Integrand size = 21, antiderivative size = 138 \[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}-\frac {24 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3} \] Output:
-2*sec(b*x+a)^3/b/d/(d*tan(b*x+a))^(1/2)+24/5*cos(b*x+a)*EllipticE(cos(a+1 /4*Pi+b*x),2^(1/2))*(d*tan(b*x+a))^(1/2)/b/d^2/sin(2*b*x+2*a)^(1/2)+24/5*c os(b*x+a)*(d*tan(b*x+a))^(3/2)/b/d^3+12/5*sec(b*x+a)*(d*tan(b*x+a))^(3/2)/ b/d^3
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.81 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {2 \csc (a+b x) \sqrt {d \tan (a+b x)} \left (-8 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \tan ^2(a+b x)+\sqrt {\sec ^2(a+b x)} \left (-5+12 \sin ^2(a+b x)+\tan ^2(a+b x)\right )\right )}{5 b d^2 \sqrt {\sec ^2(a+b x)}} \] Input:
Integrate[Sec[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]
Output:
(2*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]]*(-8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Tan[a + b*x]^2 + Sqrt[Sec[a + b*x]^2]*(-5 + 12*Sin[a + b* x]^2 + Tan[a + b*x]^2)))/(5*b*d^2*Sqrt[Sec[a + b*x]^2])
Time = 0.76 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3088, 3042, 3093, 3042, 3093, 3042, 3095, 3042, 3052, 3042, 3119}
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.
| \(\displaystyle \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (a+b x)^5}{(d \tan (a+b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3088 |
| \(\displaystyle \frac {6 \int \sec ^3(a+b x) \sqrt {d \tan (a+b x)}dx}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \int \sec (a+b x)^3 \sqrt {d \tan (a+b x)}dx}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \int \sec (a+b x) \sqrt {d \tan (a+b x)}dx+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \int \sec (a+b x) \sqrt {d \tan (a+b x)}dx+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-2 \int \cos (a+b x) \sqrt {d \tan (a+b x)}dx\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-2 \int \frac {\sqrt {d \tan (a+b x)}}{\sec (a+b x)}dx\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3095 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{\sqrt {\sin (a+b x)}}\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{\sqrt {\sin (a+b x)}}\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3052 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-\frac {2 \cos (a+b x) \sqrt {d \tan (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)}}\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-\frac {2 \cos (a+b x) \sqrt {d \tan (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)}}\right )+\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {6 \left (\frac {2 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d}+\frac {2}{5} \left (\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d}-\frac {2 \cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{b \sqrt {\sin (2 a+2 b x)}}\right )\right )}{d^2}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
Input:
Int[Sec[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]
Output:
(-2*Sec[a + b*x]^3)/(b*d*Sqrt[d*Tan[a + b*x]]) + (6*((2*Sec[a + b*x]*(d*Ta n[a + b*x])^(3/2))/(5*b*d) + (2*((-2*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x , 2]*Sqrt[d*Tan[a + b*x]])/(b*Sqrt[Sin[2*a + 2*b*x]]) + (2*Cos[a + b*x]*(d *Tan[a + b*x])^(3/2))/(b*d)))/5))/d^2
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Simp[a^2*((m - 2)/(b^2*(n + 1))) Int[(a*Sec[e + f *x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1)) Int[(a*Sec[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 2*m, 2*n]
Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[b*Tan[e + f*x]]/Sqrt[Sin[e + f*x]]) Int[ Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Time = 1.47 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {-\frac {24}{5}+\frac {24 \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \left (1+\sec \left (b x +a \right )\right )}{5}+\frac {12 \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-1-\sec \left (b x +a \right )\right )}{5}+\frac {12 \sec \left (b x +a \right )}{5}+\frac {2 \sec \left (b x +a \right )^{3}}{5}}{d \sqrt {d \tan \left (b x +a \right )}\, b}\) | \(217\) |
Input:
int(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5/b/(d*tan(b*x+a))^(1/2)*(-12+12*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(- csc(b*x+a)+cot(b*x+a))^(1/2)*EllipticE((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2 *2^(1/2))*(csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(1+sec(b*x+a))+6*(csc(b*x+a)-cot (b*x+a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x +a))^(1/2)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-1-sec( b*x+a))+6*sec(b*x+a)+sec(b*x+a)^3)/d
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {2 \, {\left (6 i \, \sqrt {i \, d} \cos \left (b x + a\right )^{2} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - 6 i \, \sqrt {-i \, d} \cos \left (b x + a\right )^{2} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - 6 i \, \sqrt {i \, d} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 i \, \sqrt {-i \, d} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (12 \, \cos \left (b x + a\right )^{4} - 6 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{5 \, b d^{2} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \] Input:
integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
Output:
-2/5*(6*I*sqrt(I*d)*cos(b*x + a)^2*elliptic_e(arcsin(cos(b*x + a) + I*sin( b*x + a)), -1)*sin(b*x + a) - 6*I*sqrt(-I*d)*cos(b*x + a)^2*elliptic_e(arc sin(cos(b*x + a) - I*sin(b*x + a)), -1)*sin(b*x + a) - 6*I*sqrt(I*d)*cos(b *x + a)^2*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + a) + 6*I*sqrt(-I*d)*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) - I*sin( b*x + a)), -1)*sin(b*x + a) + (12*cos(b*x + a)^4 - 6*cos(b*x + a)^2 - 1)*s qrt(d*sin(b*x + a)/cos(b*x + a)))/(b*d^2*cos(b*x + a)^2*sin(b*x + a))
\[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(b*x+a)**5/(d*tan(b*x+a))**(3/2),x)
Output:
Integral(sec(a + b*x)**5/(d*tan(a + b*x))**(3/2), x)
\[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
Output:
integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)
\[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="giac")
Output:
integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)
Timed out. \[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2)),x)
Output:
int(1/(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2)), x)
\[ \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sec \left (b x +a \right )^{5}}{\tan \left (b x +a \right )^{2}}d x \right )}{d^{2}} \] Input:
int(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x)
Output:
(sqrt(d)*int((sqrt(tan(a + b*x))*sec(a + b*x)**5)/tan(a + b*x)**2,x))/d**2