) \left (c^{2}+d^{2}\right )}-\frac {b^{2} \ln \left (a +b \tan \left (f x +e \right )\right )}{\left (a^{2}+b^{2}\right ) \left (a d -b c \right )}}{f}\)	\(133\)
norman	\(\frac {\left (a c -b d \right ) x}{\left (a^{2}+b^{2}\right ) \left (c^{2}+d^{2}\right )}+\frac {d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{f \left (a \,c^{2} d +a \,d^{3}-b \,c^{3}-b c \,d^{2}\right )}-\frac {b^{2} \ln \left (a +b \tan \left (f x +e \right )\right )}{\left (a d -b c \right ) f \left (a^{2}+b^{2}\right )}-\frac {\left (a d +b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 \left (a^{2}+b^{2}\right ) \left (c^{2}+d^{2}\right ) f}\)	\(154\)
risch	\(-\frac {x}{i a d +i b c -a c +b d}+\frac {2 i b^{2} x}{a^{3} d -a^{2} b c +a \,b^{2} d -b^{3} c}+\frac {2 i b^{2} e}{f \left (a^{3} d -a^{2} b c +a \,b^{2} d -b^{3} c \right )}-\frac {2 i d^{2} x}{a \,c^{2} d +a \,d^{3}-b \,c^{3}-b c \,d^{2}}-\frac {2 i d^{2} e}{f \left (a \,c^{2} d +a \,d^{3}-b \,c^{3}-b c \,d^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i b +a}{i b -a}\right )}{f \left (a^{3} d -a^{2} b c +a \,b^{2} d -b^{3} c \right )}+\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{f \left (a \,c^{2} d +a \,d^{3}-b \,c^{3}-b c \,d^{2}\right )}\)	\(294\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.55, size = 180, normalized size = 1.53 \begin {gather*} \frac {\frac {2 \, b^{2} \log \left (b \tan \left (f x + e\right ) + a\right )}{{\left (a^{2} b + b^{3}\right )} c - {\left (a^{3} + a b^{2}\right )} d} - \frac {2 \, d^{2} \log \left (d \tan \left (f x + e\right ) + c\right )}{b c^{3} - a c^{2} d + b c d^{2} - a d^{3}} + \frac {2 \, {\left (a c - b d\right )} {\left (f x + e\right )}}{{\left (a^{2} + b^{2}\right )} c^{2} + {\left (a^{2} + b^{2}\right )} d^{2}} - \frac {{\left (b c + a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{{\left (a^{2} + b^{2}\right )} c^{2} + {\left (a^{2} + b^{2}\right )} d^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.06, size = 207, normalized size = 1.75 \begin {gather*} -\frac {{\left (a^{2} + b^{2}\right )} d^{2} \log \left (\frac {d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left (a b c^{2} + a b d^{2} - {\left (a^{2} + b^{2}\right )} c d\right )} f x - {\left (b^{2} c^{2} + b^{2} d^{2}\right )} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left ({\left (a^{2} b + b^{3}\right )} c^{3} - {\left (a^{3} + a b^{2}\right )} c^{2} d + {\left (a^{2} b + b^{3}\right )} c d^{2} - {\left (a^{3} + a b^{2}\right )} d^{3}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 11.75, size = 8053, normalized size = 68.25 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e)),x)

[Out]

Piecewise(((2*c*f*x/(2*c**2*f + 2*d**2*f) + 2*d*log(c/d + tan(e + f*x))/(2*c**2*f + 2*d**2*f) - d*log(tan(e +
f*x)**2 + 1)/(2*c**2*f + 2*d**2*f))/a, Eq(b, 0)), ((2*a*f*x/(2*a**2*f + 2*b**2*f) + 2*b*log(a/b + tan(e + f*x)
)/(2*a**2*f + 2*b**2*f) - b*log(tan(e + f*x)**2 + 1)/(2*a**2*f + 2*b**2*f))/c, Eq(d, 0)), (I*c**2*f*x*tan(e +
f*x)/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e
+ f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + c**2*f*x/(2*b*c**3*f*tan(e + f*x) - 2*I*b
*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3
*f*tan(e + f*x) + 2*b*d**3*f) + I*c**2/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) +
2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - 2*c*d*f
*x*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d
**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + 2*I*c*d*f*x/(2*b*c**3*f*tan(e
+ f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*
f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + I*d**2*f*x*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f
+ 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e
+ f*x) + 2*b*d**3*f) + d**2*f*x/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c
**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - 2*d**2*log(c/
d + tan(e + f*x))*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**
2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + 2*I*d**2*log(c/
d + tan(e + f*x))/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c
*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + d**2*log(tan(e + f*x)**2 + 1
)*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d*
*2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - I*d**2*log(tan(e + f*x)**2 + 1)
/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f
*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + I*d**2/(2*b*c**3*f*tan(e + f*x) - 2*I*b*c**3*
f + 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) - 2*I*b*c*d**2*f + 2*I*b*d**3*f*tan
(e + f*x) + 2*b*d**3*f), Eq(a, -I*b)), (-I*c**2*f*x*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I
*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*tan(e + f*
x) + 2*b*d**3*f) + c**2*f*x/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d
*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - I*c**2/(2*b*c**3*f
*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*
c*d**2*f - 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - 2*c*d*f*x*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c
**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f
*tan(e + f*x) + 2*b*d**3*f) - 2*I*c*d*f*x/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x
) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - I*d*
*2*f*x*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b
*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) + d**2*f*x/(2*b*c**3*f*tan(e
+ f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2
*f - 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - 2*d**2*log(c/d + tan(e + f*x))*tan(e + f*x)/(2*b*c**3*f*tan(e +
f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f
- 2*I*b*d**3*f*tan(e + f*x) + 2*b*d**3*f) - 2*I*d**2*log(c/d + tan(e + f*x))/(2*b*c**3*f*tan(e + f*x) + 2*I*b
*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3
*f*tan(e + f*x) + 2*b*d**3*f) + d**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c*
*3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*
tan(e + f*x) + 2*b*d**3*f) + I*d**2*log(tan(e + f*x)**2 + 1)/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c
**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I*b*d**3*f*tan(e + f*x) +
2*b*d**3*f) - I*d**2/(2*b*c**3*f*tan(e + f*x) + 2*I*b*c**3*f - 2*I*b*c**2*d*f*tan(e + f*x) + 2*b*c**2*d*f + 2
*b*c*d**2*f*tan(e + f*x) + 2*I*b*c*d**2*f - 2*I...

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Giac [A]
time = 0.51, size = 201, normalized size = 1.70 \begin {gather*} \frac {\frac {2 \, b^{3} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{2} b^{2} c + b^{4} c - a^{3} b d - a b^{3} d} - \frac {2 \, d^{3} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{b c^{3} d - a c^{2} d^{2} + b c d^{3} - a d^{4}} + \frac {2 \, {\left (a c - b d\right )} {\left (f x + e\right )}}{a^{2} c^{2} + b^{2} c^{2} + a^{2} d^{2} + b^{2} d^{2}} - \frac {{\left (b c + a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} c^{2} + b^{2} c^{2} + a^{2} d^{2} + b^{2} d^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))/(c+d*tan(f*x+e)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.75, size = 173, normalized size = 1.47 \begin {gather*} \frac {d^2\,\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}{f\,\left (a\,d-b\,c\right )\,\left (c^2+d^2\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}{2\,f\,\left (a\,c\,1{}\mathrm {i}+a\,d+b\,c-b\,d\,1{}\mathrm {i}\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (\frac {d^2}{\left (a\,d-b\,c\right )\,\left (c^2+d^2\right )}-\frac {a\,d+b\,c}{\left (a^2+b^2\right )\,\left (c^2+d^2\right )}\right )}{f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}{2\,f\,\left (a\,d-a\,c\,1{}\mathrm {i}+b\,c+b\,d\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*tan(e + f*x))*(c + d*tan(e + f*x))),x)

[Out]

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

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