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

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Rubi [A]  time = 0.12, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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]

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*}

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Mathematica [C]  time = 0.16, 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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fricas [A]  time = 0.51, size = 133, normalized size = 1.29 \[ \frac {2 \, {\left (2 \, a b d^{2} + {\left (a^{2} - b^{2}\right )} c d\right )} f x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\frac {d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (b^{2} c^{2} + b^{2} d^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (c^{2} d + d^{3}\right )} f} \]

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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giac [A]  time = 1.03, size = 126, normalized size = 1.22 \[ \frac {\frac {2 \, {\left (a^{2} c - b^{2} c + 2 \, a b d\right )} {\left (f x + e\right )}}{c^{2} + d^{2}} + \frac {{\left (2 \, a b c - a^{2} d + b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}} + \frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{2} d + d^{3}}}{2 \, f} \]

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

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maple [B]  time = 0.25, size = 249, normalized size = 2.42 \[ \frac {d \ln \left (c +d \tan \left (f x +e \right )\right ) a^{2}}{f \left (c^{2}+d^{2}\right )}-\frac {2 \ln \left (c +d \tan \left (f x +e \right )\right ) a b c}{f \left (c^{2}+d^{2}\right )}+\frac {\ln \left (c +d \tan \left (f x +e \right )\right ) b^{2} c^{2}}{f \left (c^{2}+d^{2}\right ) d}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{2} d}{2 f \left (c^{2}+d^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b c}{f \left (c^{2}+d^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b^{2} d}{2 f \left (c^{2}+d^{2}\right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a^{2} c}{f \left (c^{2}+d^{2}\right )}+\frac {2 \arctan \left (\tan \left (f x +e \right )\right ) a b d}{f \left (c^{2}+d^{2}\right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2} c}{f \left (c^{2}+d^{2}\right )} \]

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

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maxima [A]  time = 0.83, size = 123, normalized size = 1.19 \[ \frac {\frac {2 \, {\left (2 \, a b d + {\left (a^{2} - b^{2}\right )} c\right )} {\left (f x + e\right )}}{c^{2} + d^{2}} + \frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{2} d + d^{3}} + \frac {{\left (2 \, a b c - {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{2 \, f} \]

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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mupad [B]  time = 5.60, size = 115, normalized size = 1.12 \[ \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}{2\,f\,\left (c+d\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}{2\,f\,\left (d+c\,1{}\mathrm {i}\right )}+\frac {\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a\,d-b\,c\right )}^2}{d\,f\,\left (c^2+d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.35, size = 1040, normalized size = 10.10 \[ \text {result too large to display} \]

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)), (a**2*f*x*tan(e + f*x)/(-2*I*d*f*tan(e + f*x) - 2*d*f) -
I*a**2*f*x/(-2*I*d*f*tan(e + f*x) - 2*d*f) + a**2/(-2*I*d*f*tan(e + f*x) - 2*d*f) - 2*I*a*b*f*x*tan(e + f*x)/
(-2*I*d*f*tan(e + f*x) - 2*d*f) - 2*a*b*f*x/(-2*I*d*f*tan(e + f*x) - 2*d*f) + 2*I*a*b/(-2*I*d*f*tan(e + f*x) -
2*d*f) + b**2*f*x*tan(e + f*x)/(-2*I*d*f*tan(e + f*x) - 2*d*f) - I*b**2*f*x/(-2*I*d*f*tan(e + f*x) - 2*d*f) -
I*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(-2*I*d*f*tan(e + f*x) - 2*d*f) - b**2*log(tan(e + f*x)**2 + 1)/
(-2*I*d*f*tan(e + f*x) - 2*d*f) - b**2/(-2*I*d*f*tan(e + f*x) - 2*d*f), Eq(c, -I*d)), (a**2*f*x*tan(e + f*x)/(
2*I*d*f*tan(e + f*x) - 2*d*f) + I*a**2*f*x/(2*I*d*f*tan(e + f*x) - 2*d*f) + a**2/(2*I*d*f*tan(e + f*x) - 2*d*f
) + 2*I*a*b*f*x*tan(e + f*x)/(2*I*d*f*tan(e + f*x) - 2*d*f) - 2*a*b*f*x/(2*I*d*f*tan(e + f*x) - 2*d*f) - 2*I*a
*b/(2*I*d*f*tan(e + f*x) - 2*d*f) + b**2*f*x*tan(e + f*x)/(2*I*d*f*tan(e + f*x) - 2*d*f) + I*b**2*f*x/(2*I*d*f
*tan(e + f*x) - 2*d*f) + I*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*I*d*f*tan(e + f*x) - 2*d*f) - b**2*lo
g(tan(e + f*x)**2 + 1)/(2*I*d*f*tan(e + f*x) - 2*d*f) - b**2/(2*I*d*f*tan(e + f*x) - 2*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 + ta
n(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*l
og(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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