\(\int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [322]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{b^{3/2} f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{b^{3/2} f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {d (d \sec (e+f x))^{3/2} \left (-2 \sin (e+f x)+\frac {\left (-\arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )\right ) \cos (e+f x) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt [4]{\sec ^2(e+f x)}}\right )}{f (b \tan (e+f x))^{3/2}} \] Input:

Integrate[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(d*(d*Sec[e + f*x])^(3/2)*(-2*Sin[e + f*x] + ((-ArcTan[Sqrt[Tan[e + f*x]]/ 
(Sec[e + f*x]^2)^(1/4)] + ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4 
)])*Cos[e + f*x]*Tan[e + f*x]^(3/2))/(Sec[e + f*x]^2)^(1/4)))/(f*(b*Tan[e 
+ f*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.75, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3088, 3042, 3096, 3042, 3044, 27, 266, 827, 216, 219}

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.

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3088

\(\displaystyle \frac {d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx}{b^2}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx}{b^2}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3096

\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{b^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{b^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3044

\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{b^3 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {2 d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 d^3 \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\)

Input:

Int[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(-2*d^2*Sqrt[d*Sec[e + f*x]])/(b*f*Sqrt[b*Tan[e + f*x]]) + (2*d^3*(-1/2*Ar 
cTan[Sqrt[b]*Sin[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sin[e + f*x]]/(2*Sqrt 
[b]))*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]] 
)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3088

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]
 

rule 3096

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a^(m + n)*((b*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b* 
Sin[e + f*x])^n))   Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x] /; 
FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]
 

Maple [A] (verified)

Time = 5.70 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.16

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \arctan \left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )+\sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )+2 \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \sqrt {b \tan \left (f x +e \right )}}\)	\(198\)

Input:

int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/f*(sin(f*x+e)*arctan((sin(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)/(co 
s(f*x+e)-1))+sin(f*x+e)*arctanh((sin(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*sin(f* 
x+e)/(cos(f*x+e)-1))+2*(sin(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(s 
in(f*x+e)/(1+cos(f*x+e))^2)^(1/2))*d^2*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e)) 
/(sin(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/b/(b*tan(f*x+e))^(1/2)
                                                                                    
                                                                                    
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (143) = 286\).

Time = 0.30 (sec) , antiderivative size = 794, normalized size of antiderivative = 4.64 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

[-1/8*(2*b*d^2*sqrt(-d/b)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - 
(cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*s 
qrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e))/(d*cos(f* 
x + e)^2 - (d*cos(f*x + e) + d)*sin(f*x + e) - d))*sin(f*x + e) - b*d^2*sq 
rt(-d/b)*log((d*cos(f*x + e)^4 - 72*d*cos(f*x + e)^2 - 8*(7*cos(f*x + e)^3 
 - (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b 
*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e)) + 28*(d*cos(f* 
x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 
4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8))*sin(f*x + e) + 16*d^2*sqrt(b*sin 
(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(b^2*f*sin(f*x 
+ e)), -1/8*(2*b*d^2*sqrt(d/b)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e) 
^2 + (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 
 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e))/(d*co 
s(f*x + e)^2 + (d*cos(f*x + e) + d)*sin(f*x + e) - d))*sin(f*x + e) - b*d^ 
2*sqrt(d/b)*log((d*cos(f*x + e)^4 - 72*d*cos(f*x + e)^2 - 8*(7*cos(f*x + e 
)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqr 
t(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e)) - 28*(d*cos( 
f*x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 
+ 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8))*sin(f*x + e) + 16*d^2*sqrt(b*s 
in(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(b^2*f*sin...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(3/2),x)
 

Output:

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {d}\, \sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}}d x \right ) d^{2}}{b^{2}} \] Input:

int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x)
 

Output:

(sqrt(d)*sqrt(b)*int((sqrt(tan(e + f*x))*sqrt(sec(e + f*x))*sec(e + f*x)** 
2)/tan(e + f*x)**2,x)*d**2)/b**2