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

Optimal. Leaf size=274 \[ \frac{a^3 (133 A+119 B+108 C) \tan ^3(c+d x)}{105 d}+\frac{a^3 (133 A+119 B+108 C) \tan (c+d x)}{35 d}+\frac{a^3 (26 A+23 B+21 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 (154 A+147 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{280 d}+\frac{(3 A+4 B+3 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{a^3 (26 A+23 B+21 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(7 B+3 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{42 a d}+\frac{C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^3}{7 d} \]

[Out]

(a^3*(26*A + 23*B + 21*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(133*A + 119*B + 108*C)*Tan[c + d*x])/(35*d) +
(a^3*(26*A + 23*B + 21*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^3*(154*A + 147*B + 129*C)*Sec[c + d*x]^3*Tan[
c + d*x])/(280*d) + (C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(7*d) + ((7*B + 3*C)*Sec[c + d*x]^3
*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(42*a*d) + ((3*A + 4*B + 3*C)*Sec[c + d*x]^3*(a^3 + a^3*Sec[c + d*x]
)*Tan[c + d*x])/(15*d) + (a^3*(133*A + 119*B + 108*C)*Tan[c + d*x]^3)/(105*d)

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Rubi [A]  time = 0.598019, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{a^3 (133 A+119 B+108 C) \tan ^3(c+d x)}{105 d}+\frac{a^3 (133 A+119 B+108 C) \tan (c+d x)}{35 d}+\frac{a^3 (26 A+23 B+21 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 (154 A+147 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{280 d}+\frac{(3 A+4 B+3 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{a^3 (26 A+23 B+21 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{(7 B+3 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{42 a d}+\frac{C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^3}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(26*A + 23*B + 21*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(133*A + 119*B + 108*C)*Tan[c + d*x])/(35*d) +
(a^3*(26*A + 23*B + 21*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^3*(154*A + 147*B + 129*C)*Sec[c + d*x]^3*Tan[
c + d*x])/(280*d) + (C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(7*d) + ((7*B + 3*C)*Sec[c + d*x]^3
*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(42*a*d) + ((3*A + 4*B + 3*C)*Sec[c + d*x]^3*(a^3 + a^3*Sec[c + d*x]
)*Tan[c + d*x])/(15*d) + (a^3*(133*A + 119*B + 108*C)*Tan[c + d*x]^3)/(105*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))^3 \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))^3 \tan (c+d x)}{7 d}+\frac{\int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (a (7 A+3 C)+a (7 B+3 C) \sec (c+d x)) \, dx}{7 a}\\ &=\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (14 A+7 B+9 C)+14 a^2 (3 A+4 B+3 C) \sec (c+d x)\right ) \, dx}{42 a}\\ &=\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{(3 A+4 B+3 C) \sec ^3(c+d x) \left (a^3+a^3 \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^3 (112 A+91 B+87 C)+3 a^3 (154 A+147 B+129 C) \sec (c+d x)\right ) \, dx}{210 a}\\ &=\frac{a^3 (154 A+147 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{(3 A+4 B+3 C) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{\int \sec ^3(c+d x) \left (105 a^4 (26 A+23 B+21 C)+24 a^4 (133 A+119 B+108 C) \sec (c+d x)\right ) \, dx}{840 a}\\ &=\frac{a^3 (154 A+147 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{(3 A+4 B+3 C) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{1}{8} \left (a^3 (26 A+23 B+21 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{35} \left (a^3 (133 A+119 B+108 C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{a^3 (26 A+23 B+21 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^3 (154 A+147 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{(3 A+4 B+3 C) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{1}{16} \left (a^3 (26 A+23 B+21 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (133 A+119 B+108 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 d}\\ &=\frac{a^3 (26 A+23 B+21 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 (133 A+119 B+108 C) \tan (c+d x)}{35 d}+\frac{a^3 (26 A+23 B+21 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{a^3 (154 A+147 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{C \sec ^3(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{7 d}+\frac{(7 B+3 C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{42 a d}+\frac{(3 A+4 B+3 C) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac{a^3 (133 A+119 B+108 C) \tan ^3(c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 6.16201, size = 402, normalized size = 1.47 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^7(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (105 (26 A+23 B+21 C) \cos ^7(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) \cos ^6(c+d x) (105 \sin (c) (26 A+23 B+21 C)+32 (133 A+119 B+108 C) \sin (d x))-\sec (c) \cos ^5(c+d x) (16 \sin (c) (133 A+119 B+108 C)+105 (26 A+23 B+21 C) \sin (d x))-2 \sec (c) \cos ^4(c+d x) (35 \sin (c) (18 A+23 B+21 C)+8 (133 A+119 B+108 C) \sin (d x))-2 \sec (c) \cos ^3(c+d x) (24 \sin (c) (7 A+21 B+27 C)+35 (18 A+23 B+21 C) \sin (d x))-8 \sec (c) \cos ^2(c+d x) (6 (7 A+21 B+27 C) \sin (d x)+35 (B+3 C) \sin (c))-40 \sec (c) \cos (c+d x) (7 (B+3 C) \sin (d x)+6 C \sin (c))-240 C \sec (c) \sin (d x)\right )}{6720 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Cos[c + d*x])^3*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^7*(105*(26*
A + 23*B + 21*C)*Cos[c + d*x]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]]) - 240*C*Sec[c]*Sin[d*x] - 40*Cos[c + d*x]*Sec[c]*(6*C*Sin[c] + 7*(B + 3*C)*Sin[d*x]) - 2*Cos[c + d*x]^
3*Sec[c]*(24*(7*A + 21*B + 27*C)*Sin[c] + 35*(18*A + 23*B + 21*C)*Sin[d*x]) - Cos[c + d*x]^5*Sec[c]*(16*(133*A
+ 119*B + 108*C)*Sin[c] + 105*(26*A + 23*B + 21*C)*Sin[d*x]) - 8*Cos[c + d*x]^2*Sec[c]*(35*(B + 3*C)*Sin[c] +
6*(7*A + 21*B + 27*C)*Sin[d*x]) - 2*Cos[c + d*x]^4*Sec[c]*(35*(18*A + 23*B + 21*C)*Sin[c] + 8*(133*A + 119*B
+ 108*C)*Sin[d*x]) - Cos[c + d*x]^6*Sec[c]*(105*(26*A + 23*B + 21*C)*Sin[c] + 32*(133*A + 119*B + 108*C)*Sin[d
*x])))/(6720*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.074, size = 455, normalized size = 1.7 \begin{align*}{\frac{23\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{23\,B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{23\,B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{3\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{2\,d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{34\,B{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{17\,B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{38\,A{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{19\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{72\,{a}^{3}C\tan \left ( dx+c \right ) }{35\,d}}+{\frac{27\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35\,d}}+{\frac{36\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35\,d}}+{\frac{3\,B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7\,d}}+{\frac{13\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{21\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{13\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{21\,{a}^{3}C\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)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

23/16/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+23/24/d*B*a^3*tan(d*x+c)*sec(d*x+c)^3+23/16/d*B*a^3*sec(d*x+c)*tan(d*x
+c)+3/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+1/2/d*a^3*C*tan(d*x+c)*sec(d*x+c)^5+1/6/d*B*a^3*tan(d*x+c)*sec(d*x+c)^
5+34/15/d*B*a^3*tan(d*x+c)+17/15/d*B*a^3*tan(d*x+c)*sec(d*x+c)^2+38/15/d*A*a^3*tan(d*x+c)+19/15/d*A*a^3*tan(d*
x+c)*sec(d*x+c)^2+72/35*a^3*C*tan(d*x+c)/d+27/35/d*a^3*C*tan(d*x+c)*sec(d*x+c)^4+36/35/d*a^3*C*tan(d*x+c)*sec(
d*x+c)^2+3/5/d*B*a^3*tan(d*x+c)*sec(d*x+c)^4+1/5/d*A*a^3*tan(d*x+c)*sec(d*x+c)^4+1/7/d*a^3*C*tan(d*x+c)*sec(d*
x+c)^6+13/8/d*A*a^3*sec(d*x+c)*tan(d*x+c)+7/8/d*a^3*C*tan(d*x+c)*sec(d*x+c)^3+21/16/d*a^3*C*sec(d*x+c)*tan(d*x
+c)+13/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+21/16/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [B]  time = 0.989039, size = 876, normalized size = 3.2 \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)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/3360*(224*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 3360*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*A*a^3 + 672*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 1120*(tan(d*x + c)^3 + 3*ta
n(d*x + c))*B*a^3 + 96*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*C*a^3 + 67
2*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^3 - 35*B*a^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)) - 105*C*a^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)) -
630*A*a^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)) - 630*B*a^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)) - 210*C*a^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)) - 840*A*a^
3*(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 = 0.556788, size = 617, normalized size = 2.25 \begin{align*} \frac{105 \,{\left (26 \, A + 23 \, B + 21 \, C\right )} a^{3} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (26 \, A + 23 \, B + 21 \, C\right )} a^{3} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \,{\left (133 \, A + 119 \, B + 108 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} + 105 \,{\left (26 \, A + 23 \, B + 21 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} + 16 \,{\left (133 \, A + 119 \, B + 108 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 70 \,{\left (18 \, A + 23 \, B + 21 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 48 \,{\left (7 \, A + 21 \, B + 27 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 280 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 240 \, C a^{3}\right )} \sin \left (d x + c\right )}{3360 \, d \cos \left (d x + c\right )^{7}} \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))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/3360*(105*(26*A + 23*B + 21*C)*a^3*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(26*A + 23*B + 21*C)*a^3*cos(d
*x + c)^7*log(-sin(d*x + c) + 1) + 2*(32*(133*A + 119*B + 108*C)*a^3*cos(d*x + c)^6 + 105*(26*A + 23*B + 21*C)
*a^3*cos(d*x + c)^5 + 16*(133*A + 119*B + 108*C)*a^3*cos(d*x + c)^4 + 70*(18*A + 23*B + 21*C)*a^3*cos(d*x + c)
^3 + 48*(7*A + 21*B + 27*C)*a^3*cos(d*x + c)^2 + 280*(B + 3*C)*a^3*cos(d*x + c) + 240*C*a^3)*sin(d*x + c))/(d*
cos(d*x + c)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{5}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{7}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{6}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{7}{\left (c + d x \right )}\, dx + \int C \sec ^{8}{\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))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a**3*(Integral(A*sec(c + d*x)**3, x) + Integral(3*A*sec(c + d*x)**4, x) + Integral(3*A*sec(c + d*x)**5, x) + I
ntegral(A*sec(c + d*x)**6, x) + Integral(B*sec(c + d*x)**4, x) + Integral(3*B*sec(c + d*x)**5, x) + Integral(3
*B*sec(c + d*x)**6, x) + Integral(B*sec(c + d*x)**7, x) + Integral(C*sec(c + d*x)**5, x) + Integral(3*C*sec(c
+ d*x)**6, x) + Integral(3*C*sec(c + d*x)**7, x) + Integral(C*sec(c + d*x)**8, x))

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Giac [A]  time = 1.3511, size = 598, normalized size = 2.18 \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)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/1680*(105*(26*A*a^3 + 23*B*a^3 + 21*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(26*A*a^3 + 23*B*a^3 + 2
1*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2730*A*a^3*tan(1/2*d*x + 1/2*c)^13 + 2415*B*a^3*tan(1/2*d*x +
1/2*c)^13 + 2205*C*a^3*tan(1/2*d*x + 1/2*c)^13 - 18200*A*a^3*tan(1/2*d*x + 1/2*c)^11 - 16100*B*a^3*tan(1/2*d*
x + 1/2*c)^11 - 14700*C*a^3*tan(1/2*d*x + 1/2*c)^11 + 51506*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 45563*B*a^3*tan(1/2
*d*x + 1/2*c)^9 + 41601*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 77952*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 72576*B*a^3*tan(1/
2*d*x + 1/2*c)^7 - 62592*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 71246*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 62853*B*a^3*tan(1
/2*d*x + 1/2*c)^5 + 63231*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40040*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 33180*B*a^3*tan(
1/2*d*x + 1/2*c)^3 - 25620*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 10710*A*a^3*tan(1/2*d*x + 1/2*c) + 11025*B*a^3*tan(1
/2*d*x + 1/2*c) + 11235*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d