\(\int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [120]

Optimal result

Integrand size = 33, antiderivative size = 101 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 a (5 A+7 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 A-2 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d} \] Output:

2/15*a*(5*A+7*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*(5*A-2*B)*(a+a*s 
ec(d*x+c))^(1/2)*tan(d*x+c)/d+2/5*B*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 (5 A+7 B+(5 A+4 B) \cos (c+d x)+(5 A+4 B) \cos (2 (c+d x))) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{15 d (1+\cos (c+d x))} \] Input:

Integrate[Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(2*(5*A + 7*B + (5*A + 4*B)*Cos[c + d*x] + (5*A + 4*B)*Cos[2*(c + d*x)])*S 
ec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(15*d*(1 + Cos[c + d* 
x]))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4498, 27, 3042, 4489, 3042, 4279}

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.

\(\displaystyle \int \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) \sqrt {\sec (c+d x) a+a} (3 a B+a (5 A-2 B) \sec (c+d x))dx}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) \sqrt {\sec (c+d x) a+a} (3 a B+a (5 A-2 B) \sec (c+d x))dx}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a B+a (5 A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {1}{3} a (5 A+7 B) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a (5 A-2 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (5 A+7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a (5 A-2 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {\frac {2 a^2 (5 A+7 B) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 A-2 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

Input:

Int[Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
 

Output:

(2*B*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((2*a^2*(5*A + 7*B 
)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*A - 2*B)*Sqrt[a + 
 a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)
 

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 3042

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

rule 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4489

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

rule 4498

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

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (10 \sin \left (d x +c \right )+5 \tan \left (d x +c \right )\right ) A +\left (3 \sec \left (d x +c \right ) \tan \left (d x +c \right )+8 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )\right ) B \right )}{15 d \left (1+\cos \left (d x +c \right )\right )}\)	\(81\)
parts	\(\frac {A \left (4 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right )}+\frac {B \left (16 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}\)	\(108\)

Input:

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

2/15/d*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((10*sin(d*x+c)+5*tan(d*x+c 
))*A+(3*sec(d*x+c)*tan(d*x+c)+8*sin(d*x+c)+4*tan(d*x+c))*B)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (2 \, {\left (5 \, A + 4 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 4 \, B\right )} \cos \left (d x + c\right ) + 3 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

2/15*(2*(5*A + 4*B)*cos(d*x + c)^2 + (5*A + 4*B)*cos(d*x + c) + 3*B)*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos( 
d*x + c)^2)
 

Sympy [F]

\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)),x)
 

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))*sec(c + d*x)**2, 
x)
 

Maxima [F]

\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \] Input:

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 

Output:

4/15*(15*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1 
)^(1/4)*((A*d*cos(2*d*x + 2*c)^2 + A*d*sin(2*d*x + 2*c)^2 + 2*A*d*cos(2*d* 
x + 2*c) + A*d)*integrate((((cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d 
*x + 6*c)*cos(2*d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d 
*x + 2*c)^2 + sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin(2 
*d*x + 2*c) + 3*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*co 
s(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin 
(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*s 
in(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*s 
in(2*d*x + 2*c) - 3*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(5/2*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 2* 
c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8* 
c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 
4*c)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) 
) - (cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d*x + 6*c)*cos(2*d*x + 2* 
c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(8*d*x 
+ 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 3*sin(4*d* 
x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(5/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x...
 

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.74 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (15 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (20 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 10 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (5 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 

Output:

2/15*(15*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*B*a^3*sgn(cos(d*x + 
c)) - (20*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 10*sqrt(2)*B*a^3*sgn(cos(d*x + 
 c)) - (5*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 7*sqrt(2)*B*a^3*sgn(cos(d*x + 
c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/ 
((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 15.51 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.10 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=-\frac {4\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (A\,5{}\mathrm {i}+B\,4{}\mathrm {i}+A\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,5{}\mathrm {i}+A\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,10{}\mathrm {i}+A\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,5{}\mathrm {i}+A\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,14{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,4{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,4{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^2,x)
 

Output:

-(4*(exp(c*1i + d*x*1i) - 1)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d 
*x*1i)/2))^(1/2)*(A*5i + B*4i + A*exp(c*1i + d*x*1i)*5i + A*exp(c*2i + d*x 
*2i)*10i + A*exp(c*3i + d*x*3i)*5i + A*exp(c*4i + d*x*4i)*5i + B*exp(c*1i 
+ d*x*1i)*4i + B*exp(c*2i + d*x*2i)*14i + B*exp(c*3i + d*x*3i)*4i + B*exp( 
c*4i + d*x*4i)*4i))/(15*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1 
)^2)
 

Reduce [F]

\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x)
 

Output:

sqrt(a)*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*b + int(sqrt(sec(c 
+ d*x) + 1)*sec(c + d*x)**2,x)*a)