\(\int \sqrt {a+b \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [817]

Optimal result

Integrand size = 34, antiderivative size = 256 \[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \] Output:

-2/3*(a-b)*(a+b)^(1/2)*(3*B*b+C*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/d+2/3*(a-b)*(a+b)^(1/2)*(3*B-C)*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-se 
c(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d+2/3*C*(a+b*sec( 
d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 12.20 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a+b \sec (c+d x)} \left (\left (6 B+\frac {2 a C}{b}\right ) \sin (c+d x)+\frac {2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-2 (a+b) (3 b B+a C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 B+C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-(3 b B+a C) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+2 C \tan (c+d x)\right )}{3 d} \] Input:

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

Output:

(Sqrt[a + b*Sec[c + d*x]]*((6*B + (2*a*C)/b)*Sin[c + d*x] + (2*Sqrt[Sec[(c 
 + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(-2*(a + b)*(3*b*B + 
 a*C)*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*b*(a + b)*(3*B + C)*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)] - (3*b*B + a*C)*Cos[c + d*x]*(b + a*Cos[c + d* 
x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(b*(b + a*Cos[c + d*x])*Sec[c + 
d*x]^(3/2)) + 2*C*Tan[c + d*x]))/(3*d)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4544, 27, 3042, 4546, 27, 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.

\(\displaystyle \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

((-2*(a - b)*Sqrt[a + b]*(3*b*B + a*C)*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]*(3*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)]*S 
qrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 + (2*C*Sqrt[a + b*Sec[c + 
 d*x]]*Tan[c + d*x])/(3*d)
 

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 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 4544

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

rule 4546

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(940\) vs. \(2(230)=460\).

Time = 36.69 (sec) , antiderivative size = 941, normalized size of antiderivative = 3.68

method	result	size
default	\(\text {Expression too large to display}\)	\(941\)
parts	\(\text {Expression too large to display}\)	\(968\)

Input:

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

Output:

-2/3/d/b*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+c)+ 
b)*((-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*B*(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*EllipticE(-csc(d*x+c)+c 
ot(d*x+c),((a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*B*(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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2 
*cos(d*x+c)-1)*C*(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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b) 
)^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(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*EllipticE(-csc(d*x+c 
)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*B*(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*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^ 
2+6*cos(d*x+c)+3)*B*(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*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a 
+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(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*EllipticF(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(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*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))-3*B*a*b*cos...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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