3.10.1 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [901]

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

[Out]

1/2*(b^2*(2*A+C)-2*a*(B*b-C*a))*arctanh(sin(d*x+c))/b^3/d-2*a*(A*b^2-a*(B*b-C*a))*arctanh((a-b)^(1/2)*tan(1/2*
d*x+1/2*c)/(a+b)^(1/2))/b^3/d/(a-b)^(1/2)/(a+b)^(1/2)+(B*b-C*a)*tan(d*x+c)/b^2/d+1/2*C*sec(d*x+c)*tan(d*x+c)/b
/d

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Rubi [A]
time = 0.33, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {4177, 4167, 4083, 3855, 3916, 2738, 214} \begin {gather*} \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} \end {gather*}

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4083

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 4167

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 4177

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]

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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.74, size = 472, normalized size = 3.08 \begin {gather*} \frac {\cos (c+d x) (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \left (2 A b^2-2 a b B+2 a^2 C+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 b^2-2 a b B+2 a^2 C+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 )+\frac {8 a \left (A b^2+a (-b B+a C)\right ) \text {ArcTan}\left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (i \cos (c)+\sin (c))}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+\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 {4 b (b B-a C) \sin \left (\frac {d x}{2}\right )}{\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 {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 {4 b (b B-a C) \sin \left (\frac {d x}{2}\right )}{\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 )}\right )}{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))} \end {gather*}

Warning: Unable to verify antiderivative.

[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]))

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Maple [A]
time = 0.30, size = 247, normalized size = 1.61

method	result	size
derivativedivides	\(\frac {\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 b B -2 a C -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+2 a b B -2 a^{2} C -b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 b B -2 a C -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-2 a b B +2 a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}-\frac {2 a \left (A \,b^{2}-a b B +a^{2} C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(247\)
default	\(\frac {\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 b B -2 a C -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+2 a b B -2 a^{2} C -b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 b B -2 a C -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-2 a b B +2 a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}-\frac {2 a \left (A \,b^{2}-a b B +a^{2} C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(247\)
risch	\(-\frac {i \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 b B +2 a C \right )}{b^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{b d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a B}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 b d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a B}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 b d}\)	\(708\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (140) = 280\).
time = 31.28, size = 657, normalized size = 4.29 \begin {gather*} \left [\frac {2 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\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 + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} 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 + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} 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 + A a b^{2}\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 + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} 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 + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} 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 {gather*}

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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (140) = 280\).
time = 0.51, size = 287, normalized size = 1.88 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 12.10, size = 2500, normalized size = 16.34 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(2*B*b - 2*C*a + C*b))/b^2 + (tan(c/2 + (d*x)/2)^3*(2*C*a - 2*B*b + C*b))/b^2)/(d*(tan(c/
2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (atan(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^
7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^
4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*
b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b
^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*
B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C
*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2
+ b^2*(A + C/2) - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2
+ b^2*(A + C/2) - B*a*b)*1i)/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 -
3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3
- 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 -
12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5
- 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*
C*a^5*b^2))/b^4 - (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*
b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B
*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) -
B*a*b)*1i)/b^3)/((16*(4*C^3*a^8 - 4*A^3*a*b^7 - 6*C^3*a^7*b + 4*A^3*a^2*b^6 + 4*B^3*a^4*b^4 - 4*B^3*a^5*b^3 -
C^3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3 + 6*C^3*a^6*b^2 - A*C^2*a*b^7 - 4*A^2*C*a*b^7 - 12*B*C^2*a^7*b - 1
2*A*B^2*a^3*b^5 + 12*A*B^2*a^4*b^4 + 12*A^2*B*a^2*b^6 - 12*A^2*B*a^3*b^5 + 2*A*C^2*a^2*b^6 - 9*A*C^2*a^3*b^5 +
12*A*C^2*a^4*b^4 - 16*A*C^2*a^5*b^3 + 12*A*C^2*a^6*b^2 + 6*A^2*C*a^2*b^6 - 14*A^2*C*a^3*b^5 + 12*A^2*C*a^4*b^
4 + B*C^2*a^2*b^6 - 2*B*C^2*a^3*b^5 + 9*B*C^2*a^4*b^4 - 12*B*C^2*a^5*b^3 + 16*B*C^2*a^6*b^2 - 4*B^2*C*a^3*b^5
+ 6*B^2*C*a^4*b^4 - 14*B^2*C*a^5*b^3 + 12*B^2*C*a^6*b^2 + 8*A*B*C*a^2*b^6 - 12*A*B*C*a^3*b^5 + 28*A*B*C*a^4*b^
4 - 24*A*B*C*a^5*b^3))/b^6 - (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a
*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2
*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C
*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A
*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b
^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4
*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B*a*b)*(
8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) - B*a*b))
/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b +
16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b
^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 +
16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^
4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 - (((8*(4*A*b
^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9
- 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B*a*b)*(8*a*b^8 - 16*a^2*b^7 +
8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3))*(C*a^2 + b^2*(A
+ C/2) - B*a*b)*2i)/(b^3*d) + (a*atan(((a*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a
^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12
*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b
^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 1
6*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^
3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 + (a*((a + b)*(a - b))^(1/2)*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*
b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 -...

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