3.2.85 \(\int \sec (e+f x) (a+a \sec (e+f x)) (c+d \sec (e+f x))^4 \, dx\) [185]

Optimal. Leaf size=236 \[ \frac {a \left (8 c^4+16 c^3 d+24 c^2 d^2+12 c d^3+3 d^4\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {a \left (12 c^4+95 c^3 d+112 c^2 d^2+80 c d^3+16 d^4\right ) \tan (e+f x)}{30 f}+\frac {a d \left (24 c^3+130 c^2 d+116 c d^2+45 d^3\right ) \sec (e+f x) \tan (e+f x)}{120 f}+\frac {a \left (12 c^2+35 c d+16 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac {a (4 c+5 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac {a (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f} \]

[Out]

1/8*a*(8*c^4+16*c^3*d+24*c^2*d^2+12*c*d^3+3*d^4)*arctanh(sin(f*x+e))/f+1/30*a*(12*c^4+95*c^3*d+112*c^2*d^2+80*
c*d^3+16*d^4)*tan(f*x+e)/f+1/120*a*d*(24*c^3+130*c^2*d+116*c*d^2+45*d^3)*sec(f*x+e)*tan(f*x+e)/f+1/60*a*(12*c^
2+35*c*d+16*d^2)*(c+d*sec(f*x+e))^2*tan(f*x+e)/f+1/20*a*(4*c+5*d)*(c+d*sec(f*x+e))^3*tan(f*x+e)/f+1/5*a*(c+d*s
ec(f*x+e))^4*tan(f*x+e)/f

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Rubi [A]
time = 0.30, antiderivative size = 236, normalized size of antiderivative = 1.00, 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 = {4087, 4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {a \left (12 c^2+35 c d+16 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{60 f}+\frac {a d \left (24 c^3+130 c^2 d+116 c d^2+45 d^3\right ) \tan (e+f x) \sec (e+f x)}{120 f}+\frac {a \left (12 c^4+95 c^3 d+112 c^2 d^2+80 c d^3+16 d^4\right ) \tan (e+f x)}{30 f}+\frac {a \left (8 c^4+16 c^3 d+24 c^2 d^2+12 c d^3+3 d^4\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {a \tan (e+f x) (c+d \sec (e+f x))^4}{5 f}+\frac {a (4 c+5 d) \tan (e+f x) (c+d \sec (e+f x))^3}{20 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*c^4 + 16*c^3*d + 24*c^2*d^2 + 12*c*d^3 + 3*d^4)*ArcTanh[Sin[e + f*x]])/(8*f) + (a*(12*c^4 + 95*c^3*d + 1
12*c^2*d^2 + 80*c*d^3 + 16*d^4)*Tan[e + f*x])/(30*f) + (a*d*(24*c^3 + 130*c^2*d + 116*c*d^2 + 45*d^3)*Sec[e +
f*x]*Tan[e + f*x])/(120*f) + (a*(12*c^2 + 35*c*d + 16*d^2)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(60*f) + (a*(4
*c + 5*d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(20*f) + (a*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(5*f)

Rule 8

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

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 3855

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

Rule 3872

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 4082

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
*Csc[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 4087

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] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 1.82, size = 153, normalized size = 0.65 \begin {gather*} \frac {a \left (15 \left (8 c^4+16 c^3 d+24 c^2 d^2+12 c d^3+3 d^4\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (120 (c+d)^4+15 d \left (16 c^3+24 c^2 d+12 c d^2+3 d^3\right ) \sec (e+f x)+30 d^3 (4 c+d) \sec ^3(e+f x)+80 d^2 \left (3 c^2+2 c d+d^2\right ) \tan ^2(e+f x)+24 d^4 \tan ^4(e+f x)\right )\right )}{120 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(15*(8*c^4 + 16*c^3*d + 24*c^2*d^2 + 12*c*d^3 + 3*d^4)*ArcTanh[Sin[e + f*x]] + Tan[e + f*x]*(120*(c + d)^4
+ 15*d*(16*c^3 + 24*c^2*d + 12*c*d^2 + 3*d^3)*Sec[e + f*x] + 30*d^3*(4*c + d)*Sec[e + f*x]^3 + 80*d^2*(3*c^2 +
2*c*d + d^2)*Tan[e + f*x]^2 + 24*d^4*Tan[e + f*x]^4)))/(120*f)

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Maple [A]
time = 0.40, size = 313, normalized size = 1.33 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a*c^4*tan(f*x+e)+4*a*c^3*d*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))-6*a*c^2*d^2*(-2/3-1/
3*sec(f*x+e)^2)*tan(f*x+e)+4*a*c*d^3*(-(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x+e)+tan(f*x
+e)))-a*d^4*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)+a*c^4*ln(sec(f*x+e)+tan(f*x+e))+4*a*c^3*d*ta
n(f*x+e)+6*a*c^2*d^2*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))-4*a*c*d^3*(-2/3-1/3*sec(f*x+e)^
2)*tan(f*x+e)+a*d^4*(-(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x+e)+tan(f*x+e))))

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Maxima [A]
time = 0.29, size = 410, normalized size = 1.74 \begin {gather*} \frac {480 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a 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 )} a d^{4} - 60 \, a 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 \, a 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 \, a c^{4} \tan \left (f x + e\right ) + 960 \, a c^{3} d \tan \left (f x + e\right )}{240 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*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))*a*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))*a*d^4 - 60*a*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*a*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*a*c^4*tan(f*x + e) + 960*a*c^3*d*tan(f*x + e))/f

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Fricas [A]
time = 2.35, size = 291, normalized size = 1.23 \begin {gather*} \frac {15 \, {\left (8 \, a c^{4} + 16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a 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 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a 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 \, a d^{4} + 8 \, {\left (15 \, a c^{4} + 60 \, a c^{3} d + 60 \, a c^{2} d^{2} + 40 \, a c d^{3} + 8 \, a d^{4}\right )} \cos \left (f x + e\right )^{4} + 15 \, {\left (16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a c d^{3} + 3 \, a d^{4}\right )} \cos \left (f x + e\right )^{3} + 16 \, {\left (15 \, a c^{2} d^{2} + 10 \, a c d^{3} + 2 \, a d^{4}\right )} \cos \left (f x + e\right )^{2} + 30 \, {\left (4 \, a 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(8*a*c^4 + 16*a*c^3*d + 24*a*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*cos(f*x + e)^5*log(sin(f*x + e) + 1) -
15*(8*a*c^4 + 16*a*c^3*d + 24*a*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(24*
a*d^4 + 8*(15*a*c^4 + 60*a*c^3*d + 60*a*c^2*d^2 + 40*a*c*d^3 + 8*a*d^4)*cos(f*x + e)^4 + 15*(16*a*c^3*d + 24*a
*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*cos(f*x + e)^3 + 16*(15*a*c^2*d^2 + 10*a*c*d^3 + 2*a*d^4)*cos(f*x + e)^2 + 30
*(4*a*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int c^{4} \sec {\left (e + f x \right )}\, dx + \int c^{4} \sec ^{2}{\left (e + f x \right )}\, dx + \int d^{4} \sec ^{5}{\left (e + f x \right )}\, dx + \int d^{4} \sec ^{6}{\left (e + f x \right )}\, dx + \int 4 c d^{3} \sec ^{4}{\left (e + f x \right )}\, dx + \int 4 c d^{3} \sec ^{5}{\left (e + f x \right )}\, dx + \int 6 c^{2} d^{2} \sec ^{3}{\left (e + f x \right )}\, dx + \int 6 c^{2} d^{2} \sec ^{4}{\left (e + f x \right )}\, dx + \int 4 c^{3} d \sec ^{2}{\left (e + f x \right )}\, dx + \int 4 c^{3} d \sec ^{3}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(c**4*sec(e + f*x), x) + Integral(c**4*sec(e + f*x)**2, x) + Integral(d**4*sec(e + f*x)**5, x) + In
tegral(d**4*sec(e + f*x)**6, x) + Integral(4*c*d**3*sec(e + f*x)**4, x) + Integral(4*c*d**3*sec(e + f*x)**5, x
) + Integral(6*c**2*d**2*sec(e + f*x)**3, x) + Integral(6*c**2*d**2*sec(e + f*x)**4, x) + Integral(4*c**3*d*se
c(e + f*x)**2, x) + Integral(4*c**3*d*sec(e + f*x)**3, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (224) = 448\).
time = 0.52, size = 566, normalized size = 2.40 \begin {gather*} \frac {15 \, {\left (8 \, a c^{4} + 16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a c d^{3} + 3 \, a d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, a c^{4} + 16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a c d^{3} + 3 \, a d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 240 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 360 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 180 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 45 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 480 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 1440 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 1200 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 1160 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 130 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 720 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2880 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2400 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 1600 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 464 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 480 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2400 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2640 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 1400 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 190 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 120 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 720 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1080 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 780 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 195 \, a 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 )}^{5}}}{120 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(8*a*c^4 + 16*a*c^3*d + 24*a*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15
*(8*a*c^4 + 16*a*c^3*d + 24*a*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(120*a*c^
4*tan(1/2*f*x + 1/2*e)^9 + 240*a*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 + 180*a*c
*d^3*tan(1/2*f*x + 1/2*e)^9 + 45*a*d^4*tan(1/2*f*x + 1/2*e)^9 - 480*a*c^4*tan(1/2*f*x + 1/2*e)^7 - 1440*a*c^3*
d*tan(1/2*f*x + 1/2*e)^7 - 1200*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^7 - 1160*a*c*d^3*tan(1/2*f*x + 1/2*e)^7 - 130*a
*d^4*tan(1/2*f*x + 1/2*e)^7 + 720*a*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*a*
c^2*d^2*tan(1/2*f*x + 1/2*e)^5 + 1600*a*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 464*a*d^4*tan(1/2*f*x + 1/2*e)^5 - 480*
a*c^4*tan(1/2*f*x + 1/2*e)^3 - 2400*a*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 2640*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 1
400*a*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 190*a*d^4*tan(1/2*f*x + 1/2*e)^3 + 120*a*c^4*tan(1/2*f*x + 1/2*e) + 720*a
*c^3*d*tan(1/2*f*x + 1/2*e) + 1080*a*c^2*d^2*tan(1/2*f*x + 1/2*e) + 780*a*c*d^3*tan(1/2*f*x + 1/2*e) + 195*a*d
^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f

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Mupad [B]
time = 5.51, size = 361, normalized size = 1.53 \begin {gather*} \frac {a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (8\,c^4+16\,c^3\,d+24\,c^2\,d^2+12\,c\,d^3+3\,d^4\right )}{2\,\left (4\,c^4+8\,c^3\,d+12\,c^2\,d^2+6\,c\,d^3+\frac {3\,d^4}{2}\right )}\right )\,\left (8\,c^4+16\,c^3\,d+24\,c^2\,d^2+12\,c\,d^3+3\,d^4\right )}{4\,f}-\frac {\left (2\,a\,c^4+4\,a\,c^3\,d+6\,a\,c^2\,d^2+3\,a\,c\,d^3+\frac {3\,a\,d^4}{4}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+\left (-8\,a\,c^4-24\,a\,c^3\,d-20\,a\,c^2\,d^2-\frac {58\,a\,c\,d^3}{3}-\frac {13\,a\,d^4}{6}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+\left (12\,a\,c^4+48\,a\,c^3\,d+40\,a\,c^2\,d^2+\frac {80\,a\,c\,d^3}{3}+\frac {116\,a\,d^4}{15}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (-8\,a\,c^4-40\,a\,c^3\,d-44\,a\,c^2\,d^2-\frac {70\,a\,c\,d^3}{3}-\frac {19\,a\,d^4}{6}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (2\,a\,c^4+12\,a\,c^3\,d+18\,a\,c^2\,d^2+13\,a\,c\,d^3+\frac {13\,a\,d^4}{4}\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atanh((tan(e/2 + (f*x)/2)*(12*c*d^3 + 16*c^3*d + 8*c^4 + 3*d^4 + 24*c^2*d^2))/(2*(6*c*d^3 + 8*c^3*d + 4*c^4
+ (3*d^4)/2 + 12*c^2*d^2)))*(12*c*d^3 + 16*c^3*d + 8*c^4 + 3*d^4 + 24*c^2*d^2))/(4*f) - (tan(e/2 + (f*x)/2)^9
*(2*a*c^4 + (3*a*d^4)/4 + 6*a*c^2*d^2 + 3*a*c*d^3 + 4*a*c^3*d) - tan(e/2 + (f*x)/2)^7*(8*a*c^4 + (13*a*d^4)/6
+ 20*a*c^2*d^2 + (58*a*c*d^3)/3 + 24*a*c^3*d) - tan(e/2 + (f*x)/2)^3*(8*a*c^4 + (19*a*d^4)/6 + 44*a*c^2*d^2 +
(70*a*c*d^3)/3 + 40*a*c^3*d) + tan(e/2 + (f*x)/2)^5*(12*a*c^4 + (116*a*d^4)/15 + 40*a*c^2*d^2 + (80*a*c*d^3)/3
+ 48*a*c^3*d) + tan(e/2 + (f*x)/2)*(2*a*c^4 + (13*a*d^4)/4 + 18*a*c^2*d^2 + 13*a*c*d^3 + 12*a*c^3*d))/(f*(5*t
an(e/2 + (f*x)/2)^2 - 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (
f*x)/2)^10 - 1))

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