3.2.83 \(\int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx\) [183]

3.2.83.1 Optimal result

Integrand size = 27, antiderivative size = 999 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=-\frac {\tan (e+f x)}{4 a^2 (c-d)^3 f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)}}-\frac {(c-4 d) \tan (e+f x)}{2 a^2 (c-d)^4 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}-\frac {3 \tan (e+f x)}{16 a^2 (c-d)^3 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {(c-4 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} a^{3/2} (c-d)^4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{16 \sqrt {2} a^{3/2} (c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} \left (c^2-5 c d+10 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{a^{3/2} (c-d)^5 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {3 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 a^{3/2} c (c-d)^3 (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(4 c-d) d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{a^{3/2} c^2 (c-d)^4 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 d^{7/2} \left (10 c^2-5 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{a^{3/2} c^3 (c-d)^5 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {d^4 \tan (e+f x)}{2 a^2 c (c-d)^3 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {3 d^4 \tan (e+f x)}{4 a^2 c (c-d)^3 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(4 c-d) d^4 \tan (e+f x)}{a^2 c^2 (c-d)^4 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]

output

-1/4*tan(f*x+e)/a^2/(c-d)^3/f/(1+sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)-1/2* 
(c-4*d)*tan(f*x+e)/a^2/(c-d)^4/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-3/1 
6*tan(f*x+e)/a^2/(c-d)^3/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+1/2*d^4*t 
an(f*x+e)/a^2/c/(c-d)^3/(c+d)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)+ 
3/4*d^4*tan(f*x+e)/a^2/c/(c-d)^3/(c+d)^2/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e 
))^(1/2)+(4*c-d)*d^4*tan(f*x+e)/a^2/c^2/(c-d)^4/(c+d)/f/(c+d*sec(f*x+e))/( 
a+a*sec(f*x+e))^(1/2)+2*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e) 
/a^(3/2)/c^3/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+3/4*d^(7/2)*a 
rctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/a^(3 
/2)/c/(c-d)^3/(c+d)^(5/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+ 
(4*c-d)*d^(7/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2) 
)*tan(f*x+e)/a^(3/2)/c^2/(c-d)^4/(c+d)^(3/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a 
*sec(f*x+e))^(1/2)-1/4*(c-4*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/ 
a^(1/2))*tan(f*x+e)/a^(3/2)/(c-d)^4/f*2^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a* 
sec(f*x+e))^(1/2)-3/32*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2)) 
*tan(f*x+e)/a^(3/2)/(c-d)^3/f*2^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+ 
e))^(1/2)-(c^2-5*c*d+10*d^2)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^ 
(1/2))*2^(1/2)*tan(f*x+e)/a^(3/2)/(c-d)^5/f/(a-a*sec(f*x+e))^(1/2)/(a+a*se 
c(f*x+e))^(1/2)+2*d^(7/2)*(10*c^2-5*c*d+d^2)*arctanh(d^(1/2)*(a-a*sec(f*x+ 
e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/a^(3/2)/c^3/(c-d)^5/f/(c+d)^(...
 

3.2.83.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2904\) vs. \(2(999)=1998\).

Time = 22.08 (sec) , antiderivative size = 2904, normalized size of antiderivative = 2.91 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\text {Result too large to show} \]

input

Integrate[1/((a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^3),x]
 

output

(Cos[(e + f*x)/2]^5*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^6*((-3*(5*c^6 - 3* 
c^5*d - 21*c^4*d^2 - 13*c^3*d^3 - 28*c^2*d^4 - 12*c*d^5 + 8*d^6)*Sin[(e + 
f*x)/2])/(2*c^3*(-c + d)^4*(c + d)^2) - (4*d^6*Sin[(e + f*x)/2])/(c^3*(-c 
+ d)^3*(c + d)*(d + c*Cos[e + f*x])^2) + (Sec[(e + f*x)/2]^2*(19*c*Sin[(e 
+ f*x)/2] - 43*d*Sin[(e + f*x)/2]))/(4*(-c + d)^4) + (2*(-23*c^2*d^5*Sin[( 
e + f*x)/2] - 9*c*d^6*Sin[(e + f*x)/2] + 8*d^7*Sin[(e + f*x)/2]))/(c^3*(-c 
 + d)^4*(c + d)^2*(d + c*Cos[e + f*x])) + (Sec[(e + f*x)/2]^3*Tan[(e + f*x 
)/2])/(2*(-c + d)^3)))/(f*(a*(1 + Sec[e + f*x]))^(5/2)*(c + d*Sec[e + f*x] 
)^3) - ((c^3*(c + d)^2*(43*c^2 - 206*c*d + 355*d^2)*ArcSin[Tan[(e + f*x)/2 
]] - 32*Sqrt[2]*(c - d)^5*(c + d)^2*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f 
*x]/(1 + Cos[e + f*x])]] + (4*Sqrt[2]*d^(7/2)*(99*c^4 + 110*c^3*d - 5*c^2* 
d^2 - 20*c*d^3 + 8*d^4)*ArcTanh[(Sqrt[d]*Tan[(e + f*x)/2])/(Sqrt[-c - d]*S 
qrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])/Sqrt[-c - d])*Cos[(e + f*x)/2]^5*( 
d + c*Cos[e + f*x])^3*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2]*((-11*c^4*Sec[ 
(e + f*x)/2])/(8*(-c + d)^4*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f* 
x]]) + (45*c^3*d*Sec[(e + f*x)/2])/(8*(-c + d)^4*(c + d)^2*(d + c*Cos[e + 
f*x])*Sqrt[Sec[e + f*x]]) - (5*c^2*d^2*Sec[(e + f*x)/2])/(8*(-c + d)^4*(c 
+ d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) - (317*c*d^3*Sec[(e + f*x) 
/2])/(8*(-c + d)^4*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) - (6 
9*d^4*Sec[(e + f*x)/2])/(2*(-c + d)^4*(c + d)^2*(d + c*Cos[e + f*x])*Sq...
 

3.2.83.3 Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 729, normalized size of antiderivative = 0.73, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 4428, 27, 198, 2009}

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.

\(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{5/2} (c+d \sec (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 4428

\(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a^3 (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\)

\(\Big \downarrow \) 198

\(\displaystyle -\frac {\tan (e+f x) \int \left (\frac {\left (10 c^2-5 d c+d^2\right ) d^4}{c^3 (c-d)^5 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(4 c-d) d^4}{c^2 (c-d)^4 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {d^4}{c (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}+\frac {\cos (e+f x)}{c^3 \sqrt {a-a \sec (e+f x)}}+\frac {-c^2+5 d c-10 d^2}{(c-d)^5 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}+\frac {4 d-c}{(c-d)^4 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d)^3 (\sec (e+f x)+1)^3 \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\tan (e+f x) \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^3}-\frac {d^{7/2} (4 c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 (c-d)^4 (c+d)^{3/2}}+\frac {\sqrt {2} \left (c^2-5 c d+10 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^5}-\frac {2 d^{7/2} \left (10 c^2-5 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^3 (c-d)^5 \sqrt {c+d}}-\frac {3 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{4 \sqrt {a} c (c-d)^3 (c+d)^{5/2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} \sqrt {a} (c-d)^3}+\frac {(c-4 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} (c-d)^4}-\frac {d^4 (4 c-d) \sqrt {a-a \sec (e+f x)}}{a c^2 (c-d)^4 (c+d) (c+d \sec (e+f x))}-\frac {3 d^4 \sqrt {a-a \sec (e+f x)}}{4 a c (c-d)^3 (c+d)^2 (c+d \sec (e+f x))}-\frac {d^4 \sqrt {a-a \sec (e+f x)}}{2 a c (c-d)^3 (c+d) (c+d \sec (e+f x))^2}+\frac {3 \sqrt {a-a \sec (e+f x)}}{16 a (c-d)^3 (\sec (e+f x)+1)}+\frac {(c-4 d) \sqrt {a-a \sec (e+f x)}}{2 a (c-d)^4 (\sec (e+f x)+1)}+\frac {\sqrt {a-a \sec (e+f x)}}{4 a (c-d)^3 (\sec (e+f x)+1)^2}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\)

input

Int[1/((a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^3),x]
 

output

-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^3) + ((c - 4 
*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*Sqrt[a 
]*(c - d)^4) + (3*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(16 
*Sqrt[2]*Sqrt[a]*(c - d)^3) + (Sqrt[2]*(c^2 - 5*c*d + 10*d^2)*ArcTanh[Sqrt 
[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]*(c - d)^5) - (3*d^(7/2)* 
ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(4*Sqrt 
[a]*c*(c - d)^3*(c + d)^(5/2)) - ((4*c - d)*d^(7/2)*ArcTanh[(Sqrt[d]*Sqrt[ 
a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^2*(c - d)^4*(c + d 
)^(3/2)) - (2*d^(7/2)*(10*c^2 - 5*c*d + d^2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*S 
ec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^3*(c - d)^5*Sqrt[c + d]) 
+ Sqrt[a - a*Sec[e + f*x]]/(4*a*(c - d)^3*(1 + Sec[e + f*x])^2) + ((c - 4* 
d)*Sqrt[a - a*Sec[e + f*x]])/(2*a*(c - d)^4*(1 + Sec[e + f*x])) + (3*Sqrt[ 
a - a*Sec[e + f*x]])/(16*a*(c - d)^3*(1 + Sec[e + f*x])) - (d^4*Sqrt[a - a 
*Sec[e + f*x]])/(2*a*c*(c - d)^3*(c + d)*(c + d*Sec[e + f*x])^2) - (3*d^4* 
Sqrt[a - a*Sec[e + f*x]])/(4*a*c*(c - d)^3*(c + d)^2*(c + d*Sec[e + f*x])) 
 - ((4*c - d)*d^4*Sqrt[a - a*Sec[e + f*x]])/(a*c^2*(c - d)^4*(c + d)*(c + 
d*Sec[e + f*x])))*Tan[e + f*x])/(a*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*S 
ec[e + f*x]]))
 

3.2.83.3.1 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 198

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_))^(q_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c 
 + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, 
 m, n}, x] && IntegersQ[p, q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4428

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e 
 + f*x]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(a + b*x)^(m - 1/2)*((c + d 
*x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, 
 f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0 
] && IntegerQ[m - 1/2]
 

3.2.83.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(139612\) vs. \(2(868)=1736\).

Time = 23.72 (sec) , antiderivative size = 139613, normalized size of antiderivative = 139.75

method	result	size
default	\(\text {Expression too large to display}\)	\(139613\)

input

int(1/(a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))^3,x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.2.83.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]

input

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))^3,x, algorithm="fricas 
")
 

output

Timed out
 

3.2.83.6 Sympy [F]

\[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]

input

integrate(1/(a+a*sec(f*x+e))**(5/2)/(c+d*sec(f*x+e))**3,x)
 

output

Integral(1/((a*(sec(e + f*x) + 1))**(5/2)*(c + d*sec(e + f*x))**3), x)
 

3.2.83.7 Maxima [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]

input

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))^3,x, algorithm="maxima 
")
 

output

Timed out
 

3.2.83.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))^3,x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

3.2.83.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3} \, dx=\text {Hanged} \]

input

int(1/((a + a/cos(e + f*x))^(5/2)*(c + d/cos(e + f*x))^3),x)
 

output

\text{Hanged}