3.255 \(\int \frac{1}{(a+b \tan ^2(c+d x))^2} \, dx\)
Optimal. Leaf size=97 \[ -\frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^2}-\frac{b \tan (c+d x)}{2 a d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{x}{(a-b)^2} \]
[Out]
x/(a - b)^2 - ((3*a - b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^2*d) - (b*Tan[c +
d*x])/(2*a*(a - b)*d*(a + b*Tan[c + d*x]^2))
________________________________________________________________________________________
Rubi [A] time = 0.0920903, antiderivative size = 97, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used =
{3661, 414, 522, 203, 205} \[ -\frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^2}-\frac{b \tan (c+d x)}{2 a d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{x}{(a-b)^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[c + d*x]^2)^(-2),x]
[Out]
x/(a - b)^2 - ((3*a - b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^2*d) - (b*Tan[c +
d*x])/(2*a*(a - b)*d*(a + b*Tan[c + d*x]^2))
Rule 3661
Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Rule 414
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{b \tan (c+d x)}{2 a (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a-b-b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 a (a-b) d}\\ &=-\frac{b \tan (c+d x)}{2 a (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{(a-b)^2 d}-\frac{((3 a-b) b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a (a-b)^2 d}\\ &=\frac{x}{(a-b)^2}-\frac{(3 a-b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} (a-b)^2 d}-\frac{b \tan (c+d x)}{2 a (a-b) d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.993233, size = 88, normalized size = 0.91 \[ \frac{\frac{\sqrt{b} (b-3 a) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{b (b-a) \tan (c+d x)}{a \left (a+b \tan ^2(c+d x)\right )}+2 \tan ^{-1}(\tan (c+d x))}{2 d (a-b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[c + d*x]^2)^(-2),x]
[Out]
(2*ArcTan[Tan[c + d*x]] + (Sqrt[b]*(-3*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2) + (b*(-a + b)*Ta
n[c + d*x])/(a*(a + b*Tan[c + d*x]^2)))/(2*(a - b)^2*d)
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Maple [A] time = 0.027, size = 160, normalized size = 1.7 \begin{align*} -{\frac{b\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{2}\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{2}a \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{3\,b}{2\,d \left ( a-b \right ) ^{2}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{2\,d \left ( a-b \right ) ^{2}a}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ( a-b \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*tan(d*x+c)^2)^2,x)
[Out]
-1/2/d/(a-b)^2*b*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2/d/(a-b)^2*b^2/a*tan(d*x+c)/(a+b*tan(d*x+c)^2)-3/2/d/(a-b)^2
*b/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+1/2/d/(a-b)^2*b^2/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2
))+1/d/(a-b)^2*arctan(tan(d*x+c))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.52268, size = 865, normalized size = 8.92 \begin{align*} \left [\frac{8 \, a b d x \tan \left (d x + c\right )^{2} + 8 \, a^{2} d x -{\left ({\left (3 \, a b - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{4} - 6 \, a b \tan \left (d x + c\right )^{2} + a^{2} + 4 \,{\left (a b \tan \left (d x + c\right )^{3} - a^{2} \tan \left (d x + c\right )\right )} \sqrt{-\frac{b}{a}}}{b^{2} \tan \left (d x + c\right )^{4} + 2 \, a b \tan \left (d x + c\right )^{2} + a^{2}}\right ) - 4 \,{\left (a b - b^{2}\right )} \tan \left (d x + c\right )}{8 \,{\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d \tan \left (d x + c\right )^{2} +{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, \frac{4 \, a b d x \tan \left (d x + c\right )^{2} + 4 \, a^{2} d x -{\left ({\left (3 \, a b - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left (b \tan \left (d x + c\right )^{2} - a\right )} \sqrt{\frac{b}{a}}}{2 \, b \tan \left (d x + c\right )}\right ) - 2 \,{\left (a b - b^{2}\right )} \tan \left (d x + c\right )}{4 \,{\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d \tan \left (d x + c\right )^{2} +{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
[1/8*(8*a*b*d*x*tan(d*x + c)^2 + 8*a^2*d*x - ((3*a*b - b^2)*tan(d*x + c)^2 + 3*a^2 - a*b)*sqrt(-b/a)*log((b^2*
tan(d*x + c)^4 - 6*a*b*tan(d*x + c)^2 + a^2 + 4*(a*b*tan(d*x + c)^3 - a^2*tan(d*x + c))*sqrt(-b/a))/(b^2*tan(d
*x + c)^4 + 2*a*b*tan(d*x + c)^2 + a^2)) - 4*(a*b - b^2)*tan(d*x + c))/((a^3*b - 2*a^2*b^2 + a*b^3)*d*tan(d*x
+ c)^2 + (a^4 - 2*a^3*b + a^2*b^2)*d), 1/4*(4*a*b*d*x*tan(d*x + c)^2 + 4*a^2*d*x - ((3*a*b - b^2)*tan(d*x + c)
^2 + 3*a^2 - a*b)*sqrt(b/a)*arctan(1/2*(b*tan(d*x + c)^2 - a)*sqrt(b/a)/(b*tan(d*x + c))) - 2*(a*b - b^2)*tan(
d*x + c))/((a^3*b - 2*a^2*b^2 + a*b^3)*d*tan(d*x + c)^2 + (a^4 - 2*a^3*b + a^2*b^2)*d)]
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Sympy [A] time = 36.543, size = 2086, normalized size = 21.51 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c)**2)**2,x)
[Out]
Piecewise((zoo*x/tan(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x + 1/(d*tan(c + d*x)) - 1/(3*d*tan(c + d*x)**3
))/b**2, Eq(a, 0)), (x/(a + b*tan(c)**2)**2, Eq(d, 0)), (x/a**2, Eq(b, 0)), (4*I*a**(5/2)*d*x*sqrt(1/b)/(4*I*a
**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b
**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**
2) + 4*I*a**(3/2)*b*d*x*sqrt(1/b)*tan(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c
+ d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*
sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) - 2*I*a**(3/2)*b*sqrt(1/b)*tan(c + d*x)/(4*I*a**(9/
2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d
*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) +
2*I*sqrt(a)*b**2*sqrt(1/b)*tan(c + d*x)/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2
- 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b)
+ 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) - 3*a**2*log(-I*sqrt(a)*sqrt(1/b) + tan(c + d*x))/(4*I*a**(9/
2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d
*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) +
3*a**2*log(I*sqrt(a)*sqrt(1/b) + tan(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c +
d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqr
t(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) - 3*a*b*log(-I*sqrt(a)*sqrt(1/b) + tan(c + d*x))*tan(c
+ d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b)
- 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/
b)*tan(c + d*x)**2) + a*b*log(-I*sqrt(a)*sqrt(1/b) + tan(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*
d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I
*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) + 3*a*b*log(I*sqrt(a)*sqrt(1/b) +
tan(c + d*x))*tan(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**
(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(
3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) - a*b*log(I*sqrt(a)*sqrt(1/b) + tan(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b)
+ 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan
(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) + b**2*log(-I*sq
rt(a)*sqrt(1/b) + tan(c + d*x))*tan(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c +
d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sq
rt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*tan(c + d*x)**2) - b**2*log(I*sqrt(a)*sqrt(1/b) + tan(c + d*x))*tan(c
+ d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*tan(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b)
- 8*I*a**(5/2)*b**2*d*sqrt(1/b)*tan(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b
)*tan(c + d*x)**2), True))
________________________________________________________________________________________
Giac [A] time = 1.32069, size = 165, normalized size = 1.7 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (d x + c\right )}{\sqrt{a b}}\right )\right )}{\left (3 \, a b - b^{2}\right )}}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt{a b}} - \frac{2 \,{\left (d x + c\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac{b \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )}{\left (a^{2} - a b\right )}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")
[Out]
-1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))*(3*a*b - b^2)/((a^3 - 2*a^2*b +
a*b^2)*sqrt(a*b)) - 2*(d*x + c)/(a^2 - 2*a*b + b^2) + b*tan(d*x + c)/((b*tan(d*x + c)^2 + a)*(a^2 - a*b)))/d