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

Optimal. Leaf size=92 \[ -\frac {\sqrt {b} (3 a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 f (a+b)^{3/2}}+\frac {x}{a^2}-\frac {b \tan (e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

x/a^2-1/2*(3*a+2*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/a^2/(a+b)^(3/2)/f-1/2*b*tan(f*x+e)/a/(a+b)/
f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {4128, 414, 522, 203, 205} \[ -\frac {\sqrt {b} (3 a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 f (a+b)^{3/2}}+\frac {x}{a^2}-\frac {b \tan (e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^(-2),x]

[Out]

x/a^2 - (Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(3/2)*f) - (b*Tan[e +
f*x])/(2*a*(a + b)*f*(a + b + b*Tan[e + f*x]^2))

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]

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 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]

Rubi steps

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

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Mathematica [C]  time = 1.98, size = 240, normalized size = 2.61 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (2 x (a \cos (2 (e+f x))+a+2 b)+\frac {b ((a+2 b) \sin (2 e)-a \sin (2 f x))}{f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+\frac {b (3 a+2 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{f (a+b)^{3/2} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{8 a^2 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(2*x*(a + 2*b + a*Cos[2*(e + f*x)]) + (b*(3*a + 2*b)*ArcTan[(Se
c[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*S
in[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/((a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e]
)^4]) + (b*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/((a + b)*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8*a^2*(a +
b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 0.79, size = 435, normalized size = 4.73 \[ \left [\frac {8 \, {\left (a^{2} + a b\right )} f x \cos \left (f x + e\right )^{2} - 4 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 8 \, {\left (a b + b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left ({\left (a^{4} + a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + a^{2} b^{2}\right )} f\right )}}, \frac {4 \, {\left (a^{2} + a b\right )} f x \cos \left (f x + e\right )^{2} - 2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 4 \, {\left (a b + b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{4} + a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(8*(a^2 + a*b)*f*x*cos(f*x + e)^2 - 4*a*b*cos(f*x + e)*sin(f*x + e) + 8*(a*b + b^2)*f*x + ((3*a^2 + 2*a*b
)*cos(f*x + e)^2 + 3*a*b + 2*b^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)))/((a^4 + a^3*b)*f*cos(f*x + e)^2 + (a^3*b +
a^2*b^2)*f), 1/4*(4*(a^2 + a*b)*f*x*cos(f*x + e)^2 - 2*a*b*cos(f*x + e)*sin(f*x + e) + 4*(a*b + b^2)*f*x + ((3
*a^2 + 2*a*b)*cos(f*x + e)^2 + 3*a*b + 2*b^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))))/((a^4 + a^3*b)*f*cos(f*x + e)^2 + (a^3*b + a^2*b^2)*f)]

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giac [A]  time = 0.23, size = 119, normalized size = 1.29 \[ -\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b + 2 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} \sqrt {a b + b^{2}}} + \frac {b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} {\left (a^{2} + a b\right )}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3*a*b + 2*b^2)/((a^3 + a
^2*b)*sqrt(a*b + b^2)) + b*tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)*(a^2 + a*b)) - 2*(f*x + e)/a^2)/f

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maple [A]  time = 0.90, size = 127, normalized size = 1.38 \[ -\frac {b \tan \left (f x +e \right )}{2 a \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f a \left (a +b \right ) \sqrt {\left (a +b \right ) b}}-\frac {b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{2} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*tan(f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)-3/2/f*b/a/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)
^(1/2))-1/f*b^2/a^2/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/f/a^2*arctan(tan(f*x+e))

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maxima [A]  time = 0.48, size = 106, normalized size = 1.15 \[ -\frac {\frac {b \tan \left (f x + e\right )}{a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{2} b + a b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*tan(f*x + e)/(a^3 + 2*a^2*b + a*b^2 + (a^2*b + a*b^2)*tan(f*x + e)^2) + (3*a*b + 2*b^2)*arctan(b*tan(f
*x + e)/sqrt((a + b)*b))/((a^3 + a^2*b)*sqrt((a + b)*b)) - 2*(f*x + e)/a^2)/f

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mupad [B]  time = 6.35, size = 2056, normalized size = 22.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cos(e + f*x)^2)^2,x)

[Out]

atan((((((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)*1i)/(2*(2*a^4*b + a^5 + a^3*b^2)) - (tan(e + f*x)*(32*a^4*b^5 + 8
0*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))/(2*a^2) + (tan(e + f*x)*(20*a*b^4 + 8
*b^5 + 13*a^2*b^3))/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2 - ((((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)*1i)/(2*(2*a^4*
b + a^5 + a^3*b^2)) + (tan(e + f*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4
+ a^2*b^2)))/(2*a^2) - (tan(e + f*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2)/(((
(((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)*1i)/(2*(2*a^4*b + a^5 + a^3*b^2)) - (tan(e + f*x)*(32*a^4*b^5 + 80*a^5*b
^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))*1i)/(2*a^2) + (tan(e + f*x)*(20*a*b^4 + 8*b^
5 + 13*a^2*b^3)*1i)/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2 + (((((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)*1i)/(2*(2*a^4
*b + a^5 + a^3*b^2)) + (tan(e + f*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^
4 + a^2*b^2)))*1i)/(2*a^2) - (tan(e + f*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3)*1i)/(4*(2*a^3*b + a^4 + a^2*b^2)))/
a^2 + ((3*a*b^3)/2 + b^4)/(2*a^4*b + a^5 + a^3*b^2)))/(a^2*f) + (atan(((((tan(e + f*x)*(20*a*b^4 + 8*b^5 + 13*
a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) - ((-b*(a + b)^3)^(1/2)*((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b
+ a^5 + a^3*b^2) - (tan(e + f*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a
^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 +
a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)) + (((ta
n(e + f*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((-b*(a + b)^3)^(1/2)*((2*a^4*b^4
+ 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) + (tan(e + f*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^
5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2))
)*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(4*(3*a^4*b + a
^5 + a^2*b^3 + 3*a^3*b^2)))/(((3*a*b^3)/2 + b^4)/(2*a^4*b + a^5 + a^3*b^2) - (((tan(e + f*x)*(20*a*b^4 + 8*b^5
+ 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) - ((-b*(a + b)^3)^(1/2)*((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*
a^4*b + a^5 + a^3*b^2) - (tan(e + f*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3
+ 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b +
a^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)) + ((
(tan(e + f*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((-b*(a + b)^3)^(1/2)*((2*a^4*b
^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) + (tan(e + f*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4
*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^
2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b))/(4*(3*a^4*b + a
^5 + a^2*b^3 + 3*a^3*b^2))))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(2*f*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2))
- (b*tan(e + f*x))/(2*a*f*(a + b)*(a + b + b*tan(e + f*x)^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**(-2), x)

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