\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [311]

Optimal result

Integrand size = 31, antiderivative size = 187 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\left (2 a^2+b^2\right ) (A b-a B) \text {arctanh}(\sin (c+d x))}{2 b^4 d}-\frac {2 a^3 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d} \] Output:

1/2*(2*a^2+b^2)*(A*b-B*a)*arctanh(sin(d*x+c))/b^4/d-2*a^3*(A*b-B*a)*arctan 
h((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(1/2)/b^4/(a+b)^(1/2)/ 
d-1/3*(3*A*a*b-3*B*a^2-2*B*b^2)*tan(d*x+c)/b^3/d+1/2*(A*b-B*a)*sec(d*x+c)* 
tan(d*x+c)/b^2/d+1/3*B*sec(d*x+c)^2*tan(d*x+c)/b/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(422\) vs. \(2(187)=374\).

Time = 2.18 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.26 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {24 a^3 (A b-a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+6 \left (2 a^2+b^2\right ) (-A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \left (2 a^2+b^2\right ) (-A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2 (3 A b+(-3 a+b) B)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 b \left (-3 a A b+3 a^2 B+2 b^2 B\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {b^2 (3 A b+(-3 a+b) B)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b \left (-3 a A b+3 a^2 B+2 b^2 B\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{12 b^4 d} \] Input:

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4521, 3042, 4580, 3042, 4570, 27, 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.

\(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4521

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (2 a^2+b^2\right ) (A b-a B) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^3 (A b-a B) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (2 a^2+b^2\right ) (A b-a B) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^3 (A b-a B) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (2 a^2+b^2\right ) (A b-a B) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^3 (A b-a B) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {\left (2 a^2+b^2\right ) (A b-a B) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^3 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 b d}}{3 b}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d}\)

Input:

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 4580

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] + Simp[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]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.79

method	result	size
derivativedivides	\(\frac {-\frac {B}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A b -B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (2 A \,a^{2} b +A \,b^{3}-2 B \,a^{3}-B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {-2 A a b -A \,b^{2}+2 B \,a^{2}+B a b +2 b^{2} B}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{3} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-A b +B a +B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-2 A \,a^{2} b -A \,b^{3}+2 B \,a^{3}+B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {-2 A a b -A \,b^{2}+2 B \,a^{2}+B a b +2 b^{2} B}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(335\)
default	\(\frac {-\frac {B}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A b -B a -B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (2 A \,a^{2} b +A \,b^{3}-2 B \,a^{3}-B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {-2 A a b -A \,b^{2}+2 B \,a^{2}+B a b +2 b^{2} B}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{3} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {B}{3 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-A b +B a +B b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-2 A \,a^{2} b -A \,b^{3}+2 B \,a^{3}+B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {-2 A a b -A \,b^{2}+2 B \,a^{2}+B a b +2 b^{2} B}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(335\)
risch	\(\frac {i \left (-3 A \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+3 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A a b \,{\mathrm e}^{2 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 A \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 B a b \,{\mathrm e}^{i \left (d x +c \right )}-6 A a b +6 B \,a^{2}+4 b^{2} B \right )}{3 d \,b^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 b d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,a^{3}}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{2 b^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,a^{3}}{b^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{2 b^{2} d}\)	\(659\)

Input:

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE 
)
                                                                                    
                                                                                    
 

Output:

1/d*(-1/3*B/b/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(A*b-B*a-B*b)/b^2/(tan(1/2*d*x+ 
1/2*c)+1)^2+1/2*(2*A*a^2*b+A*b^3-2*B*a^3-B*a*b^2)/b^4*ln(tan(1/2*d*x+1/2*c 
)+1)-1/2*(-2*A*a*b-A*b^2+2*B*a^2+B*a*b+2*B*b^2)/b^3/(tan(1/2*d*x+1/2*c)+1) 
-2*a^3*(A*b-B*a)/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))-1/3*B/b/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(-A*b+B*a+B*b)/b 
^2/(tan(1/2*d*x+1/2*c)-1)^2+1/2/b^4*(-2*A*a^2*b-A*b^3+2*B*a^3+B*a*b^2)*ln( 
tan(1/2*d*x+1/2*c)-1)-1/2*(-2*A*a*b-A*b^2+2*B*a^2+B*a*b+2*B*b^2)/b^3/(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. 343 vs. \(2 (170) = 340\).

Time = 0.50 (sec) , antiderivative size = 743, normalized size of antiderivative = 3.97 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:

integrate(sec(d*x+c)**4*(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)**4/(a + b*sec(c + d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^4*(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. 412 vs. \(2 (170) = 340\).

Time = 0.18 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.20 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A 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 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A 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 (B a^{4} - A 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 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B 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} \] Input:

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="gia 
c")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.79 (sec) , antiderivative size = 4667, normalized size of antiderivative = 24.96 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + b/cos(c + d*x))),x)
 

Output:

- ((tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 2*B*b^2 - 2*A*a*b - B*a*b))/b^3 
+ (tan(c/2 + (d*x)/2)^5*(2*B*a^2 - A*b^2 + 2*B*b^2 - 2*A*a*b + B*a*b))/b^3 
 - (4*tan(c/2 + (d*x)/2)^3*(3*B*a^2 + B*b^2 - 3*A*a*b))/(3*b^3))/(d*(3*tan 
(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) - 
(atan(((((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2 
*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 
+ 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4 
*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16 
*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5* 
b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 - (((8*(2*A*b^13 + 2*A*a^2*b^1 
1 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 
 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 - (4*tan(c/2 + (d*x)/2)*(8* 
a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/ 
b^10)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/(2*b^4))*(A*b^3 - 2*B*a^3 + 
 2*A*a^2*b - B*a*b^2)*1i)/(2*b^4) + (((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B 
^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16* 
A^2*a^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^ 
7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B 
^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 
 26*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.23 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (2 \sin \left (d x +c \right )^{2}-3\right )}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
 

Output:

(sin(c + d*x)*(2*sin(c + d*x)**2 - 3))/(3*cos(c + d*x)*d*(sin(c + d*x)**2 
- 1))