3.3.37 \(\int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [237]

3.3.37.1 Optimal result
3.3.37.2 Mathematica [A] (verified)
3.3.37.3 Rubi [A] (verified)
3.3.37.4 Maple [B] (verified)
3.3.37.5 Fricas [A] (verification not implemented)
3.3.37.6 Sympy [F(-1)]
3.3.37.7 Maxima [B] (verification not implemented)
3.3.37.8 Giac [F]
3.3.37.9 Mupad [F(-1)]

3.3.37.1 Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a^{5/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {14 a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

output
2*a^(5/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+14/3*a^3*si 
n(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/3*a^2*sin(d*x+c)*(a+a 
*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)
 
3.3.37.2 Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (3 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x)+\sqrt {1-\sec (c+d x)} (1+8 \sec (c+d x))\right ) \sin (c+d x)}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(3/2),x]
 
output
(2*a^3*(3*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(3/2) + Sqrt[1 - Sec 
[c + d*x]]*(1 + 8*Sec[c + d*x]))*Sin[c + d*x])/(3*d*Sqrt[-((-1 + Sec[c + d 
*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.3.37.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4300, 27, 3042, 4503, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4300

\(\displaystyle \frac {2}{3} a \int \frac {\sqrt {\sec (c+d x) a+a} (3 \sec (c+d x) a+7 a)}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} a \int \frac {\sqrt {\sec (c+d x) a+a} (3 \sec (c+d x) a+7 a)}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 \csc \left (c+d x+\frac {\pi }{2}\right ) a+7 a\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{3} a \left (3 a \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {14 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (3 a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {14 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {1}{3} a \left (\frac {14 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} a \left (\frac {6 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {14 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\)

input
Int[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(3/2),x]
 
output
(2*a^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ( 
a*((6*a^(3/2)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d 
+ (14*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/ 
3
 

3.3.37.3.1 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 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4300
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[b^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* 
((d*Csc[e + f*x])^n/(f*n)), x] - Simp[a/(d*n)   Int[(a + b*Csc[e + f*x])^(m 
 - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f 
*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] 
 && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2*m]
 

rule 4503
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a 
*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
 
3.3.37.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(299\) vs. \(2(100)=200\).

Time = 1.82 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.54

method result size
default \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-3 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \sin \left (d x +c \right )-3 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )+3 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )+16 \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) \(300\)

input
int((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/3/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(-3*(-1 
/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1) 
/(-1/(cos(d*x+c)+1))^(1/2))+3*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d* 
x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+2*sin(d*x+c)- 
3*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+ 
c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*sec(d*x+c)+3*(-1/(cos(d*x+c)+1))^(1/2)*ar 
ctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2 
))*sec(d*x+c)+16*tan(d*x+c))
 
3.3.37.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.08 \[ \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

input
integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")
 
output
[1/6*(3*(a^2*cos(d*x + c) + a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d 
*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c) 
^3 + cos(d*x + c)^2)) + 4*(a^2*cos(d*x + c)^2 + 8*a^2*cos(d*x + c))*sqrt(( 
a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos( 
d*x + c) + d), 1/3*(3*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*arctan(2*sqrt(-a)* 
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a 
*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(a^2*cos(d*x + c)^2 + 8*a^2*c 
os(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos 
(d*x + c)))/(d*cos(d*x + c) + d)]
 
3.3.37.6 Sympy [F(-1)]

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

input
integrate((a+a*sec(d*x+c))**(5/2)/sec(d*x+c)**(3/2),x)
 
output
Timed out
 
3.3.37.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (100) = 200\).

Time = 0.41 (sec) , antiderivative size = 593, normalized size of antiderivative = 5.03 \[ \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

input
integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")
 
output
1/12*sqrt(2)*(30*a^2*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 
/2*c)))*sin(3/2*d*x + 3/2*c) - 30*a^2*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2 
(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*log(2*cos(1/ 
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arcta 
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arct 
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arcta 
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqrt(2)*a^2*log(2 
*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/ 
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1 
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/ 
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 3*sqrt(2)*a^ 
2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2 
*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2 
)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2) 
*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqr 
t(2)*a^2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) 
)^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2 
*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2* 
sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) 
+ 4*a^2*sin(3/2*d*x + 3/2*c) + 30*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2...
 
3.3.37.8 Giac [F]

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

input
integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="giac")
 
output
sage0*x
 
3.3.37.9 Mupad [F(-1)]

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

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