3.654 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=306 \[ \frac{a \left (a^2 b^2 (30 A+17 C)+2 a^4 C+24 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b^2 d}+\frac{a \left (2 a^2 C+30 A b^2+21 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b^2 d}+\frac{\left (12 a^2 b^2 (5 A+3 C)+4 a^4 C+15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]

[Out]

(b*(6*a^2*(4*A + 3*C) + b^2*(6*A + 5*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (a*(2*a^4*C + 24*b^4*(5*A + 4*C) + a^
2*b^2*(30*A + 17*C))*Tan[c + d*x])/(60*b^2*d) + ((4*a^4*C + 12*a^2*b^2*(5*A + 3*C) + 15*b^4*(6*A + 5*C))*Sec[c
+ d*x]*Tan[c + d*x])/(240*b*d) + (a*(30*A*b^2 + 2*a^2*C + 21*b^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(120
*b^2*d) + ((2*a^2*C + 5*b^2*(6*A + 5*C))*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b^2*d) - (a*C*(a + b*Sec[c
+ d*x])^4*Tan[c + d*x])/(15*b^2*d) + (C*Sec[c + d*x]*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(6*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.719236, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4093, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{a \left (a^2 b^2 (30 A+17 C)+2 a^4 C+24 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b^2 d}+\frac{a \left (2 a^2 C+30 A b^2+21 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b^2 d}+\frac{\left (12 a^2 b^2 (5 A+3 C)+4 a^4 C+15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(b*(6*a^2*(4*A + 3*C) + b^2*(6*A + 5*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (a*(2*a^4*C + 24*b^4*(5*A + 4*C) + a^
2*b^2*(30*A + 17*C))*Tan[c + d*x])/(60*b^2*d) + ((4*a^4*C + 12*a^2*b^2*(5*A + 3*C) + 15*b^4*(6*A + 5*C))*Sec[c
+ d*x]*Tan[c + d*x])/(240*b*d) + (a*(30*A*b^2 + 2*a^2*C + 21*b^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(120
*b^2*d) + ((2*a^2*C + 5*b^2*(6*A + 5*C))*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b^2*d) - (a*C*(a + b*Sec[c
+ d*x])^4*Tan[c + d*x])/(15*b^2*d) + (C*Sec[c + d*x]*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(6*b*d)

Rule 4093

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))
^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[
1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - 2*a
*C*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 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) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, 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 ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (a C+b (6 A+5 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{6 b}\\ &=-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (-3 a b C+\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{30 b^2}\\ &=\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (30 A b^2-2 a^2 C+25 b^2 C\right )+3 a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (-3 a b \left (2 a^2 C-3 b^2 (50 A+39 C)\right )+3 \left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{360 b^2}\\ &=\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) \left (45 b^3 \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right )+12 a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right ) \sec (c+d x)\right ) \, dx}{720 b^2}\\ &=\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{1}{16} \left (b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right )\right ) \int \sec (c+d x) \, dx+\frac{\left (a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right )\right ) \int \sec ^2(c+d x) \, dx}{60 b^2}\\ &=\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}-\frac{\left (a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b^2 d}\\ &=\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}\\ \end{align*}

Mathematica [A]  time = 3.61659, size = 407, normalized size = 1.33 \[ \frac{\sec ^6(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (2 \sin (c+d x) \left (16 a \left (a^2 (75 A+80 C)+24 b^2 (10 A+11 C)\right ) \cos (c+d x)+20 b \left (18 a^2 (4 A+5 C)+5 b^2 (6 A+5 C)\right ) \cos (2 (c+d x))+360 a^2 A b \cos (4 (c+d x))+1080 a^2 A b+600 a^3 A \cos (3 (c+d x))+120 a^3 A \cos (5 (c+d x))+270 a^2 b C \cos (4 (c+d x))+1530 a^2 b C+560 a^3 C \cos (3 (c+d x))+80 a^3 C \cos (5 (c+d x))+1680 a A b^2 \cos (3 (c+d x))+240 a A b^2 \cos (5 (c+d x))+1344 a b^2 C \cos (3 (c+d x))+192 a b^2 C \cos (5 (c+d x))+90 A b^3 \cos (4 (c+d x))+510 A b^3+75 b^3 C \cos (4 (c+d x))+745 b^3 C\right )-240 b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{1920 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

((C + A*Cos[c + d*x]^2)*Sec[c + d*x]^6*(-240*b*(6*a^2*(4*A + 3*C) + b^2*(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*(1080*a^2*A*b + 510*A*b^3 + 15
30*a^2*b*C + 745*b^3*C + 16*a*(24*b^2*(10*A + 11*C) + a^2*(75*A + 80*C))*Cos[c + d*x] + 20*b*(18*a^2*(4*A + 5*
C) + 5*b^2*(6*A + 5*C))*Cos[2*(c + d*x)] + 600*a^3*A*Cos[3*(c + d*x)] + 1680*a*A*b^2*Cos[3*(c + d*x)] + 560*a^
3*C*Cos[3*(c + d*x)] + 1344*a*b^2*C*Cos[3*(c + d*x)] + 360*a^2*A*b*Cos[4*(c + d*x)] + 90*A*b^3*Cos[4*(c + d*x)
] + 270*a^2*b*C*Cos[4*(c + d*x)] + 75*b^3*C*Cos[4*(c + d*x)] + 120*a^3*A*Cos[5*(c + d*x)] + 240*a*A*b^2*Cos[5*
(c + d*x)] + 80*a^3*C*Cos[5*(c + d*x)] + 192*a*b^2*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 = 0.051, size = 430, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{2\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{3\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,{a}^{2}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+2\,{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{8\,Ca{b}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,C{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.02067, size = 521, normalized size = 1.7 \begin{align*} \frac{160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} + 96 \,{\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^{2} - 5 \, C b^{3}{\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 )} - 90 \, C a^{2} b{\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^{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 )} - 360 \, A a^{2} 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 )} + 480 \, A a^{3} \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)^2*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 + 96*(3*tan
(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a*b^2 - 5*C*b^3*(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)) - 90*C*a^2*b*(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^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)) - 360*A*a^2*b*(2*s
in(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480*A*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.576212, size = 636, normalized size = 2.08 \begin{align*} \frac{15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \,{\left (3 \, A + 2 \, C\right )} a^{3} + 6 \,{\left (5 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} + 144 \, C a b^{2} \cos \left (d x + c\right ) + 15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, C b^{3} + 16 \,{\left (5 \, C a^{3} + 3 \,{\left (5 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (18 \, C a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\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)^2*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/480*(15*(6*(4*A + 3*C)*a^2*b + (6*A + 5*C)*b^3)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(6*(4*A + 3*C)*a^2
*b + (6*A + 5*C)*b^3)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*(5*(3*A + 2*C)*a^3 + 6*(5*A + 4*C)*a*b^2)*
cos(d*x + c)^5 + 144*C*a*b^2*cos(d*x + c) + 15*(6*(4*A + 3*C)*a^2*b + (6*A + 5*C)*b^3)*cos(d*x + c)^4 + 40*C*b
^3 + 16*(5*C*a^3 + 3*(5*A + 4*C)*a*b^2)*cos(d*x + c)^3 + 10*(18*C*a^2*b + (6*A + 5*C)*b^3)*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 )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**3*sec(c + d*x)**2, x)

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Giac [B]  time = 1.27429, size = 1258, normalized size = 4.11 \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)^2*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(24*A*a^2*b + 18*C*a^2*b + 6*A*b^3 + 5*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(24*A*a^2*b +
18*C*a^2*b + 6*A*b^3 + 5*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(240*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 24
0*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 360*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 450*C*a^2*b*tan(1/2*d*x + 1/2*c)^11 +
720*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 150*A*b^3*tan(1/2*d*x + 1/2*c)^11
- 165*C*b^3*tan(1/2*d*x + 1/2*c)^11 - 1200*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 880*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 1
080*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 630*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 2640*A*a*b^2*tan(1/2*d*x + 1/2*c)^9
- 1680*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 210*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 25*C*b^3*tan(1/2*d*x + 1/2*c)^9 + 2
400*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 1440*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 18
0*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 4320*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 3744*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 -
60*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 450*C*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 1440
*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 4320
*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3744*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 60*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 450*
C*b^3*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 880*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 1080*A*a
^2*b*tan(1/2*d*x + 1/2*c)^3 + 630*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 2640*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 1680*
C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 210*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 25*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 240*A*a^
3*tan(1/2*d*x + 1/2*c) - 240*C*a^3*tan(1/2*d*x + 1/2*c) - 360*A*a^2*b*tan(1/2*d*x + 1/2*c) - 450*C*a^2*b*tan(1
/2*d*x + 1/2*c) - 720*A*a*b^2*tan(1/2*d*x + 1/2*c) - 720*C*a*b^2*tan(1/2*d*x + 1/2*c) - 150*A*b^3*tan(1/2*d*x
+ 1/2*c) - 165*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d