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

Optimal. Leaf size=252 \[ \frac{2 a^4 (56 A+49 B+44 C) \tan ^3(c+d x)}{105 d}+\frac{4 a^4 (56 A+49 B+44 C) \tan (c+d x)}{35 d}+\frac{a^4 (56 A+49 B+44 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (56 A+49 B+44 C) \tan (c+d x) \sec ^3(c+d x)}{280 d}+\frac{27 a^4 (56 A+49 B+44 C) \tan (c+d x) \sec (c+d x)}{560 d}+\frac{(42 A-7 B+8 C) \tan (c+d x) (a \sec (c+d x)+a)^4}{210 d}+\frac{(7 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^5}{42 a d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^4}{7 d} \]

[Out]

(a^4*(56*A + 49*B + 44*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a^4*(56*A + 49*B + 44*C)*Tan[c + d*x])/(35*d) + (
27*a^4*(56*A + 49*B + 44*C)*Sec[c + d*x]*Tan[c + d*x])/(560*d) + (a^4*(56*A + 49*B + 44*C)*Sec[c + d*x]^3*Tan[
c + d*x])/(280*d) + ((42*A - 7*B + 8*C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(210*d) + (C*Sec[c + d*x]^2*(a +
a*Sec[c + d*x])^4*Tan[c + d*x])/(7*d) + ((7*B + 4*C)*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(42*a*d) + (2*a^4*(5
6*A + 49*B + 44*C)*Tan[c + d*x]^3)/(105*d)

________________________________________________________________________________________

Rubi [A]  time = 0.520057, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {4088, 4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac{2 a^4 (56 A+49 B+44 C) \tan ^3(c+d x)}{105 d}+\frac{4 a^4 (56 A+49 B+44 C) \tan (c+d x)}{35 d}+\frac{a^4 (56 A+49 B+44 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (56 A+49 B+44 C) \tan (c+d x) \sec ^3(c+d x)}{280 d}+\frac{27 a^4 (56 A+49 B+44 C) \tan (c+d x) \sec (c+d x)}{560 d}+\frac{(42 A-7 B+8 C) \tan (c+d x) (a \sec (c+d x)+a)^4}{210 d}+\frac{(7 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^5}{42 a d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^4}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(56*A + 49*B + 44*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a^4*(56*A + 49*B + 44*C)*Tan[c + d*x])/(35*d) + (
27*a^4*(56*A + 49*B + 44*C)*Sec[c + d*x]*Tan[c + d*x])/(560*d) + (a^4*(56*A + 49*B + 44*C)*Sec[c + d*x]^3*Tan[
c + d*x])/(280*d) + ((42*A - 7*B + 8*C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(210*d) + (C*Sec[c + d*x]^2*(a +
a*Sec[c + d*x])^4*Tan[c + d*x])/(7*d) + ((7*B + 4*C)*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(42*a*d) + (2*a^4*(5
6*A + 49*B + 44*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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

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[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{\int \sec ^2(c+d x) (a+a \sec (c+d x))^4 (a (7 A+2 C)+a (7 B+4 C) \sec (c+d x)) \, dx}{7 a}\\ &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^4 \left (5 a^2 (7 B+4 C)+a^2 (42 A-7 B+8 C) \sec (c+d x)\right ) \, dx}{42 a^2}\\ &=\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{1}{70} (56 A+49 B+44 C) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{1}{70} (56 A+49 B+44 C) \int \left (a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+a^4 \sec ^5(c+d x)\right ) \, dx\\ &=\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{1}{70} \left (a^4 (56 A+49 B+44 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{70} \left (a^4 (56 A+49 B+44 C)\right ) \int \sec ^5(c+d x) \, dx+\frac{1}{35} \left (2 a^4 (56 A+49 B+44 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{35} \left (2 a^4 (56 A+49 B+44 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{35} \left (3 a^4 (56 A+49 B+44 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^4 (56 A+49 B+44 C) \tanh ^{-1}(\sin (c+d x))}{70 d}+\frac{3 a^4 (56 A+49 B+44 C) \sec (c+d x) \tan (c+d x)}{70 d}+\frac{a^4 (56 A+49 B+44 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{1}{280} \left (3 a^4 (56 A+49 B+44 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{70} \left (3 a^4 (56 A+49 B+44 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (2 a^4 (56 A+49 B+44 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{35 d}-\frac{\left (2 a^4 (56 A+49 B+44 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 d}\\ &=\frac{2 a^4 (56 A+49 B+44 C) \tanh ^{-1}(\sin (c+d x))}{35 d}+\frac{4 a^4 (56 A+49 B+44 C) \tan (c+d x)}{35 d}+\frac{27 a^4 (56 A+49 B+44 C) \sec (c+d x) \tan (c+d x)}{560 d}+\frac{a^4 (56 A+49 B+44 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{2 a^4 (56 A+49 B+44 C) \tan ^3(c+d x)}{105 d}+\frac{1}{560} \left (3 a^4 (56 A+49 B+44 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{a^4 (56 A+49 B+44 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{4 a^4 (56 A+49 B+44 C) \tan (c+d x)}{35 d}+\frac{27 a^4 (56 A+49 B+44 C) \sec (c+d x) \tan (c+d x)}{560 d}+\frac{a^4 (56 A+49 B+44 C) \sec ^3(c+d x) \tan (c+d x)}{280 d}+\frac{(42 A-7 B+8 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{210 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{(7 B+4 C) (a+a \sec (c+d x))^5 \tan (c+d x)}{42 a d}+\frac{2 a^4 (56 A+49 B+44 C) \tan ^3(c+d x)}{105 d}\\ \end{align*}

Mathematica [B]  time = 6.45561, size = 1087, normalized size = 4.31 \[ \frac{(-56 A-49 B-44 C) \cos ^6(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{128 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{(56 A+49 B+44 C) \cos ^6(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{128 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{C \sec (c) \sec (c+d x) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sin (d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{56 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (6 C \sin (c)+7 B \sin (d x)+28 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{336 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\cos (c+d x) \sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (35 B \sin (c)+140 C \sin (c)+42 A \sin (d x)+168 B \sin (d x)+288 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{1680 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\cos ^2(c+d x) \sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (168 A \sin (c)+672 B \sin (c)+1152 C \sin (c)+840 A \sin (d x)+1435 B \sin (d x)+1540 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{6720 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\cos ^3(c+d x) \sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (840 A \sin (c)+1435 B \sin (c)+1540 C \sin (c)+1904 A \sin (d x)+2016 B \sin (d x)+1816 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{6720 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\cos ^4(c+d x) \sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (3808 A \sin (c)+4032 B \sin (c)+3632 C \sin (c)+5880 A \sin (d x)+5145 B \sin (d x)+4620 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{13440 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\cos ^5(c+d x) \sec (c) (\sec (c+d x) a+a)^4 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (5880 A \sin (c)+5145 B \sin (c)+4620 C \sin (c)+9296 A \sin (d x)+8064 B \sin (d x)+7264 C \sin (d x)) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right )}{13440 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-56*A - 49*B - 44*C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a
*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(128*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d
*x])) + ((56*A + 49*B + 44*C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8
*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(128*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*
c + 2*d*x])) + (C*Sec[c]*Sec[c/2 + (d*x)/2]^8*Sec[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[
c + d*x]^2)*Sin[d*x])/(56*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (Sec[c]*Sec[c/2 + (d*x)/2]^8*
(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(6*C*Sin[c] + 7*B*Sin[d*x] + 28*C*Sin[d*x]))/(3
36*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (Cos[c + d*x]*Sec[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec
[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(35*B*Sin[c] + 140*C*Sin[c] + 42*A*Sin[d*x] + 168*B*Sin[d
*x] + 288*C*Sin[d*x]))/(1680*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*Sec[c]*Sec
[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(168*A*Sin[c] + 672*B*Sin[c]
+ 1152*C*Sin[c] + 840*A*Sin[d*x] + 1435*B*Sin[d*x] + 1540*C*Sin[d*x]))/(6720*d*(A + 2*C + 2*B*Cos[c + d*x] + A
*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^3*Sec[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] +
C*Sec[c + d*x]^2)*(840*A*Sin[c] + 1435*B*Sin[c] + 1540*C*Sin[c] + 1904*A*Sin[d*x] + 2016*B*Sin[d*x] + 1816*C*
Sin[d*x]))/(6720*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^4*Sec[c]*Sec[c/2 + (d*x)
/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(3808*A*Sin[c] + 4032*B*Sin[c] + 3632*C*S
in[c] + 5880*A*Sin[d*x] + 5145*B*Sin[d*x] + 4620*C*Sin[d*x]))/(13440*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c
+ 2*d*x])) + (Cos[c + d*x]^5*Sec[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*(5880*A*Sin[c] + 5145*B*Sin[c] + 4620*C*Sin[c] + 9296*A*Sin[d*x] + 8064*B*Sin[d*x] + 7264*C*Sin[d*x
]))/(13440*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [A]  time = 0.077, size = 454, normalized size = 1.8 \begin{align*}{\frac{B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{454\,{a}^{4}C\tan \left ( dx+c \right ) }{105\,d}}+{\frac{227\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{105\,d}}+{\frac{24\,B{a}^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{12\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{7\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{11\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{6\,d}}+{\frac{11\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{41\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{2\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{3\,d}}+{\frac{49\,B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{34\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{48\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35\,d}}+{\frac{4\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7\,d}}+{\frac{11\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{83\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{49\,B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{7\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d*B*a^4*tan(d*x+c)*sec(d*x+c)^5+454/105/d*a^4*C*tan(d*x+c)+227/105/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+24/5/d*
B*a^4*tan(d*x+c)+12/5/d*B*a^4*tan(d*x+c)*sec(d*x+c)^2+7/2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+11/6/d*a^4*C*tan(d*x+c
)*sec(d*x+c)^3+11/4/d*a^4*C*sec(d*x+c)*tan(d*x+c)+41/24/d*B*a^4*tan(d*x+c)*sec(d*x+c)^3+1/d*A*a^4*tan(d*x+c)*s
ec(d*x+c)^3+2/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^5+49/16/d*B*a^4*sec(d*x+c)*tan(d*x+c)+34/15/d*A*a^4*tan(d*x+c)*s
ec(d*x+c)^2+48/35/d*a^4*C*tan(d*x+c)*sec(d*x+c)^4+4/5/d*B*a^4*tan(d*x+c)*sec(d*x+c)^4+1/5/d*A*a^4*tan(d*x+c)*s
ec(d*x+c)^4+1/7/d*a^4*C*tan(d*x+c)*sec(d*x+c)^6+11/4/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+83/15/d*A*a^4*tan(d*x+c
)+49/16/d*B*a^4*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [B]  time = 1.00252, size = 987, normalized size = 3.92 \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+a*sec(d*x+c))^4*(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^4 + 6720*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*A*a^4 + 896*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^4 + 4480*(tan(d*x + c)^3 + 3*ta
n(d*x + c))*B*a^4 + 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^4 + 13
44*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 1120*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a
^4 - 35*B*a^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)) - 140*C*a^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)) - 840*A*a^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)) - 1260*B*a^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)) - 840*C*a^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)) - 3360*A*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)) - 840*B*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)) + 3360*A*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.55581, size = 617, normalized size = 2.45 \begin{align*} \frac{105 \,{\left (56 \, A + 49 \, B + 44 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (56 \, A + 49 \, B + 44 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (581 \, A + 504 \, B + 454 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} + 105 \,{\left (56 \, A + 49 \, B + 44 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 16 \,{\left (238 \, A + 252 \, B + 227 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \,{\left (24 \, A + 41 \, B + 44 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 48 \,{\left (7 \, A + 28 \, B + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 280 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 240 \, C a^{4}\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)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/3360*(105*(56*A + 49*B + 44*C)*a^4*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(56*A + 49*B + 44*C)*a^4*cos(d
*x + c)^7*log(-sin(d*x + c) + 1) + 2*(16*(581*A + 504*B + 454*C)*a^4*cos(d*x + c)^6 + 105*(56*A + 49*B + 44*C)
*a^4*cos(d*x + c)^5 + 16*(238*A + 252*B + 227*C)*a^4*cos(d*x + c)^4 + 70*(24*A + 41*B + 44*C)*a^4*cos(d*x + c)
^3 + 48*(7*A + 28*B + 48*C)*a^4*cos(d*x + c)^2 + 280*(B + 4*C)*a^4*cos(d*x + c) + 240*C*a^4)*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^{4} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 A \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{4}{\left (c + d x \right )}\, dx + \int 6 B \sec ^{5}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{7}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{5}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{6}{\left (c + d x \right )}\, dx + \int 4 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)**2*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a**4*(Integral(A*sec(c + d*x)**2, x) + Integral(4*A*sec(c + d*x)**3, x) + Integral(6*A*sec(c + d*x)**4, x) + I
ntegral(4*A*sec(c + d*x)**5, x) + Integral(A*sec(c + d*x)**6, x) + Integral(B*sec(c + d*x)**3, x) + Integral(4
*B*sec(c + d*x)**4, x) + Integral(6*B*sec(c + d*x)**5, x) + Integral(4*B*sec(c + d*x)**6, x) + Integral(B*sec(
c + d*x)**7, x) + Integral(C*sec(c + d*x)**4, x) + Integral(4*C*sec(c + d*x)**5, x) + Integral(6*C*sec(c + d*x
)**6, x) + Integral(4*C*sec(c + d*x)**7, x) + Integral(C*sec(c + d*x)**8, x))

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Giac [A]  time = 1.30492, size = 598, normalized size = 2.37 \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+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/1680*(105*(56*A*a^4 + 49*B*a^4 + 44*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(56*A*a^4 + 49*B*a^4 + 4
4*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5880*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 5145*B*a^4*tan(1/2*d*x +
1/2*c)^13 + 4620*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 39200*A*a^4*tan(1/2*d*x + 1/2*c)^11 - 34300*B*a^4*tan(1/2*d*
x + 1/2*c)^11 - 30800*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 110936*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 97069*B*a^4*tan(1/
2*d*x + 1/2*c)^9 + 87164*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 172032*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 150528*B*a^4*tan
(1/2*d*x + 1/2*c)^7 - 135168*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 159656*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 134099*B*a^4
*tan(1/2*d*x + 1/2*c)^5 + 126084*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 86240*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 73220*B*a
^4*tan(1/2*d*x + 1/2*c)^3 - 58800*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 21000*A*a^4*tan(1/2*d*x + 1/2*c) + 21735*B*a^
4*tan(1/2*d*x + 1/2*c) + 22260*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d