\(\int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx\) [311]

Optimal result

Integrand size = 25, antiderivative size = 168 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \] Output:

-b^(5/2)*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/d/f/(d* 
sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)+b^(5/2)*arctanh((b*sin(f*x+e))^(1/2 
)/b^(1/2))*(b*tan(f*x+e))^(1/2)/d/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1 
/2)-2/3*b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {\sqrt {d \sec (e+f x)} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )-3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\sqrt [4]{\sec ^2(e+f x)} \sin (2 (e+f x)) \sqrt {\tan (e+f x)}\right ) (b \tan (e+f x))^{5/2}}{3 d^2 f \sqrt [4]{\sec ^2(e+f x)} \tan ^{\frac {5}{2}}(e+f x)} \] Input:

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

Output:

-1/3*(Sqrt[d*Sec[e + f*x]]*(3*ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^( 
1/4)] - 3*ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)] + (Sec[e + f* 
x]^2)^(1/4)*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*(b*Tan[e + f*x])^(5/2))/( 
d^2*f*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.74, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3090, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3090

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3096

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 3090

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)) 
, x] - Simp[b^2*((n - 1)/(a^2*m))   Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + 
 f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[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 = 5.66 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\left (\frac {\operatorname {arctanh}\left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (-3 \cot \left (f x +e \right )-3 \csc \left (f x +e \right )\right )}{3}+\frac {\arctan \left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (-3 \cot \left (f x +e \right )-3 \csc \left (f x +e \right )\right )}{3}-\frac {2 \sin \left (f x +e \right )}{3}\right ) b^{2} \sqrt {b \tan \left (f x +e \right )}}{f \sqrt {d \sec \left (f x +e \right )}\, d}\)	\(190\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (138) = 276\).

Time = 0.58 (sec) , antiderivative size = 766, normalized size of antiderivative = 4.56 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/24*(16*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos( 
f*x + e)*sin(f*x + e) + 6*b^2*d*sqrt(-b/d)*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*co 
s(f*x + e) + 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/d)*sqrt(d/cos(f* 
x + e))/(b*cos(f*x + e)^2 - (b*cos(f*x + e) + b)*sin(f*x + e) - b)) - 3*b^ 
2*d*sqrt(-b/d)*log((b*cos(f*x + e)^4 - 72*b*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(-b/d)*sqrt(d/cos(f*x + e)) + 28*(b* 
cos(f*x + e)^2 - 2*b)*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f*x + e 
)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/(d^2*f), -1/24*(16*b^2*sq 
rt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)*sin(f*x 
+ e) + 6*b^2*d*sqrt(b/d)*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)*sq 
rt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*x 
+ e)^2 + (b*cos(f*x + e) + b)*sin(f*x + e) - b)) - 3*b^2*d*sqrt(b/d)*log(( 
b*cos(f*x + e)^4 - 72*b*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(b/d)*sqrt(d/cos(f*x + e)) - 28*(b*cos(f*x + e)^2 - 2*b) 
*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e) 
^2 - 2)*sin(f*x + e) + 8)))/(d^2*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[4,14]%%%}+%%%{6,[4,12]%%%}+%%%{15,[4,10]%%%}+%%%{20 
,[4,8]%%%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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