3.256 \(\int \frac{1}{(a+b \tan ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=150 \[ -\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a-b)^3}-\frac{b (7 a-3 b) \tan (c+d x)}{8 a^2 d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )}-\frac{b \tan (c+d x)}{4 a d (a-b) \left (a+b \tan ^2(c+d x)\right )^2}+\frac{x}{(a-b)^3} \]

[Out]

x/(a - b)^3 - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(8*a^(5/2)*(a - b)^3*
d) - (b*Tan[c + d*x])/(4*a*(a - b)*d*(a + b*Tan[c + d*x]^2)^2) - ((7*a - 3*b)*b*Tan[c + d*x])/(8*a^2*(a - b)^2
*d*(a + b*Tan[c + d*x]^2))

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Rubi [A]  time = 0.144264, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3661, 414, 527, 522, 203, 205} \[ -\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a-b)^3}-\frac{b (7 a-3 b) \tan (c+d x)}{8 a^2 d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )}-\frac{b \tan (c+d x)}{4 a d (a-b) \left (a+b \tan ^2(c+d x)\right )^2}+\frac{x}{(a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^(-3),x]

[Out]

x/(a - b)^3 - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(8*a^(5/2)*(a - b)^3*
d) - (b*Tan[c + d*x])/(4*a*(a - b)*d*(a + b*Tan[c + d*x]^2)^2) - ((7*a - 3*b)*b*Tan[c + d*x])/(8*a^2*(a - b)^2
*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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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 )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a-3 b-3 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 a (a-b) d}\\ &=-\frac{b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}-\frac{(7 a-3 b) b \tan (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2-7 a b+3 b^2-(7 a-3 b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 a^2 (a-b)^2 d}\\ &=-\frac{b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}-\frac{(7 a-3 b) b \tan (c+d x)}{8 a^2 (a-b)^2 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)^3 d}-\frac{\left (b \left (15 a^2-10 a b+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{8 a^2 (a-b)^3 d}\\ &=\frac{x}{(a-b)^3}-\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} (a-b)^3 d}-\frac{b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}-\frac{(7 a-3 b) b \tan (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.87812, size = 138, normalized size = 0.92 \[ -\frac{\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2}}+\frac{b (7 a-3 b) (a-b) \tan (c+d x)}{a^2 \left (a+b \tan ^2(c+d x)\right )}+\frac{2 b (a-b)^2 \tan (c+d x)}{a \left (a+b \tan ^2(c+d x)\right )^2}-8 \tan ^{-1}(\tan (c+d x))}{8 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-8*ArcTan[Tan[c + d*x]] + (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(5/2)
+ (2*(a - b)^2*b*Tan[c + d*x])/(a*(a + b*Tan[c + d*x]^2)^2) + ((7*a - 3*b)*(a - b)*b*Tan[c + d*x])/(a^2*(a +
b*Tan[c + d*x]^2)))/(8*(a - b)^3*d)

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Maple [B]  time = 0.027, size = 350, normalized size = 2.3 \begin{align*} -{\frac{7\,{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{5\,{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}a}}-{\frac{3\,{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}{a}^{2}}}-{\frac{9\,ab\tan \left ( dx+c \right ) }{8\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{7\,{b}^{2}\tan \left ( dx+c \right ) }{4\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{5\,{b}^{3}\tan \left ( dx+c \right ) }{8\,d \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}a}}-{\frac{15\,b}{8\,d \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}}{4\,d \left ( a-b \right ) ^{3}a}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,{b}^{3}}{8\,d \left ( a-b \right ) ^{3}{a}^{2}}\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 ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/d*b^2/(a-b)^3/(a+b*tan(d*x+c)^2)^2*tan(d*x+c)^3+5/4/d*b^3/(a-b)^3/(a+b*tan(d*x+c)^2)^2/a*tan(d*x+c)^3-3/8
/d*b^4/(a-b)^3/(a+b*tan(d*x+c)^2)^2/a^2*tan(d*x+c)^3-9/8/d*b/(a-b)^3/(a+b*tan(d*x+c)^2)^2*a*tan(d*x+c)+7/4/d*b
^2/(a-b)^3/(a+b*tan(d*x+c)^2)^2*tan(d*x+c)-5/8/d*b^3/(a-b)^3/(a+b*tan(d*x+c)^2)^2/a*tan(d*x+c)-15/8/d*b/(a-b)^
3/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+5/4/d*b^2/(a-b)^3/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)
)-3/8/d*b^3/(a-b)^3/a^2/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+1/d/(a-b)^3*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.68687, size = 1643, normalized size = 10.95 \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)^3,x, algorithm="fricas")

[Out]

[1/32*(32*a^2*b^2*d*x*tan(d*x + c)^4 + 64*a^3*b*d*x*tan(d*x + c)^2 + 32*a^4*d*x - 4*(7*a^2*b^2 - 10*a*b^3 + 3*
b^4)*tan(d*x + c)^3 - ((15*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*
a^3*b - 10*a^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*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
*(9*a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d*x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*d*tan(d*x + c)^4
+ 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d), 1/1
6*(16*a^2*b^2*d*x*tan(d*x + c)^4 + 32*a^3*b*d*x*tan(d*x + c)^2 + 16*a^4*d*x - 2*(7*a^2*b^2 - 10*a*b^3 + 3*b^4)
*tan(d*x + c)^3 - ((15*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*a^3*
b - 10*a^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*sqrt(b/a)*arctan(1/2*(b*tan(d*x + c)^2 - a)*sqrt(b/a)/(b*tan(d*x + c
))) - 2*(9*a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d*x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*d*tan(d*x
+ c)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*
d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.31306, size = 277, normalized size = 1.85 \begin{align*} -\frac{\frac{{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )}{\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 (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt{a b}} - \frac{8 \,{\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{7 \, a b^{2} \tan \left (d x + c\right )^{3} - 3 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a^{2} b \tan \left (d x + c\right ) - 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/(
(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*sqrt(a*b)) - 8*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (7*a*b^2*tan(
d*x + c)^3 - 3*b^3*tan(d*x + c)^3 + 9*a^2*b*tan(d*x + c) - 5*a*b^2*tan(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*(b
*tan(d*x + c)^2 + a)^2))/d