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

Optimal. Leaf size=82 \[ \frac{a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{2 a b x}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0947094, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3529, 3531, 3530} \[ \frac{a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{2 a b x}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3529

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

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

Mathematica [C]  time = 0.437586, size = 181, normalized size = 2.21 \[ \frac{a \left (2 \left (\left (b^2-a^2\right ) \log (a+b \tan (c+d x))+a^2+b^2\right )+(a-i b)^2 \log (-\tan (c+d x)+i)+(a+i b)^2 \log (\tan (c+d x)+i)\right )+b \tan (c+d x) \left (2 \left (b^2-a^2\right ) \log (a+b \tan (c+d x))+(a-i b)^2 \log (-\tan (c+d x)+i)+(a+i b)^2 \log (\tan (c+d x)+i)\right )}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.024, size = 162, normalized size = 2. \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{ab\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ){b}^{2}}{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)/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.67063, size = 188, normalized size = 2.29 \begin{align*} \frac{\frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (a^{2} - b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \, a}{a^{3} + a b^{2} +{\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02968, size = 351, normalized size = 4.28 \begin{align*} \frac{4 \, a^{2} b d x + 2 \, a b^{2} -{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \tan \left (d x + c\right )\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 ) + 2 \,{\left (2 \, a b^{2} d x - a^{2} b\right )} \tan \left (d x + c\right )}{2 \,{\left ({\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\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.27912, size = 234, normalized size = 2.85 \begin{align*} \frac{\frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac{2 \,{\left (a^{2} b \tan \left (d x + c\right ) - b^{3} \tan \left (d x + c\right ) + 2 \, a^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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