\(\int (a+b \sec (c+d x))^m (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [1074]

Optimal result

Integrand size = 48, antiderivative size = 48 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} b C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{1+m} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-1-m} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)}}+(b B-a C) \text {Int}\left ((a+b \sec (c+d x))^{1+m},x\right ) \] Output:

2^(1/2)*b*C*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d 
*x+c))*(a+b*sec(d*x+c))^(1+m)*((a+b*sec(d*x+c))/(a+b))^(-1-m)*tan(d*x+c)/d 
/(1+sec(d*x+c))^(1/2)+(B*b-C*a)*Defer(Int)((a+b*sec(d*x+c))^(1+m),x)
                                                                                    
                                                                                    
 

Mathematica [N/A]

Not integrable

Time = 29.98 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \] Input:

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

Output:

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

Rubi [N/A]

Not integrable

Time = 0.62 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 4529, 3042, 4412, 3042, 4273, 4321, 156, 155}

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.

\(\displaystyle \int (a+b \sec (c+d x))^m \left (a^2 (-C)+a b B+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4529

\(\displaystyle \frac {\int (a+b \sec (c+d x))^{m+1} \left (C \sec (c+d x) b^3+(b B-a C) b^2\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{m+1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2\right )dx}{b^2}\)

\(\Big \downarrow \) 4412

\(\displaystyle \frac {b^3 C \int \sec (c+d x) (a+b \sec (c+d x))^{m+1}dx+b^2 (b B-a C) \int (a+b \sec (c+d x))^{m+1}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{m+1}dx+b^2 (b B-a C) \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{m+1}dx}{b^2}\)

\(\Big \downarrow \) 4273

\(\displaystyle \frac {b^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{m+1}dx+b^2 (b B-a C) \int (a+b \sec (c+d x))^{m+1}dx}{b^2}\)

\(\Big \downarrow \) 4321

\(\displaystyle \frac {b^2 (b B-a C) \int (a+b \sec (c+d x))^{m+1}dx-\frac {b^3 C \tan (c+d x) \int \frac {(a+b \sec (c+d x))^{m+1}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{b^2}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {b^2 (b B-a C) \int (a+b \sec (c+d x))^{m+1}dx-\frac {b^3 C (a+b) \tan (c+d x) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{m+1}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{b^2}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {b^2 (b B-a C) \int (a+b \sec (c+d x))^{m+1}dx+\frac {\sqrt {2} b^3 C (a+b) \tan (c+d x) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt {\sec (c+d x)+1}}}{b^2}\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 4273

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Unintegrable[ 
(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0 
] &&  !IntegerQ[2*n]
 

rule 4321

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x 
]])   Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f*x]] 
, x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]
 

rule 4412

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[c   Int[(a + b*Csc[e + f*x])^m, x], x] + Sim 
p[d   Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[2*m]
 

rule 4529

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

Maple [N/A]

Not integrable

Time = 0.47 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00

\[\int \left (a +b \sec \left (d x +c \right )\right )^{m} \left (B a b -C \,a^{2}+b^{2} B \sec \left (d x +c \right )+b^{2} C \sec \left (d x +c \right )^{2}\right )d x\]

Input:

int((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x 
)
 

Output:

int((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x 
)
 

Fricas [N/A]

Not integrable

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c 
)^2),x, algorithm="fricas")
 

Output:

integral((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*se 
c(d*x + c) + a)^m, x)
 

Sympy [N/A]

Not integrable

Time = 10.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=- \int C a^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m}\, dx - \int \left (- B a b \left (a + b \sec {\left (c + d x \right )}\right )^{m}\right )\, dx - \int \left (- B b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- C b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\right )\, dx \] Input:

integrate((a+b*sec(d*x+c))**m*(B*a*b-C*a**2+b**2*B*sec(d*x+c)+b**2*C*sec(d 
*x+c)**2),x)
 

Output:

-Integral(C*a**2*(a + b*sec(c + d*x))**m, x) - Integral(-B*a*b*(a + b*sec( 
c + d*x))**m, x) - Integral(-B*b**2*(a + b*sec(c + d*x))**m*sec(c + d*x), 
x) - Integral(-C*b**2*(a + b*sec(c + d*x))**m*sec(c + d*x)**2, x)
 

Maxima [N/A]

Not integrable

Time = 19.62 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c 
)^2),x, algorithm="maxima")
 

Output:

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*s 
ec(d*x + c) + a)^m, x)
 

Giac [N/A]

Not integrable

Time = 0.82 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:

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

Output:

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*s 
ec(d*x + c) + a)^m, x)
 

Mupad [N/A]

Not integrable

Time = 16.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^m\,\left (\frac {B\,b^2}{\cos \left (c+d\,x\right )}-C\,a^2+\frac {C\,b^2}{{\cos \left (c+d\,x\right )}^2}+B\,a\,b\right ) \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.94 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\left (\int \left (\sec \left (d x +c \right ) b +a \right )^{m}d x \right ) a^{2} c +\left (\int \left (\sec \left (d x +c \right ) b +a \right )^{m}d x \right ) a \,b^{2}+\left (\int \left (\sec \left (d x +c \right ) b +a \right )^{m} \sec \left (d x +c \right )^{2}d x \right ) b^{2} c +\left (\int \left (\sec \left (d x +c \right ) b +a \right )^{m} \sec \left (d x +c \right )d x \right ) b^{3} \] Input:

int((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x 
)
 

Output:

 - int((sec(c + d*x)*b + a)**m,x)*a**2*c + int((sec(c + d*x)*b + a)**m,x)* 
a*b**2 + int((sec(c + d*x)*b + a)**m*sec(c + d*x)**2,x)*b**2*c + int((sec( 
c + d*x)*b + a)**m*sec(c + d*x),x)*b**3