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

Optimal. Leaf size=312 \[ \frac{\tan (c+d x) \left (2 a^2 b B-3 a^3 C-a b^2 (A-2 C)-b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b^2 d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.2334, antiderivative size = 312, 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, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\tan (c+d x) \left (2 a^2 b B-3 a^3 C-a b^2 (A-2 C)-b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b^2 d \left (a^2-b^2\right )} \]

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])^2,x]

[Out]

((2*A*b^2 - 4*a*b*B + 6*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (2*a*(a^2*A*b^2 - 2*A*b^4 - 2*a^3*b*
B + 3*a*b^3*B + 3*a^4*C - 4*a^2*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^4
*(a + b)^(3/2)*d) + ((2*a^2*b*B - b^3*B - a*b^2*(A - 2*C) - 3*a^3*C)*Tan[c + d*x])/(b^3*(a^2 - b^2)*d) + ((2*A
*b^2 - 2*a*b*B + 3*a^2*C - b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*b^2*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a*C))*
Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^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 4092

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[(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 - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, 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 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))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec ^2(c+d x) \left (2 \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \sec (c+d x)-\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a \left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right )+b \left (2 A b^2-2 a b B+a^2 C+b^2 C\right ) \sec (c+d x)-2 \left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a b \left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right )-\left (a^2-b^2\right ) \left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (a \left (2 A b^4+2 a^3 b B-3 a b^3 B-a^2 b^2 (A-4 C)-3 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^4 \left (a^2-b^2\right )}+\frac{\left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \int \sec (c+d x) \, dx}{2 b^4}\\ &=\frac{\left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (a \left (2 A b^4+2 a^3 b B-3 a b^3 B-a^2 b^2 (A-4 C)-3 a^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=\frac{\left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 a \left (2 A b^4+2 a^3 b B-3 a b^3 B-a^2 b^2 (A-4 C)-3 a^4 C\right )\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 ) d}\\ &=\frac{\left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a \left (a^2 A b^2-2 A b^4-2 a^3 b B+3 a b^3 B+3 a^4 C-4 a^2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac{\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \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)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.85377, size = 519, normalized size = 1.66 \[ \frac{(a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 a^2 b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(b-a) (a+b)}-2 \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 a \left (a^2 b^2 (A-4 C)-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b^2 C (a \cos (c+d x)+b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C (a \cos (c+d x)+b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{4 b (b B-2 a C) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b (b B-2 a C) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{2 b^4 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

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])^2,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.105, size = 926, normalized size = 3. \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))^2,x)

[Out]

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

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 160.142, size = 3341, normalized size = 10.71 \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))^2,x, algorithm="fricas")

[Out]

[1/4*(2*((3*C*a^6 - 2*B*a^5*b + (A - 4*C)*a^4*b^2 + 3*B*a^3*b^3 - 2*A*a^2*b^4)*cos(d*x + c)^3 + (3*C*a^5*b - 2
*B*a^4*b^2 + (A - 4*C)*a^3*b^3 + 3*B*a^2*b^4 - 2*A*a*b^5)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x +
c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*c
os(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + ((6*C*a^7 - 4*B*a^6*b + (2*A - 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A
- C)*a^3*b^4 - 4*B*a^2*b^5 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 + (2*A - 11*C)*a^4*b^
3 + 8*B*a^3*b^4 - 4*(A - C)*a^2*b^5 - 4*B*a*b^6 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((6*C
*a^7 - 4*B*a^6*b + (2*A - 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A - C)*a^3*b^4 - 4*B*a^2*b^5 + (2*A + C)*a*b^6)*cos
(d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 + (2*A - 11*C)*a^4*b^3 + 8*B*a^3*b^4 - 4*(A - C)*a^2*b^5 - 4*B*a*b^6 +
(2*A + C)*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7 - 2*(3*C*a^6*b - 2*
B*a^5*b^2 + (A - 5*C)*a^4*b^3 + 3*B*a^3*b^4 - (A - 2*C)*a^2*b^5 - B*a*b^6)*cos(d*x + c)^2 - (3*C*a^5*b^2 - 2*B
*a^4*b^3 - 6*C*a^3*b^4 + 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))*sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6
+ a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2), -1/4*(4*((3*C*a^6 - 2*B*a^5*b + (A
- 4*C)*a^4*b^2 + 3*B*a^3*b^3 - 2*A*a^2*b^4)*cos(d*x + c)^3 + (3*C*a^5*b - 2*B*a^4*b^2 + (A - 4*C)*a^3*b^3 + 3*
B*a^2*b^4 - 2*A*a*b^5)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 -
b^2)*sin(d*x + c))) - ((6*C*a^7 - 4*B*a^6*b + (2*A - 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A - C)*a^3*b^4 - 4*B*a^2
*b^5 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 + (2*A - 11*C)*a^4*b^3 + 8*B*a^3*b^4 - 4*(A
- C)*a^2*b^5 - 4*B*a*b^6 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((6*C*a^7 - 4*B*a^6*b + (2*A
- 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A - C)*a^3*b^4 - 4*B*a^2*b^5 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*
b - 4*B*a^5*b^2 + (2*A - 11*C)*a^4*b^3 + 8*B*a^3*b^4 - 4*(A - C)*a^2*b^5 - 4*B*a*b^6 + (2*A + C)*b^7)*cos(d*x
+ c)^2)*log(-sin(d*x + c) + 1) - 2*(C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7 - 2*(3*C*a^6*b - 2*B*a^5*b^2 + (A - 5*C)*a
^4*b^3 + 3*B*a^3*b^4 - (A - 2*C)*a^2*b^5 - B*a*b^6)*cos(d*x + c)^2 - (3*C*a^5*b^2 - 2*B*a^4*b^3 - 6*C*a^3*b^4
+ 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))*sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)
^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2)]

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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 )^{2}}\, 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))**2,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))**2, x)

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Giac [A]  time = 1.38434, size = 578, normalized size = 1.85 \begin{align*} -\frac{\frac{4 \,{\left (3 \, C a^{5} - 2 \, B a^{4} b + A a^{3} b^{2} - 4 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{4 \,{\left (C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}} - \frac{{\left (6 \, C a^{2} - 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac{{\left (6 \, C a^{2} - 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac{2 \,{\left (4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C b \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 )}^{2} b^{3}}}{2 \, d} \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))^2,x, algorithm="giac")

[Out]

-1/2*(4*(3*C*a^5 - 2*B*a^4*b + A*a^3*b^2 - 4*C*a^3*b^2 + 3*B*a^2*b^3 - 2*A*a*b^4)*(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^2*b^4
- b^6)*sqrt(-a^2 + b^2)) - 4*(C*a^4*tan(1/2*d*x + 1/2*c) - B*a^3*b*tan(1/2*d*x + 1/2*c) + A*a^2*b^2*tan(1/2*d
*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (6*C*a^2 - 4*B*
a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*C*a^2 - 4*B*a*b + 2*A*b^2 + C*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*
d*x + 1/2*c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1/2*c))/((tan(1/2
*d*x + 1/2*c)^2 - 1)^2*b^3))/d