3.7.36 \(\int (a+a \sec (c+d x))^m (B-C+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [636]

3.7.36.1 Optimal result

Integrand size = 36, antiderivative size = 171 \[ \int (a+a \sec (c+d x))^m \left (B-C+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} (B-C) \operatorname {AppellF1}\left (\frac {3}{2}+m,\frac {1}{2},1,\frac {5}{2}+m,\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) (a+a \sec (c+d x))^m \tan (c+d x)}{d (3+2 m) \sqrt {1-\sec (c+d x)}}+\frac {2^{\frac {3}{2}+m} C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-m} (a+a \sec (c+d x))^m \tan (c+d x)}{d} \]

output

2^(3/2+m)*C*hypergeom([1/2, -1/2-m],[3/2],1/2-1/2*sec(d*x+c))*(1+sec(d*x+c 
))^(-1/2-m)*(a+a*sec(d*x+c))^m*tan(d*x+c)/d+(B-C)*AppellF1(3/2+m,1,1/2,5/2 
+m,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(1+sec(d*x+c))*(a+a*sec(d*x+c))^m*2^(1 
/2)*tan(d*x+c)/d/(3+2*m)/(1-sec(d*x+c))^(1/2)
 

3.7.36.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1817\) vs. \(2(171)=342\).

Time = 12.89 (sec) , antiderivative size = 1817, normalized size of antiderivative = 10.63 \[ \int (a+a \sec (c+d x))^m \left (B-C+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \]

input

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

output

((C*(1 + Cos[c + d*x] + 4*m*Cos[c + d*x]*Hypergeometric2F1[1/2, 2 + m, 3/2 
, Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m)*(a*(1 + Sec[c + 
 d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*Tan[(c + d*x)/2])/(d*(1 + 2*m)*(C 
 + B*Cos[c + d*x] - C*Cos[c + d*x])*(1 + Sec[c + d*x])) + (2^(1 + m)*B*Cos 
[c + d*x]*Hypergeometric2F1[1/2, 1 + m, 3/2, Tan[(c + d*x)/2]^2]*(Cos[c + 
d*x]*Sec[(c + d*x)/2]^2)^m*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^m*(1 + Sec[c 
+ d*x])^(-1 - m)*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x])*T 
an[(c + d*x)/2])/(d*(C + B*Cos[c + d*x] - C*Cos[c + d*x])) + (30*B*AppellF 
1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2 
]^2*Cos[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(1 + m)*(B - C + C*Sec[c + d*x]) 
*Sin[c + d*x]*(3*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d* 
x)/2]^2] - 2*(AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/ 
2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/ 
2]^2])*Tan[(c + d*x)/2]^2))/(d*(C + B*Cos[c + d*x] - C*Cos[c + d*x])*(1 + 
Sec[c + d*x])*(45*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d 
*x)/2]^2]^2*Cos[(c + d*x)/2]^2*(1 + 2*m - 2*m*Cos[c + d*x] + Cos[2*(c + d* 
x)]) + 6*AppellF1[1/2, m, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] 
*Sin[(c + d*x)/2]^2*(-5*AppellF1[3/2, m, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[ 
(c + d*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[c + d*x] + Cos[2*(c + d*x)]) + 5* 
m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]...
 

3.7.36.3 Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4529, 3042, 4412, 3042, 4266, 3042, 4265, 153, 4315, 3042, 4314, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4412

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4266

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4265

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

\(\Big \downarrow \) 153

\(\displaystyle \frac {a C \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{m+1}dx+\frac {\sqrt {2} a^2 (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},1,m+\frac {5}{2},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}}{a^2}\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {a^2 C (\sec (c+d x)+1)^{-m} (a \sec (c+d x)+a)^m \int \sec (c+d x) (\sec (c+d x)+1)^{m+1}dx+\frac {\sqrt {2} a^2 (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},1,m+\frac {5}{2},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 C (\sec (c+d x)+1)^{-m} (a \sec (c+d x)+a)^m \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{m+1}dx+\frac {\sqrt {2} a^2 (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},1,m+\frac {5}{2},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}}{a^2}\)

\(\Big \downarrow \) 4314

\(\displaystyle \frac {\frac {\sqrt {2} a^2 (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},1,m+\frac {5}{2},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}-\frac {a^2 C \tan (c+d x) (\sec (c+d x)+1)^{-m-\frac {1}{2}} (a \sec (c+d x)+a)^m \int \frac {(\sec (c+d x)+1)^{m+\frac {1}{2}}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)}}}{a^2}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\frac {\sqrt {2} a^2 (B-C) \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},1,m+\frac {5}{2},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d (2 m+3) \sqrt {1-\sec (c+d x)}}+\frac {a^2 C 2^{m+\frac {3}{2}} \tan (c+d x) (\sec (c+d x)+1)^{-m-\frac {1}{2}} (a \sec (c+d x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m-\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )}{d}}{a^2}\)

input

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

output

((Sqrt[2]*a^2*(B - C)*AppellF1[3/2 + m, 1/2, 1, 5/2 + m, (1 + Sec[c + d*x] 
)/2, 1 + Sec[c + d*x]]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^m*Tan[c + d 
*x])/(d*(3 + 2*m)*Sqrt[1 - Sec[c + d*x]]) + (2^(3/2 + m)*a^2*C*Hypergeomet 
ric2F1[1/2, -1/2 - m, 3/2, (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 
- m)*(a + a*Sec[c + d*x])^m*Tan[c + d*x])/d)/a^2
 

3.7.36.3.1 Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 153

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

rule 3042

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

rule 4265

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot 
[c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]]))   Subst[Int[(1 
 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0 
]
 

rule 4266

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 4314

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

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m 
]/(1 + (b/a)*Csc[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Csc[e + f*x])^m*(d 
*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

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]
 

3.7.36.4 Maple [F]

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

input

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

output

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

3.7.36.5 Fricas [F]

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

input

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

output

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

3.7.36.6 Sympy [F]

\[ \int (a+a \sec (c+d x))^m \left (B-C+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{m} \left (\sec {\left (c + d x \right )} + 1\right ) \left (B + C \sec {\left (c + d x \right )} - C\right )\, dx \]

input

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

output

Integral((a*(sec(c + d*x) + 1))**m*(sec(c + d*x) + 1)*(B + C*sec(c + d*x) 
- C), x)
 

3.7.36.7 Maxima [F]

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

input

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

output

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

3.7.36.8 Giac [F]

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

input

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

output

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

3.7.36.9 Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^m \left (B-C+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^m\,\left (B-C+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

input

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

output

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