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

Optimal. Leaf size=66 \[ \frac{a^2 \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}-\frac{a x}{a^2+b^2}-\frac{\log (\cos (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.099732, antiderivative size = 77, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3541, 3475, 3484, 3530} \[ \frac{a^2 \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}+\frac{a^3 x}{b^2 \left (a^2+b^2\right )}-\frac{a x}{b^2}-\frac{\log (\cos (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x)/b^2) + (a^3*x)/(b^2*(a^2 + b^2)) - Log[Cos[c + d*x]]/(b*d) + (a^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]
)/(b*(a^2 + b^2)*d)

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{\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx &=-\frac{a x}{b^2}+\frac{a^2 \int \frac{1}{a+b \tan (c+d x)} \, dx}{b^2}+\frac{\int \tan (c+d x) \, dx}{b}\\ &=-\frac{a x}{b^2}+\frac{a^3 x}{b^2 \left (a^2+b^2\right )}-\frac{\log (\cos (c+d x))}{b d}+\frac{a^2 \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{a x}{b^2}+\frac{a^3 x}{b^2 \left (a^2+b^2\right )}-\frac{\log (\cos (c+d x))}{b d}+\frac{a^2 \log (a \cos (c+d x)+b \sin (c+d x))}{b \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.0828734, size = 78, normalized size = 1.18 \[ \frac{2 a^2 \log (a+b \tan (c+d x))+b (b+i a) \log (-\tan (c+d x)+i)+b (b-i a) \log (\tan (c+d x)+i)}{2 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 80, normalized size = 1.2 \begin{align*}{\frac{b\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) b}} \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)),x)

[Out]

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

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Maxima [A]  time = 1.58623, size = 97, normalized size = 1.47 \begin{align*} \frac{\frac{2 \, a^{2} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.1047, size = 213, normalized size = 3.23 \begin{align*} -\frac{2 \, a b d x - a^{2} \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^{2} + b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b + b^{3}\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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.94669, size = 405, normalized size = 6.14 \begin{align*} \begin{cases} \tilde{\infty } x \tan{\left (c \right )} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac{i d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{\log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = - i b \\- \frac{i d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{\log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i}{2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = i b \\\frac{- x + \frac{\tan{\left (c + d x \right )}}{d}}{a} & \text{for}\: b = 0 \\\frac{x \tan ^{2}{\left (c \right )}}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\\frac{2 a^{2} \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} - \frac{2 a b d x}{2 a^{2} b d + 2 b^{3} d} + \frac{b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} & \text{otherwise} \end{cases} \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)),x)

[Out]

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

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Giac [A]  time = 1.52533, size = 99, normalized size = 1.5 \begin{align*} \frac{\frac{2 \, a^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} - \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, 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)),x, algorithm="giac")

[Out]

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