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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.526849, antiderivative size = 297, normalized size of antiderivative = 1., 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 = {3988, 2952, 2664, 12, 2659, 208, 3770, 3767, 8, 3768} \[ \frac{d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}-\frac{(b c-a d)^4 \sin (e+f x)}{b^3 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{2 d^3 (2 b c-a d) \tan (e+f x)}{b^3 f}+\frac{2 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^2 f (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 (b c-a d)^3 (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^4 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^4 \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{d^4 \tan (e+f x) \sec (e+f x)}{2 b^2 f} \]

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 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))^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 ) \operatorname{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 ) \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^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 ) \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^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 = 3.46889, size = 511, normalized size = 1.72 \[ \frac{\cos ^2(e+f x) (a \cos (e+f x)+b) (c+d \sec (e+f x))^4 \left (-2 d^2 \left (6 a^2 d^2-16 a b c d+b^2 \left (12 c^2+d^2\right )\right ) (a \cos (e+f x)+b) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+2 d^2 \left (6 a^2 d^2-16 a b c d+b^2 \left (12 c^2+d^2\right )\right ) (a \cos (e+f x)+b) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{8 (a d-b c)^3 \left (3 a^2 d+a b c-4 b^2 d\right ) (a \cos (e+f x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b^2 d^4 (a \cos (e+f x)+b)}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{b^2 d^4 (a \cos (e+f x)+b)}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{8 b d^3 (2 b c-a d) \sin \left (\frac{1}{2} (e+f x)\right ) (a \cos (e+f x)+b)}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{8 b d^3 (2 b c-a d) \sin \left (\frac{1}{2} (e+f x)\right ) (a \cos (e+f x)+b)}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}+\frac{4 b (b c-a d)^4 \sin (e+f x)}{(b-a) (a+b)}\right )}{4 b^4 f (a+b \sec (e+f x))^2 (c \cos (e+f x)+d)^4} \]

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 [B]  time = 0.118, size = 1249, normalized size = 4.2 \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))^4/(a+b*sec(f*x+e))^2,x)

[Out]

8/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*d^4-8/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^3*d+24/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^2*d^2-6/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*d^4-8/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*d^3+12/f/b/(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^2*c^2*d^2+6/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)*c^2-1/2/f*d^4/b^2/(ta
n(1/2*f*x+1/2*e)+1)^2+1/2/f*d^4/b^2*ln(tan(1/2*f*x+1/2*e)+1)+1/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)+1/2/f*d^4/b^
2/(tan(1/2*f*x+1/2*e)-1)^2-1/2/f*d^4/b^2*ln(tan(1/2*f*x+1/2*e)-1)+1/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)-1)-24/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*d^3+16/f/b^3/(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^4*c*d^3-12/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^2*d^2+2/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))*c^4*a+2/f/b^3/(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^4*d^4-8/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^3*d+2/f*d^4/b^3/(tan(1/2*f*x+1/2*e)+1)*a-4/f*d^3/b^2/(tan(1/2*f*x+1/2*
e)+1)*c-3/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)-1)*a^2-6/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*c^2+2/f*d^4/b^3/(tan(1/2
*f*x+1/2*e)-1)*a-4/f*d^3/b^2/(tan(1/2*f*x+1/2*e)-1)*c+3/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)+1)*a^2+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-tan(1/2*f*x+1/2*e)^2*b-a-b)*c^4-8/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)+
1)*a*c+8/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)-1)*a*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))^4/(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))^4/(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 )^{4} \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))**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)

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Giac [B]  time = 1.32983, size = 770, normalized size = 2.59 \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))^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