\(\int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx\) [328]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{b^{5/2} f \sqrt {b \tan (e+f x)}}+\frac {d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{b^{5/2} f \sqrt {b \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \cos (e+f x) (d \sec (e+f x))^{7/2} \sin (e+f x)}{3 f (b \tan (e+f x))^{5/2}}+\frac {\left (\arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )\right ) \cos ^2(e+f x) (d \sec (e+f x))^{7/2} \tan ^{\frac {5}{2}}(e+f x)}{f \sec ^2(e+f x)^{3/4} (b \tan (e+f x))^{5/2}} \] Input:

Integrate[(d*Sec[e + f*x])^(7/2)/(b*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*Cos[e + f*x]*(d*Sec[e + f*x])^(7/2)*Sin[e + f*x])/(3*f*(b*Tan[e + f*x] 
)^(5/2)) + ((ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)] + ArcTanh[S 
qrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)])*Cos[e + f*x]^2*(d*Sec[e + f*x]) 
^(7/2)*Tan[e + f*x]^(5/2))/(f*(Sec[e + f*x]^2)^(3/4)*(b*Tan[e + f*x])^(5/2 
))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76, 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, 756, 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.

\(\displaystyle \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3088

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3096

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[(d*Sec[e + f*x])^(7/2)/(b*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*d^2*(d*Sec[e + f*x])^(3/2))/(3*b*f*(b*Tan[e + f*x])^(3/2)) + (2*d^3*(A 
rcTan[Sqrt[b]*Sin[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Sin[e + f*x]]/(2 
*b^(3/2)))*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(b*f*Sqrt[b*Tan[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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   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 = 6.11 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\csc \left (f x +e \right ) \left (\left (3 \cos \left (f x +e \right )-3\right ) \operatorname {arctanh}\left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )+\left (-3 \cos \left (f x +e \right )+3\right ) \arctan \left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )-2 \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) d^{3} \sqrt {d \sec \left (f x +e \right )}}{3 f \sqrt {b \tan \left (f x +e \right )}\, \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}}\)	\(175\)

Input:

int((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (142) = 284\).

Time = 0.33 (sec) , antiderivative size = 850, normalized size of antiderivative = 4.94 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

[1/24*(16*d^3*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f 
*x + e) - 6*(b*d^3*cos(f*x + e)^2 - b*d^3)*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)*sq 
rt(d/cos(f*x + e))/(d*cos(f*x + e)^2 - (d*cos(f*x + e) + d)*sin(f*x + e) - 
 d)) + 3*(b*d^3*cos(f*x + e)^2 - b*d^3)*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))*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) + 7 
2*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + 
 e) + 8)))/(b^3*f*cos(f*x + e)^2 - b^3*f), 1/24*(16*d^3*sqrt(b*sin(f*x + e 
)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e) + 6*(b*d^3*cos(f*x + e)^ 
2 - b*d^3)*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*cos(f*x + e) 
^2 + (d*cos(f*x + e) + d)*sin(f*x + e) - d)) + 3*(b*d^3*cos(f*x + e)^2 - b 
*d^3)*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)) 
*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 ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(7/2)/(b*tan(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e))^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((d/cos(e + f*x))^(7/2)/(b*tan(e + f*x))^(5/2),x)
 

Output:

int((d/cos(e + f*x))^(7/2)/(b*tan(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3}}d x \right ) d^{3}}{b^{3}} \] Input:

int((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),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)**3,x)*d**3)/b**3