3.1218 \(\int \frac{(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=126 \[ -\frac{(b c-a d)^2}{d f \left (c^2+d^2\right ) (c+d \tan (e+f x))}-\frac{2 (a c+b d) (b c-a d) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^2}-\frac{x (b (c-d)-a (c+d)) (a (c-d)+b (c+d))}{\left (c^2+d^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.222707, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3542, 3531, 3530} \[ -\frac{(b c-a d)^2}{d f \left (c^2+d^2\right ) (c+d \tan (e+f x))}-\frac{2 (a c+b d) (b c-a d) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^2}-\frac{x (b (c-d)-a (c+d)) (a (c-d)+b (c+d))}{\left (c^2+d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3542

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

Rule 3531

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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]

Rubi steps

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

Mathematica [C]  time = 1.83363, size = 320, normalized size = 2.54 \[ \frac{(a+b \tan (e+f x))^2 (c \cos (e+f x)+d \sin (e+f x)) \left (2 i (e+f x) \left (a^2 c d+a b \left (d^2-c^2\right )-b^2 c d\right ) (c \cos (e+f x)+d \sin (e+f x))+\left (a^2 c d+a b \left (d^2-c^2\right )-b^2 c d\right ) (c \cos (e+f x)+d \sin (e+f x)) \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )+2 i \left (a^2 (-c) d+a b \left (c^2-d^2\right )+b^2 c d\right ) \tan ^{-1}(\tan (e+f x)) (c \cos (e+f x)+d \sin (e+f x))+\frac{\left (c^2+d^2\right ) (b c-a d)^2 \sin (e+f x)}{c}+(e+f x) (a (c+d)+b (d-c)) (a (c-d)+b (c+d)) (c \cos (e+f x)+d \sin (e+f x))\right )}{f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))^2 (a \cos (e+f x)+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.035, size = 465, normalized size = 3.7 \begin{align*} -{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ab{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ab{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{2}cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+4\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) abcd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{{a}^{2}d}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+2\,{\frac{abc}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{{b}^{2}{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) d \left ( c+d\tan \left ( fx+e \right ) \right ) }}+2\,{\frac{{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) ab{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+2\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) ab{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ){b}^{2}cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*a^2/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*c*d+1/f/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*a*b*c^2-1/f/(c^2+d^2)^2*ln(1+ta
n(f*x+e)^2)*a*b*d^2+1/f/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*b^2*c*d+1/f*a^2/(c^2+d^2)^2*arctan(tan(f*x+e))*c^2-1/f*
a^2/(c^2+d^2)^2*arctan(tan(f*x+e))*d^2+4/f/(c^2+d^2)^2*arctan(tan(f*x+e))*a*b*c*d-1/f/(c^2+d^2)^2*arctan(tan(f
*x+e))*b^2*c^2+1/f/(c^2+d^2)^2*arctan(tan(f*x+e))*b^2*d^2-1/f*a^2/(c^2+d^2)*d/(c+d*tan(f*x+e))+2/f/(c^2+d^2)/(
c+d*tan(f*x+e))*a*b*c-1/f/(c^2+d^2)/d/(c+d*tan(f*x+e))*b^2*c^2+2/f*a^2/(c^2+d^2)^2*ln(c+d*tan(f*x+e))*c*d-2/f/
(c^2+d^2)^2*ln(c+d*tan(f*x+e))*a*b*c^2+2/f/(c^2+d^2)^2*ln(c+d*tan(f*x+e))*a*b*d^2-2/f/(c^2+d^2)^2*ln(c+d*tan(f
*x+e))*b^2*c*d

________________________________________________________________________________________

Maxima [A]  time = 1.78939, size = 309, normalized size = 2.45 \begin{align*} \frac{\frac{{\left (4 \, a b c d +{\left (a^{2} - b^{2}\right )} c^{2} -{\left (a^{2} - b^{2}\right )} d^{2}\right )}{\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac{2 \,{\left (a b c^{2} - a b d^{2} -{\left (a^{2} - b^{2}\right )} c d\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{{\left (a b c^{2} - a b d^{2} -{\left (a^{2} - b^{2}\right )} c d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{3} d + c d^{3} +{\left (c^{2} d^{2} + d^{4}\right )} \tan \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.55563, size = 612, normalized size = 4.86 \begin{align*} -\frac{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3} -{\left (4 \, a b c^{2} d +{\left (a^{2} - b^{2}\right )} c^{3} -{\left (a^{2} - b^{2}\right )} c d^{2}\right )} f x +{\left (a b c^{3} - a b c d^{2} -{\left (a^{2} - b^{2}\right )} c^{2} d +{\left (a b c^{2} d - a b d^{3} -{\left (a^{2} - b^{2}\right )} c d^{2}\right )} \tan \left (f x + e\right )\right )} \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 ) -{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (4 \, a b c d^{2} +{\left (a^{2} - b^{2}\right )} c^{2} d -{\left (a^{2} - b^{2}\right )} d^{3}\right )} f x\right )} \tan \left (f x + e\right )}{{\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f \tan \left (f x + e\right ) +{\left (c^{5} + 2 \, c^{3} d^{2} + c d^{4}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3 - (4*a*b*c^2*d + (a^2 - b^2)*c^3 - (a^2 - b^2)*c*d^2)*f*x + (a*b*c^3 - a*b
*c*d^2 - (a^2 - b^2)*c^2*d + (a*b*c^2*d - a*b*d^3 - (a^2 - b^2)*c*d^2)*tan(f*x + e))*log((d^2*tan(f*x + e)^2 +
2*c*d*tan(f*x + e) + c^2)/(tan(f*x + e)^2 + 1)) - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (4*a*b*c*d^2 + (a^2 -
b^2)*c^2*d - (a^2 - b^2)*d^3)*f*x)*tan(f*x + e))/((c^4*d + 2*c^2*d^3 + d^5)*f*tan(f*x + e) + (c^5 + 2*c^3*d^2
+ c*d^4)*f)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [B]  time = 1.59364, size = 447, normalized size = 3.55 \begin{align*} \frac{\frac{{\left (a^{2} c^{2} - b^{2} c^{2} + 4 \, a b c d - a^{2} d^{2} + b^{2} d^{2}\right )}{\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{{\left (a b c^{2} - a^{2} c d + b^{2} c d - a b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac{2 \,{\left (a b c^{2} d - a^{2} c d^{2} + b^{2} c d^{2} - a b d^{3}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{4} d + 2 \, c^{2} d^{3} + d^{5}} + \frac{2 \, a b c^{2} d^{2} \tan \left (f x + e\right ) - 2 \, a^{2} c d^{3} \tan \left (f x + e\right ) + 2 \, b^{2} c d^{3} \tan \left (f x + e\right ) - 2 \, a b d^{4} \tan \left (f x + e\right ) - b^{2} c^{4} + 4 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2} - a^{2} d^{4}}{{\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )}{\left (d \tan \left (f x + e\right ) + c\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*c^2 - b^2*c^2 + 4*a*b*c*d - a^2*d^2 + b^2*d^2)*(f*x + e)/(c^4 + 2*c^2*d^2 + d^4) + (a*b*c^2 - a^2*c*d +
b^2*c*d - a*b*d^2)*log(tan(f*x + e)^2 + 1)/(c^4 + 2*c^2*d^2 + d^4) - 2*(a*b*c^2*d - a^2*c*d^2 + b^2*c*d^2 - a*
b*d^3)*log(abs(d*tan(f*x + e) + c))/(c^4*d + 2*c^2*d^3 + d^5) + (2*a*b*c^2*d^2*tan(f*x + e) - 2*a^2*c*d^3*tan(
f*x + e) + 2*b^2*c*d^3*tan(f*x + e) - 2*a*b*d^4*tan(f*x + e) - b^2*c^4 + 4*a*b*c^3*d - 3*a^2*c^2*d^2 + b^2*c^2
*d^2 - a^2*d^4)/((c^4*d + 2*c^2*d^3 + d^5)*(d*tan(f*x + e) + c)))/f