3.917 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=323 \[ \frac{\tan (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)+5 a^3 b^3 B-15 a^4 b^2 C-2 a^5 b B+6 a^6 C-6 a b^5 B+2 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a \tan (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

[Out]

((b*B - 3*a*C)*ArcTanh[Sin[c + d*x]])/(b^4*d) + ((2*A*b^6 - 2*a^5*b*B + 5*a^3*b^3*B - 6*a*b^5*B + 6*a^6*C - 15
*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a +
b)^(5/2)*d) + ((A*b^2 - a*b*B + 3*a^2*C - 2*b^2*C)*Tan[c + d*x])/(2*b^3*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a*
C))*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - (a*(2*A*b^4 + a^3*b*B - 4*a*b^3*
B - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 2.98692, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {4098, 4090, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\tan (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)+5 a^3 b^3 B-15 a^4 b^2 C-2 a^5 b B+6 a^6 C-6 a b^5 B+2 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a \tan (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - 3*a*C)*ArcTanh[Sin[c + d*x]])/(b^4*d) + ((2*A*b^6 - 2*a^5*b*B + 5*a^3*b^3*B - 6*a*b^5*B + 6*a^6*C - 15
*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a +
b)^(5/2)*d) + ((A*b^2 - a*b*B + 3*a^2*C - 2*b^2*C)*Tan[c + d*x])/(2*b^3*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a*
C))*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - (a*(2*A*b^4 + a^3*b*B - 4*a*b^3*
B - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4098

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[(d*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(
a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(a^2 - b^2)*(m + 1)), x] + Dist[d/(b*(a^2 - b^2)*(m
+ 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) +
b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]
^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 4090

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*(-(a*(b*B - a*C)) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, 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 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) \left (2 \left (A b^2-a (b B-a C)\right )+2 b (b B-a (A+C)) \sec (c+d x)-\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-b \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right )-\left (a^2-b^2\right ) \left (a^2 b B-2 b^3 B-3 a^3 C+a b^2 (A+4 C)\right ) \sec (c+d x)-b \left (a^2-b^2\right ) \left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-b^2 \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right )-2 b \left (a^2-b^2\right )^2 (b B-3 a C) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{(b B-3 a C) \int \sec (c+d x) \, dx}{b^4}+\frac{\left (2 A b^6-2 a^5 b B+5 a^3 b^3 B-6 a b^5 B+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 A b^6-2 a^5 b B+5 a^3 b^3 B-6 a b^5 B+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 A b^6-2 a^5 b B+5 a^3 b^3 B-6 a b^5 B+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right )^2 d}\\ &=\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (a^2 A b^4+2 A b^6-2 a^5 b B+5 a^3 b^3 B-6 a b^5 B+6 a^6 C-15 a^4 b^2 C+12 a^2 b^4 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac{\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.9103, size = 492, normalized size = 1.52 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{2 b \sin (c+d x) \left (a^2 \cos (2 (c+d x)) \left (-11 a^2 b^2 C-2 a^3 b B+6 a^4 C+5 a b^3 B-3 A b^4+2 b^4 C\right )+2 a b \cos (c+d x) \left (a^2 b^2 (A-16 C)-3 a^3 b B+9 a^4 C+6 a b^3 B+4 b^4 (C-A)\right )-3 a^2 A b^4+5 a^3 b^3 B-7 a^4 b^2 C-6 a^2 b^4 C-2 a^5 b B+6 a^6 C+4 b^6 C\right )}{\left (a^2-b^2\right )^2}-\frac{8 \cos (c+d x) \left (a^2 b^4 (A+12 C)+5 a^3 b^3 B-15 a^4 b^2 C-2 a^5 b B+6 a^6 C-6 a b^5 B+2 A b^6\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-8 (b B-3 a C) \cos (c+d x) (a \cos (c+d x)+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 (b B-3 a C) \cos (c+d x) (a \cos (c+d x)+b)^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 b^4 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-8*(2*A*b^6 - 2*a^5*b*B + 5*a^3
*b^3*B - 6*a*b^5*B + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2
- b^2]]*Cos[c + d*x]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) - 8*(b*B - 3*a*C)*Cos[c + d*x]*(b + a*Cos[c +
d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*(b*B - 3*a*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])^2*Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*b*(-3*a^2*A*b^4 - 2*a^5*b*B + 5*a^3*b^3*B + 6*a^6*C - 7*a^4*b^2*C - 6*a
^2*b^4*C + 4*b^6*C + 2*a*b*(-3*a^3*b*B + 6*a*b^3*B + a^2*b^2*(A - 16*C) + 9*a^4*C + 4*b^4*(-A + C))*Cos[c + d*
x] + a^2*(-3*A*b^4 - 2*a^3*b*B + 5*a*b^3*B + 6*a^4*C - 11*a^2*b^2*C + 2*b^4*C)*Cos[2*(c + d*x)])*Sin[c + d*x])
/(a^2 - b^2)^2))/(4*b^4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.106, size = 1813, normalized size = 5.6 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

-1/d/b^3/(tan(1/2*d*x+1/2*c)-1)*C-1/d/b^3/(tan(1/2*d*x+1/2*c)+1)*C+4/d*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1
/2*c)^2*b-a-b)^2*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-4/d*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*
c)^2*b-a-b)^2*a/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A-8/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-
b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C+2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b
)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)
/(a-b)^2*tan(1/2*d*x+1/2*c)*B-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*
tan(1/2*d*x+1/2*c)*B-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan
(1/2*d*x+1/2*c)^3*B+1/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*
x+1/2*c)*C+1/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x
+1/2*c)^3*C-4/d*a^5/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*
x+1/2*c)^3*C+4/d*a^5/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c
)*C+8/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3
*C+3/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)*a*C-3/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a*C+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*
B-1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*B+6/d/b^4/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x
+1/2*c)/((a+b)*(a-b))^(1/2))*a^6*C-15/d/b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+
1/2*c)/((a+b)*(a-b))^(1/2))*a^4*C-2/d/b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/
2*c)/((a+b)*(a-b))^(1/2))*B*a^5+5/d/b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)
/((a+b)*(a-b))^(1/2))*B*a^3-6/d*b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a
+b)*(a-b))^(1/2))*a*B-6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(
1/2*d*x+1/2*c)^3*B+6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2
*c)*B+1/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+1/d*a^2
/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+2/d*b^2/(a
^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+1/d/(a^4-2*a^2*b
^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*a^2+12/d/(a^4-2*a^2*b^2+b^
4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.53882, size = 952, normalized size = 2.95 \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

((6*C*a^6 - 2*B*a^5*b - 15*C*a^4*b^2 + 5*B*a^3*b^3 + A*a^2*b^4 + 12*C*a^2*b^4 - 6*B*a*b^5 + 2*A*b^6)*(pi*floor
(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2
+ b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(-a^2 + b^2)) - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^5*b*tan(1/2
*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*
d*x + 1/2*c)^3 - A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d
*x + 1/2*c)^3 - 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*A*a*b^5*tan(1/2*d*
x + 1/2*c)^3 - 4*C*a^6*tan(1/2*d*x + 1/2*c) + 2*B*a^5*b*tan(1/2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)
+ 3*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c) - A*a^3*b^3*tan(1/2*d*x + 1/2*c) - 5*B*a
^3*b^3*tan(1/2*d*x + 1/2*c) + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c) - 6*B*a^2*b^
4*tan(1/2*d*x + 1/2*c) + 4*A*a*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^
2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) - (3*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (3*C*a - B*b
)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3))/d