3.271 \(\int \frac {1}{(a \sec (x)+b \tan (x))^4} \, dx\)

Optimal. Leaf size=156 \[ \frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {x}{b^4} \]

[Out]

x/b^4-a*(2*a^2-3*b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/b^4/(a^2-b^2)^(3/2)-1/3*cos(x)^3/b/(a+b*sin(x))
^3+1/2*a*cos(x)^3/b/(a^2-b^2)/(a+b*sin(x))^2+1/2*cos(x)*(2*a^2-2*b^2+a*b*sin(x))/b^3/(a^2-b^2)/(a+b*sin(x))

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {4391, 2693, 2864, 2863, 2735, 2660, 618, 204} \[ -\frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {x}{b^4} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x] + b*Tan[x])^(-4),x]

[Out]

x/b^4 - (a*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^(3/2)) - Cos[x]^3/(3*b*(
a + b*Sin[x])^3) + (a*Cos[x]^3)/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2) + (Cos[x]*(2*(a^2 - b^2) + a*b*Sin[x]))/(2*
b^3*(a^2 - b^2)*(a + b*Sin[x]))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2863

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2864

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a
^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*Sim
p[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {1}{(a \sec (x)+b \tan (x))^4} \, dx &=\int \frac {\cos ^4(x)}{(a+b \sin (x))^4} \, dx\\ &=-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}-\frac {\int \frac {\cos ^2(x) \sin (x)}{(a+b \sin (x))^3} \, dx}{b}\\ &=-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\int \frac {\cos ^2(x) (2 b+a \sin (x))}{(a+b \sin (x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {-a b-2 \left (a^2-b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac {x}{b^4}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (a \left (2 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{2 b^4 \left (a^2-b^2\right )}\\ &=\frac {x}{b^4}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (a \left (2 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )}\\ &=\frac {x}{b^4}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\left (2 a \left (2 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )}\\ &=\frac {x}{b^4}-\frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}-\frac {\cos ^3(x)}{3 b (a+b \sin (x))^3}+\frac {a \cos ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\cos (x) \left (2 \left (a^2-b^2\right )+a b \sin (x)\right )}{2 b^3 \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.36, size = 2661, normalized size = 17.06 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a*Sec[x] + b*Tan[x])^(-4),x]

[Out]

(Sec[x]*(a + b*Sin[x])^4*(-1/3*(b*(-(b/(a - b)) - (b*Sin[x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[x])/(a + b))^(
5/2))/(((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[x])^3) - ((a*b^3*(-(b/(a - b)) -
(b*Sin[x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[x])/(a + b))^(5/2))/(2*(a^2 - b^2)*((a*b)/(a - b) - b^2/(a - b)
)*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[x])^2) - (-((((-3*a^2*b^5)/((a - b)^2*(a + b)^2) + (2*b^5*(3*a^2 -
2*b^2))/((a - b)^2*(a + b)^2))*(-(b/(a - b)) - (b*Sin[x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[x])/(a + b))^(5/2
))/(((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[x]))) - ((16*Sqrt[2]*b^6*(3*a^2 - 4
*b^2)*(-(b/(a - b)) - (b*Sin[x])/(a - b))^(5/2)*Sqrt[b/(a + b) - (b*Sin[x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b
)) - (b*Sin[x])/(a - b)))/(2*b))^(5/2)*((5*(1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^2)
+ (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^(-1)))/8 - (15*b^3*(((a - b)*(-(b/(a - b)) - (b*Si
n[x])/(a - b)))/b - ((a - b)^2*(-(b/(a - b)) - (b*Sin[x])/(a - b))^2)/(3*b^2) - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(
Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)
])/(Sqrt[b]*Sqrt[1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b)])))/(32*(a - b)^3*(-(b/(a - b)) - (b*
Sin[x])/(a - b))^3*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^2)))/(5*(a - b)^2*(a + b)^4*Sqrt[
((a + b)*(b/(a + b) - (b*Sin[x])/(a + b)))/b]) + ((-((a*b^7*(6*a^2 - 7*b^2))/((a - b)^3*(a + b)^3)) + (4*a*b^7
*(3*a^2 - 4*b^2))/((a - b)^3*(a + b)^3))*((4*Sqrt[2]*(-(b/(a - b)) - (b*Sin[x])/(a - b))^(3/2)*Sqrt[b/(a + b)
- (b*Sin[x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^(5/2)*((3/(4*(1 + ((a - b)*(-(
b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^(-1))/
2 + (3*b^2*(((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/b - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-
(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[b]*Sqrt[1
+ ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b)])))/(8*(a - b)^2*(-(b/(a - b)) - (b*Sin[x])/(a - b))^2*
(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^2)))/(3*(a + b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[x]
)/(a + b)))/b]) - ((-((a*b)/(a - b)) + b^2/(a - b))*(-(((-((a*b)/(a - b)) + b^2/(a - b))*(-(((-((a*b)/(a + b))
- b^2/(a + b))*((2*Sqrt[a - b]*ArcTanh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[a + b]*Sqr
t[b/(a + b) - (b*Sin[x])/(a + b)])])/(b*Sqrt[a + b]) - (2*Sqrt[-((a*b)/(a + b)) - b^2/(a + b)]*ArcTanh[(Sqrt[-
((a*b)/(a + b)) - b^2/(a + b)]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[-((a*b)/(a - b)) + b^2/(a - b)]*
Sqrt[b/(a + b) - (b*Sin[x])/(a + b)])])/(b*Sqrt[-((a*b)/(a - b)) + b^2/(a - b)])))/b) + (2*Sqrt[2]*(a - b)*Sqr
t[-(b/(a - b)) - (b*Sin[x])/(a - b)]*Sqrt[b/(a + b) - (b*Sin[x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin
[x])/(a - b)))/(2*b))^(3/2)*((Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[2]*S
qrt[b])])/(Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x]
)/(a - b)))/(2*b))^(3/2)) + 1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b)))))/(b*(a + b)*Sqrt[
((a + b)*(b/(a + b) - (b*Sin[x])/(a + b)))/b])))/b) + (4*Sqrt[2]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)]*Sqrt[
b/(a + b) - (b*Sin[x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^(5/2)*((3*Sqrt[b]*Ar
cSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[2]*Sqrt[b])])/(4*Sqrt[2]*Sqrt[a - b]*Sqrt[-(
b/(a - b)) - (b*Sin[x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^(5/2)) + (3/(2*(1 +
((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[x])/(a - b)))/
(2*b))^(-1))/4))/((a + b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[x])/(a + b)))/b])))/b))/b)/(((a*b)/(a - b) - b^2/(
a - b))*((a*b)/(a + b) + b^2/(a + b))))/(2*((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))))/(3*((
a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b)))))/((1 - (a + b*Sin[x])/(a - b))^(3/2)*(1 - (a + b*S
in[x])/(a + b))^(3/2)*(a*Sec[x] + b*Tan[x])^4)

________________________________________________________________________________________

fricas [B]  time = 1.40, size = 931, normalized size = 5.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^4,x, algorithm="fricas")

[Out]

[-1/12*(36*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*x*cos(x)^2 + 2*(11*a^4*b^3 - 19*a^2*b^5 + 8*b^7)*cos(x)^3 + 3*(2*a^6
+ 3*a^4*b^2 - 9*a^2*b^4 - 3*(2*a^4*b^2 - 3*a^2*b^4)*cos(x)^2 + (6*a^5*b - 7*a^3*b^3 - 3*a*b^5 - (2*a^3*b^3 - 3
*a*b^5)*cos(x)^2)*sin(x))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(
x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 12*(a^7 + a^5*b^2 - 5*a^3
*b^4 + 3*a*b^6)*x - 12*(a^6*b - 2*a^2*b^5 + b^7)*cos(x) + 6*(2*(a^4*b^3 - 2*a^2*b^5 + b^7)*x*cos(x)^2 - 2*(3*a
^6*b - 5*a^4*b^3 + a^2*b^5 + b^7)*x - (5*a^5*b^2 - 8*a^3*b^4 + 3*a*b^6)*cos(x))*sin(x))/(a^7*b^4 + a^5*b^6 - 5
*a^3*b^8 + 3*a*b^10 - 3*(a^5*b^6 - 2*a^3*b^8 + a*b^10)*cos(x)^2 + (3*a^6*b^5 - 5*a^4*b^7 + a^2*b^9 + b^11 - (a
^4*b^7 - 2*a^2*b^9 + b^11)*cos(x)^2)*sin(x)), -1/6*(18*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*x*cos(x)^2 + (11*a^4*b^3
- 19*a^2*b^5 + 8*b^7)*cos(x)^3 - 3*(2*a^6 + 3*a^4*b^2 - 9*a^2*b^4 - 3*(2*a^4*b^2 - 3*a^2*b^4)*cos(x)^2 + (6*a^
5*b - 7*a^3*b^3 - 3*a*b^5 - (2*a^3*b^3 - 3*a*b^5)*cos(x)^2)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sq
rt(a^2 - b^2)*cos(x))) - 6*(a^7 + a^5*b^2 - 5*a^3*b^4 + 3*a*b^6)*x - 6*(a^6*b - 2*a^2*b^5 + b^7)*cos(x) + 3*(2
*(a^4*b^3 - 2*a^2*b^5 + b^7)*x*cos(x)^2 - 2*(3*a^6*b - 5*a^4*b^3 + a^2*b^5 + b^7)*x - (5*a^5*b^2 - 8*a^3*b^4 +
3*a*b^6)*cos(x))*sin(x))/(a^7*b^4 + a^5*b^6 - 5*a^3*b^8 + 3*a*b^10 - 3*(a^5*b^6 - 2*a^3*b^8 + a*b^10)*cos(x)^
2 + (3*a^6*b^5 - 5*a^4*b^7 + a^2*b^9 + b^11 - (a^4*b^7 - 2*a^2*b^9 + b^11)*cos(x)^2)*sin(x))]

________________________________________________________________________________________

giac [B]  time = 0.17, size = 369, normalized size = 2.37 \[ -\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{6} b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x\right )^{5} + 6 \, a^{7} \tan \left (\frac {1}{2} \, x\right )^{4} + 9 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} - 12 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x\right )^{4} + 12 \, a b^{6} \tan \left (\frac {1}{2} \, x\right )^{4} + 36 \, a^{6} b \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 8 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x\right )^{3} + 8 \, b^{7} \tan \left (\frac {1}{2} \, x\right )^{3} + 12 \, a^{7} \tan \left (\frac {1}{2} \, x\right )^{2} + 48 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 42 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a b^{6} \tan \left (\frac {1}{2} \, x\right )^{2} + 33 \, a^{6} b \tan \left (\frac {1}{2} \, x\right ) - 24 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, x\right ) + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x\right ) + 6 \, a^{7} - 5 \, a^{5} b^{2} + 2 \, a^{3} b^{4}}{3 \, {\left (a^{5} b^{3} - a^{3} b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{3}} + \frac {x}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^4,x, algorithm="giac")

[Out]

-(2*a^3 - 3*a*b^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^2*b^4 -
b^6)*sqrt(a^2 - b^2)) + 1/3*(3*a^6*b*tan(1/2*x)^5 - 6*a^4*b^3*tan(1/2*x)^5 + 6*a^2*b^5*tan(1/2*x)^5 + 6*a^7*ta
n(1/2*x)^4 + 9*a^5*b^2*tan(1/2*x)^4 - 12*a^3*b^4*tan(1/2*x)^4 + 12*a*b^6*tan(1/2*x)^4 + 36*a^6*b*tan(1/2*x)^3
- 6*a^4*b^3*tan(1/2*x)^3 - 8*a^2*b^5*tan(1/2*x)^3 + 8*b^7*tan(1/2*x)^3 + 12*a^7*tan(1/2*x)^2 + 48*a^5*b^2*tan(
1/2*x)^2 - 42*a^3*b^4*tan(1/2*x)^2 + 12*a*b^6*tan(1/2*x)^2 + 33*a^6*b*tan(1/2*x) - 24*a^4*b^3*tan(1/2*x) + 6*a
^2*b^5*tan(1/2*x) + 6*a^7 - 5*a^5*b^2 + 2*a^3*b^4)/((a^5*b^3 - a^3*b^5)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^
3) + x/b^4

________________________________________________________________________________________

maple [B]  time = 0.21, size = 967, normalized size = 6.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)+b*tan(x))^4,x)

[Out]

1/b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3*a^3/(a^2-b^2)*tan(1/2*x)^5-2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3*a/(
a^2-b^2)*tan(1/2*x)^5+2*b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/a/(a^2-b^2)*tan(1/2*x)^5+2/b^3/(a*tan(1/2*x)^2
+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*a^4*tan(1/2*x)^4+3/b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*a^2*tan(1/2*
x)^4-4*b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*tan(1/2*x)^4+4*b^3/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/
(a^2-b^2)/a^2*tan(1/2*x)^4+12/b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3*a^3/(a^2-b^2)*tan(1/2*x)^3-2/(a*tan(1/2*
x)^2+2*b*tan(1/2*x)+a)^3*a/(a^2-b^2)*tan(1/2*x)^3-8/3*b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/a/(a^2-b^2)*tan(
1/2*x)^3+8/3*b^4/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/a^3/(a^2-b^2)*tan(1/2*x)^3+4/b^3/(a*tan(1/2*x)^2+2*b*tan(
1/2*x)+a)^3/(a^2-b^2)*a^4*tan(1/2*x)^2+16/b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*a^2*tan(1/2*x)^2-14*
b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*tan(1/2*x)^2+4*b^3/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^
2)/a^2*tan(1/2*x)^2+11/b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3*a^3/(a^2-b^2)*tan(1/2*x)-8/(a*tan(1/2*x)^2+2*b*
tan(1/2*x)+a)^3*a/(a^2-b^2)*tan(1/2*x)+2*b^2/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/a/(a^2-b^2)*tan(1/2*x)+2/b^3/
(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*a^4-5/3/b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)*a^2+2/3*
b/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^3/(a^2-b^2)-2/b^4*a^3/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2
-b^2)^(1/2))+3/b^2*a/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+2/b^4*arctan(tan(1/2*x))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 6.37, size = 2782, normalized size = 17.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(x) + a/cos(x))^4,x)

[Out]

((6*a^4 + 2*b^4 - 5*a^2*b^2)/(3*b^3*(a^2 - b^2)) + (tan(x/2)*(11*a^4 + 2*b^4 - 8*a^2*b^2))/(a*b^2*(a^2 - b^2))
+ (tan(x/2)^5*(a^4 + 2*b^4 - 2*a^2*b^2))/(a*b^2*(a^2 - b^2)) + (tan(x/2)^4*(2*a^6 + 4*b^6 - 4*a^2*b^4 + 3*a^4
*b^2))/(a^2*b^3*(a^2 - b^2)) + (2*tan(x/2)^2*(2*a^6 + 2*b^6 - 7*a^2*b^4 + 8*a^4*b^2))/(a^2*b^3*(a^2 - b^2)) +
(2*tan(x/2)^3*(3*a^2 + 2*b^2)*(6*a^4 + 2*b^4 - 5*a^2*b^2))/(3*a^3*b^2*(a^2 - b^2)))/(tan(x/2)^2*(12*a*b^2 + 3*
a^3) + tan(x/2)^4*(12*a*b^2 + 3*a^3) + tan(x/2)^3*(12*a^2*b + 8*b^3) + a^3 + a^3*tan(x/2)^6 + 6*a^2*b*tan(x/2)
^5 + 6*a^2*b*tan(x/2)) + (2*atan((48*a^3*b^3*tan(x/2))/((176*a^3*b^15)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (160*a^
5*b^13)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (48*a^7*b^11)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (64*a*b^17)/(b^12 - 2*a^
2*b^10 + a^4*b^8)) - (64*a*b^5*tan(x/2))/((176*a^3*b^15)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (160*a^5*b^13)/(b^12
- 2*a^2*b^10 + a^4*b^8) + (48*a^7*b^11)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (64*a*b^17)/(b^12 - 2*a^2*b^10 + a^4*b
^8))))/b^4 + (a*atan(((a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^7 - 8*a^4*b^5 + 4*a^6*b^3))
/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(8*a*b^9 - 29*a^3*b^7 + 28*a^5*b^5 - 8*a^7*b^3))/(b^13 - 2*a^2*b^
11 + a^4*b^9) - (a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a*b^12 - 6*a^3*b^10 + 2*a^5*b^8))/(b^12
- 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a^2*b^12 - 20*a^4*b^10 + 8*a^6*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)
- (a*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 - 32*a
^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/
2))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))*1i)/(2*(b^10
- 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)) + (a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^7 - 8*a^4*b
^5 + 4*a^6*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(8*a*b^9 - 29*a^3*b^7 + 28*a^5*b^5 - 8*a^7*b^3))/
(b^13 - 2*a^2*b^11 + a^4*b^9) + (a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a*b^12 - 6*a^3*b^10 + 2
*a^5*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a^2*b^12 - 20*a^4*b^10 + 8*a^6*b^8))/(b^13 - 2*a^2*
b^11 + a^4*b^9) + (a*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(
12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(2*a^2 - 3*b^2)*(-(a + b)^
3*(a - b)^3)^(1/2))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)
))*1i)/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/((16*(2*a^5 - 3*a^3*b^2))/(b^12 - 2*a^2*b^10 + a^4*b^8) +
(16*tan(x/2)*(8*a^6 + 12*a^2*b^4 - 20*a^4*b^2))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (a*(2*a^2 - 3*b^2)*(-(a + b)^
3*(a - b)^3)^(1/2)*((8*(4*a^2*b^7 - 8*a^4*b^5 + 4*a^6*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(8*a*b
^9 - 29*a^3*b^7 + 28*a^5*b^5 - 8*a^7*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a -
b)^3)^(1/2)*((8*(4*a*b^12 - 6*a^3*b^10 + 2*a^5*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a^2*b^12
- 20*a^4*b^10 + 8*a^6*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (a*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b
^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*
b^11 + a^4*b^9))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))/
(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)) + (a*(2*a^2 - 3*b^
2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^7 - 8*a^4*b^5 + 4*a^6*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*ta
n(x/2)*(8*a*b^9 - 29*a^3*b^7 + 28*a^5*b^5 - 8*a^7*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + (a*(2*a^2 - 3*b^2)*(-(
a + b)^3*(a - b)^3)^(1/2)*((8*(4*a*b^12 - 6*a^3*b^10 + 2*a^5*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)
*(12*a^2*b^12 - 20*a^4*b^10 + 8*a^6*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) + (a*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*
a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(
b^13 - 2*a^2*b^11 + a^4*b^9))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 -
a^6*b^4))))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))/(2*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))))*(2*
a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*1i)/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sec {\relax (x )} + b \tan {\relax (x )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))**4,x)

[Out]

Integral((a*sec(x) + b*tan(x))**(-4), x)

________________________________________________________________________________________