\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [380]

Optimal result

Integrand size = 31, antiderivative size = 254 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 (A b-a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 \sqrt {a+b} d}+\frac {2 (A+B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b \sqrt {a+b} d}-\frac {2 (A b-a B) \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:

-2*(A*b-B*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b 
)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^( 
1/2)/b^2/(a+b)^(1/2)/d+2*(A+B)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2) 
/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+se 
c(d*x+c))/(a-b))^(1/2)/b/(a+b)^(1/2)/d-2*(A*b-B*a)*tan(d*x+c)/(a^2-b^2)/d/ 
(a+b*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 12.31 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.36 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 (b+a \cos (c+d x)) \sqrt {\sec (c+d x)} (A+B \sec (c+d x)) \left (\frac {a (-A b+a B) \sin (c+d x)}{\sqrt {\sec (c+d x)}}+\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 (a+b) (-A b+a B) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-2 b (a+b) (A-B) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+(A b-a B) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}\right )}{\left (-a^2 b+b^3\right ) d (B+A \cos (c+d x)) (a+b \sec (c+d x))^{3/2}} \] Input:

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x 
]
 

Output:

(2*(b + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x])*((a*(-(A*b 
) + a*B)*Sin[c + d*x])/Sqrt[Sec[c + d*x]] + (Sqrt[Cos[(c + d*x)/2]^2*Sec[c 
 + d*x]]*(-2*(a + b)*(-(A*b) + a*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a 
 - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a 
 + b)*(1 + Sec[c + d*x]))] - 2*b*(a + b)*(A - B)*EllipticF[ArcSin[Tan[(c + 
 d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[ 
c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + (A*b - a*B)*Cos[c + d*x]*(b + a* 
Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/Sqrt[Sec[(c + d*x)/2]^ 
2]))/((-(a^2*b) + b^3)*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4491, 27, 3042, 4493, 3042, 4319, 4492}

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.

\(\displaystyle \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {(A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (A+B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 (A b-a B) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {2 (a-b) \sqrt {a+b} (A+B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (A b-a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {2 (A b-a B) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

Input:

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x]
 

Output:

((-2*(a - b)*Sqrt[a + b]*(A*b - a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a 
+ b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x] 
))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)* 
Sqrt[a + b]*(A + B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]] 
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[ 
-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/(a^2 - b^2) - (2*(A*b - a*B)*Ta 
n[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])
 

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 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4491

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

rule 4492

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

rule 4493

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(903\) vs. \(2(234)=468\).

Time = 19.18 (sec) , antiderivative size = 904, normalized size of antiderivative = 3.56

method	result	size
default	\(\text {Expression too large to display}\)	\(904\)
parts	\(\text {Expression too large to display}\)	\(970\)

Input:

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

2/d/b/(a+b)/(a-b)*(((1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*A 
*a*b+((1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*B*a*b+(-(1-cos( 
d*x+c))^3*csc(d*x+c)^3-cot(d*x+c)+csc(d*x+c))*b^2*A+(-(1-cos(d*x+c))^3*csc 
(d*x+c)^3-cot(d*x+c)+csc(d*x+c))*a^2*B+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+c 
ot(d*x+c),((a-b)/(a+b))^(1/2))*a*b+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*( 
1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d 
*x+c),((a-b)/(a+b))^(1/2))*b^2-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a 
+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c 
),((a-b)/(a+b))^(1/2))*a*b-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* 
(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c),(( 
a-b)/(a+b))^(1/2))*b^2-2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a 
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b) 
/(a+b))^(1/2))*a*b-2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos 
(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+ 
b))^(1/2))*b^2+2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x 
+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^ 
(1/2))*a^2+2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)) 
/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2 
))*a*b)*(a+b*sec(d*x+c))^(1/2)/((1-cos(d*x+c))^2*a*csc(d*x+c)^2-b*(1-co...
 

Fricas [F]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm= 
"fricas")
 

Output:

integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^2 
*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(3/2),x)
 

Output:

Integral((A + B*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm= 
"maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm= 
"giac")
 

Output:

integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)*(a + b/cos(c + d*x))^(3/2)),x)
 

Output:

int((A + B/cos(c + d*x))/(cos(c + d*x)*(a + b/cos(c + d*x))^(3/2)), x)
 

Reduce [F]

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

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x)
 

Output:

int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x))/(sec(c + d*x)*b + a),x)