3.244 \(\int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=250 \[ \frac{\left (95 a c^3 d+80 a c d^3+112 b c^2 d^2+12 b c^4+16 b d^4\right ) \tan (e+f x)}{30 f}+\frac{\left (24 a c^2 d^2+8 a c^4+3 a d^4+16 b c^3 d+12 b c d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{\left (35 a c d+12 b c^2+16 b d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{60 f}+\frac{d \left (130 a c^2 d+45 a d^3+24 b c^3+116 b c d^2\right ) \tan (e+f x) \sec (e+f x)}{120 f}+\frac{(5 a d+4 b c) \tan (e+f x) (c+d \sec (e+f x))^3}{20 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^4}{5 f} \]

[Out]

((8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*ArcTanh[Sin[e + f*x]])/(8*f) + ((12*b*c^4 + 95*a
*c^3*d + 112*b*c^2*d^2 + 80*a*c*d^3 + 16*b*d^4)*Tan[e + f*x])/(30*f) + (d*(24*b*c^3 + 130*a*c^2*d + 116*b*c*d^
2 + 45*a*d^3)*Sec[e + f*x]*Tan[e + f*x])/(120*f) + ((12*b*c^2 + 35*a*c*d + 16*b*d^2)*(c + d*Sec[e + f*x])^2*Ta
n[e + f*x])/(60*f) + ((4*b*c + 5*a*d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(20*f) + (b*(c + d*Sec[e + f*x])^4*
Tan[e + f*x])/(5*f)

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Rubi [A]  time = 0.500991, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (95 a c^3 d+80 a c d^3+112 b c^2 d^2+12 b c^4+16 b d^4\right ) \tan (e+f x)}{30 f}+\frac{\left (24 a c^2 d^2+8 a c^4+3 a d^4+16 b c^3 d+12 b c d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{\left (35 a c d+12 b c^2+16 b d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{60 f}+\frac{d \left (130 a c^2 d+45 a d^3+24 b c^3+116 b c d^2\right ) \tan (e+f x) \sec (e+f x)}{120 f}+\frac{(5 a d+4 b c) \tan (e+f x) (c+d \sec (e+f x))^3}{20 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^4}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*ArcTanh[Sin[e + f*x]])/(8*f) + ((12*b*c^4 + 95*a
*c^3*d + 112*b*c^2*d^2 + 80*a*c*d^3 + 16*b*d^4)*Tan[e + f*x])/(30*f) + (d*(24*b*c^3 + 130*a*c^2*d + 116*b*c*d^
2 + 45*a*d^3)*Sec[e + f*x]*Tan[e + f*x])/(120*f) + ((12*b*c^2 + 35*a*c*d + 16*b*d^2)*(c + d*Sec[e + f*x])^2*Ta
n[e + f*x])/(60*f) + ((4*b*c + 5*a*d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(20*f) + (b*(c + d*Sec[e + f*x])^4*
Tan[e + f*x])/(5*f)

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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]

Rubi steps

\begin{align*} \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x))^4 \, dx &=\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}+\frac{1}{5} \int \sec (e+f x) (c+d \sec (e+f x))^3 (5 a c+4 b d+(4 b c+5 a d) \sec (e+f x)) \, dx\\ &=\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}+\frac{1}{20} \int \sec (e+f x) (c+d \sec (e+f x))^2 \left (20 a c^2+28 b c d+15 a d^2+\left (12 b c^2+35 a c d+16 b d^2\right ) \sec (e+f x)\right ) \, dx\\ &=\frac{\left (12 b c^2+35 a c d+16 b d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}+\frac{1}{60} \int \sec (e+f x) (c+d \sec (e+f x)) \left (60 a c^3+108 b c^2 d+115 a c d^2+32 b d^3+\left (24 b c^3+130 a c^2 d+116 b c d^2+45 a d^3\right ) \sec (e+f x)\right ) \, dx\\ &=\frac{d \left (24 b c^3+130 a c^2 d+116 b c d^2+45 a d^3\right ) \sec (e+f x) \tan (e+f x)}{120 f}+\frac{\left (12 b c^2+35 a c d+16 b d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}+\frac{1}{120} \int \sec (e+f x) \left (15 \left (8 a c^4+16 b c^3 d+24 a c^2 d^2+12 b c d^3+3 a d^4\right )+4 \left (12 b c^4+95 a c^3 d+112 b c^2 d^2+80 a c d^3+16 b d^4\right ) \sec (e+f x)\right ) \, dx\\ &=\frac{d \left (24 b c^3+130 a c^2 d+116 b c d^2+45 a d^3\right ) \sec (e+f x) \tan (e+f x)}{120 f}+\frac{\left (12 b c^2+35 a c d+16 b d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}+\frac{1}{8} \left (8 a c^4+16 b c^3 d+24 a c^2 d^2+12 b c d^3+3 a d^4\right ) \int \sec (e+f x) \, dx+\frac{1}{30} \left (12 b c^4+95 a c^3 d+112 b c^2 d^2+80 a c d^3+16 b d^4\right ) \int \sec ^2(e+f x) \, dx\\ &=\frac{\left (8 a c^4+16 b c^3 d+24 a c^2 d^2+12 b c d^3+3 a d^4\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{d \left (24 b c^3+130 a c^2 d+116 b c d^2+45 a d^3\right ) \sec (e+f x) \tan (e+f x)}{120 f}+\frac{\left (12 b c^2+35 a c d+16 b d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}-\frac{\left (12 b c^4+95 a c^3 d+112 b c^2 d^2+80 a c d^3+16 b d^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{30 f}\\ &=\frac{\left (8 a c^4+16 b c^3 d+24 a c^2 d^2+12 b c d^3+3 a d^4\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{\left (12 b c^4+95 a c^3 d+112 b c^2 d^2+80 a c d^3+16 b d^4\right ) \tan (e+f x)}{30 f}+\frac{d \left (24 b c^3+130 a c^2 d+116 b c d^2+45 a d^3\right ) \sec (e+f x) \tan (e+f x)}{120 f}+\frac{\left (12 b c^2+35 a c d+16 b d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{(4 b c+5 a d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{b (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 4.42882, size = 201, normalized size = 0.8 \[ \frac{15 \left (a \left (24 c^2 d^2+8 c^4+3 d^4\right )+4 b c d \left (4 c^2+3 d^2\right )\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (8 \left (10 d^2 \left (2 a c d+b \left (3 c^2+d^2\right )\right ) \tan ^2(e+f x)+15 \left (4 a c d \left (c^2+d^2\right )+b \left (6 c^2 d^2+c^4+d^4\right )\right )+3 b d^4 \tan ^4(e+f x)\right )+15 d \left (3 a d \left (8 c^2+d^2\right )+4 b \left (4 c^3+3 c d^2\right )\right ) \sec (e+f x)+30 d^3 (a d+4 b c) \sec ^3(e+f x)\right )}{120 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(4*b*c*d*(4*c^2 + 3*d^2) + a*(8*c^4 + 24*c^2*d^2 + 3*d^4))*ArcTanh[Sin[e + f*x]] + Tan[e + f*x]*(15*d*(3*a
*d*(8*c^2 + d^2) + 4*b*(4*c^3 + 3*c*d^2))*Sec[e + f*x] + 30*d^3*(4*b*c + a*d)*Sec[e + f*x]^3 + 8*(15*(4*a*c*d*
(c^2 + d^2) + b*(c^4 + 6*c^2*d^2 + d^4)) + 10*d^2*(2*a*c*d + b*(3*c^2 + d^2))*Tan[e + f*x]^2 + 3*b*d^4*Tan[e +
f*x]^4)))/(120*f)

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Maple [A]  time = 0.048, size = 431, normalized size = 1.7 \begin{align*}{\frac{a{c}^{4}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+4\,{\frac{a{c}^{3}d\tan \left ( fx+e \right ) }{f}}+3\,{\frac{a{c}^{2}{d}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+3\,{\frac{a{c}^{2}{d}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{8\,ac{d}^{3}\tan \left ( fx+e \right ) }{3\,f}}+{\frac{4\,ac{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{a{d}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{3\,a{d}^{4}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,a{d}^{4}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{b{c}^{4}\tan \left ( fx+e \right ) }{f}}+2\,{\frac{b{c}^{3}d\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+2\,{\frac{b{c}^{3}d\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+4\,{\frac{b{c}^{2}{d}^{2}\tan \left ( fx+e \right ) }{f}}+2\,{\frac{b{c}^{2}{d}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{bc{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{f}}+{\frac{3\,bc{d}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{3\,bc{d}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{8\,b{d}^{4}\tan \left ( fx+e \right ) }{15\,f}}+{\frac{b{d}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}}+{\frac{4\,b{d}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+b*sec(f*x+e))*(c+d*sec(f*x+e))^4,x)

[Out]

1/f*a*c^4*ln(sec(f*x+e)+tan(f*x+e))+4/f*a*c^3*d*tan(f*x+e)+3/f*a*c^2*d^2*sec(f*x+e)*tan(f*x+e)+3/f*a*c^2*d^2*l
n(sec(f*x+e)+tan(f*x+e))+8/3/f*a*c*d^3*tan(f*x+e)+4/3/f*a*c*d^3*tan(f*x+e)*sec(f*x+e)^2+1/4/f*a*d^4*tan(f*x+e)
*sec(f*x+e)^3+3/8/f*a*d^4*sec(f*x+e)*tan(f*x+e)+3/8/f*a*d^4*ln(sec(f*x+e)+tan(f*x+e))+1/f*b*c^4*tan(f*x+e)+2/f
*b*c^3*d*sec(f*x+e)*tan(f*x+e)+2/f*b*c^3*d*ln(sec(f*x+e)+tan(f*x+e))+4/f*b*c^2*d^2*tan(f*x+e)+2/f*b*c^2*d^2*ta
n(f*x+e)*sec(f*x+e)^2+1/f*b*c*d^3*tan(f*x+e)*sec(f*x+e)^3+3/2/f*b*c*d^3*sec(f*x+e)*tan(f*x+e)+3/2/f*b*c*d^3*ln
(sec(f*x+e)+tan(f*x+e))+8/15/f*b*d^4*tan(f*x+e)+1/5/f*b*d^4*tan(f*x+e)*sec(f*x+e)^4+4/15/f*b*d^4*tan(f*x+e)*se
c(f*x+e)^2

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Maxima [A]  time = 1.15576, size = 512, normalized size = 2.05 \begin{align*} \frac{480 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} b c^{2} d^{2} + 320 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a c d^{3} + 16 \,{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} b d^{4} - 60 \, b c d^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 15 \, a d^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 240 \, b c^{3} d{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 360 \, a c^{2} d^{2}{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 240 \, a c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 240 \, b c^{4} \tan \left (f x + e\right ) + 960 \, a c^{3} d \tan \left (f x + e\right )}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(480*(tan(f*x + e)^3 + 3*tan(f*x + e))*b*c^2*d^2 + 320*(tan(f*x + e)^3 + 3*tan(f*x + e))*a*c*d^3 + 16*(3
*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*b*d^4 - 60*b*c*d^3*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e
))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 15*a*d^4*(2*
(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(
sin(f*x + e) - 1)) - 240*b*c^3*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x +
e) - 1)) - 360*a*c^2*d^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1))
+ 240*a*c^4*log(sec(f*x + e) + tan(f*x + e)) + 240*b*c^4*tan(f*x + e) + 960*a*c^3*d*tan(f*x + e))/f

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Fricas [A]  time = 0.616146, size = 687, normalized size = 2.75 \begin{align*} \frac{15 \,{\left (8 \, a c^{4} + 16 \, b c^{3} d + 24 \, a c^{2} d^{2} + 12 \, b c d^{3} + 3 \, a d^{4}\right )} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (8 \, a c^{4} + 16 \, b c^{3} d + 24 \, a c^{2} d^{2} + 12 \, b c d^{3} + 3 \, a d^{4}\right )} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (24 \, b d^{4} + 8 \,{\left (15 \, b c^{4} + 60 \, a c^{3} d + 60 \, b c^{2} d^{2} + 40 \, a c d^{3} + 8 \, b d^{4}\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (16 \, b c^{3} d + 24 \, a c^{2} d^{2} + 12 \, b c d^{3} + 3 \, a d^{4}\right )} \cos \left (f x + e\right )^{3} + 16 \,{\left (15 \, b c^{2} d^{2} + 10 \, a c d^{3} + 2 \, b d^{4}\right )} \cos \left (f x + e\right )^{2} + 30 \,{\left (4 \, b c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*cos(f*x + e)^5*log(sin(f*x + e) + 1) -
15*(8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(24*
b*d^4 + 8*(15*b*c^4 + 60*a*c^3*d + 60*b*c^2*d^2 + 40*a*c*d^3 + 8*b*d^4)*cos(f*x + e)^4 + 15*(16*b*c^3*d + 24*a
*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*cos(f*x + e)^3 + 16*(15*b*c^2*d^2 + 10*a*c*d^3 + 2*b*d^4)*cos(f*x + e)^2 + 30
*(4*b*c*d^3 + a*d^4)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.45158, size = 1207, normalized size = 4.83 \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)*(a+b*sec(f*x+e))*(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

1/120*(15*(8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15
*(8*a*c^4 + 16*b*c^3*d + 24*a*c^2*d^2 + 12*b*c*d^3 + 3*a*d^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(120*b*c^
4*tan(1/2*f*x + 1/2*e)^9 + 480*a*c^3*d*tan(1/2*f*x + 1/2*e)^9 - 240*b*c^3*d*tan(1/2*f*x + 1/2*e)^9 - 360*a*c^2
*d^2*tan(1/2*f*x + 1/2*e)^9 + 720*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^9 + 480*a*c*d^3*tan(1/2*f*x + 1/2*e)^9 - 300*
b*c*d^3*tan(1/2*f*x + 1/2*e)^9 - 75*a*d^4*tan(1/2*f*x + 1/2*e)^9 + 120*b*d^4*tan(1/2*f*x + 1/2*e)^9 - 480*b*c^
4*tan(1/2*f*x + 1/2*e)^7 - 1920*a*c^3*d*tan(1/2*f*x + 1/2*e)^7 + 480*b*c^3*d*tan(1/2*f*x + 1/2*e)^7 + 720*a*c^
2*d^2*tan(1/2*f*x + 1/2*e)^7 - 1920*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^7 - 1280*a*c*d^3*tan(1/2*f*x + 1/2*e)^7 + 1
20*b*c*d^3*tan(1/2*f*x + 1/2*e)^7 + 30*a*d^4*tan(1/2*f*x + 1/2*e)^7 - 160*b*d^4*tan(1/2*f*x + 1/2*e)^7 + 720*b
*c^4*tan(1/2*f*x + 1/2*e)^5 + 2880*a*c^3*d*tan(1/2*f*x + 1/2*e)^5 + 2400*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 + 16
00*a*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 464*b*d^4*tan(1/2*f*x + 1/2*e)^5 - 480*b*c^4*tan(1/2*f*x + 1/2*e)^3 - 1920
*a*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 480*b*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 720*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 -
1920*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 1280*a*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 120*b*c*d^3*tan(1/2*f*x + 1/2*e)
^3 - 30*a*d^4*tan(1/2*f*x + 1/2*e)^3 - 160*b*d^4*tan(1/2*f*x + 1/2*e)^3 + 120*b*c^4*tan(1/2*f*x + 1/2*e) + 480
*a*c^3*d*tan(1/2*f*x + 1/2*e) + 240*b*c^3*d*tan(1/2*f*x + 1/2*e) + 360*a*c^2*d^2*tan(1/2*f*x + 1/2*e) + 720*b*
c^2*d^2*tan(1/2*f*x + 1/2*e) + 480*a*c*d^3*tan(1/2*f*x + 1/2*e) + 300*b*c*d^3*tan(1/2*f*x + 1/2*e) + 75*a*d^4*
tan(1/2*f*x + 1/2*e) + 120*b*d^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f