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

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

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.336986, antiderivative size = 156, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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 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])

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 = 6.36915, size = 2677, normalized size = 17.16 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[x]*(a + b*Sin[x])^4*(-(b*(-(b/(a - b)) - (b*Sin[x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[x])/(a + b))^(5/2)
)/(3*((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*Sin[
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[(Sq
rt[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*Si
n[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*(-((a*b)/(a + b)) - b^2/(a + b))*ArcTan[(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)]*Sqrt[(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]*Sqrt[b/(a + b) - (b*Sin[x])/(a + b)])])/(b*Sqrt[a + b])))/b)
+ (2*Sqrt[2]*(a - b)*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))^(3/2)*((Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin
[x])/(a - b)])/(Sqrt[2]*Sqrt[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]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[x])/(a - b)])/(Sqrt[2]*Sqrt[b])])/(4*Sq
rt[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*Sin[x])/(a + b))^(3/2)*(a*Sec[x] + b*Tan[x])^4)

________________________________________________________________________________________

Maple [B]  time = 0.113, size = 967, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

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))

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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*sec(x)+b*tan(x))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.96195, size = 2061, normalized size = 13.21 \begin{align*} \text{result too large to display} \end{align*}

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))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sec{\left (x \right )} + b \tan{\left (x \right )}\right )^{4}}\, dx \end{align*}

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)

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Giac [B]  time = 1.16997, size = 498, normalized size = 3.19 \begin{align*} -\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}\left (a\right ) + \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}} \end{align*}

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