3.372 \(\int \frac{1}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{5/2}}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x}{a^3}-\frac{b \tan (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

x/a^3 - (Sqrt[b]*(15*a^2 + 20*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^3*(a + b)^(5/2)*f)
- (b*Tan[e + f*x])/(4*a*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(7*a + 4*b)*Tan[e + f*x])/(8*a^2*(a + b)
^2*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.177944, antiderivative size = 144, 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 = {4128, 414, 527, 522, 203, 205} \[ -\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{5/2}}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x}{a^3}-\frac{b \tan (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^(-3),x]

[Out]

x/a^3 - (Sqrt[b]*(15*a^2 + 20*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^3*(a + b)^(5/2)*f)
- (b*Tan[e + f*x])/(4*a*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(7*a + 4*b)*Tan[e + f*x])/(8*a^2*(a + b)
^2*f*(a + b + b*Tan[e + f*x]^2))

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

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 \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a+b) f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2+9 a b+4 b^2-b (7 a+4 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^2 f}\\ &=-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}-\frac{\left (b \left (15 a^2+20 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b)^2 f}\\ &=\frac{x}{a^3}-\frac{\sqrt{b} \left (15 a^2+20 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 (a+b)^{5/2} f}-\frac{b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 5.82632, size = 332, normalized size = 2.31 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{b \left (\left (9 a^2+28 a b+16 b^2\right ) \sin (2 e)-3 a (3 a+2 b) \sin (2 f x)\right ) (a \cos (2 (e+f x))+a+2 b)}{f (a+b)^2 (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+\frac{b \left (15 a^2+20 a b+8 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{f (a+b)^{5/2} \sqrt{b (\cos (e)-i \sin (e))^4}}-\frac{4 b^2 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+8 x (a \cos (2 (e+f x))+a+2 b)^2\right )}{64 a^3 \left (a+b \sec ^2(e+f x)\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^(-3),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(8*x*(a + 2*b + a*Cos[2*(e + f*x)])^2 + (b*(15*a^2 + 20*a*b + 8
*b^2)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt
[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])^2*(Cos[2*e] - I*Sin[2*e]))/((a + b)^(5/2)*f*Sqrt[b*
(Cos[e] - I*Sin[e])^4]) - (4*b^2*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/((a + b)*f*(Cos[e] - Sin[e])*(Cos[e] + S
in[e])) + (b*(a + 2*b + a*Cos[2*(e + f*x)])*((9*a^2 + 28*a*b + 16*b^2)*Sin[2*e] - 3*a*(3*a + 2*b)*Sin[2*f*x]))
/((a + b)^2*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(64*a^3*(a + b*Sec[e + f*x]^2)^3)

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Maple [B]  time = 0.089, size = 321, normalized size = 2.2 \begin{align*}{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{3}}}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,fa \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{9\,b\tan \left ( fx+e \right ) }{8\, \left ( a+b \right ) af \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{2}\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}-{\frac{15\,b}{8\,fa \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{5\,{b}^{2}}{2\,f{a}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{{b}^{3}}{f{a}^{3} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sec(f*x+e)^2)^3,x)

[Out]

1/f/a^3*arctan(tan(f*x+e))-7/8/f/a*b^2/(a+b+b*tan(f*x+e)^2)^2/(a^2+2*a*b+b^2)*tan(f*x+e)^3-1/2/f/a^2*b^3/(a+b+
b*tan(f*x+e)^2)^2/(a^2+2*a*b+b^2)*tan(f*x+e)^3-9/8*b*tan(f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2-1/2/f/a^2*b^2
/(a+b+b*tan(f*x+e)^2)^2/(a+b)*tan(f*x+e)-15/8/f/a*b/(a^2+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)
*b)^(1/2))-5/2/f/a^2*b^2/(a^2+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/f/a^3*b^3/(a^2
+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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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*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.723028, size = 1854, normalized size = 12.88 \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*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(32*(a^4 + 2*a^3*b + a^2*b^2)*f*x*cos(f*x + e)^4 + 64*(a^3*b + 2*a^2*b^2 + a*b^3)*f*x*cos(f*x + e)^2 + 3
2*(a^2*b^2 + 2*a*b^3 + b^4)*f*x + ((15*a^4 + 20*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2 + 20*a*b^3 + 8*
b^4 + 2*(15*a^3*b + 20*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x
+ e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e)
)*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 4*(3*(3*a^3*b + 2*
a^2*b^2)*cos(f*x + e)^3 + (7*a^2*b^2 + 4*a*b^3)*cos(f*x + e))*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f
*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f), 1/16*(16*(a
^4 + 2*a^3*b + a^2*b^2)*f*x*cos(f*x + e)^4 + 32*(a^3*b + 2*a^2*b^2 + a*b^3)*f*x*cos(f*x + e)^2 + 16*(a^2*b^2 +
2*a*b^3 + b^4)*f*x + ((15*a^4 + 20*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2 + 20*a*b^3 + 8*b^4 + 2*(15*
a^3*b + 20*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b
/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) - 2*(3*(3*a^3*b + 2*a^2*b^2)*cos(f*x + e)^3 + (7*a^2*b^2 + 4*a*b^3)*c
os(f*x + e))*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos
(f*x + e)^2 + (a^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f)]

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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*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.41983, size = 277, normalized size = 1.92 \begin{align*} -\frac{\frac{{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt{a b + b^{2}}} + \frac{7 \, a b^{2} \tan \left (f x + e\right )^{3} + 4 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 13 \, a b^{2} \tan \left (f x + e\right ) + 4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac{8 \,{\left (f x + e\right )}}{a^{3}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/8*((15*a^2*b + 20*a*b^2 + 8*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^
2)))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(a*b + b^2)) + (7*a*b^2*tan(f*x + e)^3 + 4*b^3*tan(f*x + e)^3 + 9*a^2*b*ta
n(f*x + e) + 13*a*b^2*tan(f*x + e) + 4*b^3*tan(f*x + e))/((a^4 + 2*a^3*b + a^2*b^2)*(b*tan(f*x + e)^2 + a + b)
^2) - 8*(f*x + e)/a^3)/f