3.8.56 \(\int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [756]

3.8.56.1 Optimal result

Integrand size = 32, antiderivative size = 200 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d} \]

output

-2*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^( 
1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b)) 
^(1/2)/d-2*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/ 
a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+s 
ec(d*x+c))/(a-b))^(1/2)/d
 

3.8.56.2 Mathematica [A] (verified)

Time = 5.61 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.72 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left ((a+b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-2 a \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec (c+d x)}{d \sqrt {a+b \sec (c+d x)}} \]

input

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

output

(-4*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*C 
os[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*((a + b)*EllipticF[ArcSin[Tan[( 
c + d*x)/2]], (a - b)/(a + b)] - 2*a*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2 
]], (a - b)/(a + b)])*Sec[c + d*x])/(d*Sqrt[a + b*Sec[c + d*x]])
 

3.8.56.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4530, 25, 3042, 4409, 3042, 4271, 4319}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4530

\(\displaystyle -\int -\frac {a-b \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {a-b \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

\(\displaystyle a \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

\(\displaystyle -\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\)

input

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

output

(-2*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqr 
t[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b 
*(1 + Sec[c + d*x]))/(a - b))])/d - (2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi 
[(a + b)/a, 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 
))])/d
 

3.8.56.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4409

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

rule 4530

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

3.8.56.4 Maple [A] (verified)

Time = 8.03 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {2 \left (\cos \left (d x +c \right )+1\right ) \left (\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a +\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b -2 a \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a +b \sec \left (d x +c \right )}}{d \left (b +a \cos \left (d x +c \right )\right )}\)	\(181\)
parts	\(\text {Expression too large to display}\)	\(2901\)

input

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

output

2/d*(cos(d*x+c)+1)*(EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a 
+EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b-2*a*EllipticPi(cot 
(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2)))*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(a+b*sec(d*x+c))^(1/2) 
/(b+a*cos(d*x+c))
 

3.8.56.5 Fricas [F]

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

input

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fric 
as")
 

output

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

3.8.56.6 Sympy [F]

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

input

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

output

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

3.8.56.7 Maxima [F]

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

input

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxi 
ma")
 

output

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

3.8.56.8 Giac [F]

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

input

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

output

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

3.8.56.9 Mupad [F(-1)]

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

input

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

output

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