3.3.59 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx\) [259]

Optimal. Leaf size=297 \[ \frac {d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac {2 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} b^2 (a+b)^{3/2} f}+\frac {2 (b c-a d)^3 (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^4 \sqrt {a+b} f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f} \]

[Out]

1/2*d^4*arctanh(sin(f*x+e))/b^2/f+d^2*(3*a^2*d^2-8*a*b*c*d+6*b^2*c^2)*arctanh(sin(f*x+e))/b^4/f+2*(-a*d+b*c)^4
*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/(a-b)^(3/2)/b^2/(a+b)^(3/2)/f-(-a*d+b*c)^4*sin(f*x+e)/b
^3/(a^2-b^2)/f/(b+a*cos(f*x+e))+2*(-a*d+b*c)^3*(3*a*d+b*c)*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))
/a/b^4/f/(a-b)^(1/2)/(a+b)^(1/2)+2*d^3*(-a*d+2*b*c)*tan(f*x+e)/b^3/f+1/2*d^4*sec(f*x+e)*tan(f*x+e)/b^2/f

________________________________________________________________________________________

Rubi [A]
time = 0.38, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {4073, 3031, 2743, 12, 2738, 214, 3855, 3852, 8, 3853} \begin {gather*} \frac {d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac {2 (b c-a d)^3 (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^4 f \sqrt {a-b} \sqrt {a+b}}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {2 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^2 f (a-b)^{3/2} (a+b)^{3/2}}+\frac {d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {d^4 \tan (e+f x) \sec (e+f x)}{2 b^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + b*Sec[e + f*x])^2,x]

[Out]

(d^4*ArcTanh[Sin[e + f*x]])/(2*b^2*f) + (d^2*(6*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*ArcTanh[Sin[e + f*x]])/(b^4*f
) + (2*(b*c - a*d)^4*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*(a - b)^(3/2)*b^2*(a + b)^(3/2)*f
) + (2*(b*c - a*d)^3*(b*c + 3*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^4*Sqr
t[a + b]*f) - ((b*c - a*d)^4*Sin[e + f*x])/(b^3*(a^2 - b^2)*f*(b + a*Cos[e + f*x])) + (2*d^3*(2*b*c - a*d)*Tan
[e + f*x])/(b^3*f) + (d^4*Sec[e + f*x]*Tan[e + f*x])/(2*b^2*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 214

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 3031

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4073

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[1/g^(m + n), Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d
+ c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && Inte
gerQ[n]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+b \sec (e+f x))^2} \, dx &=\int \frac {(d+c \cos (e+f x))^4 \sec ^3(e+f x)}{(b+a \cos (e+f x))^2} \, dx\\ &=\int \left (-\frac {(-b c+a d)^4}{a b^3 (b+a \cos (e+f x))^2}-\frac {(-b c+a d)^3 (b c+3 a d)}{a b^4 (b+a \cos (e+f x))}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \sec (e+f x)}{b^4}+\frac {2 d^3 (2 b c-a d) \sec ^2(e+f x)}{b^3}+\frac {d^4 \sec ^3(e+f x)}{b^2}\right ) \, dx\\ &=\frac {d^4 \int \sec ^3(e+f x) \, dx}{b^2}-\frac {(b c-a d)^4 \int \frac {1}{(b+a \cos (e+f x))^2} \, dx}{a b^3}+\frac {\left (2 d^3 (2 b c-a d)\right ) \int \sec ^2(e+f x) \, dx}{b^3}+\frac {\left ((b c-a d)^3 (b c+3 a d)\right ) \int \frac {1}{b+a \cos (e+f x)} \, dx}{a b^4}+\frac {\left (d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right )\right ) \int \sec (e+f x) \, dx}{b^4}\\ &=\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f}+\frac {d^4 \int \sec (e+f x) \, dx}{2 b^2}+\frac {(b c-a d)^4 \int \frac {b}{b+a \cos (e+f x)} \, dx}{a b^3 \left (a^2-b^2\right )}-\frac {\left (2 d^3 (2 b c-a d)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{b^3 f}+\frac {\left (2 (b c-a d)^3 (b c+3 a d)\right ) \text {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a b^4 f}\\ &=\frac {d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac {2 (b c-a d)^3 (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^4 \sqrt {a+b} f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f}+\frac {(b c-a d)^4 \int \frac {1}{b+a \cos (e+f x)} \, dx}{a b^2 \left (a^2-b^2\right )}\\ &=\frac {d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac {2 (b c-a d)^3 (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^4 \sqrt {a+b} f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f}+\frac {\left (2 (b c-a d)^4\right ) \text {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a b^2 \left (a^2-b^2\right ) f}\\ &=\frac {d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac {2 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} b^2 (a+b)^{3/2} f}+\frac {2 (b c-a d)^3 (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^4 \sqrt {a+b} f}-\frac {(b c-a d)^4 \sin (e+f x)}{b^3 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac {d^4 \sec (e+f x) \tan (e+f x)}{2 b^2 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.01, size = 511, normalized size = 1.72 \begin {gather*} \frac {\cos ^2(e+f x) (b+a \cos (e+f x)) (c+d \sec (e+f x))^4 \left (\frac {8 (-b c+a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (e+f x))}{\left (a^2-b^2\right )^{3/2}}-2 d^2 \left (-16 a b c d+6 a^2 d^2+b^2 \left (12 c^2+d^2\right )\right ) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 d^2 \left (-16 a b c d+6 a^2 d^2+b^2 \left (12 c^2+d^2\right )\right ) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {b^2 d^4 (b+a \cos (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {8 b d^3 (2 b c-a d) (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}-\frac {b^2 d^4 (b+a \cos (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {8 b d^3 (2 b c-a d) (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {4 b (b c-a d)^4 \sin (e+f x)}{(-a+b) (a+b)}\right )}{4 b^4 f (d+c \cos (e+f x))^4 (a+b \sec (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + b*Sec[e + f*x])^2,x]

[Out]

(Cos[e + f*x]^2*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^4*((8*(-(b*c) + a*d)^3*(a*b*c + 3*a^2*d - 4*b^2*d)*A
rcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[e + f*x]))/(a^2 - b^2)^(3/2) - 2*d^2*(-16*a*b*c
*d + 6*a^2*d^2 + b^2*(12*c^2 + d^2))*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 2*d^2*(-1
6*a*b*c*d + 6*a^2*d^2 + b^2*(12*c^2 + d^2))*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (b
^2*d^4*(b + a*Cos[e + f*x]))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (8*b*d^3*(2*b*c - a*d)*(b + a*Cos[e + f
*x])*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (b^2*d^4*(b + a*Cos[e + f*x]))/(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2])^2 + (8*b*d^3*(2*b*c - a*d)*(b + a*Cos[e + f*x])*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Si
n[(e + f*x)/2]) + (4*b*(b*c - a*d)^4*Sin[e + f*x])/((-a + b)*(a + b))))/(4*b^4*f*(d + c*Cos[e + f*x])^4*(a + b
*Sec[e + f*x])^2)

________________________________________________________________________________________

Maple [A]
time = 1.09, size = 465, normalized size = 1.57

method	result	size
derivativedivides	\(\frac {\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d^{2} \left (6 a^{2} d^{2}-16 a b d c +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +d b \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d^{2} \left (6 a^{2} d^{2}-16 a b d c +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +d b \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\frac {2 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-a -b \right )}-\frac {2 \left (3 a^{5} d^{4}-8 a^{4} b c \,d^{3}+6 a^{3} b^{2} c^{2} d^{2}-4 a^{3} b^{2} d^{4}+12 a^{2} b^{3} c \,d^{3}-b^{4} c^{4} a -12 a \,b^{4} c^{2} d^{2}+4 c^{3} d \,b^{5}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{4}}}{f}\)	\(465\)
default	\(\frac {\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d^{2} \left (6 a^{2} d^{2}-16 a b d c +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +d b \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d^{4}}{2 b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d^{2} \left (6 a^{2} d^{2}-16 a b d c +12 b^{2} c^{2}+b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 b^{4}}+\frac {d^{3} \left (4 a d -8 b c +d b \right )}{2 b^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\frac {2 b \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-a -b \right )}-\frac {2 \left (3 a^{5} d^{4}-8 a^{4} b c \,d^{3}+6 a^{3} b^{2} c^{2} d^{2}-4 a^{3} b^{2} d^{4}+12 a^{2} b^{3} c \,d^{3}-b^{4} c^{4} a -12 a \,b^{4} c^{2} d^{2}+4 c^{3} d \,b^{5}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{4}}}{f}\)	\(465\)
risch	\(\text {Expression too large to display}\)	\(2483\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+b*sec(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/2*d^4/b^2/(tan(1/2*f*x+1/2*e)-1)^2-1/2*d^2*(6*a^2*d^2-16*a*b*c*d+12*b^2*c^2+b^2*d^2)/b^4*ln(tan(1/2*f*x
+1/2*e)-1)+1/2*d^3*(4*a*d-8*b*c+b*d)/b^3/(tan(1/2*f*x+1/2*e)-1)-1/2*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)^2+1/2*d^2*(
6*a^2*d^2-16*a*b*c*d+12*b^2*c^2+b^2*d^2)/b^4*ln(tan(1/2*f*x+1/2*e)+1)+1/2*d^3*(4*a*d-8*b*c+b*d)/b^3/(tan(1/2*f
*x+1/2*e)+1)+2/b^4*(b*(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)/(a^2-b^2)*tan(1/2*f*x+1/
2*e)/(a*tan(1/2*f*x+1/2*e)^2-b*tan(1/2*f*x+1/2*e)^2-a-b)-(3*a^5*d^4-8*a^4*b*c*d^3+6*a^3*b^2*c^2*d^2-4*a^3*b^2*
d^4+12*a^2*b^3*c*d^3-a*b^4*c^4-12*a*b^4*c^2*d^2+4*b^5*c^3*d)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan
(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+b*sec(f*x+e))^2,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*a^2-4*b^2>0)', see `assume?`
for more de

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \sec {\left (e + f x \right )}\right )^{4} \sec {\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**4/(a+b*sec(f*x+e))**2,x)

[Out]

Integral((c + d*sec(e + f*x))**4*sec(e + f*x)/(a + b*sec(e + f*x))**2, x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 551 vs. \(2 (275) = 550\).
time = 0.55, size = 551, normalized size = 1.86 \begin {gather*} -\frac {\frac {4 \, {\left (a b^{4} c^{4} - 4 \, b^{5} c^{3} d - 6 \, a^{3} b^{2} c^{2} d^{2} + 12 \, a b^{4} c^{2} d^{2} + 8 \, a^{4} b c d^{3} - 12 \, a^{2} b^{3} c d^{3} - 3 \, a^{5} d^{4} + 4 \, a^{3} b^{2} d^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (b^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a b^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a^{2} b^{2} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a^{3} b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{4} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a - b\right )}} - \frac {{\left (12 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} + 6 \, a^{2} d^{4} + b^{2} d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{b^{4}} + \frac {{\left (12 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} + 6 \, a^{2} d^{4} + b^{2} d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{b^{4}} + \frac {2 \, {\left (8 \, b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

-1/2*(4*(a*b^4*c^4 - 4*b^5*c^3*d - 6*a^3*b^2*c^2*d^2 + 12*a*b^4*c^2*d^2 + 8*a^4*b*c*d^3 - 12*a^2*b^3*c*d^3 - 3
*a^5*d^4 + 4*a^3*b^2*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f*x + 1/2*e) -
b*tan(1/2*f*x + 1/2*e))/sqrt(-a^2 + b^2)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - 4*(b^4*c^4*tan(1/2*f*x + 1/2*e
) - 4*a*b^3*c^3*d*tan(1/2*f*x + 1/2*e) + 6*a^2*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) - 4*a^3*b*c*d^3*tan(1/2*f*x +
1/2*e) + a^4*d^4*tan(1/2*f*x + 1/2*e))/((a^2*b^3 - b^5)*(a*tan(1/2*f*x + 1/2*e)^2 - b*tan(1/2*f*x + 1/2*e)^2 -
a - b)) - (12*b^2*c^2*d^2 - 16*a*b*c*d^3 + 6*a^2*d^4 + b^2*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b^4 + (12*
b^2*c^2*d^2 - 16*a*b*c*d^3 + 6*a^2*d^4 + b^2*d^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^4 + 2*(8*b*c*d^3*tan(1/
2*f*x + 1/2*e)^3 - 4*a*d^4*tan(1/2*f*x + 1/2*e)^3 - b*d^4*tan(1/2*f*x + 1/2*e)^3 - 8*b*c*d^3*tan(1/2*f*x + 1/2
*e) + 4*a*d^4*tan(1/2*f*x + 1/2*e) - b*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*b^3))/f

________________________________________________________________________________________

Mupad [B]
time = 14.37, size = 2500, normalized size = 8.42 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/cos(e + f*x))^4/(cos(e + f*x)*(a + b/cos(e + f*x))^2),x)

[Out]

(atan(((((((8*(2*b^15*d^4 - 4*a*b^14*c^4 + 16*b^15*c^3*d + 4*a^2*b^13*c^4 + 4*a^3*b^12*c^4 - 4*a^4*b^11*c^4 +
6*a^2*b^13*d^4 - 16*a^3*b^12*d^4 - 14*a^4*b^11*d^4 + 28*a^5*b^10*d^4 + 6*a^6*b^9*d^4 - 12*a^7*b^8*d^4 + 24*b^1
5*c^2*d^2 - 48*a*b^14*c^2*d^2 + 48*a^2*b^13*c*d^3 - 16*a^2*b^13*c^3*d + 48*a^3*b^12*c*d^3 + 16*a^3*b^12*c^3*d
- 80*a^4*b^11*c*d^3 - 16*a^5*b^10*c*d^3 + 32*a^6*b^9*c*d^3 - 24*a^2*b^13*c^2*d^2 + 72*a^3*b^12*c^2*d^2 - 24*a^
5*b^10*c^2*d^2 - 32*a*b^14*c*d^3 - 16*a*b^14*c^3*d))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (8*tan(e/2 + (f*x)
/2)*(b^2*(d^4/2 + 6*c^2*d^2) + 3*a^2*d^4 - 8*a*b*c*d^3)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8
*a^5*b^9 - 8*a^6*b^8))/(b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(b^2*(d^4/2 + 6*c^2*d^2) + 3*a^2*d^4 - 8*a*b*c
*d^3))/b^4 - (8*tan(e/2 + (f*x)/2)*(72*a^10*d^8 + b^10*d^8 - 2*a*b^9*d^8 - 72*a^9*b*d^8 + 4*a^2*b^8*c^8 + 11*a
^2*b^8*d^8 - 20*a^3*b^7*d^8 + 23*a^4*b^6*d^8 - 26*a^5*b^5*d^8 + 17*a^6*b^4*d^8 + 120*a^7*b^3*d^8 - 120*a^8*b^2
*d^8 + 24*b^10*c^2*d^6 + 144*b^10*c^4*d^4 + 64*b^10*c^6*d^2 - 48*a*b^9*c^2*d^6 - 384*a*b^9*c^3*d^5 - 288*a*b^9
*c^4*d^4 - 384*a*b^9*c^5*d^3 + 64*a^2*b^8*c*d^7 - 160*a^3*b^7*c*d^7 + 256*a^4*b^6*c*d^7 - 160*a^5*b^5*c*d^7 -
704*a^6*b^4*c*d^7 + 704*a^7*b^3*c*d^7 + 384*a^8*b^2*c*d^7 + 376*a^2*b^8*c^2*d^6 + 768*a^2*b^8*c^3*d^5 + 816*a^
2*b^8*c^4*d^4 + 96*a^2*b^8*c^6*d^2 - 704*a^3*b^7*c^2*d^6 - 896*a^3*b^7*c^3*d^5 + 576*a^3*b^7*c^4*d^4 + 96*a^3*
b^7*c^5*d^3 + 536*a^4*b^6*c^2*d^6 - 1536*a^4*b^6*c^3*d^5 - 944*a^4*b^6*c^4*d^4 - 48*a^4*b^6*c^6*d^2 + 1552*a^5
*b^5*c^2*d^6 + 1824*a^5*b^5*c^3*d^5 - 288*a^5*b^5*c^4*d^4 + 64*a^5*b^5*c^5*d^3 - 1624*a^6*b^4*c^2*d^6 + 768*a^
6*b^4*c^3*d^5 + 264*a^6*b^4*c^4*d^4 - 800*a^7*b^3*c^2*d^6 - 768*a^7*b^3*c^3*d^5 + 800*a^8*b^2*c^2*d^6 - 32*a*b
^9*c*d^7 - 32*a*b^9*c^7*d - 384*a^9*b*c*d^7))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6))*(b^2*(d^4/2 + 6*c^2*d^2) + 3*
a^2*d^4 - 8*a*b*c*d^3)*1i)/b^4 - (((((8*(2*b^15*d^4 - 4*a*b^14*c^4 + 16*b^15*c^3*d + 4*a^2*b^13*c^4 + 4*a^3*b^
12*c^4 - 4*a^4*b^11*c^4 + 6*a^2*b^13*d^4 - 16*a^3*b^12*d^4 - 14*a^4*b^11*d^4 + 28*a^5*b^10*d^4 + 6*a^6*b^9*d^4
- 12*a^7*b^8*d^4 + 24*b^15*c^2*d^2 - 48*a*b^14*c^2*d^2 + 48*a^2*b^13*c*d^3 - 16*a^2*b^13*c^3*d + 48*a^3*b^12*
c*d^3 + 16*a^3*b^12*c^3*d - 80*a^4*b^11*c*d^3 - 16*a^5*b^10*c*d^3 + 32*a^6*b^9*c*d^3 - 24*a^2*b^13*c^2*d^2 + 7
2*a^3*b^12*c^2*d^2 - 24*a^5*b^10*c^2*d^2 - 32*a*b^14*c*d^3 - 16*a*b^14*c^3*d))/(a*b^11 + b^12 - a^2*b^10 - a^3
*b^9) + (8*tan(e/2 + (f*x)/2)*(b^2*(d^4/2 + 6*c^2*d^2) + 3*a^2*d^4 - 8*a*b*c*d^3)*(8*a*b^13 - 8*a^2*b^12 - 16*
a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/(b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(b^2*(d^4/2 + 6*c^2*
d^2) + 3*a^2*d^4 - 8*a*b*c*d^3))/b^4 + (8*tan(e/2 + (f*x)/2)*(72*a^10*d^8 + b^10*d^8 - 2*a*b^9*d^8 - 72*a^9*b*
d^8 + 4*a^2*b^8*c^8 + 11*a^2*b^8*d^8 - 20*a^3*b^7*d^8 + 23*a^4*b^6*d^8 - 26*a^5*b^5*d^8 + 17*a^6*b^4*d^8 + 120
*a^7*b^3*d^8 - 120*a^8*b^2*d^8 + 24*b^10*c^2*d^6 + 144*b^10*c^4*d^4 + 64*b^10*c^6*d^2 - 48*a*b^9*c^2*d^6 - 384
*a*b^9*c^3*d^5 - 288*a*b^9*c^4*d^4 - 384*a*b^9*c^5*d^3 + 64*a^2*b^8*c*d^7 - 160*a^3*b^7*c*d^7 + 256*a^4*b^6*c*
d^7 - 160*a^5*b^5*c*d^7 - 704*a^6*b^4*c*d^7 + 704*a^7*b^3*c*d^7 + 384*a^8*b^2*c*d^7 + 376*a^2*b^8*c^2*d^6 + 76
8*a^2*b^8*c^3*d^5 + 816*a^2*b^8*c^4*d^4 + 96*a^2*b^8*c^6*d^2 - 704*a^3*b^7*c^2*d^6 - 896*a^3*b^7*c^3*d^5 + 576
*a^3*b^7*c^4*d^4 + 96*a^3*b^7*c^5*d^3 + 536*a^4*b^6*c^2*d^6 - 1536*a^4*b^6*c^3*d^5 - 944*a^4*b^6*c^4*d^4 - 48*
a^4*b^6*c^6*d^2 + 1552*a^5*b^5*c^2*d^6 + 1824*a^5*b^5*c^3*d^5 - 288*a^5*b^5*c^4*d^4 + 64*a^5*b^5*c^5*d^3 - 162
4*a^6*b^4*c^2*d^6 + 768*a^6*b^4*c^3*d^5 + 264*a^6*b^4*c^4*d^4 - 800*a^7*b^3*c^2*d^6 - 768*a^7*b^3*c^3*d^5 + 80
0*a^8*b^2*c^2*d^6 - 32*a*b^9*c*d^7 - 32*a*b^9*c^7*d - 384*a^9*b*c*d^7))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6))*(b^
2*(d^4/2 + 6*c^2*d^2) + 3*a^2*d^4 - 8*a*b*c*d^3)*1i)/b^4)/((16*(108*a^11*d^12 - 54*a^10*b*d^12 + 4*a^3*b^8*d^1
2 - 4*a^4*b^7*d^12 + 41*a^5*b^6*d^12 - 9*a^6*b^5*d^12 + 63*a^7*b^4*d^12 + 81*a^8*b^3*d^12 - 216*a^9*b^2*d^12 -
4*b^11*c^3*d^9 - 96*b^11*c^5*d^7 + 32*b^11*c^6*d^6 - 576*b^11*c^7*d^5 + 384*b^11*c^8*d^4 + 12*a*b^10*c^2*d^10
+ 4*a*b^10*c^3*d^9 + 417*a*b^10*c^4*d^8 - 96*a*b^10*c^5*d^7 + 3288*a*b^10*c^6*d^6 - 2256*a*b^10*c^7*d^5 + 144
*a*b^10*c^8*d^4 - 192*a*b^10*c^9*d^3 - 12*a^2*b^9*c*d^11 + 12*a^3*b^8*c*d^11 - 252*a^4*b^7*c*d^11 + 60*a^5*b^6
*c*d^11 - 744*a^6*b^5*c*d^11 - 648*a^7*b^4*c*d^11 + 1872*a^8*b^3*c*d^11 + 432*a^9*b^2*c*d^11 - 12*a^2*b^9*c^2*
d^10 - 716*a^2*b^9*c^3*d^9 + 63*a^2*b^9*c^4*d^8 - 7872*a^2*b^9*c^5*d^7 + 5784*a^2*b^9*c^6*d^6 + 192*a^2*b^9*c^
7*d^5 + 690*a^2*b^9*c^8*d^4 + 24*a^2*b^9*c^10*d^2 + 606*a^3*b^8*c^2*d^10 + 76*a^3*b^8*c^3*d^9 + 10203*a^3*b^8*
c^4*d^8 - 8592*a^3*b^8*c^5*d^7 - 3752*a^3*b^8*c^6*d^6 - 480*a^3*b^8*c^7*d^5 - 144*a^3*b^8*c^8*d^4 - 32*a^3*b^8
*c^9*d^3 - 126*a^4*b^7*c^2*d^10 - 7680*a^4*b^7*c^3*d^9 + 8277*a^4*b^7*c^4*d^8 + 11232*a^4*b^7*c^5*d^7 - 1552*a
^4*b^7*c^6*d^6 + 384*a^4*b^7*c^7*d^5 - 132*a^4*b^7*c^8*d^4 + 3318*a^5*b^6*c^2*d^10 - 5424*a^5*b^6*c^3*d^9 - 16
488*a^5*b^6*c^4*d^8 + 4128*a^5*b^6*c^5*d^7 + 46...

________________________________________________________________________________________