3.3.26 \(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [226]

3.3.26.1 Optimal result
3.3.26.2 Mathematica [A] (verified)
3.3.26.3 Rubi [A] (verified)
3.3.26.4 Maple [B] (verified)
3.3.26.5 Fricas [A] (verification not implemented)
3.3.26.6 Sympy [F(-1)]
3.3.26.7 Maxima [B] (verification not implemented)
3.3.26.8 Giac [F]
3.3.26.9 Mupad [F(-1)]

3.3.26.1 Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {7 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {7 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \]

output
7/4*a^(3/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+7/4*a^2*s 
ec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*a^2*sec(d*x+c)^(5/ 
2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 
3.3.26.2 Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {a^2 \left (7 \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+7 \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2),x]
 
output
(a^2*(7*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 2*Sqrt[1 - Sec[c + d*x]]*Sec[c + 
d*x]^(3/2) + 7*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Tan[c + d*x])/(4 
*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.3.26.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4301, 27, 3042, 4290, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

\(\displaystyle \frac {1}{2} a \int \frac {7}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7}{4} a \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{4} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {7}{4} a \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{4} a \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {7}{4} a \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {7}{4} a \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\)

input
Int[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2),x]
 
output
(a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (7* 
a*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + 
(a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4
 

3.3.26.3.1 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 4290
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
 

rule 4301
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*(m + n - 1))), x] + Simp[b/(m + n - 1)   Int[(a + 
 b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n 
 - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 
2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
 
3.3.26.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(100)=200\).

Time = 1.57 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.72

method result size
default \(\frac {a \left (14 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )-7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-7 \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+4 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(206\)

input
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
1/8/d*a*(14*(-1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*cos(d*x+c)-7*cos(d*x+c)^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))-7*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))*cos(d*x+c)^2+4*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2))*(a*(1+s 
ec(d*x+c)))^(1/2)*sec(d*x+c)^(3/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2 
)
 
3.3.26.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.08 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {7 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (7 \, a \cos \left (d x + c\right ) + 2 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {7 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (7 \, a \cos \left (d x + c\right ) + 2 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]

input
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 
output
[1/16*(7*(a*cos(d*x + c)^2 + a*cos(d*x + c))*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*(7*a*cos(d*x + c) + 2*a)*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)^2 + d*cos(d*x + c)), 1/8*(7*(a*cos(d*x + c)^2 + a*cos(d*x + c))*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*(7*a*cos 
(d*x + c) + 2*a)*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)^2 + d*cos(d*x + c))]
 
3.3.26.6 Sympy [F(-1)]

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

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

Leaf count of result is larger than twice the leaf count of optimal. 2244 vs. \(2 (100) = 200\).

Time = 0.45 (sec) , antiderivative size = 2244, normalized size of antiderivative = 18.70 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 
output
-1/16*(56*sqrt(2)*a*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/ 
2*c)))*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 24*s 
qrt(2)*a*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin( 
4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*sqrt(2)*a*si 
n(3/2*d*x + 3/2*c) + 28*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co 
s(3/2*d*x + 3/2*c))) - 4*(3*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 7*sqrt(2)*a*s 
in(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*a* 
sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 7*sqrt(2)*a 
*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(8/3*arc 
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*(3*sqrt(2)*a*sin(3/2 
*d*x + 3/2*c) - 7*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
)) - 7*(a*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 
 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a*si 
n(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 4*a*sin(8/3 
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2(sin( 
3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*a*sin(4/3*arctan2(sin(3/2*d*x 
 + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*(2*a*cos(4/3*arctan2(sin(3/2*d*x + 
 3/2*c), cos(3/2*d*x + 3/2*c))) + a)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), 
 cos(3/2*d*x + 3/2*c))) + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos...
 
3.3.26.8 Giac [F]

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

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

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

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