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

Optimal. Leaf size=143 \[ -\frac{\left (-2 a^2 C+2 a b B-b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{2 a^2 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.450333, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4072, 4033, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{\left (-2 a^2 C+2 a b B-b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{2 a^2 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*b*B - 2*a^2*C - b^2*C)*ArcTanh[Sin[c + d*x]])/(2*b^3*d) + (2*a^2*(b*B - a*C)*ArcTanh[(Sqrt[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 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4033

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2))/(
b*f*(m + n)), x] + Dist[d^2/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2)
+ B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] &&  !IGtQ[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 (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\int \frac{\sec ^3(c+d x) (B+C \sec (c+d x))}{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 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 (2 a b B-2 a^2 C-b^2 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 (a^2 (b B-a C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^3}-\frac{\left (2 a b B-2 a^2 C-b^2 C\right ) \int \sec (c+d x) \, dx}{2 b^3}\\ &=-\frac{\left (2 a b B-2 a^2 C-b^2 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^2 (b B-a C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^4}\\ &=-\frac{\left (2 a b B-2 a^2 C-b^2 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^2 (b B-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 )}{b^4 d}\\ &=-\frac{\left (2 a b B-2 a^2 C-b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{2 a^2 (b B-a C) \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 [B]  time = 1.73591, size = 300, normalized size = 2.1 \[ \frac{\frac{8 a^2 (a C-b B) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-2 \left (2 a^2 C-2 a b B+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (2 a^2 C-2 a b B+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 b (b B-a C) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b (b B-a C) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{b^2 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.08, size = 410, normalized size = 2.9 \begin{align*} 2\,{\frac{B{a}^{2}}{d{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{{a}^{3}C}{d{b}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{B}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{aC}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{Ba}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{a}^{2}C}{d{b}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{2\,db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{B}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{aC}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{Ba}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{a}^{2}C}{d{b}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{C}{2\,db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*tan(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/(ta
n(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^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+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^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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 11.3992, size = 1353, normalized size = 9.46 \begin{align*} \left [-\frac{2 \,{\left (C a^{3} - B a^{2} b\right )} \sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )^{2} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) -{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (C a^{2} b^{2} - C b^{4} - 2 \,{\left (C a^{3} b - B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}, -\frac{4 \,{\left (C a^{3} - B a^{2} b\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} -{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (C a^{2} b^{2} - C b^{4} - 2 \,{\left (C a^{3} b - B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)*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*cos(d*x +
c) + b^2)) - (2*C*a^4 - 2*B*a^3*b - C*a^2*b^2 + 2*B*a*b^3 - C*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*C
*a^4 - 2*B*a^3*b - C*a^2*b^2 + 2*B*a*b^3 - 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)*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 - C*a^2*b^2 + 2*B*a*b^3 - C*b^4)*cos(d*x + c)^2*log(sin(d*x + c) +
1) + (2*C*a^4 - 2*B*a^3*b - C*a^2*b^2 + 2*B*a*b^3 - 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)]

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

[Out]

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

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Giac [B]  time = 1.21725, size = 363, normalized size = 2.54 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} - 2 \, B a b + 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 + 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\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*(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 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 - (2*C*a^2 - 2*B*a*b + C*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1))/b^3 - 4*(C*a^3 - B*a^2*b)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arcta
n(-(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