3.2.63 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx\) [163]

3.2.63.1 Optimal result

Integrand size = 27, antiderivative size = 203 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 a^{7/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c d^{3/2} \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]

output

2*a^3*tan(f*x+e)/d/f/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*arctanh((a-a*sec(f*x 
+e))^(1/2)/a^(1/2))*tan(f*x+e)/c/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e)) 
^(1/2)-2*a^(7/2)*(c-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)/c/d^(3/2)/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a* 
sec(f*x+e))^(1/2)
 

3.2.63.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.71 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.69 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\frac {\cos ^{\frac {3}{2}}(e+f x) (d+c \cos (e+f x)) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (1+\sec (e+f x)))^{5/2} \left (\frac {10 (c-d)^2 (c+3 d+2 c \cos (e+f x)) \csc \left (\frac {1}{2} (e+f x)\right ) \left (-\text {arctanh}\left (\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )+\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )}{d (c+d) \sqrt {\cos (e+f x)} \sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}}+\frac {20 (3 c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}-\frac {16 (c-d)^2 d (d+c \cos (e+f x)) \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\frac {2 d \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )}{c+d}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right )}{(c+d)^3 \cos ^{\frac {5}{2}}(e+f x)}+10 c \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right )-\frac {2 \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )\right )}{40 c^2 f (c+d \sec (e+f x))} \]

input

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

output

(Cos[e + f*x]^(3/2)*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^5*(a*(1 + Sec[e 
+ f*x]))^(5/2)*((10*(c - d)^2*(c + 3*d + 2*c*Cos[e + f*x])*Csc[(e + f*x)/2 
]*(-ArcTanh[Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]] + Sqrt[-((d*(-1 + Se 
c[e + f*x]))/(c + d))]))/(d*(c + d)*Sqrt[Cos[e + f*x]]*Sqrt[-((d*(-1 + Sec 
[e + f*x]))/(c + d))]) + (20*(3*c - d)*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x] 
] - (16*(c - d)^2*d*(d + c*Cos[e + f*x])*Hypergeometric2F1[2, 5/2, 7/2, (- 
2*d*Sec[e + f*x]*Sin[(e + f*x)/2]^2)/(c + d)]*Sin[(e + f*x)/2]^3)/((c + d) 
^3*Cos[e + f*x]^(5/2)) + 10*c*(Sqrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]] - 
(2*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x]])))/(40*c^2*f*(c + d*Sec[e + f*x]))
 

3.2.63.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.80, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4428

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 198

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

\(\Big \downarrow \) 2009

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

input

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

output

-((a^4*((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c) + (2*(c 
 - d)^2*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]) 
/(Sqrt[a]*c*d^(3/2)*Sqrt[c + d]) - (2*Sqrt[a - a*Sec[e + f*x]])/(a*d))*Tan 
[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
 

3.2.63.3.1 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]
 

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

Leaf count of result is larger than twice the leaf count of optimal. \(1401\) vs. \(2(173)=346\).

Time = 23.45 (sec) , antiderivative size = 1402, normalized size of antiderivative = 6.91

method	result	size
default	\(\text {Expression too large to display}\)	\(1402\)

input

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

output

1/2/f*a^2/c/((c+d)*(c-d))^(1/2)/d/(d/(c-d))^(1/2)*(2*((c+d)*(c-d))^(1/2)*2 
^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(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 
^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln(-2*(((1-cos(f*x+e))^2*cs 
c(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c-2^(1/2)*(d/(c-d))^(1/2)*((1- 
cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc 
(f*x+e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot(f*x+e)+csc(f*x+e))*d+((c+ 
d)*(c-d))^(1/2)))*c^2+2*2^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln 
(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c-2^( 
1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d 
))^(1/2)*(-cot(f*x+e)+csc(f*x+e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot( 
f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*c*d-2^(1/2)*((1-cos(f*x+e))^2*c 
sc(f*x+e)^2-1)^(1/2)*ln(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2 
)*(d/(c-d))^(1/2)*c-2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2 
-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))-c+d)/(-c*(-cot(f* 
x+e)+csc(f*x+e))+(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*d^2+2^(1 
/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln(-2*(-((1-cos(f*x+e))^2*csc( 
f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c+2^(1/2)*(d/(c-d))^(1/2)*((1-co 
s(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f 
*x+e))+c-d)/(c*(-cot(f*x+e)+csc(f*x+e))-(-cot(f*x+e)+csc(f*x+e))*d+((c+...
 

3.2.63.5 Fricas [A] (verification not implemented)

Time = 2.93 (sec) , antiderivative size = 1140, normalized size of antiderivative = 5.62 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\text {Too large to display} \]

input

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

output

[(2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + (a^2*c^2 
- 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt 
(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + 
 e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + 
 e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f* 
x + e) + d)) + (a^2*d*cos(f*x + e) + a^2*d)*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)))/(c*d*f*cos(f*x + e) + c 
*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 2*( 
a^2*d*cos(f*x + e) + a^2*d)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f 
*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (a^2*c^2 - 2*a^2*c*d + a^2 
*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(-a/(c*d + d^2))* 
log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x 
+ e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a* 
c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)))/(c* 
d*f*cos(f*x + e) + c*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e) 
)*sin(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + 
 a^2*d^2)*cos(f*x + e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqrt(a/(c*d + d 
^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*sin(f*x + e)) 
) + (a^2*d*cos(f*x + e) + a^2*d)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*s...
 

3.2.63.6 Sympy [F]

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

input

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

output

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

3.2.63.7 Maxima [F]

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

input

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

output

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

3.2.63.8 Giac [F]

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

input

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

output

sage0*x
 

3.2.63.9 Mupad [F(-1)]

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

input

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

output

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