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

Optimal. Leaf size=187 \[ -\frac{\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \tan (c+d x)}{3 b^3 d}+\frac{\left (2 a^2+b^2\right ) (b B-a C) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a^3 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.716545, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.225, Rules used = {4072, 4033, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \tan (c+d x)}{3 b^3 d}+\frac{\left (2 a^2+b^2\right ) (b B-a C) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a^3 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 2.34835, size = 422, normalized size = 2.26 \[ \frac{\frac{4 b \left (3 a^2 C-3 a b B+2 b^2 C\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b \left (3 a^2 C-3 a b B+2 b^2 C\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{24 a^3 (b B-a C) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+6 \left (2 a^2+b^2\right ) (a C-b B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 \left (2 a^2+b^2\right ) (a C-b B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b^2 (b (3 B+C)-3 a C)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 (b (3 B+C)-3 a C)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((24*a^3*(b*B - a*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 6*(2*a^2 + b^2)*(
-(b*B) + a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*(2*a^2 + b^2)*(-(b*B) + a*C)*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] + (b^2*(-3*a*C + b*(3*B + C)))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*C*Sin[(c +
d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*b*(-3*a*b*B + 3*a^2*C + 2*b^2*C)*Sin[(c + d*x)/2])/(Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (b^2*(
-3*a*C + b*(3*B + C)))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b*(-3*a*b*B + 3*a^2*C + 2*b^2*C)*Sin[(c +
d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(12*b^4*d)

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Maple [B]  time = 0.095, size = 688, normalized size = 3.7 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

-1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^2*B+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^2*C-1/2/d/b*ln(tan(1/2*d*x+1/2*c)-1)*B+1/2/
d/b/(tan(1/2*d*x+1/2*c)-1)*B+1/2/d/b*ln(tan(1/2*d*x+1/2*c)+1)*B+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)*B-1/3/d*C/b/(ta
n(1/2*d*x+1/2*c)-1)^3+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2*B-1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2*C-1/3/d*C/b/(tan(1/2
*d*x+1/2*c)+1)^3-2/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))*B-1/d/b
/(tan(1/2*d*x+1/2*c)+1)*C-1/d/b/(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*a*C+1/d/b^3*ln(tan
(1/2*d*x+1/2*c)+1)*B*a^2-1/2/d*a/b^2*ln(tan(1/2*d*x+1/2*c)+1)*C+1/2/d*a/b^2*ln(tan(1/2*d*x+1/2*c)-1)*C+2/d*a^4
/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-1/d/b^3*ln(tan(1/2*d*x+1/2*c)
-1)*B*a^2+1/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)*a^3*C+1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*B*a-1/d/b^3/(tan(1/2*d*x+1/2*c
)-1)*a^2*C-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a*C-1/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a^2*C-1/2/d/b^2/(tan(1/2*d*x+1/
2*c)+1)*a*C-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2*a*C-1/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a^3*C+1/d/b^2/(tan(1/2*d*x
+1/2*c)+1)*B*a

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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*(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 = 2.29042, size = 1650, normalized size = 8.82 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[-1/12*(6*(C*a^4 - B*a^3*b)*sqrt(a^2 - b^2)*cos(d*x + c)^3*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)) + 3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c)^3*log(sin(d*x + c
) + 1) - 3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) +
1) - 2*(2*C*a^2*b^3 - 2*C*b^5 + 2*(3*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5)*cos(d*x + c)^2 -
3*(C*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c)^3), 1
/12*(12*(C*a^4 - B*a^3*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)^3 - 3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c)^3*log(s
in(d*x + c) + 1) + 3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 2*(2*C*a^2*b^3 - 2*C*b^5 + 2*(3*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5)*cos(d*
x + c)^2 - 3*(C*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x
+ c)^3)]

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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 ^{4}{\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)**3*(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)**4/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 1.23036, size = 556, normalized size = 2.97 \begin{align*} -\frac{\frac{3 \,{\left (2 \, C a^{3} - 2 \, B a^{2} b + C a b^{2} - B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac{3 \,{\left (2 \, C a^{3} - 2 \, B a^{2} b + C a b^{2} - B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac{12 \,{\left (C a^{4} - B a^{3} 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^{4}} + \frac{2 \,{\left (6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{2} \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 )}^{3} b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(2*C*a^3 - 2*B*a^2*b + C*a*b^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*C*a^3 - 2*B*a^2*
b + C*a*b^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 12*(C*a^4 - B*a^3*b)*(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^4) + 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/
2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 1
2*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1
/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/
((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^3))/d