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

Optimal. Leaf size=153 \[ \frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.453871, antiderivative size = 153, 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 = {4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

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]),x]

[Out]

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

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

Mathematica [C]  time = 2.5958, size = 472, normalized size = 3.08 \[ \frac{\cos (c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 a (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}+\frac{4 b \sin \left (\frac{d x}{2}\right ) (b B-a C)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 b \sin \left (\frac{d x}{2}\right ) (b B-a C)}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{b^2 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{2 b^3 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[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]),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.083, size = 499, normalized size = 3.3 \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)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

-2/d*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^2/b^2/((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^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^2*C-1/d/b/(tan(1/2*d*x+1/2*c)+1)*B+1/d
/b^2/(tan(1/2*d*x+1/2*c)+1)*a*C+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)*C+1/d/b*ln(tan(1/2*d*x+1/2*c)+1)*A-1/d/b^2*ln(t
an(1/2*d*x+1/2*c)+1)*B*a+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*a^2*C+1/2/d/b*ln(tan(1/2*d*x+1/2*c)+1)*C+1/2/d/b/(ta
n(1/2*d*x+1/2*c)-1)^2*C-1/d/b/(tan(1/2*d*x+1/2*c)-1)*B+1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a*C+1/2/d/b/(tan(1/2*d*x
+1/2*c)-1)*C-1/d/b*ln(tan(1/2*d*x+1/2*c)-1)*A+1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*B*a-1/d/b^3*ln(tan(1/2*d*x+1/2*
c)-1)*a^2*C-1/2/d/b*ln(tan(1/2*d*x+1/2*c)-1)*C

________________________________________________________________________________________

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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 37.0882, size = 1465, normalized size = 9.58 \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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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 )}}{a + b \sec{\left (c + d x \right )}}\, 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)),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)), x)

________________________________________________________________________________________

Giac [B]  time = 1.28547, size = 387, normalized size = 2.53 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} - 2 \, 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^{3}} - \frac{{\left (2 \, C a^{2} - 2 \, 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^{3}} - \frac{4 \,{\left (C a^{3} - B a^{2} b + A a b^{2}\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 )}}{\sqrt{-a^{2} + b^{2}} b^{3}} + \frac{2 \,{\left (2 \, 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} - 2 \, 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^{2}}}{2 \, d} \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)),x, algorithm="giac")

[Out]

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