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

Optimal. Leaf size=310 \[ -\frac{a \left (-a^2 b^2 (190 A+121 C)+4 a^4 C-32 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b d}+\frac{\left (12 a^2 b^2 (4 A+3 C)+8 a^4 (2 A+C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}+\frac{a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac{\left (-2 a^2 b^2 (130 A+89 C)+8 a^4 C-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

[Out]

((8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*ArcTanh[Sin[c + d*x]])/(16*d) - (a*(4*a^4*C - 32
*b^4*(5*A + 4*C) - a^2*b^2*(190*A + 121*C))*Tan[c + d*x])/(60*b*d) - ((8*a^4*C - 15*b^4*(6*A + 5*C) - 2*a^2*b^
2*(130*A + 89*C))*Sec[c + d*x]*Tan[c + d*x])/(240*d) + (a*(70*A*b^2 - 4*a^2*C + 53*b^2*C)*(a + b*Sec[c + d*x])
^2*Tan[c + d*x])/(120*b*d) - ((4*a^2*C - 5*b^2*(6*A + 5*C))*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b*d) - (
a*C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + (C*(a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*d)

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Rubi [A]  time = 0.704779, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4083, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac{a \left (-a^2 b^2 (190 A+121 C)+4 a^4 C-32 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b d}+\frac{\left (12 a^2 b^2 (4 A+3 C)+8 a^4 (2 A+C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}+\frac{a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac{\left (-2 a^2 b^2 (130 A+89 C)+8 a^4 C-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*ArcTanh[Sin[c + d*x]])/(16*d) - (a*(4*a^4*C - 32
*b^4*(5*A + 4*C) - a^2*b^2*(190*A + 121*C))*Tan[c + d*x])/(60*b*d) - ((8*a^4*C - 15*b^4*(6*A + 5*C) - 2*a^2*b^
2*(130*A + 89*C))*Sec[c + d*x]*Tan[c + d*x])/(240*d) + (a*(70*A*b^2 - 4*a^2*C + 53*b^2*C)*(a + b*Sec[c + d*x])
^2*Tan[c + d*x])/(120*b*d) - ((4*a^2*C - 5*b^2*(6*A + 5*C))*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b*d) - (
a*C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + (C*(a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*d)

Rule 4083

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + 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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C*Csc[e + f*x], x], x], x] /; FreeQ
[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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 (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^4 (b (6 A+5 C)-a C \sec (c+d x)) \, dx}{6 b}\\ &=-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (3 a b (10 A+7 C)-\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{30 b}\\ &=-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (8 a^2 (5 A+3 C)+5 b^2 (6 A+5 C)\right )+3 a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac{a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 a b \left (8 a^2 (15 A+8 C)+b^2 (230 A+181 C)\right )-3 \left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x)\right ) \, dx}{360 b}\\ &=-\frac{\left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) \left (45 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right )-12 a \left (4 a^4 C-32 b^4 (5 A+4 C)-a^2 b^2 (190 A+121 C)\right ) \sec (c+d x)\right ) \, dx}{720 b}\\ &=-\frac{\left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{1}{16} \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a \left (4 a^4 C-32 b^4 (5 A+4 C)-a^2 b^2 (190 A+121 C)\right )\right ) \int \sec ^2(c+d x) \, dx}{60 b}\\ &=\frac{\left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\left (a \left (4 a^4 C-32 b^4 (5 A+4 C)-a^2 b^2 (190 A+121 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b d}\\ &=\frac{\left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{a \left (4 a^4 C-32 b^4 (5 A+4 C)-a^2 b^2 (190 A+121 C)\right ) \tan (c+d x)}{60 b d}-\frac{\left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}\\ \end{align*}

Mathematica [A]  time = 2.90493, size = 460, normalized size = 1.48 \[ -\frac{\sec ^6(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (240 \left (12 a^2 b^2 (4 A+3 C)+8 a^4 (2 A+C)+b^4 (6 A+5 C)\right ) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-2 \sin (c+d x) \left (64 a b \left (a^2 (75 A+80 C)+8 b^2 (10 A+11 C)\right ) \cos (c+d x)+20 \left (36 a^2 b^2 (4 A+5 C)+24 a^4 C+5 b^4 (6 A+5 C)\right ) \cos (2 (c+d x))+720 a^2 A b^2 \cos (4 (c+d x))+2160 a^2 A b^2+2400 a^3 A b \cos (3 (c+d x))+480 a^3 A b \cos (5 (c+d x))+540 a^2 b^2 C \cos (4 (c+d x))+3060 a^2 b^2 C+2240 a^3 b C \cos (3 (c+d x))+320 a^3 b C \cos (5 (c+d x))+120 a^4 C \cos (4 (c+d x))+360 a^4 C+2240 a A b^3 \cos (3 (c+d x))+320 a A b^3 \cos (5 (c+d x))+1792 a b^3 C \cos (3 (c+d x))+256 a b^3 C \cos (5 (c+d x))+90 A b^4 \cos (4 (c+d x))+510 A b^4+75 b^4 C \cos (4 (c+d x))+745 b^4 C\right )\right )}{1920 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C + A*Cos[c + d*x]^2)*Sec[c + d*x]^6*(240*(8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*Cos[
c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 2*(2160*a^2
*A*b^2 + 510*A*b^4 + 360*a^4*C + 3060*a^2*b^2*C + 745*b^4*C + 64*a*b*(8*b^2*(10*A + 11*C) + a^2*(75*A + 80*C))
*Cos[c + d*x] + 20*(24*a^4*C + 36*a^2*b^2*(4*A + 5*C) + 5*b^4*(6*A + 5*C))*Cos[2*(c + d*x)] + 2400*a^3*A*b*Cos
[3*(c + d*x)] + 2240*a*A*b^3*Cos[3*(c + d*x)] + 2240*a^3*b*C*Cos[3*(c + d*x)] + 1792*a*b^3*C*Cos[3*(c + d*x)]
+ 720*a^2*A*b^2*Cos[4*(c + d*x)] + 90*A*b^4*Cos[4*(c + d*x)] + 120*a^4*C*Cos[4*(c + d*x)] + 540*a^2*b^2*C*Cos[
4*(c + d*x)] + 75*b^4*C*Cos[4*(c + d*x)] + 480*a^3*A*b*Cos[5*(c + d*x)] + 320*a*A*b^3*Cos[5*(c + d*x)] + 320*a
^3*b*C*Cos[5*(c + d*x)] + 256*a*b^3*C*Cos[5*(c + d*x)])*Sin[c + d*x]))/(1920*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.059, size = 511, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^4*C*sec(d*x+c)*tan(d*x+c)+1/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4/
d*A*a^3*b*tan(d*x+c)+8/3/d*a^3*b*C*tan(d*x+c)+4/3/d*a^3*b*C*tan(d*x+c)*sec(d*x+c)^2+3/d*A*a^2*b^2*sec(d*x+c)*t
an(d*x+c)+3/d*A*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*C*a^2*b^2*tan(d*x+c)*sec(d*x+c)^3+9/4/d*C*a^2*b^2*sec(
d*x+c)*tan(d*x+c)+9/4/d*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+8/3/d*A*a*b^3*tan(d*x+c)+4/3/d*A*a*b^3*tan(d*x+c)*
sec(d*x+c)^2+32/15/d*C*a*b^3*tan(d*x+c)+4/5/d*C*a*b^3*tan(d*x+c)*sec(d*x+c)^4+16/15/d*C*a*b^3*tan(d*x+c)*sec(d
*x+c)^2+1/4/d*A*b^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*b^4*sec(d*x+c)*tan(d*x+c)+3/8/d*A*b^4*ln(sec(d*x+c)+tan(d*
x+c))+1/6/d*C*b^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*C*b^4*tan(d*x+c)*sec(d*x+c)^3+5/16/d*C*b^4*sec(d*x+c)*tan(d*x
+c)+5/16/d*C*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.02264, size = 620, normalized size = 2. \begin{align*} \frac{640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} b + 640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} + 128 \,{\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 a b^{3} - 5 \, C b^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{2} b^{2}{\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 )} - 30 \, A 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 )} - 120 \, C a^{4}{\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 )} - 720 \, A 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 )} + 480 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 1920 \, A a^{3} b \tan \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3*b + 640*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^3 + 128*(3*
tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a*b^3 - 5*C*b^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c
)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) +
15*log(sin(d*x + c) - 1)) - 180*C*a^2*b^2*(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)) - 30*A*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)) - 120*C*a^4*
(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 720*A*a^2*b^2*(2*sin(d
*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480*A*a^4*log(sec(d*x + c) + t
an(d*x + c)) + 1920*A*a^3*b*tan(d*x + c))/d

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Fricas [A]  time = 0.592328, size = 716, normalized size = 2.31 \begin{align*} \frac{15 \,{\left (8 \,{\left (2 \, A + C\right )} a^{4} + 12 \,{\left (4 \, A + 3 \, C\right )} a^{2} b^{2} +{\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \,{\left (2 \, A + C\right )} a^{4} + 12 \,{\left (4 \, A + 3 \, C\right )} a^{2} b^{2} +{\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (192 \, C a b^{3} \cos \left (d x + c\right ) + 64 \,{\left (5 \,{\left (3 \, A + 2 \, C\right )} a^{3} b + 2 \,{\left (5 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} + 40 \, C b^{4} + 15 \,{\left (8 \, C a^{4} + 12 \,{\left (4 \, A + 3 \, C\right )} a^{2} b^{2} +{\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 64 \,{\left (5 \, C a^{3} b +{\left (5 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (36 \, C a^{2} b^{2} +{\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(8*(2*A + C)*a^4 + 12*(4*A + 3*C)*a^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) -
15*(8*(2*A + C)*a^4 + 12*(4*A + 3*C)*a^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(192
*C*a*b^3*cos(d*x + c) + 64*(5*(3*A + 2*C)*a^3*b + 2*(5*A + 4*C)*a*b^3)*cos(d*x + c)^5 + 40*C*b^4 + 15*(8*C*a^4
+ 12*(4*A + 3*C)*a^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^4 + 64*(5*C*a^3*b + (5*A + 4*C)*a*b^3)*cos(d*x + c)^
3 + 10*(36*C*a^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.23601, size = 1485, normalized size = 4.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(16*A*a^4 + 8*C*a^4 + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 6*A*b^4 + 5*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c)
+ 1)) - 15*(16*A*a^4 + 8*C*a^4 + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 6*A*b^4 + 5*C*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)^11 - 960*A*a^3*b*tan(1/2*d*x + 1/2*c)^11 - 960*C*a^3*b*tan(1/2*d*x
+ 1/2*c)^11 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 900*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*A*a*b^3*tan(
1/2*d*x + 1/2*c)^11 - 960*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 150*A*b^4*tan(1/2*d*x + 1/2*c)^11 + 165*C*b^4*tan(
1/2*d*x + 1/2*c)^11 - 360*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 4800*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 3520*C*a^3*b*ta
n(1/2*d*x + 1/2*c)^9 - 2160*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 1260*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 3520*A*
a*b^3*tan(1/2*d*x + 1/2*c)^9 + 2240*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 210*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 25*C*b
^4*tan(1/2*d*x + 1/2*c)^9 + 240*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 9600*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 5760*C*a^
3*b*tan(1/2*d*x + 1/2*c)^7 + 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 57
60*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 4992*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 60*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 45
0*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 240*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 9600*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 5760
*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5
+ 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4992*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 60*A*b^4*tan(1/2*d*x + 1/2*c)^5
+ 450*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 360*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 4800*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 -
3520*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2160*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 1260*C*a^2*b^2*tan(1/2*d*x + 1/
2*c)^3 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 2240*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 210*A*b^4*tan(1/2*d*x + 1
/2*c)^3 + 25*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*C*a^4*tan(1/2*d*x + 1/2*c) + 960*A*a^3*b*tan(1/2*d*x + 1/2*c)
+ 960*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) + 900*C*a^2*b^2*tan(1/2*d*x + 1/2*c) +
960*A*a*b^3*tan(1/2*d*x + 1/2*c) + 960*C*a*b^3*tan(1/2*d*x + 1/2*c) + 150*A*b^4*tan(1/2*d*x + 1/2*c) + 165*C*
b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d