3.889 \(\int (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=290 \[ \frac{\tan (c+d x) \left (2 a^2 b^2 (85 A+56 C)+95 a^3 b B+12 a^4 C+80 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 d}+\frac{\left (16 a^3 b (2 A+C)+24 a^2 b^2 B+8 a^4 B+4 a b^3 (4 A+3 C)+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \left (12 a^2 C+35 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 d}+\frac{b \tan (c+d x) \sec (c+d x) \left (130 a^2 b B+24 a^3 C+4 a b^2 (40 A+29 C)+45 b^3 B\right )}{120 d}+a^4 A x+\frac{(4 a C+5 b B) \tan (c+d x) (a+b \sec (c+d x))^3}{20 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]

[Out]

a^4*A*x + ((8*a^4*B + 24*a^2*b^2*B + 3*b^4*B + 16*a^3*b*(2*A + C) + 4*a*b^3*(4*A + 3*C))*ArcTanh[Sin[c + d*x]]
)/(8*d) + ((95*a^3*b*B + 80*a*b^3*B + 12*a^4*C + 4*b^4*(5*A + 4*C) + 2*a^2*b^2*(85*A + 56*C))*Tan[c + d*x])/(3
0*d) + (b*(130*a^2*b*B + 45*b^3*B + 24*a^3*C + 4*a*b^2*(40*A + 29*C))*Sec[c + d*x]*Tan[c + d*x])/(120*d) + ((2
0*A*b^2 + 35*a*b*B + 12*a^2*C + 16*b^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*d) + ((5*b*B + 4*a*C)*(a +
b*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.542844, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4056, 4048, 3770, 3767, 8} \[ \frac{\tan (c+d x) \left (2 a^2 b^2 (85 A+56 C)+95 a^3 b B+12 a^4 C+80 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 d}+\frac{\left (16 a^3 b (2 A+C)+24 a^2 b^2 B+8 a^4 B+4 a b^3 (4 A+3 C)+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \left (12 a^2 C+35 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 d}+\frac{b \tan (c+d x) \sec (c+d x) \left (130 a^2 b B+24 a^3 C+4 a b^2 (40 A+29 C)+45 b^3 B\right )}{120 d}+a^4 A x+\frac{(4 a C+5 b B) \tan (c+d x) (a+b \sec (c+d x))^3}{20 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^4*A*x + ((8*a^4*B + 24*a^2*b^2*B + 3*b^4*B + 16*a^3*b*(2*A + C) + 4*a*b^3*(4*A + 3*C))*ArcTanh[Sin[c + d*x]]
)/(8*d) + ((95*a^3*b*B + 80*a*b^3*B + 12*a^4*C + 4*b^4*(5*A + 4*C) + 2*a^2*b^2*(85*A + 56*C))*Tan[c + d*x])/(3
0*d) + (b*(130*a^2*b*B + 45*b^3*B + 24*a^3*C + 4*a*b^2*(40*A + 29*C))*Sec[c + d*x]*Tan[c + d*x])/(120*d) + ((2
0*A*b^2 + 35*a*b*B + 12*a^2*C + 16*b^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*d) + ((5*b*B + 4*a*C)*(a +
b*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

Rule 4056

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

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, 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 (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \sec (c+d x))^3 \left (5 a A+(5 A b+5 a B+4 b C) \sec (c+d x)+(5 b B+4 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+b \sec (c+d x))^2 \left (20 a^2 A+\left (40 a A b+20 a^2 B+15 b^2 B+28 a b C\right ) \sec (c+d x)+\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{60} \int (a+b \sec (c+d x)) \left (60 a^3 A+\left (60 a^3 B+115 a b^2 B+36 a^2 b (5 A+3 C)+8 b^3 (5 A+4 C)\right ) \sec (c+d x)+\left (130 a^2 b B+45 b^3 B+24 a^3 C+4 a b^2 (40 A+29 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b \left (130 a^2 b B+45 b^3 B+24 a^3 C+4 a b^2 (40 A+29 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{120} \int \left (120 a^4 A+15 \left (8 a^4 B+24 a^2 b^2 B+3 b^4 B+16 a^3 b (2 A+C)+4 a b^3 (4 A+3 C)\right ) \sec (c+d x)+4 \left (95 a^3 b B+80 a b^3 B+12 a^4 C+4 b^4 (5 A+4 C)+2 a^2 b^2 (85 A+56 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 A x+\frac{b \left (130 a^2 b B+45 b^3 B+24 a^3 C+4 a b^2 (40 A+29 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{8} \left (8 a^4 B+24 a^2 b^2 B+3 b^4 B+16 a^3 b (2 A+C)+4 a b^3 (4 A+3 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{30} \left (95 a^3 b B+80 a b^3 B+12 a^4 C+4 b^4 (5 A+4 C)+2 a^2 b^2 (85 A+56 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=a^4 A x+\frac{\left (8 a^4 B+24 a^2 b^2 B+3 b^4 B+16 a^3 b (2 A+C)+4 a b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (130 a^2 b B+45 b^3 B+24 a^3 C+4 a b^2 (40 A+29 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}-\frac{\left (95 a^3 b B+80 a b^3 B+12 a^4 C+4 b^4 (5 A+4 C)+2 a^2 b^2 (85 A+56 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 d}\\ &=a^4 A x+\frac{\left (8 a^4 B+24 a^2 b^2 B+3 b^4 B+16 a^3 b (2 A+C)+4 a b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (95 a^3 b B+80 a b^3 B+12 a^4 C+4 b^4 (5 A+4 C)+2 a^2 b^2 (85 A+56 C)\right ) \tan (c+d x)}{30 d}+\frac{b \left (130 a^2 b B+45 b^3 B+24 a^3 C+4 a b^2 (40 A+29 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (20 A b^2+35 a b B+12 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 b B+4 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 4.10403, size = 690, normalized size = 2.38 \[ \frac{\sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (-120 \cos ^5(c+d x) \left (16 a^3 b (2 A+C)+24 a^2 b^2 B+8 a^4 B+4 a b^3 (4 A+3 C)+3 b^4 B\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+720 a^2 A b^2 \sin (c+d x)+1080 a^2 A b^2 \sin (3 (c+d x))+360 a^2 A b^2 \sin (5 (c+d x))+600 a^4 A (c+d x) \cos (c+d x)+300 a^4 A (c+d x) \cos (3 (c+d x))+60 a^4 A c \cos (5 (c+d x))+60 a^4 A d x \cos (5 (c+d x))+720 a^2 b^2 B \sin (2 (c+d x))+360 a^2 b^2 B \sin (4 (c+d x))+960 a^2 b^2 C \sin (c+d x)+1200 a^2 b^2 C \sin (3 (c+d x))+240 a^2 b^2 C \sin (5 (c+d x))+480 a^3 b B \sin (c+d x)+720 a^3 b B \sin (3 (c+d x))+240 a^3 b B \sin (5 (c+d x))+480 a^3 b C \sin (2 (c+d x))+240 a^3 b C \sin (4 (c+d x))+120 a^4 C \sin (c+d x)+180 a^4 C \sin (3 (c+d x))+60 a^4 C \sin (5 (c+d x))+480 a A b^3 \sin (2 (c+d x))+240 a A b^3 \sin (4 (c+d x))+640 a b^3 B \sin (c+d x)+800 a b^3 B \sin (3 (c+d x))+160 a b^3 B \sin (5 (c+d x))+840 a b^3 C \sin (2 (c+d x))+180 a b^3 C \sin (4 (c+d x))+160 A b^4 \sin (c+d x)+200 A b^4 \sin (3 (c+d x))+40 A b^4 \sin (5 (c+d x))+210 b^4 B \sin (2 (c+d x))+45 b^4 B \sin (4 (c+d x))+320 b^4 C \sin (c+d x)+160 b^4 C \sin (3 (c+d x))+32 b^4 C \sin (5 (c+d x))\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[c + d*x]^5*(600*a^4*A*(c + d*x)*Cos[c + d*x] + 300*a^4*A*(c + d*x
)*Cos[3*(c + d*x)] + 60*a^4*A*c*Cos[5*(c + d*x)] + 60*a^4*A*d*x*Cos[5*(c + d*x)] - 120*(8*a^4*B + 24*a^2*b^2*B
+ 3*b^4*B + 16*a^3*b*(2*A + C) + 4*a*b^3*(4*A + 3*C))*Cos[c + d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 720*a^2*A*b^2*Sin[c + d*x] + 160*A*b^4*Sin[c + d*x] + 480*a^3*
b*B*Sin[c + d*x] + 640*a*b^3*B*Sin[c + d*x] + 120*a^4*C*Sin[c + d*x] + 960*a^2*b^2*C*Sin[c + d*x] + 320*b^4*C*
Sin[c + d*x] + 480*a*A*b^3*Sin[2*(c + d*x)] + 720*a^2*b^2*B*Sin[2*(c + d*x)] + 210*b^4*B*Sin[2*(c + d*x)] + 48
0*a^3*b*C*Sin[2*(c + d*x)] + 840*a*b^3*C*Sin[2*(c + d*x)] + 1080*a^2*A*b^2*Sin[3*(c + d*x)] + 200*A*b^4*Sin[3*
(c + d*x)] + 720*a^3*b*B*Sin[3*(c + d*x)] + 800*a*b^3*B*Sin[3*(c + d*x)] + 180*a^4*C*Sin[3*(c + d*x)] + 1200*a
^2*b^2*C*Sin[3*(c + d*x)] + 160*b^4*C*Sin[3*(c + d*x)] + 240*a*A*b^3*Sin[4*(c + d*x)] + 360*a^2*b^2*B*Sin[4*(c
+ d*x)] + 45*b^4*B*Sin[4*(c + d*x)] + 240*a^3*b*C*Sin[4*(c + d*x)] + 180*a*b^3*C*Sin[4*(c + d*x)] + 360*a^2*A
*b^2*Sin[5*(c + d*x)] + 40*A*b^4*Sin[5*(c + d*x)] + 240*a^3*b*B*Sin[5*(c + d*x)] + 160*a*b^3*B*Sin[5*(c + d*x)
] + 60*a^4*C*Sin[5*(c + d*x)] + 240*a^2*b^2*C*Sin[5*(c + d*x)] + 32*b^4*C*Sin[5*(c + d*x)]))/(480*d*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.066, size = 572, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

a^4*A*x+1/d*A*a^4*c+4/d*A*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^3*b*C*ln(sec(d*x+c)+tan(d*x+c))+4/d*C*a^2*b^2*
tan(d*x+c)+1/3/d*A*b^4*tan(d*x+c)*sec(d*x+c)^2+1/5/d*C*b^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*C*b^4*tan(d*x+c)*sec
(d*x+c)^2+8/3/d*a*b^3*B*tan(d*x+c)+4/d*B*a^3*b*tan(d*x+c)+6/d*A*a^2*b^2*tan(d*x+c)+3/2/d*C*a*b^3*ln(sec(d*x+c)
+tan(d*x+c))+1/4/d*B*b^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*B*b^4*sec(d*x+c)*tan(d*x+c)+2/d*A*a*b^3*ln(sec(d*x+c)+t
an(d*x+c))+1/d*a^4*C*tan(d*x+c)+1/d*B*a^4*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*C*b^4*tan(d*x+c)+2/3/d*A*b^4*tan(d*
x+c)+3/8/d*B*b^4*ln(sec(d*x+c)+tan(d*x+c))+4/3/d*a*b^3*B*tan(d*x+c)*sec(d*x+c)^2+1/d*C*a*b^3*tan(d*x+c)*sec(d*
x+c)^3+3/2/d*C*a*b^3*sec(d*x+c)*tan(d*x+c)+3/d*a^2*b^2*B*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^3*b*C*sec(d*x+c)*tan(
d*x+c)+3/d*a^2*b^2*B*sec(d*x+c)*tan(d*x+c)+2/d*A*a*b^3*sec(d*x+c)*tan(d*x+c)+2/d*C*a^2*b^2*tan(d*x+c)*sec(d*x+
c)^2

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Maxima [A]  time = 1.05134, size = 671, normalized size = 2.31 \begin{align*} \frac{240 \,{\left (d x + c\right )} A a^{4} + 480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b^{2} + 320 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{3} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} + 16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{4} - 60 \, C a b^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, C a^{3} b{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{2} b^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a b^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 960 \, A a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 240 \, C a^{4} \tan \left (d x + c\right ) + 960 \, B a^{3} b \tan \left (d x + c\right ) + 1440 \, A a^{2} b^{2} \tan \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(240*(d*x + c)*A*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b^2 + 320*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*B*a*b^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^4 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*ta
n(d*x + c))*C*b^4 - 60*C*a*b^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1)
- 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 15*B*b^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*
x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 240*C*a^3*b*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 360*B*a^2*b^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 240*A*a*b^3*(2*sin(d*x + c)/(sin(d*x + c
)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*B*a^4*log(sec(d*x + c) + tan(d*x + c)) + 960*A
*a^3*b*log(sec(d*x + c) + tan(d*x + c)) + 240*C*a^4*tan(d*x + c) + 960*B*a^3*b*tan(d*x + c) + 1440*A*a^2*b^2*t
an(d*x + c))/d

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Fricas [A]  time = 0.625407, size = 824, normalized size = 2.84 \begin{align*} \frac{240 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \,{\left (8 \, B a^{4} + 16 \,{\left (2 \, A + C\right )} a^{3} b + 24 \, B a^{2} b^{2} + 4 \,{\left (4 \, A + 3 \, C\right )} a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \, B a^{4} + 16 \,{\left (2 \, A + C\right )} a^{3} b + 24 \, B a^{2} b^{2} + 4 \,{\left (4 \, A + 3 \, C\right )} a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, C b^{4} + 8 \,{\left (15 \, C a^{4} + 60 \, B a^{3} b + 30 \,{\left (3 \, A + 2 \, C\right )} a^{2} b^{2} + 40 \, B a b^{3} + 2 \,{\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (16 \, C a^{3} b + 24 \, B a^{2} b^{2} + 4 \,{\left (4 \, A + 3 \, C\right )} a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (30 \, C a^{2} b^{2} + 20 \, B a b^{3} +{\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(240*A*a^4*d*x*cos(d*x + c)^5 + 15*(8*B*a^4 + 16*(2*A + C)*a^3*b + 24*B*a^2*b^2 + 4*(4*A + 3*C)*a*b^3 +
3*B*b^4)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(8*B*a^4 + 16*(2*A + C)*a^3*b + 24*B*a^2*b^2 + 4*(4*A + 3*C
)*a*b^3 + 3*B*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(24*C*b^4 + 8*(15*C*a^4 + 60*B*a^3*b + 30*(3*A +
2*C)*a^2*b^2 + 40*B*a*b^3 + 2*(5*A + 4*C)*b^4)*cos(d*x + c)^4 + 15*(16*C*a^3*b + 24*B*a^2*b^2 + 4*(4*A + 3*C)*
a*b^3 + 3*B*b^4)*cos(d*x + c)^3 + 8*(30*C*a^2*b^2 + 20*B*a*b^3 + (5*A + 4*C)*b^4)*cos(d*x + c)^2 + 30*(4*C*a*b
^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{4} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a + b*sec(c + d*x))**4*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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Giac [B]  time = 1.41148, size = 1539, normalized size = 5.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(120*(d*x + c)*A*a^4 + 15*(8*B*a^4 + 32*A*a^3*b + 16*C*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 12*C*a*b^3 +
3*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*B*a^4 + 32*A*a^3*b + 16*C*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^
3 + 12*C*a*b^3 + 3*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 480*B*a^3
*b*tan(1/2*d*x + 1/2*c)^9 - 240*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 360*B*
a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 240*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 4
80*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 300*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 75
*B*b^4*tan(1/2*d*x + 1/2*c)^9 + 120*C*b^4*tan(1/2*d*x + 1/2*c)^9 - 480*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 1920*B*a
^3*b*tan(1/2*d*x + 1/2*c)^7 + 480*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 2880*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 720
*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 1920*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^7
- 1280*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 120*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 320*A*b^4*tan(1/2*d*x + 1/2*c)^7
+ 30*B*b^4*tan(1/2*d*x + 1/2*c)^7 - 160*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 720*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 288
0*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 4320*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 2400*C*a^2*b^2*tan(1/2*d*x + 1/2*c)
^5 + 1600*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 400*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 464*C*b^4*tan(1/2*d*x + 1/2*c)^5
- 480*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 1920*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 480*C*a^3*b*tan(1/2*d*x + 1/2*c)^3
- 2880*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 720*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 1920*C*a^2*b^2*tan(1/2*d*x +
1/2*c)^3 - 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 1280*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 120*C*a*b^3*tan(1/2*d*x
+ 1/2*c)^3 - 320*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 30*B*b^4*tan(1/2*d*x + 1/2*c)^3 - 160*C*b^4*tan(1/2*d*x + 1/2
*c)^3 + 120*C*a^4*tan(1/2*d*x + 1/2*c) + 480*B*a^3*b*tan(1/2*d*x + 1/2*c) + 240*C*a^3*b*tan(1/2*d*x + 1/2*c) +
720*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)
+ 240*A*a*b^3*tan(1/2*d*x + 1/2*c) + 480*B*a*b^3*tan(1/2*d*x + 1/2*c) + 300*C*a*b^3*tan(1/2*d*x + 1/2*c) + 120
*A*b^4*tan(1/2*d*x + 1/2*c) + 75*B*b^4*tan(1/2*d*x + 1/2*c) + 120*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1
/2*c)^2 - 1)^5)/d