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

Optimal. Leaf size=88 \[ -\frac{a^2}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{2 a b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.115915, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3542, 3531, 3530} \[ -\frac{a^2}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{2 a b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)*x)/(a^2 + b^2)^2) - (2*a*b*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^2*d) - a^2/(b*(a^
2 + b^2)*d*(a + b*Tan[c + d*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{\tan ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=-\frac{a^2}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{-a+b \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{\left (a^2-b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac{a^2}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{(2 a b) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{\left (a^2-b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac{2 a b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a^2}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.661689, size = 161, normalized size = 1.83 \[ \frac{\tan (c+d x) \left (-a b^2 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+(a+i b) \left (a^2-a b (c+d x+i)-i b^2 (c+d x)\right )\right )+2 i a b \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))-a \left (a b \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+(a+i b)^2 (c+d x)\right )}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.024, size = 134, normalized size = 1.5 \begin{align*}{\frac{ab\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{2}}{b \left ({a}^{2}+{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-2\,{\frac{ab\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.64388, size = 185, normalized size = 2.1 \begin{align*} -\frac{\frac{2 \, a b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{a^{2}}{a^{3} b + a b^{3} +{\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.12379, size = 336, normalized size = 3.82 \begin{align*} -\frac{a^{2} b +{\left (a^{3} - a b^{2}\right )} d x +{\left (a b^{2} \tan \left (d x + c\right ) + a^{2} b\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (a^{3} -{\left (a^{2} b - b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.55569, size = 223, normalized size = 2.53 \begin{align*} -\frac{\frac{2 \, a b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac{a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, a b^{3} \tan \left (d x + c\right ) - a^{4} + a^{2} b^{2}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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