3.6.40 \(\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \, dx\) [540]

3.6.40.1 Optimal result

Integrand size = 21, antiderivative size = 249 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \, dx=-\frac {8 a (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 (a-b) (3 a-b) \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 b \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]

output

-8/3*a*(a-b)*cot(d*x+c)*EllipticE((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)/b/d+2/3*(a-b)*(3*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))*(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/d+2/3*b*(a+b*sec(d*x+c))^(1/2 
)*tan(d*x+c)/d
 

3.6.40.2 Mathematica [A] (verified)

Time = 7.49 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.22 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \, dx=-\frac {2 \sqrt {a+b \sec (c+d x)} \left (8 a (a+b) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 \left (3 a^2+4 a b+b^2\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-5 a b \sin (c+d x)-2 a^2 \sin (2 (c+d x))+4 a b \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+4 a^2 \cos ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )-b^2 \tan (c+d x)\right )}{3 d (b+a \cos (c+d x))} \]

input

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

output

(-2*Sqrt[a + b*Sec[c + d*x]]*(8*a*(a + b)*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + 
d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d 
*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*(3*a^2 + 4 
*a*b + b^2)*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[ 
(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c 
 + d*x)/2]], (a - b)/(a + b)] - 5*a*b*Sin[c + d*x] - 2*a^2*Sin[2*(c + d*x) 
] + 4*a*b*Cos[c + d*x]*Tan[(c + d*x)/2] + 4*a^2*Cos[c + d*x]^2*Tan[(c + d* 
x)/2] - b^2*Tan[c + d*x]))/(3*d*(b + a*Cos[c + d*x]))
 

3.6.40.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4317, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4317

\(\displaystyle \frac {2}{3} \int \frac {\sec (c+d x) \left (3 a^2+4 b \sec (c+d x) a+b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {\sec (c+d x) \left (3 a^2+4 b \sec (c+d x) a+b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a^2+4 b \csc \left (c+d x+\frac {\pi }{2}\right ) a+b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {1}{3} \left ((a-b) (3 a-b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+4 a b \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left ((a-b) (3 a-b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+4 a b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{3} \left (4 a b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) (3 a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{3} \left (\frac {2 (a-b) (3 a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {8 a (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

input

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

output

((-8*a*(a - b)*Sqrt[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*d) + (2*(a - b)*(3*a - b)*Sq 
rt[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))/3 + (2*b*Sqrt[a + b*Sec[c + d*x]]*Tan[c + 
 d*x])/(3*d)
 

3.6.40.3.1 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 4317

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] 
+ Simp[1/m   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(b^2*(m - 1) + a 
^2*m + a*b*(2*m - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ 
[a^2 - b^2, 0] && GtQ[m, 1] && IntegerQ[2*m]
 

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 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]
 

3.6.40.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1382\) vs. \(2(223)=446\).

Time = 11.09 (sec) , antiderivative size = 1383, normalized size of antiderivative = 5.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1383\)

input

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

output

-2/3/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(3*(cos(d*x+ 
c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*E 
llipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*cos(d*x+c)^2+4*Ell 
ipticF(cot(d*x+c)-csc(d*x+c),((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*cos(d*x+c)^2 
+EllipticF(cot(d*x+c)-csc(d*x+c),((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)*b^2*cos(d*x+ 
c)^2-4*EllipticE(cot(d*x+c)-csc(d*x+c),((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^2*co 
s(d*x+c)^2-4*EllipticE(cot(d*x+c)-csc(d*x+c),((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*cos(d*x+c)^2+6*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x 
+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^( 
1/2))*a^2*cos(d*x+c)+8*EllipticF(cot(d*x+c)-csc(d*x+c),((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*cos(d*x+c)+2*EllipticF(cot(d*x+c)-csc(d*x+c),((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)*b^2*cos(d*x+c)-8*EllipticE(cot(d*x+c)-csc(d*x+c),((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^2*cos(d*x+c)-8*EllipticE(cot(d*x+c)-csc(d*x+c),((a-...
 

3.6.40.5 Fricas [F]

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

input

integrate(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), x)
 

3.6.40.6 Sympy [F]

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

input

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

output

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

3.6.40.7 Maxima [F]

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

input

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

output

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

3.6.40.8 Giac [F]

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

input

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

output

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

3.6.40.9 Mupad [F(-1)]

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

input

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

output

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