Integrand size = 23, antiderivative size = 292 \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (2 a^2-9 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}+\frac {2 (a-b) \sqrt {a+b} (2 a+9 b) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}-\frac {4 a \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac {2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d} \]
2/15*(a-b)*(2*a^2-9*b^2)*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+2/15*(a-b)*(2*a+9*b)*cot(d*x+c)*Ellipti cF((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+2/5*(a+b *sec(d*x+c))^(3/2)*tan(d*x+c)/b/d-4/15*a*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c) /b/d
Time = 27.52 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\frac {2 \sqrt {a+b \sec (c+d x)} \left (\left (-2 a^2+9 b^2\right ) \sin (c+d x)+\frac {\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (2 \left (2 a^3+2 a^2 b-9 a b^2-9 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b \left (-2 a^2+7 a b+9 b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (2 a^2-9 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{(b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+b (a+3 b \sec (c+d x)) \tan (c+d x)\right )}{15 b^2 d} \]
(2*Sqrt[a + b*Sec[c + d*x]]*((-2*a^2 + 9*b^2)*Sin[c + d*x] + (Sqrt[Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(2*(2*a^3 + 2*a^2*b - 9*a*b^2 - 9*b^3)*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*(-2*a^2 + 7*a*b + 9*b^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]))]*Ellipt icF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (2*a^2 - 9*b^2)*Cos[c + d *x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/((b + a*Cos [c + d*x])*Sec[c + d*x]^(3/2)) + b*(a + 3*b*Sec[c + d*x])*Tan[c + d*x]))/( 15*b^2*d)
Time = 1.08 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4327, 27, 3042, 4490, 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.
\(\displaystyle \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4327 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (3 b-2 a \sec (c+d x)) \sqrt {a+b \sec (c+d x)}dx}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec (c+d x) (3 b-2 a \sec (c+d x)) \sqrt {a+b \sec (c+d x)}dx}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {\sec (c+d x) \left (7 a b-\left (2 a^2-9 b^2\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {\sec (c+d x) \left (7 a b-\left (2 a^2-9 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (7 a b+\left (9 b^2-2 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {\frac {1}{3} \left ((a-b) (2 a+9 b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (2 a^2-9 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left ((a-b) (2 a+9 b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (2 a^2-9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (2 a+9 b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (2 a^2-9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (2 a^2-9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 (a-b) \sqrt {a+b} (2 a+9 b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {4 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
(2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d) + (((2*(a - b)*Sqrt[a + b]*(2*a^2 - 9*b^2)*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]*(2*a + 9*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 + Se c[c + d*x]))/(a - b))])/(b*d))/3 - (4*a*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d *x])/(3*d))/(5*b)
3.6.31.3.1 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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(1981\) vs. \(2(262)=524\).
Time = 11.20 (sec) , antiderivative size = 1982, normalized size of antiderivative = 6.79
-2/15/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-4*a*b ^2*sin(d*x+c)+a^2*b*sin(d*x+c)+2*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)/(c os(d*x+c)+1))^(1/2)*a^2*b-9*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^2+2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(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))*a^3*cos(d*x+c)^2-9*(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b )/(a+b))^(1/2))*b^3*cos(d*x+c)^2-2*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)*a^2*b+7*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^2+18*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^3*cos(d*x+c)+4*(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a- b)/(a+b))^(1/2))*a^3*cos(d*x+c)+9*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^3*cos(d*x+c)^2-18*(1/(a+b)*(b+a*cos(d*x+c))/(cos...
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\int { \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\int \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\int { \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\int { \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx=\int \frac {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^3} \,d x \]