Integrand size = 33, antiderivative size = 253 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {4 a (a-b) \sqrt {a+b} C \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 d}+\frac {2 \sqrt {a+b} (3 A b+(2 a+b) C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b d} \]
4/3*a*(a-b)*C*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^3/d+2/3*(3*A*b+(2*a+b)*C)*cot(d*x+c)*EllipticF((a+b*se c(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*C*(a+b*sec(d*x +c))^(1/2)*tan(d*x+c)/b/d
Time = 18.18 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {8 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (2 a (a+b) C \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b (3 A b+(-2 a+b) C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a C \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^2 d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {\cos (c+d x) (b+a \cos (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {8 a C \sin (c+d x)}{3 b^2}+\frac {4 C \tan (c+d x)}{3 b}\right )}{d (A+2 C+A \cos (2 c+2 d x)) \sqrt {a+b \sec (c+d x)}} \]
(8*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*(2*a*(a + 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]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(3*A*b + (-2*a + 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)] + 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^2*d*(A + 2*C + A*Cos[2*c + 2*d*x])* Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) + (C os[c + d*x]*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*((-8*a*C*Sin[c + d *x])/(3*b^2) + (4*C*Tan[c + d*x])/(3*b)))/(d*(A + 2*C + A*Cos[2*c + 2*d*x] )*Sqrt[a + b*Sec[c + d*x]])
Time = 0.84 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4571, 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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4571 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x) (b (3 A+C)-2 a C \sec (c+d x))}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec (c+d x) (b (3 A+C)-2 a C \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (3 A+C)-2 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {(C (2 a+b)+3 A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-2 a C \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(C (2 a+b)+3 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a C \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b} (C (2 a+b)+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-2 a C \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {\frac {2 \sqrt {a+b} (C (2 a+b)+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {4 a C (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}\) |
((4*a*(a - b)*Sqrt[a + b]*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*Sqrt[a + b]*(3*A* b + (2*a + b)*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq rt[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*b) + (2*C*Sqrt[a + b*Sec[c + d* x]]*Tan[c + d*x])/(3*b*d)
3.8.37.3.1 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[(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]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + 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 + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) In t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1243\) vs. \(2(227)=454\).
Time = 16.32 (sec) , antiderivative size = 1244, normalized size of antiderivative = 4.92
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1244\) |
default | \(\text {Expression too large to display}\) | \(1405\) |
-2*A/d*(cos(d*x+c)+1)*(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)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b) )^(1/2))*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))-2/3*C/d/b^2*(a+b*sec(d*x+ c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(2*EllipticE(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*co s(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*cos(d*x+c)^2+2*EllipticE(cot(d*x+c)-cs c(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*b*cos(d*x+c)^2-2*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)*a*b*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^2*cos(d*x+c)^2+4*Ellipti cE(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^2*cos(d*x+c)+4*Ell ipticE(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*b*cos(d*x+c)-4 *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*b*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^2*c...
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]