3.10.78 \(\int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [978]

3.10.78.1 Optimal result

Integrand size = 50, antiderivative size = 212 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a+b} C \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 \sqrt {a+b} (b B-a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d} \]

output

2*C*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*(B*b-C*a)*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+sec(d*x+c))/(a-b))^(1/2)/a/d
 

3.10.78.2 Mathematica [A] (verified)

Time = 7.18 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.75 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left ((a C+b (-B+C)) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+2 (b B-a C) \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec (c+d x)}{d \sqrt {a+b \sec (c+d x)}} \]

input

Integrate[(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*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*Co 
s[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*((a*C + b*(-B + C))*EllipticF[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*(b*B - a*C)*EllipticPi[-1, Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x])/(d*Sqrt[a + b*Sec[ 
c + d*x]])
 

3.10.78.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2014, 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.

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

\(\Big \downarrow \) 2014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

\(\displaystyle \frac {b^3 C \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{b^2}\)

\(\Big \downarrow \) 4319

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

input

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

output

((2*b^2*Sqrt[a + b]*C*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)]*Sqr 
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*b^2*Sqrt[a + b]*(b*B - a*C)*C 
ot[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))])/(a*d))/b^2
 

3.10.78.3.1 Defintions of rubi rules used

rule 2014

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S 
ymbol] :> Simp[1/b^2   Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], 
x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ 
[m, -1]
 

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]
 

3.10.78.4 Maple [A] (verified)

Time = 12.45 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {2 \left (1+\cos \left (d x +c \right )\right ) \left (B \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b -2 B \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right ) b -C \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a -C \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b +2 C \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right ) a \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {a +b \sec \left (d x +c \right )}}{d \left (b +a \cos \left (d x +c \right )\right )}\)	\(252\)
parts	\(\text {Expression too large to display}\)	\(3994\)

input

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

output

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

3.10.78.5 Fricas [F]

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

input

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

output

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

3.10.78.6 Sympy [F]

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

input

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

output

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

3.10.78.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.10.78.8 Giac [F]

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

input

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

3.10.78.9 Mupad [F(-1)]

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

input

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

output

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