Integrand size = 34, antiderivative size = 256 \[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]
-2/3*(a-b)*(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))*(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)*(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))*(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
Time = 12.66 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.40 \[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (-2 (a+b) (3 b B+a C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 B+C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-(3 b B+a C) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b d (b+a \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 (3 b B+a C) \sin (c+d x)}{3 b}+\frac {2}{3} C \tan (c+d x)\right )}{d} \]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(-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[Arc Sin[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]))/(3*b*d*(b + a*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Sec[c + d*x]]*((2*(3*b*B + a*C)*Sin[c + d*x])/(3*b) + (2*C*Tan[c + d*x])/3))/d
Time = 0.80 (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 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (a C+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\) |
((-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)
3.9.17.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(2193\) vs. \(2(230)=460\).
Time = 12.30 (sec) , antiderivative size = 2194, normalized size of antiderivative = 8.57
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2194\) |
| default | \(\text {Expression too large to display}\) | \(2203\) |
-2*B/d*(EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a*cos (d*x+c)^2+EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b*c os(d*x+c)^2-(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(c os(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a *cos(d*x+c)^2-(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *b*cos(d*x+c)^2+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(co s(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)*a*cos(d*x+c)+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*( 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)*b*cos(d*x+c)-2*(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b) )^(1/2))*a*cos(d*x+c)-2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ b))^(1/2))*b*cos(d*x+c)+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co s(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+(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...
\[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \]
\[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]