Integrand size = 25, antiderivative size = 72 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}+\frac {8 \sqrt {b \tan (e+f x)}}{5 b d^2 f \sqrt {d \sec (e+f x)}} \] Output:
2/5*(b*tan(f*x+e))^(1/2)/b/f/(d*sec(f*x+e))^(5/2)+8/5*(b*tan(f*x+e))^(1/2) /b/d^2/f/(d*sec(f*x+e))^(1/2)
Time = 0.61 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {(9+\cos (2 (e+f x))) \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 d^3 f \sqrt {b \tan (e+f x)}} \] Input:
Integrate[1/((d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]),x]
Output:
((9 + Cos[2*(e + f*x)])*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/(5*d^3*f*Sqrt[b *Tan[e + f*x]])
Time = 0.35 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3092, 3042, 3085}
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.
\(\displaystyle \int \frac {1}{\sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3092 |
\(\displaystyle \frac {4 \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}dx}{5 d^2}+\frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}dx}{5 d^2}+\frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3085 |
\(\displaystyle \frac {8 \sqrt {b \tan (e+f x)}}{5 b d^2 f \sqrt {d \sec (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}\) |
Input:
Int[1/((d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]),x]
Output:
(2*Sqrt[b*Tan[e + f*x]])/(5*b*f*(d*Sec[e + f*x])^(5/2)) + (8*Sqrt[b*Tan[e + f*x]])/(5*b*d^2*f*Sqrt[d*Sec[e + f*x]])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[(-(a*Sec[e + f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b* f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(a*Sec[e + f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f* m)), x] + Simp[(m + n + 1)/(a^2*m) Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1 ] && EqQ[n, -2^(-1)])) && IntegersQ[2*m, 2*n]
Time = 1.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62
method | result | size |
default | \(\frac {2 \tan \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+4\right )}{5 f \sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, d^{2}}\) | \(45\) |
Input:
int(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/5/f*tan(f*x+e)*(cos(f*x+e)^2+4)/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2 )/d^2
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {2 \, {\left (\cos \left (f x + e\right )^{3} + 4 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, b d^{3} f} \] Input:
integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas ")
Output:
2/5*(cos(f*x + e)^3 + 4*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sq rt(d/cos(f*x + e))/(b*d^3*f)
Time = 55.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\begin {cases} \frac {8 \tan ^{3}{\left (e + f x \right )}}{5 f \sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} + \frac {2 \tan {\left (e + f x \right )}}{f \sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x}{\sqrt {b \tan {\left (e \right )}} \left (d \sec {\left (e \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(1/2),x)
Output:
Piecewise((8*tan(e + f*x)**3/(5*f*sqrt(b*tan(e + f*x))*(d*sec(e + f*x))**( 5/2)) + 2*tan(e + f*x)/(f*sqrt(b*tan(e + f*x))*(d*sec(e + f*x))**(5/2)), N e(f, 0)), (x/(sqrt(b*tan(e))*(d*sec(e))**(5/2)), True))
\[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima ")
Output:
integrate(1/((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e))), x)
\[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate(1/((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e))), x)
Time = 0.91 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {\left (17\,\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{10\,d^3\,f\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:
int(1/((b*tan(e + f*x))^(1/2)*(d/cos(e + f*x))^(5/2)),x)
Output:
((17*sin(e + f*x) + sin(3*e + 3*f*x))*(d/cos(e + f*x))^(1/2))/(10*d^3*f*(( b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))
\[ \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {\sqrt {d}\, \sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{3} \tan \left (f x +e \right )}d x \right )}{b \,d^{3}} \] Input:
int(1/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x)
Output:
(sqrt(d)*sqrt(b)*int((sqrt(tan(e + f*x))*sqrt(sec(e + f*x)))/(sec(e + f*x) **3*tan(e + f*x)),x))/(b*d**3)