3.1211 \(\int \frac{(a+b \tan (e+f x))^2}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=103 \[ \frac{(b c-a d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d f \left (c^2+d^2\right )}+\frac{c x (b c-a d)^2}{d^2 \left (c^2+d^2\right )}-\frac{b x (b c-2 a d)}{d^2}-\frac{b^2 \log (\cos (e+f x))}{d f} \]

[Out]

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

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Rubi [A]  time = 0.117847, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3541, 3475, 3484, 3530} \[ \frac{(b c-a d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d f \left (c^2+d^2\right )}+\frac{c x (b c-a d)^2}{d^2 \left (c^2+d^2\right )}-\frac{b x (b c-2 a d)}{d^2}-\frac{b^2 \log (\cos (e+f x))}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3541

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]

Rule 3475

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

Rule 3484

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

Mathematica [C]  time = 0.151737, size = 108, normalized size = 1.05 \[ \frac{\frac{2 (b c-a d)^2 \log (c+d \tan (e+f x))}{d \left (c^2+d^2\right )}-\frac{(a-i b)^2 \log (\tan (e+f x)+i)}{d+i c}+\frac{(a+i b)^2 \log (-\tan (e+f x)+i)}{-d+i c}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.025, size = 249, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) abc}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{2}d}{2\,f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) abd}{f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}c}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{a}^{2}d\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{f \left ({c}^{2}+{d}^{2} \right ) }}-2\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) abc}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}{b}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) d}} \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)),x)

[Out]

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

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Maxima [A]  time = 1.81975, size = 166, normalized size = 1.61 \begin{align*} \frac{\frac{2 \,{\left (2 \, a b d +{\left (a^{2} - b^{2}\right )} c\right )}{\left (f x + e\right )}}{c^{2} + d^{2}} + \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b 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 (2 \, a b c -{\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58334, size = 296, normalized size = 2.87 \begin{align*} \frac{2 \,{\left (2 \, a b d^{2} +{\left (a^{2} - b^{2}\right )} c d\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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^{2} + b^{2} d^{2}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (c^{2} d + d^{3}\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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.91608, size = 1025, normalized size = 9.95 \begin{align*} \text{result too large to display} \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)),x)

[Out]

Piecewise((zoo*x*(a + b*tan(e))**2/tan(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)), ((a**2*x + a*b*log(tan(e + f*x)**2
+ 1)/f - b**2*x + b**2*tan(e + f*x)/f)/c, Eq(d, 0)), (-I*a**2*f*x*tan(e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f
) - a**2*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f) - I*a**2/(-2*d*f*tan(e + f*x) + 2*I*d*f) - 2*a*b*f*x*tan(e + f*x)
/(-2*d*f*tan(e + f*x) + 2*I*d*f) + 2*I*a*b*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f) + 2*a*b/(-2*d*f*tan(e + f*x) +
2*I*d*f) - I*b**2*f*x*tan(e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f) - b**2*f*x/(-2*d*f*tan(e + f*x) + 2*I*d*f)
- b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(-2*d*f*tan(e + f*x) + 2*I*d*f) + I*b**2*log(tan(e + f*x)**2 + 1)
/(-2*d*f*tan(e + f*x) + 2*I*d*f) + I*b**2/(-2*d*f*tan(e + f*x) + 2*I*d*f), Eq(c, -I*d)), (-I*a**2*f*x*tan(e +
f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + a**2*f*x/(2*d*f*tan(e + f*x) + 2*I*d*f) - I*a**2/(2*d*f*tan(e + f*x) + 2
*I*d*f) + 2*a*b*f*x*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + 2*I*a*b*f*x/(2*d*f*tan(e + f*x) + 2*I*d*f) -
2*a*b/(2*d*f*tan(e + f*x) + 2*I*d*f) - I*b**2*f*x*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + b**2*f*x/(2*d
*f*tan(e + f*x) + 2*I*d*f) + b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*d*f*tan(e + f*x) + 2*I*d*f) + I*b**
2*log(tan(e + f*x)**2 + 1)/(2*d*f*tan(e + f*x) + 2*I*d*f) + I*b**2/(2*d*f*tan(e + f*x) + 2*I*d*f), Eq(c, I*d))
, (x*(a + b*tan(e))**2/(c + d*tan(e)), Eq(f, 0)), (2*a**2*c*d*f*x/(2*c**2*d*f + 2*d**3*f) + 2*a**2*d**2*log(c/
d + tan(e + f*x))/(2*c**2*d*f + 2*d**3*f) - a**2*d**2*log(tan(e + f*x)**2 + 1)/(2*c**2*d*f + 2*d**3*f) - 4*a*b
*c*d*log(c/d + tan(e + f*x))/(2*c**2*d*f + 2*d**3*f) + 2*a*b*c*d*log(tan(e + f*x)**2 + 1)/(2*c**2*d*f + 2*d**3
*f) + 4*a*b*d**2*f*x/(2*c**2*d*f + 2*d**3*f) + 2*b**2*c**2*log(c/d + tan(e + f*x))/(2*c**2*d*f + 2*d**3*f) - 2
*b**2*c*d*f*x/(2*c**2*d*f + 2*d**3*f) + b**2*d**2*log(tan(e + f*x)**2 + 1)/(2*c**2*d*f + 2*d**3*f), True))

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Giac [A]  time = 1.50123, size = 170, normalized size = 1.65 \begin{align*} \frac{\frac{2 \,{\left (a^{2} c - b^{2} c + 2 \, a b d\right )}{\left (f x + e\right )}}{c^{2} + d^{2}} + \frac{{\left (2 \, a b c - a^{2} d + b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}} + \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{2} d + d^{3}}}{2 \, 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)),x, algorithm="giac")

[Out]

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