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

Optimal. Leaf size=222 \[ \frac{a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}+\frac{a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac{a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^2 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{a^2 (14 A+12 B+11 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d}+\frac{C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[Out]

(a^2*(14*A + 12*B + 11*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^2*(10*A + 9*B + 8*C)*Tan[c + d*x])/(5*d) + (a^2*(
14*A + 12*B + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^2*(10*A + 12*B + 9*C)*Sec[c + d*x]^3*Tan[c + d*x])/
(40*d) + (C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((3*B + C)*Sec[c + d*x]^3*(a^2 + a^2*S
ec[c + d*x])*Tan[c + d*x])/(15*d) + (a^2*(10*A + 9*B + 8*C)*Tan[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.429762, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4088, 4018, 3997, 3787, 3768, 3770, 3767} \[ \frac{a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}+\frac{a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac{a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^2 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{a^2 (14 A+12 B+11 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d}+\frac{C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(14*A + 12*B + 11*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^2*(10*A + 9*B + 8*C)*Tan[c + d*x])/(5*d) + (a^2*(
14*A + 12*B + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^2*(10*A + 12*B + 9*C)*Sec[c + d*x]^3*Tan[c + d*x])/
(40*d) + (C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((3*B + C)*Sec[c + d*x]^3*(a^2 + a^2*S
ec[c + d*x])*Tan[c + d*x])/(15*d) + (a^2*(10*A + 9*B + 8*C)*Tan[c + d*x]^3)/(15*d)

Rule 4088

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

Rule 4018

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (3 a (2 A+C)+2 a (3 B+C) \sec (c+d x)) \, dx}{6 a}\\ &=\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{\int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (10 A+6 B+7 C)+3 a^2 (10 A+12 B+9 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac{a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{\int \sec ^3(c+d x) \left (15 a^3 (14 A+12 B+11 C)+24 a^3 (10 A+9 B+8 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac{a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{1}{5} \left (a^2 (10 A+9 B+8 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{8} \left (a^2 (14 A+12 B+11 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^2 (14 A+12 B+11 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{1}{16} \left (a^2 (14 A+12 B+11 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (10 A+9 B+8 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac{a^2 (14 A+12 B+11 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 3.29793, size = 359, normalized size = 1.62 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (15 (14 A+12 B+11 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) \cos ^5(c+d x) (15 \sin (c) (14 A+12 B+11 C)+32 (10 A+9 B+8 C) \sin (d x))-\sec (c) \cos ^4(c+d x) (16 \sin (c) (10 A+9 B+8 C)+15 (14 A+12 B+11 C) \sin (d x))-2 \sec (c) \cos ^3(c+d x) (5 \sin (c) (6 A+12 B+11 C)+8 (10 A+9 B+8 C) \sin (d x))-2 \sec (c) \cos ^2(c+d x) (5 (6 A+12 B+11 C) \sin (d x)+24 (B+2 C) \sin (c))-8 \sec (c) \cos (c+d x) (6 (B+2 C) \sin (d x)+5 C \sin (c))-40 C \sec (c) \sin (d x)\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[c + d*x])^2*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^4*Sec[c + d*x]^6*(15*(14*A
+ 12*B + 11*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]]) - 40*C*Sec[c]*Sin[d*x] - 8*Cos[c + d*x]*Sec[c]*(5*C*Sin[c] + 6*(B + 2*C)*Sin[d*x]) - 2*Cos[c + d*x]^3*S
ec[c]*(5*(6*A + 12*B + 11*C)*Sin[c] + 8*(10*A + 9*B + 8*C)*Sin[d*x]) - Cos[c + d*x]^5*Sec[c]*(15*(14*A + 12*B
+ 11*C)*Sin[c] + 32*(10*A + 9*B + 8*C)*Sin[d*x]) - 2*Cos[c + d*x]^2*Sec[c]*(24*(B + 2*C)*Sin[c] + 5*(6*A + 12*
B + 11*C)*Sin[d*x]) - Cos[c + d*x]^4*Sec[c]*(16*(10*A + 9*B + 8*C)*Sin[c] + 15*(14*A + 12*B + 11*C)*Sin[d*x]))
)/(480*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.068, size = 386, normalized size = 1.7 \begin{align*}{\frac{7\,{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{6\,B{a}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{11\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{11\,{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{11\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{4\,{a}^{2}A\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{16\,{a}^{2}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{8\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8/d*a^2*A*sec(d*x+c)*tan(d*x+c)+7/8/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+6/5/d*B*a^2*tan(d*x+c)+3/5/d*B*a^2*tan
(d*x+c)*sec(d*x+c)^2+11/24/d*a^2*C*tan(d*x+c)*sec(d*x+c)^3+11/16/d*a^2*C*sec(d*x+c)*tan(d*x+c)+11/16/d*a^2*C*l
n(sec(d*x+c)+tan(d*x+c))+4/3/d*a^2*A*tan(d*x+c)+2/3/d*a^2*A*tan(d*x+c)*sec(d*x+c)^2+1/2/d*B*a^2*tan(d*x+c)*sec
(d*x+c)^3+3/4/d*B*a^2*sec(d*x+c)*tan(d*x+c)+3/4/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+16/15/d*a^2*C*tan(d*x+c)+2/5
/d*a^2*C*tan(d*x+c)*sec(d*x+c)^4+8/15/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^2*A*tan(d*x+c)*sec(d*x+c)^3+1/5/
d*B*a^2*tan(d*x+c)*sec(d*x+c)^4+1/6/d*a^2*C*tan(d*x+c)*sec(d*x+c)^5

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Maxima [B]  time = 0.977191, size = 644, normalized size = 2.9 \begin{align*} \frac{320 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 32 \,{\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 )} B a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 64 \,{\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^{2} - 5 \, C a^{2}{\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 )} - 30 \, A a^{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 )} - 60 \, B a^{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 \, C a^{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 )} - 120 \, A a^{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 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(320*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 64*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x
+ c))*C*a^2 - 5*C*a^2*(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)) - 30*A*a^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)) - 60*B*a^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(s
in(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*C*a^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)) - 120*A*a^2*(2*sin(d*x + c)/(sin(
d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 0.555197, size = 535, normalized size = 2.41 \begin{align*} \frac{15 \,{\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \,{\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 15 \,{\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 16 \,{\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \,{\left (6 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 40 \, C a^{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)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/480*(15*(14*A + 12*B + 11*C)*a^2*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(14*A + 12*B + 11*C)*a^2*cos(d*x
+ c)^6*log(-sin(d*x + c) + 1) + 2*(32*(10*A + 9*B + 8*C)*a^2*cos(d*x + c)^5 + 15*(14*A + 12*B + 11*C)*a^2*cos(
d*x + c)^4 + 16*(10*A + 9*B + 8*C)*a^2*cos(d*x + c)^3 + 10*(6*A + 12*B + 11*C)*a^2*cos(d*x + c)^2 + 48*(B + 2*
C)*a^2*cos(d*x + c) + 40*C*a^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*} a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Int
egral(B*sec(c + d*x)**4, x) + Integral(2*B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x) + Integral(C*s
ec(c + d*x)**5, x) + Integral(2*C*sec(c + d*x)**6, x) + Integral(C*sec(c + d*x)**7, x))

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Giac [A]  time = 1.36714, size = 529, normalized size = 2.38 \begin{align*} \frac{15 \,{\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (210 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 180 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 165 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 1190 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 1020 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 935 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 2580 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2568 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1986 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 3180 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2808 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3006 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2330 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1860 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1305 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 750 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 780 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 795 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(14*A*a^2 + 12*B*a^2 + 11*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(14*A*a^2 + 12*B*a^2 + 11*C
*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(210*A*a^2*tan(1/2*d*x + 1/2*c)^11 + 180*B*a^2*tan(1/2*d*x + 1/2*
c)^11 + 165*C*a^2*tan(1/2*d*x + 1/2*c)^11 - 1190*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 1020*B*a^2*tan(1/2*d*x + 1/2*c
)^9 - 935*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 2580*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 2568*B*a^2*tan(1/2*d*x + 1/2*c)^7
+ 1986*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 3180*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 2808*B*a^2*tan(1/2*d*x + 1/2*c)^5 -
3006*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 2330*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 1860*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 1
305*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 750*A*a^2*tan(1/2*d*x + 1/2*c) - 780*B*a^2*tan(1/2*d*x + 1/2*c) - 795*C*a^2
*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d