3.2.75 \(\int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))} \, dx\) [175]

3.2.75.1 Optimal result

Integrand size = 27, antiderivative size = 394 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))} \, dx=-\frac {\tan (e+f x)}{2 a (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-2 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} \sqrt {a} (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c (c-d)^2 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]

output

-1/2*tan(f*x+e)/a/(c-d)/f/(1+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)/c/f/a^(1/2)/(a-a*sec(f*x+e))^(1 
/2)/(a+a*sec(f*x+e))^(1/2)-1/4*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/ 
a^(1/2))*tan(f*x+e)/(c-d)/f*2^(1/2)/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*se 
c(f*x+e))^(1/2)-(c-2*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2) 
)*2^(1/2)*tan(f*x+e)/(c-d)^2/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x 
+e))^(1/2)-2*d^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^ 
(1/2))*tan(f*x+e)/c/(c-d)^2/f/a^(1/2)/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/( 
a+a*sec(f*x+e))^(1/2)
 

3.2.75.2 Mathematica [A] (warning: unable to verify)

Time = 6.10 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))} \, dx=\frac {\cos ^2\left (\frac {1}{2} (e+f x)\right ) (d+c \cos (e+f x)) \sec ^{\frac {5}{2}}(e+f x) \left (\frac {\left (-c (5 c-9 d) \sqrt {c+d} \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )+4 \sqrt {2} \left ((c-d)^2 \sqrt {c+d} \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )-d^{5/2} \arctan \left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right )\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {1+\sec (e+f x)}}{c \sqrt {c+d} \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}+(c-d) \sqrt {\sec (e+f x)} \left (-\sin (e+f x)+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{(c-d)^2 f (a (1+\sec (e+f x)))^{3/2} (c+d \sec (e+f x))} \]

input

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

output

(Cos[(e + f*x)/2]^2*(d + c*Cos[e + f*x])*Sec[e + f*x]^(5/2)*(((-(c*(5*c - 
9*d)*Sqrt[c + d]*ArcSin[Tan[(e + f*x)/2]]) + 4*Sqrt[2]*((c - d)^2*Sqrt[c + 
 d]*ArcTan[Tan[(e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] - d^(5/ 
2)*ArcTan[(Sqrt[d]*Tan[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]/(1 + C 
os[e + f*x])])]))*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]*Sqrt[1 + Sec[e + f 
*x]])/(c*Sqrt[c + d]*Sqrt[Sec[(e + f*x)/2]^2]) + (c - d)*Sqrt[Sec[e + f*x] 
]*(-Sin[e + f*x] + Tan[(e + f*x)/2])))/((c - d)^2*f*(a*(1 + Sec[e + f*x])) 
^(3/2)*(c + d*Sec[e + f*x]))
 

3.2.75.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.69, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4428

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 198

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

\(\Big \downarrow \) 2009

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

input

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

output

-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c) + (Sqrt[2]* 
(c - 2*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]*(c 
 - d)^2) + ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]* 
Sqrt[a]*(c - d)) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/( 
Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c*(c - d)^2*Sqrt[c + d]) + Sqrt[a - a*Sec[ 
e + f*x]]/(2*a*(c - d)*(1 + Sec[e + f*x])))*Tan[e + f*x])/(f*Sqrt[a - a*Se 
c[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
 

3.2.75.3.1 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]
 

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

Leaf count of result is larger than twice the leaf count of optimal. \(1418\) vs. \(2(334)=668\).

Time = 16.33 (sec) , antiderivative size = 1419, normalized size of antiderivative = 3.60

method	result	size
default	\(\text {Expression too large to display}\)	\(1419\)

input

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

output

1/16/f/(d/(c-d))^(1/2)/(c-d)^2/c/((c+d)*(c-d))^(1/2)/a^2*(((1-cos(f*x+e))^ 
2*csc(f*x+e)^2-1)^(3/2)*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c^2*(-cot(f*x+ 
e)+csc(f*x+e))-2*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(3/2)*((c+d)*(c-d))^(1/ 
2)*(d/(c-d))^(1/2)*c*d*(-cot(f*x+e)+csc(f*x+e))+((1-cos(f*x+e))^2*csc(f*x+ 
e)^2-1)^(3/2)*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*d^2*(-cot(f*x+e)+csc(f*x 
+e))-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*((c+d)*(c-d))^(1/2)*(d/(c-d)) 
^(1/2)*c^2*(1-cos(f*x+e))^3*csc(f*x+e)^3+2*((1-cos(f*x+e))^2*csc(f*x+e)^2- 
1)^(1/2)*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c*d*(1-cos(f*x+e))^3*csc(f*x+ 
e)^3-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*((c+d)*(c-d))^(1/2)*(d/(c-d)) 
^(1/2)*d^2*(1-cos(f*x+e))^3*csc(f*x+e)^3+5*((1-cos(f*x+e))^2*csc(f*x+e)^2- 
1)^(1/2)*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c^2*(-cot(f*x+e)+csc(f*x+e))- 
6*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1 
/2)*c*d*(-cot(f*x+e)+csc(f*x+e))+((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*( 
(c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*d^2*(-cot(f*x+e)+csc(f*x+e))+16*((c+d)* 
(c-d))^(1/2)*2^(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/ 
2)*(-cot(f*x+e)+csc(f*x+e)))*(d/(c-d))^(1/2)*c^2-32*((c+d)*(c-d))^(1/2)*2^ 
(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(-cot(f*x+e) 
+csc(f*x+e)))*(d/(c-d))^(1/2)*c*d+16*((c+d)*(c-d))^(1/2)*2^(1/2)*arctanh(2 
^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(-cot(f*x+e)+csc(f*x+e)))*( 
d/(c-d))^(1/2)*d^2-20*((c+d)*(c-d))^(1/2)*ln(csc(f*x+e)-cot(f*x+e)+((1-...
 

3.2.75.5 Fricas [A] (verification not implemented)

Time = 215.92 (sec) , antiderivative size = 2033, normalized size of antiderivative = 5.16 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))} \, dx=\text {Too large to display} \]

input

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

output

[-1/8*(4*(c^2 - c*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)* 
sin(f*x + e) - sqrt(2)*((5*c^2 - 9*c*d)*cos(f*x + e)^2 + 5*c^2 - 9*c*d + 2 
*(5*c^2 - 9*c*d)*cos(f*x + e))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*co 
s(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 3*a*cos(f*x + e) 
^2 + 2*a*cos(f*x + e) - a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 8*(a*d 
^2*cos(f*x + e)^2 + 2*a*d^2*cos(f*x + e) + a*d^2)*sqrt(-d/(a*c + a*d))*log 
((2*(c + d)*sqrt(-d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c 
os(f*x + e)*sin(f*x + e) + (c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*x + e) 
 - d)/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + 8*((c^2 - 2*c*d + d 
^2)*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + 2*(c^2 - 2*c*d + d^2)*cos(f*x + e 
))*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a) 
/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + 
e) + 1)))/((a^2*c^3 - 2*a^2*c^2*d + a^2*c*d^2)*f*cos(f*x + e)^2 + 2*(a^2*c 
^3 - 2*a^2*c^2*d + a^2*c*d^2)*f*cos(f*x + e) + (a^2*c^3 - 2*a^2*c^2*d + a^ 
2*c*d^2)*f), -1/8*(4*(c^2 - c*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c 
os(f*x + e)*sin(f*x + e) - sqrt(2)*((5*c^2 - 9*c*d)*cos(f*x + e)^2 + 5*c^2 
 - 9*c*d + 2*(5*c^2 - 9*c*d)*cos(f*x + e))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a 
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 3*a* 
cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 
1)) - 16*(a*d^2*cos(f*x + e)^2 + 2*a*d^2*cos(f*x + e) + a*d^2)*sqrt(d/(...
 

3.2.75.6 Sympy [F]

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

input

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

output

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

3.2.75.7 Maxima [F]

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

input

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

output

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

3.2.75.8 Giac [F(-2)]

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

input

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

3.2.75.9 Mupad [F(-1)]

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

input

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

output

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