3.258 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=379 \[ \frac{d^3 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right ) \tan (e+f x)}{b^4 f}+\frac{d^2 \left (15 a^2 b c d^2-4 a^3 d^3-20 a b^2 c^2 d+10 b^3 c^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{d^4 (5 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{2 b^3 f}+\frac{d^4 (5 b c-2 a d) \tan (e+f x) \sec (e+f x)}{2 b^3 f}+\frac{2 (b c-a d)^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^3 f (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 (b c-a d)^4 (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^5 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^5 \tan ^3(e+f x)}{3 b^2 f}+\frac{d^5 \tan (e+f x)}{b^2 f} \]

[Out]

(d^4*(5*b*c - 2*a*d)*ArcTanh[Sin[e + f*x]])/(2*b^3*f) + (d^2*(10*b^3*c^3 - 20*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 4
*a^3*d^3)*ArcTanh[Sin[e + f*x]])/(b^5*f) + (2*(b*c - a*d)^5*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]
])/(a*(a - b)^(3/2)*b^3*(a + b)^(3/2)*f) + (2*(b*c - a*d)^4*(b*c + 4*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2
])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^5*Sqrt[a + b]*f) - ((b*c - a*d)^5*Sin[e + f*x])/(b^4*(a^2 - b^2)*f*(b + a*Co
s[e + f*x])) + (d^5*Tan[e + f*x])/(b^2*f) + (d^3*(10*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*Tan[e + f*x])/(b^4*f) +
(d^4*(5*b*c - 2*a*d)*Sec[e + f*x]*Tan[e + f*x])/(2*b^3*f) + (d^5*Tan[e + f*x]^3)/(3*b^2*f)

________________________________________________________________________________________

Rubi [A]  time = 0.667993, antiderivative size = 379, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.323, Rules used = {3988, 2952, 2664, 12, 2659, 208, 3770, 3767, 8, 3768} \[ \frac{d^3 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right ) \tan (e+f x)}{b^4 f}+\frac{d^2 \left (15 a^2 b c d^2-4 a^3 d^3-20 a b^2 c^2 d+10 b^3 c^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{d^4 (5 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{2 b^3 f}+\frac{d^4 (5 b c-2 a d) \tan (e+f x) \sec (e+f x)}{2 b^3 f}+\frac{2 (b c-a d)^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^3 f (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 (b c-a d)^4 (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^5 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^5 \tan ^3(e+f x)}{3 b^2 f}+\frac{d^5 \tan (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^5)/(a + b*Sec[e + f*x])^2,x]

[Out]

(d^4*(5*b*c - 2*a*d)*ArcTanh[Sin[e + f*x]])/(2*b^3*f) + (d^2*(10*b^3*c^3 - 20*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 4
*a^3*d^3)*ArcTanh[Sin[e + f*x]])/(b^5*f) + (2*(b*c - a*d)^5*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]
])/(a*(a - b)^(3/2)*b^3*(a + b)^(3/2)*f) + (2*(b*c - a*d)^4*(b*c + 4*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2
])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^5*Sqrt[a + b]*f) - ((b*c - a*d)^5*Sin[e + f*x])/(b^4*(a^2 - b^2)*f*(b + a*Co
s[e + f*x])) + (d^5*Tan[e + f*x])/(b^2*f) + (d^3*(10*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*Tan[e + f*x])/(b^4*f) +
(d^4*(5*b*c - 2*a*d)*Sec[e + f*x]*Tan[e + f*x])/(2*b^3*f) + (d^5*Tan[e + f*x]^3)/(3*b^2*f)

Rule 3988

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]

Rule 2952

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 2664

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] && Integer
Q[2*n]

Rule 12

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

Rule 2659

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 208

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

Rule 3770

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

Rule 3767

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 8

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

Rule 3768

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]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^5}{(a+b \sec (e+f x))^2} \, dx &=\int \frac{(d+c \cos (e+f x))^5 \sec ^4(e+f x)}{(b+a \cos (e+f x))^2} \, dx\\ &=\int \left (\frac{(-b c+a d)^5}{a b^4 (b+a \cos (e+f x))^2}+\frac{(-b c+a d)^4 (b c+4 a d)}{a b^5 (b+a \cos (e+f x))}+\frac{d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \sec (e+f x)}{b^5}+\frac{d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \sec ^2(e+f x)}{b^4}+\frac{d^4 (5 b c-2 a d) \sec ^3(e+f x)}{b^3}+\frac{d^5 \sec ^4(e+f x)}{b^2}\right ) \, dx\\ &=\frac{d^5 \int \sec ^4(e+f x) \, dx}{b^2}+\frac{\left (d^4 (5 b c-2 a d)\right ) \int \sec ^3(e+f x) \, dx}{b^3}-\frac{(b c-a d)^5 \int \frac{1}{(b+a \cos (e+f x))^2} \, dx}{a b^4}+\frac{\left ((b c-a d)^4 (b c+4 a d)\right ) \int \frac{1}{b+a \cos (e+f x)} \, dx}{a b^5}+\frac{\left (d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \int \sec ^2(e+f x) \, dx}{b^4}+\frac{\left (d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right )\right ) \int \sec (e+f x) \, dx}{b^5}\\ &=\frac{d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac{d^4 (5 b c-2 a d) \sec (e+f x) \tan (e+f x)}{2 b^3 f}+\frac{\left (d^4 (5 b c-2 a d)\right ) \int \sec (e+f x) \, dx}{2 b^3}+\frac{(b c-a d)^5 \int \frac{b}{b+a \cos (e+f x)} \, dx}{a b^4 \left (a^2-b^2\right )}-\frac{d^5 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{b^2 f}+\frac{\left (2 (b c-a d)^4 (b c+4 a d)\right ) \operatorname{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^5 f}-\frac{\left (d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{b^4 f}\\ &=\frac{d^4 (5 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{2 b^3 f}+\frac{d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}+\frac{2 (b c-a d)^4 (b c+4 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^5 \sqrt{a+b} f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac{d^5 \tan (e+f x)}{b^2 f}+\frac{d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tan (e+f x)}{b^4 f}+\frac{d^4 (5 b c-2 a d) \sec (e+f x) \tan (e+f x)}{2 b^3 f}+\frac{d^5 \tan ^3(e+f x)}{3 b^2 f}+\frac{(b c-a d)^5 \int \frac{1}{b+a \cos (e+f x)} \, dx}{a b^3 \left (a^2-b^2\right )}\\ &=\frac{d^4 (5 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{2 b^3 f}+\frac{d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}+\frac{2 (b c-a d)^4 (b c+4 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^5 \sqrt{a+b} f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac{d^5 \tan (e+f x)}{b^2 f}+\frac{d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tan (e+f x)}{b^4 f}+\frac{d^4 (5 b c-2 a d) \sec (e+f x) \tan (e+f x)}{2 b^3 f}+\frac{d^5 \tan ^3(e+f x)}{3 b^2 f}+\frac{\left (2 (b c-a d)^5\right ) \operatorname{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^3 \left (a^2-b^2\right ) f}\\ &=\frac{d^4 (5 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{2 b^3 f}+\frac{d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{b^5 f}+\frac{2 (b c-a d)^5 \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^3 (a+b)^{3/2} f}+\frac{2 (b c-a d)^4 (b c+4 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^5 \sqrt{a+b} f}-\frac{(b c-a d)^5 \sin (e+f x)}{b^4 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac{d^5 \tan (e+f x)}{b^2 f}+\frac{d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tan (e+f x)}{b^4 f}+\frac{d^4 (5 b c-2 a d) \sec (e+f x) \tan (e+f x)}{2 b^3 f}+\frac{d^5 \tan ^3(e+f x)}{3 b^2 f}\\ \end{align*}

Mathematica [B]  time = 6.54493, size = 1137, normalized size = 3. \[ \frac{(b+a \cos (e+f x)) \left (12 d^5 \sin (e+f x) b^5+60 c^2 d^3 \sin (e+f x) b^5+6 c^5 \sin (2 (e+f x)) b^5+30 c d^4 \sin (2 (e+f x)) b^5+4 d^5 \sin (3 (e+f x)) b^5+60 c^2 d^3 \sin (3 (e+f x)) b^5+3 c^5 \sin (4 (e+f x)) b^5-45 a c d^4 \sin (e+f x) b^4-4 a d^5 \sin (2 (e+f x)) b^4+60 a c^2 d^3 \sin (2 (e+f x)) b^4-30 a c^4 d \sin (2 (e+f x)) b^4-45 a c d^4 \sin (3 (e+f x)) b^4+2 a d^5 \sin (4 (e+f x)) b^4+30 a c^2 d^3 \sin (4 (e+f x)) b^4-15 a c^4 d \sin (4 (e+f x)) b^4-60 a^2 c^2 d^3 \sin (e+f x) b^3-90 a^2 c d^4 \sin (2 (e+f x)) b^3+60 a^2 c^3 d^2 \sin (2 (e+f x)) b^3+8 a^2 d^5 \sin (3 (e+f x)) b^3-60 a^2 c^2 d^3 \sin (3 (e+f x)) b^3-30 a^2 c d^4 \sin (4 (e+f x)) b^3+30 a^2 c^3 d^2 \sin (4 (e+f x)) b^3+45 a^3 c d^4 \sin (e+f x) b^2+22 a^3 d^5 \sin (2 (e+f x)) b^2-120 a^3 c^2 d^3 \sin (2 (e+f x)) b^2+45 a^3 c d^4 \sin (3 (e+f x)) b^2+7 a^3 d^5 \sin (4 (e+f x)) b^2-60 a^3 c^2 d^3 \sin (4 (e+f x)) b^2-12 a^4 d^5 \sin (e+f x) b+90 a^4 c d^4 \sin (2 (e+f x)) b-12 a^4 d^5 \sin (3 (e+f x)) b+45 a^4 c d^4 \sin (4 (e+f x)) b-24 a^5 d^5 \sin (2 (e+f x))-12 a^5 d^5 \sin (4 (e+f x))\right ) (c+d \sec (e+f x))^5}{24 b^4 \left (b^2-a^2\right ) f (d+c \cos (e+f x))^5 (a+b \sec (e+f x))^2}+\frac{\left (8 a^3 d^5+2 a b^2 d^5-5 b^3 c d^4-30 a^2 b c d^4+40 a b^2 c^2 d^3-20 b^3 c^3 d^2\right ) \cos ^3(e+f x) (b+a \cos (e+f x))^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sec (e+f x))^5}{2 b^5 f (d+c \cos (e+f x))^5 (a+b \sec (e+f x))^2}+\frac{\left (-8 a^3 d^5-2 a b^2 d^5+5 b^3 c d^4+30 a^2 b c d^4-40 a b^2 c^2 d^3+20 b^3 c^3 d^2\right ) \cos ^3(e+f x) (b+a \cos (e+f x))^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sec (e+f x))^5}{2 b^5 f (d+c \cos (e+f x))^5 (a+b \sec (e+f x))^2}-\frac{2 (b c-a d)^4 \left (-4 d a^2-b c a+5 b^2 d\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right ) \cos ^3(e+f x) (b+a \cos (e+f x))^2 (c+d \sec (e+f x))^5}{b^5 \sqrt{a^2-b^2} \left (b^2-a^2\right ) f (d+c \cos (e+f x))^5 (a+b \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^5)/(a + b*Sec[e + f*x])^2,x]

[Out]

(-2*(b*c - a*d)^4*(-(a*b*c) - 4*a^2*d + 5*b^2*d)*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]]*Cos[e +
f*x]^3*(b + a*Cos[e + f*x])^2*(c + d*Sec[e + f*x])^5)/(b^5*Sqrt[a^2 - b^2]*(-a^2 + b^2)*f*(d + c*Cos[e + f*x])
^5*(a + b*Sec[e + f*x])^2) + ((-20*b^3*c^3*d^2 + 40*a*b^2*c^2*d^3 - 30*a^2*b*c*d^4 - 5*b^3*c*d^4 + 8*a^3*d^5 +
2*a*b^2*d^5)*Cos[e + f*x]^3*(b + a*Cos[e + f*x])^2*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*(c + d*Sec[e + f*
x])^5)/(2*b^5*f*(d + c*Cos[e + f*x])^5*(a + b*Sec[e + f*x])^2) + ((20*b^3*c^3*d^2 - 40*a*b^2*c^2*d^3 + 30*a^2*
b*c*d^4 + 5*b^3*c*d^4 - 8*a^3*d^5 - 2*a*b^2*d^5)*Cos[e + f*x]^3*(b + a*Cos[e + f*x])^2*Log[Cos[(e + f*x)/2] +
Sin[(e + f*x)/2]]*(c + d*Sec[e + f*x])^5)/(2*b^5*f*(d + c*Cos[e + f*x])^5*(a + b*Sec[e + f*x])^2) + ((b + a*Co
s[e + f*x])*(c + d*Sec[e + f*x])^5*(-60*a^2*b^3*c^2*d^3*Sin[e + f*x] + 60*b^5*c^2*d^3*Sin[e + f*x] + 45*a^3*b^
2*c*d^4*Sin[e + f*x] - 45*a*b^4*c*d^4*Sin[e + f*x] - 12*a^4*b*d^5*Sin[e + f*x] + 12*b^5*d^5*Sin[e + f*x] + 6*b
^5*c^5*Sin[2*(e + f*x)] - 30*a*b^4*c^4*d*Sin[2*(e + f*x)] + 60*a^2*b^3*c^3*d^2*Sin[2*(e + f*x)] - 120*a^3*b^2*
c^2*d^3*Sin[2*(e + f*x)] + 60*a*b^4*c^2*d^3*Sin[2*(e + f*x)] + 90*a^4*b*c*d^4*Sin[2*(e + f*x)] - 90*a^2*b^3*c*
d^4*Sin[2*(e + f*x)] + 30*b^5*c*d^4*Sin[2*(e + f*x)] - 24*a^5*d^5*Sin[2*(e + f*x)] + 22*a^3*b^2*d^5*Sin[2*(e +
f*x)] - 4*a*b^4*d^5*Sin[2*(e + f*x)] - 60*a^2*b^3*c^2*d^3*Sin[3*(e + f*x)] + 60*b^5*c^2*d^3*Sin[3*(e + f*x)]
+ 45*a^3*b^2*c*d^4*Sin[3*(e + f*x)] - 45*a*b^4*c*d^4*Sin[3*(e + f*x)] - 12*a^4*b*d^5*Sin[3*(e + f*x)] + 8*a^2*
b^3*d^5*Sin[3*(e + f*x)] + 4*b^5*d^5*Sin[3*(e + f*x)] + 3*b^5*c^5*Sin[4*(e + f*x)] - 15*a*b^4*c^4*d*Sin[4*(e +
f*x)] + 30*a^2*b^3*c^3*d^2*Sin[4*(e + f*x)] - 60*a^3*b^2*c^2*d^3*Sin[4*(e + f*x)] + 30*a*b^4*c^2*d^3*Sin[4*(e
+ f*x)] + 45*a^4*b*c*d^4*Sin[4*(e + f*x)] - 30*a^2*b^3*c*d^4*Sin[4*(e + f*x)] - 12*a^5*d^5*Sin[4*(e + f*x)] +
7*a^3*b^2*d^5*Sin[4*(e + f*x)] + 2*a*b^4*d^5*Sin[4*(e + f*x)]))/(24*b^4*(-a^2 + b^2)*f*(d + c*Cos[e + f*x])^5
*(a + b*Sec[e + f*x])^2)

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Maple [B]  time = 0.145, size = 1870, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^5/(a+b*sec(f*x+e))^2,x)

[Out]

40/f/(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))*a*c^3*d^2-10/f*b/(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))*c^4*d+10/f/b^3/(a^2-b^2)*t
an(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^4*c*d^4-20/f/b^2/(a^2-b^2)*tan(1/2*f*x
+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^3*c^2*d^3+20/f/b/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(t
an(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^2*c^3*d^2+8/f/b^5/(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))*a^6*d^5-10/f/b^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^4*d^5-1/2/f*d^5/b^2/(tan(1/2*f*x+1/2*e)-1)^2-1/3/f*d^5/b^2/(tan(1/2*f
*x+1/2*e)+1)^3-1/f*d^5/b^2/(tan(1/2*f*x+1/2*e)+1)+1/2/f*d^5/b^2/(tan(1/2*f*x+1/2*e)+1)^2-1/3/f*d^5/b^2/(tan(1/
2*f*x+1/2*e)-1)^3-1/f*d^5/b^2/(tan(1/2*f*x+1/2*e)-1)+40/f/b^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*ta
n(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^4*c^2*d^3-20/f/b^2/(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))*a^3*c^3*d^2+40/f/b^2/(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))*a^3*c*d^4-60/f/b/(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))*a^2*c^2*d^3-30/f/b^4/(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))*a^5*c*d^4-15/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)-1)*a^2*c+20/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)
-1)*a*c^2+10/f*d^4/b^3/(tan(1/2*f*x+1/2*e)-1)*a*c+2/f*b/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-t
an(1/2*f*x+1/2*e)^2*b-a-b)*c^5+15/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)+1)*a^2*c-20/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)+
1)*a*c^2+10/f*d^4/b^3/(tan(1/2*f*x+1/2*e)+1)*a*c+1/f*d^5/b^3/(tan(1/2*f*x+1/2*e)+1)^2*a-10/f/(a^2-b^2)*tan(1/2
*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a*c^4*d-2/f/b^4/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(
tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^5*d^5+2/f/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*c^5*a-5/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)^2*c+4/f*d^5/b^5*ln(tan(1/2*f
*x+1/2*e)-1)*a^3+1/f*d^5/b^3*ln(tan(1/2*f*x+1/2*e)-1)*a-10/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*c^3-5/2/f*d^4/b^
2*ln(tan(1/2*f*x+1/2*e)-1)*c-3/f*d^5/b^4/(tan(1/2*f*x+1/2*e)-1)*a^2-1/f*d^5/b^3/(tan(1/2*f*x+1/2*e)-1)*a-10/f*
d^3/b^2/(tan(1/2*f*x+1/2*e)-1)*c^2+5/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)-1)*c-1/f*d^5/b^3/(tan(1/2*f*x+1/2*e)-1)^2
*a+5/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)-1)^2*c-4/f*d^5/b^5*ln(tan(1/2*f*x+1/2*e)+1)*a^3-1/f*d^5/b^3*ln(tan(1/2*f*
x+1/2*e)+1)*a+10/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)*c^3+5/2/f*d^4/b^2*ln(tan(1/2*f*x+1/2*e)+1)*c-3/f*d^5/b^4/(
tan(1/2*f*x+1/2*e)+1)*a^2-1/f*d^5/b^3/(tan(1/2*f*x+1/2*e)+1)*a-10/f*d^3/b^2/(tan(1/2*f*x+1/2*e)+1)*c^2+5/2/f*d
^4/b^2/(tan(1/2*f*x+1/2*e)+1)*c

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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(sec(f*x+e)*(c+d*sec(f*x+e))^5/(a+b*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^5/(a+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d \sec{\left (e + f x \right )}\right )^{5} \sec{\left (e + f x \right )}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**5/(a+b*sec(f*x+e))**2,x)

[Out]

Integral((c + d*sec(e + f*x))**5*sec(e + f*x)/(a + b*sec(e + f*x))**2, x)

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Giac [B]  time = 1.44164, size = 1197, normalized size = 3.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^5/(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*(12*(a*b^5*c^5 - 5*b^6*c^4*d - 10*a^3*b^3*c^3*d^2 + 20*a*b^5*c^3*d^2 + 20*a^4*b^2*c^2*d^3 - 30*a^2*b^4*c^
2*d^3 - 15*a^5*b*c*d^4 + 20*a^3*b^3*c*d^4 + 4*a^6*d^5 - 5*a^4*b^2*d^5)*(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^5 - b^7)*sqrt(
-a^2 + b^2)) - 12*(b^5*c^5*tan(1/2*f*x + 1/2*e) - 5*a*b^4*c^4*d*tan(1/2*f*x + 1/2*e) + 10*a^2*b^3*c^3*d^2*tan(
1/2*f*x + 1/2*e) - 10*a^3*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e) + 5*a^4*b*c*d^4*tan(1/2*f*x + 1/2*e) - a^5*d^5*tan(
1/2*f*x + 1/2*e))/((a^2*b^4 - b^6)*(a*tan(1/2*f*x + 1/2*e)^2 - b*tan(1/2*f*x + 1/2*e)^2 - a - b)) - 3*(20*b^3*
c^3*d^2 - 40*a*b^2*c^2*d^3 + 30*a^2*b*c*d^4 + 5*b^3*c*d^4 - 8*a^3*d^5 - 2*a*b^2*d^5)*log(abs(tan(1/2*f*x + 1/2
*e) + 1))/b^5 + 3*(20*b^3*c^3*d^2 - 40*a*b^2*c^2*d^3 + 30*a^2*b*c*d^4 + 5*b^3*c*d^4 - 8*a^3*d^5 - 2*a*b^2*d^5)
*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^5 + 2*(60*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 - 60*a*b*c*d^4*tan(1/2*f*x
+ 1/2*e)^5 - 15*b^2*c*d^4*tan(1/2*f*x + 1/2*e)^5 + 18*a^2*d^5*tan(1/2*f*x + 1/2*e)^5 + 6*a*b*d^5*tan(1/2*f*x +
1/2*e)^5 + 6*b^2*d^5*tan(1/2*f*x + 1/2*e)^5 - 120*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 + 120*a*b*c*d^4*tan(1/2*
f*x + 1/2*e)^3 - 36*a^2*d^5*tan(1/2*f*x + 1/2*e)^3 - 4*b^2*d^5*tan(1/2*f*x + 1/2*e)^3 + 60*b^2*c^2*d^3*tan(1/2
*f*x + 1/2*e) - 60*a*b*c*d^4*tan(1/2*f*x + 1/2*e) + 15*b^2*c*d^4*tan(1/2*f*x + 1/2*e) + 18*a^2*d^5*tan(1/2*f*x
+ 1/2*e) - 6*a*b*d^5*tan(1/2*f*x + 1/2*e) + 6*b^2*d^5*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*b
^4))/f