\(\int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx\) [332]

Optimal result

Integrand size = 25, antiderivative size = 69 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \]

[Out]

-8/3*(b*tan(f*x+e))^(1/2)/b^3/f/(d*sec(f*x+e))^(1/2)-2/3/b/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2689, 2685} \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}} \]

[In]

Int[1/(Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)),x]

[Out]

-2/(3*b*f*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2)) - (8*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*Sqrt[d*Sec[e + f*x]
])

Rule 2685

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]

Rule 2689

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*(n + 1))), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {4 \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx}{3 b^2} \\ & = -\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \left (3+\csc ^2(e+f x)\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \]

[In]

Integrate[1/(Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)),x]

[Out]

(-2*(3 + Csc[e + f*x]^2)*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*Sqrt[d*Sec[e + f*x]])

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {2 \cot \left (f x +e \right )-\frac {8 \sec \left (f x +e \right ) \csc \left (f x +e \right )}{3}}{f \sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b^{2}}\)	\(52\)

[In]

int(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)/b^2*(3*cot(f*x+e)-4*sec(f*x+e)*csc(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, {\left (b^{3} d f \cos \left (f x + e\right )^{2} - b^{3} d f\right )}} \]

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/3*(3*cos(f*x + e)^3 - 4*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b^3*d*f*cos(f
*x + e)^2 - b^3*d*f)

Sympy [A] (verification not implemented)

Time = 56.80 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\begin {cases} - \frac {8 \tan ^{3}{\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} - \frac {2 \tan {\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} & \text {for}\: f \neq 0 \\\frac {x}{\left (b \tan {\left (e \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e \right )}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Piecewise((-8*tan(e + f*x)**3/(3*f*(b*tan(e + f*x))**(5/2)*sqrt(d*sec(e + f*x))) - 2*tan(e + f*x)/(3*f*(b*tan(
e + f*x))**(5/2)*sqrt(d*sec(e + f*x))), Ne(f, 0)), (x/((b*tan(e))**(5/2)*sqrt(d*sec(e))), True))

Maxima [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(5/2)), x)

Giac [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(5/2)), x)

Mupad [B] (verification not implemented)

Time = 4.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\frac {\left (\frac {13\,\sin \left (e+f\,x\right )}{3}-\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{b^2\,d\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int(1/((b*tan(e + f*x))^(5/2)*(d/cos(e + f*x))^(1/2)),x)

[Out]

(((13*sin(e + f*x))/3 - sin(3*e + 3*f*x))*(d/cos(e + f*x))^(1/2))/(b^2*d*f*(cos(2*e + 2*f*x) - 1)*((b*sin(2*e
+ 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))