\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [312]

Optimal result

Integrand size = 31, antiderivative size = 143 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {\left (2 a A b-2 a^2 B-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 b^3 d}+\frac {2 a^2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}+\frac {(A b-a B) \tan (c+d x)}{b^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 b d} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(143)=286\).

Time = 1.79 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.10 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {8 a^2 (-A b+a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-2 \left (-2 a A b+2 a^2 B+b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (-2 a A b+2 a^2 B+b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (A b-a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {b^2 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (A b-a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{4 b^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4521, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4521

\(\displaystyle \frac {\int \frac {\sec (c+d x) \left (2 (A b-a B) \sec ^2(c+d x)+b B \sec (c+d x)+a B\right )}{a+b \sec (c+d x)}dx}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b B \csc \left (c+d x+\frac {\pi }{2}\right )+a B\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (a b B-\left (-2 B a^2+2 A b a-b^2 B\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b B+\left (2 B a^2-2 A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 4486

\(\displaystyle \frac {\frac {\frac {2 a^2 (A b-a B) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \int \sec (c+d x)dx}{b}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\frac {2 a^2 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {2 a^2 (A b-a B) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (A b-a B) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {4 a^2 (A b-a B) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {4 a^2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}-\frac {\left (-2 a^2 B+2 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 (A b-a B) \tan (c+d x)}{b d}}{2 b}+\frac {B \tan (c+d x) \sec (c+d x)}{2 b d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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 4257

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

rule 4318

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

rule 4486

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( 
e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b   Int[Csc[e + f*x], 
 x], x] + Simp[(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 4521

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] + Simp[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 4570

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_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[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]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {B}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A b -2 B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 A a b +2 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {B}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A b -2 B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 A a b -2 B \,a^{2}-b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\)	\(229\)
default	\(\frac {-\frac {B}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A b -2 B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 A a b +2 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {B}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A b -2 B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 A a b -2 B \,a^{2}-b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\)	\(229\)
risch	\(-\frac {i \left (B b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-B b \,{\mathrm e}^{i \left (d x +c \right )}-2 A b +2 B a \right )}{b^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A a}{d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,a^{2}}{d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A a}{d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,a^{2}}{d \,b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d b}\)	\(520\)

Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (130) = 260\).

Time = 2.48 (sec) , antiderivative size = 609, normalized size of antiderivative = 4.26 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [-\frac {2 \, {\left (B a^{3} - A 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 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{2} b^{2} - B b^{4} - 2 \, {\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A 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 (B a^{3} - A 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 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{2} b^{2} - B b^{4} - 2 \, {\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A 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 ] \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="max 
ima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (130) = 260\).

Time = 0.18 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.88 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (2 \, B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac {{\left (2 \, B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {4 \, {\left (B a^{3} - A a^{2} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="gia 
c")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 15.78 (sec) , antiderivative size = 4047, normalized size of antiderivative = 28.30 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + b/cos(c + d*x))),x)
 

Output:

(B*a*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b 
*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*b*sin 
(c + d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*a*atan((sin(c 
/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(d*(a^2 - b^2)*(cos(2*c + 2*d*x) 
/2 + 1/2)) + (B*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2* 
d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*a^3*atan((sin(c/2 + (d*x)/2 
)*1i)/cos(c/2 + (d*x)/2))*1i)/(b^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2 
)) + (B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*b*d*(a 
^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*a^4*atan((sin(c/2 + (d*x)/2)*1i 
)/cos(c/2 + (d*x)/2))*1i)/(b^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + 
 (A*a^2*sin(2*c + 2*d*x))/(2*b*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - 
 (B*a^3*sin(2*c + 2*d*x))/(2*b^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) 
 - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)* 
1i)/(d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B*b*atan((sin(c/2 + (d*x 
)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(2*d*(a^2 - b^2)*(cos(2* 
c + 2*d*x)/2 + 1/2)) + (B*a^2*sin(c + d*x))/(2*b*d*(a^2 - b^2)*(cos(2*c + 
2*d*x)/2 + 1/2)) + (A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2)) 
*cos(2*c + 2*d*x)*1i)/(b^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B* 
a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/ 
(2*b*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*a^4*atan((sin(c/2 +...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.66 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
 

Output:

( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + log(tan((c + d*x)/2) - 1) 
+ log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - log(tan((c + d*x)/2) + 1) - 
sin(c + d*x))/(2*d*(sin(c + d*x)**2 - 1))