3.481 \(\int \frac{\tan ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ -\frac{a^2}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.206934, antiderivative size = 129, normalized size of antiderivative = 1., 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 = {3542, 3529, 3531, 3530} \[ -\frac{a^2}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3) - (b*(3*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d
) - a^2/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (2*a*b)/((a^2 + b^2)^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 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 ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac{a^2}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{-a+b \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx}{a^2+b^2}\\ &=-\frac{a^2}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{-a^2+b^2+2 a b \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{a^2}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (b \left (3 a^2-b^2\right )\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{a^2}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.45006, size = 200, normalized size = 1.55 \[ -\frac{a \left (\frac{2 a (a-b) (a+b)}{b \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{(b+i a)^3 \log (-\tan (c+d x)+i)}{\left (a^2+b^2\right )^2}-\frac{2 b \left (b^2-3 a^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}+\frac{i (a+i b) \log (\tan (c+d x)+i)}{(a-i b)^2}\right )-\frac{b^2 \tan ^3(c+d x)}{(a+b \tan (c+d x))^2}+\frac{b \tan ^2(c+d x)}{a+b \tan (c+d x)}}{2 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-(-((b^2*Tan[c + d*x]^3)/(a + b*Tan[c + d*x])^2) + (b*Tan[c + d*x]^2)/(a + b*Tan[c + d*x]) + a*(((I*a + b)^3*L
og[I - Tan[c + d*x]])/(a^2 + b^2)^2 + (I*(a + I*b)*Log[I + Tan[c + d*x]])/(a - I*b)^2 - (2*b*(-3*a^2 + b^2)*Lo
g[a + b*Tan[c + d*x]])/(a^2 + b^2)^2 + (2*a*(a - b)*(a + b))/(b*(a^2 + b^2)*(a + b*Tan[c + d*x]))))/(2*a*(a^2
+ b^2)*d)

________________________________________________________________________________________

Maple [A]  time = 0.028, size = 224, normalized size = 1.7 \begin{align*}{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) b{a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{a}^{2}}{2\,b \left ({a}^{2}+{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{ab}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-3\,{\frac{b{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \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))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.63135, size = 346, normalized size = 2.68 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{4 \, a b^{3} \tan \left (d x + c\right ) - a^{4} + 3 \, a^{2} b^{2}}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} +{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\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)^2/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b - b^3)*log(b*tan(d*x + c) +
a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4
+ b^6) - (4*a*b^3*tan(d*x + c) - a^4 + 3*a^2*b^2)/(a^6*b + 2*a^4*b^3 + a^2*b^5 + (a^4*b^3 + 2*a^2*b^5 + b^7)*
tan(d*x + c)^2 + 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*tan(d*x + c)))/d

________________________________________________________________________________________

Fricas [B]  time = 1.79416, size = 705, normalized size = 5.47 \begin{align*} -\frac{3 \, a^{4} b - 3 \, a^{2} b^{3} + 2 \,{\left (a^{5} - 3 \, a^{3} b^{2}\right )} d x -{\left (a^{4} b - 5 \, a^{2} b^{3} - 2 \,{\left (a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{2} +{\left (3 \, a^{4} b - a^{2} b^{3} +{\left (3 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b^{2} - a b^{4}\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 (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4} - 2 \,{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \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))^3,x, algorithm="fricas")

[Out]

-1/2*(3*a^4*b - 3*a^2*b^3 + 2*(a^5 - 3*a^3*b^2)*d*x - (a^4*b - 5*a^2*b^3 - 2*(a^3*b^2 - 3*a*b^4)*d*x)*tan(d*x
+ c)^2 + (3*a^4*b - a^2*b^3 + (3*a^2*b^3 - b^5)*tan(d*x + c)^2 + 2*(3*a^3*b^2 - a*b^4)*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*(a^5 - 3*a^3*b^2 + 2*a*b^4 - 2*(a^4*b - 3
*a^2*b^3)*d*x)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3
+ 3*a^3*b^5 + a*b^7)*d*tan(d*x + c) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*d)

________________________________________________________________________________________

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))**3,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [B]  time = 1.65099, size = 355, normalized size = 2.75 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{9 \, a^{2} b^{4} \tan \left (d x + c\right )^{2} - 3 \, b^{6} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{3} \tan \left (d x + c\right ) - 2 \, a b^{5} \tan \left (d x + c\right ) - a^{6} + 11 \, a^{4} b^{2}}{{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{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))^3,x, algorithm="giac")

[Out]

-1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1
)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b^2 - b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 +
3*a^2*b^5 + b^7) - (9*a^2*b^4*tan(d*x + c)^2 - 3*b^6*tan(d*x + c)^2 + 22*a^3*b^3*tan(d*x + c) - 2*a*b^5*tan(d*
x + c) - a^6 + 11*a^4*b^2)/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*(b*tan(d*x + c) + a)^2))/d