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

Optimal. Leaf size=202 \[ \frac{b \left (5 a^2 A b^2-3 a^4 (2 A+C)-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (5 A+2 C)+a^4 (-C)+2 A b^4\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{A x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.445455, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4061, 4060, 3919, 3831, 2659, 208} \[ \frac{b \left (5 a^2 A b^2-3 a^4 (2 A+C)-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (5 A+2 C)+a^4 (-C)+2 A b^4\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{A x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4061

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(
(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(
a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b*(A + C)*(m + 1)*Csc[e + f*x] +
(A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Int
egerQ[2*m] && LtQ[m, -1]

Rule 4060

Int[((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))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 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 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

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

Mathematica [C]  time = 4.72937, size = 642, normalized size = 3.18 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{\sec (c) \left (6 a^4 A b^2 \sin (c+2 d x)-7 a^3 A b^3 \sin (2 c+d x)-3 a^2 A b^4 \sin (c+2 d x)-2 a^4 A b^2 d x \cos (c+2 d x)-2 a^4 A b^2 d x \cos (3 c+2 d x)-8 a^3 A b^3 d x \cos (2 c+d x)+a^2 A b^4 d x \cos (c+2 d x)+a^2 A b^4 d x \cos (3 c+2 d x)+2 A d x \left (a^2-b^2\right )^2 \left (a^2+2 b^2\right ) \cos (c)-6 a^4 A b^2 \sin (c)-9 a^2 A b^4 \sin (c)+17 a^3 A b^3 \sin (d x)+4 a A b d x \left (a^2-b^2\right )^2 \cos (d x)+4 a^5 A b d x \cos (2 c+d x)+a^6 A d x \cos (c+2 d x)+a^6 A d x \cos (3 c+2 d x)+a^4 b^2 C \sin (c+2 d x)-5 a^4 b^2 C \sin (c)-2 a^2 b^4 C \sin (c)+4 a^3 b^3 C \sin (d x)-3 a^5 b C \sin (2 c+d x)+5 a^5 b C \sin (d x)+2 a^6 C \sin (c+2 d x)-2 a^6 C \sin (c)+4 a A b^5 \sin (2 c+d x)+4 a A b^5 d x \cos (2 c+d x)-8 a A b^5 \sin (d x)+6 A b^6 \sin (c)\right )}{\left (a^2-b^2\right )^2}+\frac{4 b (\sin (c)+i \cos (c)) \left (-5 a^2 A b^2+3 a^4 (2 A+C)+2 A b^4\right ) (a \cos (c+d x)+b)^2 \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{5/2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{2 a^3 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + 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 + C*Sec[c + d*x]^2)*((4*b*(-5*a^2*A*b^2 + 2*A*b^4 + 3*a^4*(2*A + C))*Arc
Tan[((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]) + (Sec[c]*(2*
A*(a^2 - b^2)^2*(a^2 + 2*b^2)*d*x*Cos[c] + 4*a*A*b*(a^2 - b^2)^2*d*x*Cos[d*x] + 4*a^5*A*b*d*x*Cos[2*c + d*x] -
8*a^3*A*b^3*d*x*Cos[2*c + d*x] + 4*a*A*b^5*d*x*Cos[2*c + d*x] + a^6*A*d*x*Cos[c + 2*d*x] - 2*a^4*A*b^2*d*x*Co
s[c + 2*d*x] + a^2*A*b^4*d*x*Cos[c + 2*d*x] + a^6*A*d*x*Cos[3*c + 2*d*x] - 2*a^4*A*b^2*d*x*Cos[3*c + 2*d*x] +
a^2*A*b^4*d*x*Cos[3*c + 2*d*x] - 6*a^4*A*b^2*Sin[c] - 9*a^2*A*b^4*Sin[c] + 6*A*b^6*Sin[c] - 2*a^6*C*Sin[c] - 5
*a^4*b^2*C*Sin[c] - 2*a^2*b^4*C*Sin[c] + 17*a^3*A*b^3*Sin[d*x] - 8*a*A*b^5*Sin[d*x] + 5*a^5*b*C*Sin[d*x] + 4*a
^3*b^3*C*Sin[d*x] - 7*a^3*A*b^3*Sin[2*c + d*x] + 4*a*A*b^5*Sin[2*c + d*x] - 3*a^5*b*C*Sin[2*c + d*x] + 6*a^4*A
*b^2*Sin[c + 2*d*x] - 3*a^2*A*b^4*Sin[c + 2*d*x] + 2*a^6*C*Sin[c + 2*d*x] + a^4*b^2*C*Sin[c + 2*d*x]))/(a^2 -
b^2)^2))/(2*a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.101, size = 1143, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*A/a^3*arctan(tan(1/2*d*x+1/2*c))-6/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/(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*b^3+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*b^4-2/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-b)/(a^2+2*a*b+b
^2)*tan(1/2*d*x+1/2*c)^3*C*a^2-1/d*a/(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*C-2/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-b)/(a^2+2*a*b+b^2)*ta
n(1/2*d*x+1/2*c)^3*C*b^2+6/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/(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*b^3
-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*b^4+2/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C*a^2-1/d*a/(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*C+2/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C*b^2-6/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+5/d/a*b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arc
tanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-2/d/a^3*b^5/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arcta
nh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-3/d*b*a/(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

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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((A+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 = 0.712418, size = 2295, normalized size = 11.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.24462, size = 653, normalized size = 3.23 \begin{align*} -\frac{\frac{{\left (6 \, A a^{4} b + 3 \, C a^{4} b - 5 \, A a^{2} b^{3} + 2 \, A b^{5}\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^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{{\left (d x + c\right )} A}{a^{3}} + \frac{2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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 )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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