\(\int (a \sec (x)+b \tan (x))^4 \, dx\) [264]

Optimal result

Integrand size = 11, antiderivative size = 100 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=b^4 x+\frac {4}{3} a b \left (a^2-2 b^2\right ) \cos (x)+\frac {1}{3} b^2 \left (2 a^2-3 b^2\right ) \cos (x) \sin (x)+\frac {1}{3} \sec ^3(x) (b+a \sin (x)) (a+b \sin (x))^3-\frac {1}{3} \sec (x) (a+b \sin (x))^2 \left (a b-\left (2 a^2-3 b^2\right ) \sin (x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4476, 2770, 2940, 2813} \[ \int (a \sec (x)+b \tan (x))^4 \, dx=\frac {4}{3} a b \left (a^2-2 b^2\right ) \cos (x)+\frac {1}{3} b^2 \left (2 a^2-3 b^2\right ) \sin (x) \cos (x)-\frac {1}{3} \sec (x) (a+b \sin (x))^2 \left (a b-\left (2 a^2-3 b^2\right ) \sin (x)\right )+\frac {1}{3} \sec ^3(x) (a \sin (x)+b) (a+b \sin (x))^3+b^4 x \]

[In]

Int[(a*Sec[x] + b*Tan[x])^4,x]

[Out]

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

Rule 2770

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

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2940

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

Rule 4476

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*} \text {integral}& = \int \sec ^4(x) (a+b \sin (x))^4 \, dx \\ & = \frac {1}{3} \sec ^3(x) (b+a \sin (x)) (a+b \sin (x))^3-\frac {1}{3} \int \sec ^2(x) (a+b \sin (x))^2 \left (-2 a^2+3 b^2+a b \sin (x)\right ) \, dx \\ & = \frac {1}{3} \sec ^3(x) (b+a \sin (x)) (a+b \sin (x))^3-\frac {1}{3} \sec (x) (a+b \sin (x))^2 \left (a b-\left (2 a^2-3 b^2\right ) \sin (x)\right )+\frac {1}{3} \int (a+b \sin (x)) \left (2 a b^2-2 b \left (2 a^2-3 b^2\right ) \sin (x)\right ) \, dx \\ & = b^4 x+\frac {4}{3} a b \left (a^2-2 b^2\right ) \cos (x)+\frac {1}{3} b^2 \left (2 a^2-3 b^2\right ) \cos (x) \sin (x)+\frac {1}{3} \sec ^3(x) (b+a \sin (x)) (a+b \sin (x))^3-\frac {1}{3} \sec (x) (a+b \sin (x))^2 \left (a b-\left (2 a^2-3 b^2\right ) \sin (x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=\frac {1}{12} \sec ^3(x) \left (16 a^3 b-8 a b^3+9 b^4 x \cos (x)-24 a b^3 \cos (2 x)+3 b^4 x \cos (3 x)+6 a^4 \sin (x)+18 a^2 b^2 \sin (x)+2 a^4 \sin (3 x)-6 a^2 b^2 \sin (3 x)-4 b^4 \sin (3 x)\right ) \]

[In]

Integrate[(a*Sec[x] + b*Tan[x])^4,x]

[Out]

(Sec[x]^3*(16*a^3*b - 8*a*b^3 + 9*b^4*x*Cos[x] - 24*a*b^3*Cos[2*x] + 3*b^4*x*Cos[3*x] + 6*a^4*Sin[x] + 18*a^2*
b^2*Sin[x] + 2*a^4*Sin[3*x] - 6*a^2*b^2*Sin[3*x] - 4*b^4*Sin[3*x]))/12

Maple [A] (verified)

Time = 17.59 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74

method	result	size
parts	\(-a^{4} \left (-\frac {2}{3}-\frac {\sec \left (x \right )^{2}}{3}\right ) \tan \left (x \right )+b^{4} \left (\frac {\tan \left (x \right )^{3}}{3}-\tan \left (x \right )+\arctan \left (\tan \left (x \right )\right )\right )+\frac {4 a^{3} b \sec \left (x \right )^{3}}{3}+2 a^{2} b^{2} \tan \left (x \right )^{3}+4 a \,b^{3} \left (\frac {\sec \left (x \right )^{3}}{3}-\sec \left (x \right )\right )\)	\(74\)
default	\(-a^{4} \left (-\frac {2}{3}-\frac {\sec \left (x \right )^{2}}{3}\right ) \tan \left (x \right )+\frac {4 a^{3} b}{3 \cos \left (x \right )^{3}}+\frac {2 a^{2} b^{2} \sin \left (x \right )^{3}}{\cos \left (x \right )^{3}}+4 a \,b^{3} \left (\frac {\sin \left (x \right )^{4}}{3 \cos \left (x \right )^{3}}-\frac {\sin \left (x \right )^{4}}{3 \cos \left (x \right )}-\frac {\left (2+\sin \left (x \right )^{2}\right ) \cos \left (x \right )}{3}\right )+b^{4} \left (\frac {\tan \left (x \right )^{3}}{3}-\tan \left (x \right )+x \right )\)	\(96\)
risch	\(b^{4} x -\frac {4 \left (9 i a^{2} b^{2} {\mathrm e}^{4 i x}+3 i b^{4} {\mathrm e}^{4 i x}+6 a \,b^{3} {\mathrm e}^{5 i x}-3 i a^{4} {\mathrm e}^{2 i x}+3 i b^{4} {\mathrm e}^{2 i x}-8 a^{3} b \,{\mathrm e}^{3 i x}+4 a \,b^{3} {\mathrm e}^{3 i x}-i a^{4}+3 i a^{2} b^{2}+2 i b^{4}+6 a \,b^{3} {\mathrm e}^{i x}\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{3}}\)	\(131\)

[In]

int((a*sec(x)+b*tan(x))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=\frac {3 \, b^{4} x \cos \left (x\right )^{3} - 12 \, a b^{3} \cos \left (x\right )^{2} + 4 \, a^{3} b + 4 \, a b^{3} + {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} - 3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \, \cos \left (x\right )^{3}} \]

[In]

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

[Out]

1/3*(3*b^4*x*cos(x)^3 - 12*a*b^3*cos(x)^2 + 4*a^3*b + 4*a*b^3 + (a^4 + 6*a^2*b^2 + b^4 + 2*(a^4 - 3*a^2*b^2 -
2*b^4)*cos(x)^2)*sin(x))/cos(x)^3

Sympy [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=\frac {a^{4} \tan ^{3}{\left (x \right )}}{3} + a^{4} \tan {\left (x \right )} + \frac {4 a^{3} b \sec ^{3}{\left (x \right )}}{3} + 2 a^{2} b^{2} \tan ^{3}{\left (x \right )} + \frac {4 a b^{3} \sec ^{3}{\left (x \right )}}{3} - 4 a b^{3} \sec {\left (x \right )} + b^{4} x + \frac {b^{4} \sin ^{3}{\left (x \right )}}{3 \cos ^{3}{\left (x \right )}} - \frac {b^{4} \sin {\left (x \right )}}{\cos {\left (x \right )}} \]

[In]

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

[Out]

a**4*tan(x)**3/3 + a**4*tan(x) + 4*a**3*b*sec(x)**3/3 + 2*a**2*b**2*tan(x)**3 + 4*a*b**3*sec(x)**3/3 - 4*a*b**
3*sec(x) + b**4*x + b**4*sin(x)**3/(3*cos(x)**3) - b**4*sin(x)/cos(x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.72 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=2 \, a^{2} b^{2} \tan \left (x\right )^{3} + \frac {1}{3} \, {\left (\tan \left (x\right )^{3} + 3 \, \tan \left (x\right )\right )} a^{4} + \frac {1}{3} \, {\left (\tan \left (x\right )^{3} + 3 \, x - 3 \, \tan \left (x\right )\right )} b^{4} - \frac {4 \, {\left (3 \, \cos \left (x\right )^{2} - 1\right )} a b^{3}}{3 \, \cos \left (x\right )^{3}} + \frac {4 \, a^{3} b}{3 \, \cos \left (x\right )^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.31 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=b^{4} x - \frac {2 \, {\left (3 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 3 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{4} - 2 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 24 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 10 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 24 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, x\right ) - 3 \, b^{4} \tan \left (\frac {1}{2} \, x\right ) + 4 \, a^{3} b - 8 \, a b^{3}\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{3}} \]

[In]

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

[Out]

b^4*x - 2/3*(3*a^4*tan(1/2*x)^5 - 3*b^4*tan(1/2*x)^5 + 12*a^3*b*tan(1/2*x)^4 - 2*a^4*tan(1/2*x)^3 + 24*a^2*b^2
*tan(1/2*x)^3 + 10*b^4*tan(1/2*x)^3 + 24*a*b^3*tan(1/2*x)^2 + 3*a^4*tan(1/2*x) - 3*b^4*tan(1/2*x) + 4*a^3*b -
8*a*b^3)/(tan(1/2*x)^2 - 1)^3

Mupad [B] (verification not implemented)

Time = 28.72 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.15 \[ \int (a \sec (x)+b \tan (x))^4 \, dx=b^4\,x-\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^4-2\,b^4\right )-\frac {16\,a\,b^3}{3}+\frac {8\,a^3\,b}{3}+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (-\frac {4\,a^4}{3}+16\,a^2\,b^2+\frac {20\,b^4}{3}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (2\,a^4-2\,b^4\right )+16\,a\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+8\,a^3\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2-1\right )}^3} \]

[In]

int((b*tan(x) + a/cos(x))^4,x)

[Out]

b^4*x - (tan(x/2)*(2*a^4 - 2*b^4) - (16*a*b^3)/3 + (8*a^3*b)/3 + tan(x/2)^3*((20*b^4)/3 - (4*a^4)/3 + 16*a^2*b
^2) + tan(x/2)^5*(2*a^4 - 2*b^4) + 16*a*b^3*tan(x/2)^2 + 8*a^3*b*tan(x/2)^4)/(tan(x/2)^2 - 1)^3