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

Optimal. Leaf size=215 \[ \frac{\tan (c+d x) \left (3 a^2 C-3 a b B+3 A b^2+2 b^2 C\right )}{3 b^3 d}+\frac{\left (2 a^2 b B-2 a^3 C-a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.740864, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {4102, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\tan (c+d x) \left (3 a^2 C-3 a b B+3 A b^2+2 b^2 C\right )}{3 b^3 d}+\frac{\left (2 a^2 b B-2 a^3 C-a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*b*B + b^3*B - 2*a^3*C - a*b^2*(2*A + C))*ArcTanh[Sin[c + d*x]])/(2*b^4*d) + (2*a^2*(A*b^2 - a*(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^2 - 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 4102

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

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 (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{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+b (3 A+2 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^2-3 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^2-3 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 B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^3}\\ &=\frac{\left (3 A b^2-3 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^2 b B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) \int \sec (c+d x) \, dx}{2 b^4}+\frac{\left (a^2 \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^4}\\ &=\frac{\left (2 a^2 b B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (3 A b^2-3 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^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^5}\\ &=\frac{\left (2 a^2 b B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\left (3 A b^2-3 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^2 \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^5 d}\\ &=\frac{\left (2 a^2 b B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{2 a^2 \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^4 \sqrt{a+b} d}+\frac{\left (3 A b^2-3 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 [C]  time = 3.58884, size = 512, normalized size = 2.38 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (b \sec (c) \left (-6 \sin (2 c+d x) \left (a^2 C-a b B+A b^2\right )+12 \sin (d x) \left (a^2 C-a b B+A b^2+b^2 C\right )+6 a^2 C \sin (2 c+3 d x)-6 a b B \sin (2 c+3 d x)-3 a b C \sin (c+2 d x)-3 a b C \sin (3 c+2 d x)+6 A b^2 \sin (2 c+3 d x)+3 b^2 B \sin (c+2 d x)+3 b^2 B \sin (3 c+2 d x)+4 b^2 C \sin (2 c+3 d x)\right )+12 \cos ^3(c+d x) \left (-2 a^2 b B+2 a^3 C+a b^2 (2 A+C)-b^3 B\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-12 \cos ^3(c+d x) \left (-2 a^2 b B+2 a^3 C+a b^2 (2 A+C)-b^3 B\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{48 i a^2 (\cos (c)-i \sin (c)) \cos ^3(c+d x) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{12 b^4 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(12*(-2*a^2*b*B - b^3*B + 2*a^3*C
+ a*b^2*(2*A + C))*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 12*(-2*a^2*b*B - b^3*B + 2*a^3*C
+ a*b^2*(2*A + C))*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((48*I)*a^2*(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])]*Cos[c + d*x]^3*(Cos[c] - I*Sin[c]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]) + b*Sec[c]*(1
2*(A*b^2 - a*b*B + a^2*C + b^2*C)*Sin[d*x] - 6*(A*b^2 - a*b*B + a^2*C)*Sin[2*c + d*x] + 3*b^2*B*Sin[c + 2*d*x]
- 3*a*b*C*Sin[c + 2*d*x] + 3*b^2*B*Sin[3*c + 2*d*x] - 3*a*b*C*Sin[3*c + 2*d*x] + 6*A*b^2*Sin[2*c + 3*d*x] - 6
*a*b*B*Sin[2*c + 3*d*x] + 6*a^2*C*Sin[2*c + 3*d*x] + 4*b^2*C*Sin[2*c + 3*d*x])))/(12*b^4*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 = 0.091, size = 825, normalized size = 3.8 \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*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

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*(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 [A]  time = 54.4325, size = 1810, normalized size = 8.42 \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*(A+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 + A*a^2*b^2)*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 + (2*A - C)*a^3*b^2 + B*a^2*b^3 - (2*A + 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 + (2*A - C)*a^3*b^2 + B*a^2*b^3 - (2*A + 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 + (3*A -
C)*a^2*b^3 + 3*B*a*b^4 - (3*A + 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 + A*a^2*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)^3 - 3*(2*C*a^5 - 2*
B*a^4*b + (2*A - C)*a^3*b^2 + B*a^2*b^3 - (2*A + 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 + (2*A - C)*a^3*b^2 + B*a^2*b^3 - (2*A + 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 + (3*A - C)*a^2*b^3 + 3*B*a*b^4 - (3*A + 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)]

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [B]  time = 1.31899, size = 652, normalized size = 3.03 \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*(A+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 + 2*A*a*b^2 + 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 + 2*A*a*b^2 + 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 + A*
a^2*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^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 + 6*A*b^2*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 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2
*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) + 6*A*b^2*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