Integrand size = 21, antiderivative size = 296 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (23 a^2+9 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {2 (a-b) \sqrt {a+b} \left (15 a^2-8 a b+9 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {16 a b \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
-2/15*(a-b)*(23*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/d+2/15*(a-b)*(15*a^2-8*a*b+9*b^2)*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 +2/5*b*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+16/15*a*b*(a+b*sec(d*x+c))^(1/2 )*tan(d*x+c)/d
Time = 13.45 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.49 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=-\frac {2 (a+b \sec (c+d x))^{5/2} \left (-2 \left (23 a^3+23 a^2 b+9 a b^2+9 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 \left (15 a^3+23 a^2 b+17 a b^2+9 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-\left (23 a^2+9 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2}{15} \left (23 a^2+9 b^2\right ) \sin (c+d x)+\frac {22}{15} a b \tan (c+d x)+\frac {2}{5} b^2 \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \]
(-2*(a + b*Sec[c + d*x])^(5/2)*(-2*(23*a^3 + 23*a^2*b + 9*a*b^2 + 9*b^3)*S qrt[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*(15*a^3 + 23*a^2*b + 17*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]))]*EllipticF[Arc Sin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - (23*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]))/(15*d*(b + a*Cos[ c + d*x])^3*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x) /2]^2*Sec[c + d*x]]*(-1 + Tan[(c + d*x)/2]^2)) + (Cos[c + d*x]^2*(a + b*Se c[c + d*x])^(5/2)*((2*(23*a^2 + 9*b^2)*Sin[c + d*x])/15 + (22*a*b*Tan[c + d*x])/15 + (2*b^2*Sec[c + d*x]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x])^2 )
Time = 1.08 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4317, 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.
\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
\(\Big \downarrow \) 4317 |
\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (5 a^2+8 b \sec (c+d x) a+3 b^2\right )dx+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (5 a^2+8 b \sec (c+d x) a+3 b^2\right )dx+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 a^2+8 b \csc \left (c+d x+\frac {\pi }{2}\right ) a+3 b^2\right )dx+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (a \left (15 a^2+17 b^2\right )+b \left (23 a^2+9 b^2\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (a \left (15 a^2+17 b^2\right )+b \left (23 a^2+9 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \left (15 a^2+17 b^2\right )+b \left (23 a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (15 a^2-8 a b+9 b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b \left (23 a^2+9 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (15 a^2-8 a b+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (23 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b \left (23 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (15 a^2-8 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (15 a^2-8 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (23 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}\right )+\frac {16 a b \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
(2*b*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b]*(23*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)]*Sq rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2*(a - b)*Sqrt[a + b]*(15* a^2 - 8*a*b + 9*b^2)*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 + (16*a*b*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5
3.6.47.3.1 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_)]*(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]
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_))^(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. \(2236\) vs. \(2(266)=532\).
Time = 14.61 (sec) , antiderivative size = 2237, normalized size of antiderivative = 7.56
2/15/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(14*a*b^2*si n(d*x+c)+34*a^2*b*sin(d*x+c)-15*(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^3*cos(d*x+c)^2+23*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*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+23*(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)^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)^2-23*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-17*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)+46*(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*...
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
integral((b^2*sec(d*x + c)^3 + 2*a*b*sec(d*x + c)^2 + a^2*sec(d*x + c))*sq rt(b*sec(d*x + c) + a), x)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]