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

Optimal. Leaf size=242 \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-3 a b^4 (A+2 C)+2 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\tan (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.918053, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4090, 4080, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-3 a b^4 (A+2 C)+2 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\tan (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4080

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

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

Mathematica [C]  time = 6.00229, size = 514, normalized size = 2.12 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 (\sin (c)+i \cos (c)) \left (-a^2 b^3 B-5 a^3 b^2 C+2 a^5 C+3 a b^4 (A+2 C)-2 b^5 B\right ) (a \cos (c+d x)+b)^2 \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{5/2} \sqrt{(\cos (c)-i \sin (c))^2}}+\frac{b \left (a \sec (c) \left (a \left (b \sin (2 c+d x) \left (a^2 b B+a^3 C-a b^2 (3 A+4 C)+2 b^3 B\right )+\sin (c+2 d x) \left (a^2 b^2 (2 A+5 C)-2 a^4 C-3 a b^3 B+A b^4\right )\right )+b \sin (d x) \left (a^2 b^2 (5 A+16 C)+a^3 b B-7 a^4 C-10 a b^3 B+4 A b^4\right )\right )+\left (a^2+2 b^2\right ) \tan (c) \left (-a^2 b^2 (2 A+5 C)+2 a^4 C+3 a b^3 B-A b^4\right )\right )}{a \left (a^2-b^2\right )^2}-4 C (a \cos (c+d x)+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 C (a \cos (c+d x)+b)^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 b^3 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^2*(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]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-4*C*(b + a*Cos[c + d*x])^2*Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*C*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*
(-(a^2*b^3*B) - 2*b^5*B + 2*a^5*C - 5*a^3*b^2*C + 3*a*b^4*(A + 2*C))*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (
-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^2*(I*Cos[c]
+ Sin[c]))/((a^2 - b^2)^(5/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*(a*Sec[c]*(b*(4*A*b^4 + a^3*b*B - 10*a*b^3*B -
7*a^4*C + a^2*b^2*(5*A + 16*C))*Sin[d*x] + a*(b*(a^2*b*B + 2*b^3*B + a^3*C - a*b^2*(3*A + 4*C))*Sin[2*c + d*x
] + (A*b^4 - 3*a*b^3*B - 2*a^4*C + a^2*b^2*(2*A + 5*C))*Sin[c + 2*d*x])) + (a^2 + 2*b^2)*(-(A*b^4) + 3*a*b^3*B
+ 2*a^4*C - a^2*b^2*(2*A + 5*C))*Tan[c]))/(a*(a^2 - b^2)^2)))/(2*b^3*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 = 0.11, size = 1572, normalized size = 6.5 \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)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

-2/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-1/
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-2/d*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*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*B+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*B+2/d*a^4/b^2/(t
an(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-1/d*a^3/b/(ta
n(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-6/d/(tan(1/2*d
*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+2/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*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+2/d*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)*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-b)^2*tan(1/2*d*x+1/2*c)*B-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)*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*ta
n(1/2*d*x+1/2*c)*C-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)*C+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C-3/d*a
*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+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))*B*a^2+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))*B-2/d*a^5/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))*C+5/d*a^3/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))*C-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))*C*a+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*C-
1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*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)^2*(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 [B]  time = 78.2604, size = 3240, normalized size = 13.39 \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*((2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*b^6 - 2*B*b^7 + (2*C*a^7 - 5*C*a^5*b^2 - B*a^4*b
^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 3*(A + 2*C)*
a^2*b^5 - 2*B*a*b^6)*cos(d*x + c))*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*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^
2)) - 2*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*cos(d
*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(C*a^6*b
^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*cos(d*x + c)^2 + 2*(C
*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(3*C*a^6*b^2 - B*a^5*b^
3 - (A + 9*C)*a^4*b^4 + 5*B*a^3*b^5 - (A - 6*C)*a^2*b^6 - 4*B*a*b^7 + 2*A*b^8 + (2*C*a^7*b - (2*A + 7*C)*a^5*b
^3 + 3*B*a^4*b^4 + (A + 5*C)*a^3*b^5 - 3*B*a^2*b^6 + A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^
5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^6
*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d), -1/2*((2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*b^6 - 2*
B*b^7 + (2*C*a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5)*cos(d*x + c)^2 + 2*(2*C*a^6*b
- 5*C*a^4*b^3 - B*a^3*b^4 + 3*(A + 2*C)*a^2*b^5 - 2*B*a*b^6)*cos(d*x + c))*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))) - (C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*
a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7
)*cos(d*x + c))*log(sin(d*x + c) + 1) + (C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2
+ 3*C*a^4*b^4 - C*a^2*b^6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*cos(d*x + c))*lo
g(-sin(d*x + c) + 1) + (3*C*a^6*b^2 - B*a^5*b^3 - (A + 9*C)*a^4*b^4 + 5*B*a^3*b^5 - (A - 6*C)*a^2*b^6 - 4*B*a*
b^7 + 2*A*b^8 + (2*C*a^7*b - (2*A + 7*C)*a^5*b^3 + 3*B*a^4*b^4 + (A + 5*C)*a^3*b^5 - 3*B*a^2*b^6 + A*a*b^7)*co
s(d*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b
^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d)]

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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 ^{2}{\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)**2*(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)**2/(a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.42163, size = 853, normalized size = 3.52 \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-((2*C*a^5 - 5*C*a^3*b^2 - B*a^2*b^3 + 3*A*a*b^4 + 6*C*a*b^4 - 2*B*b^5)*(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^3 - 2*a^2*b
^5 + b^7)*sqrt(-a^2 + b^2)) - C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
b^3 - (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3
+ B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 3
*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*B*
a*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^5*tan(1/2*d*x + 1/2*c) - 3*C*a^4*b*tan(1
/2*d*x + 1/2*c) + 2*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 5*C*a^3*b^2*tan(1/2*d*x
+ 1/2*c) + A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*C*a^2*b^3*tan(1/2*d*x + 1/2*c
) + A*a*b^4*tan(1/2*d*x + 1/2*c) - 4*B*a*b^4*tan(1/2*d*x + 1/2*c) + 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^2 -
2*a^2*b^4 + b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2))/d