\(\int \frac {x^{3/2}}{a+b \sec (c+d \sqrt {x})} \, dx\) [61]

Optimal result

Integrand size = 22, antiderivative size = 653 \[ \int \frac {x^{3/2}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{5/2}}{5 a}+\frac {2 i b x^2 \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x^2 \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {24 i b x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {24 i b x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}-\frac {48 i b \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5}+\frac {48 i b \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5} \] Output:

2/5*x^(5/2)/a+2*I*b*x^2*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/ 
a/(-a^2+b^2)^(1/2)/d-2*I*b*x^2*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^( 
1/2)))/a/(-a^2+b^2)^(1/2)/d+8*b*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/ 
(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2-8*b*x^(3/2)*polylog(2,-a*exp( 
I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2+24*I*b*x*pol 
ylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^ 
3-24*I*b*x*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2 
+b^2)^(1/2)/d^3-48*b*x^(1/2)*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^ 
2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^4+48*b*x^(1/2)*polylog(4,-a*exp(I*(c+d*x^( 
1/2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^4-48*I*b*polylog(5,-a*ex 
p(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^5+48*I*b*pol 
ylog(5,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^ 
5
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 513, normalized size of antiderivative = 0.79 \[ \int \frac {x^{3/2}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {-a^2+b^2} d^5 x^{5/2}+5 i b d^4 x^2 \log \left (1-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-5 i b d^4 x^2 \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+60 i b d^2 x \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-60 i b d^2 x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )-120 b d \sqrt {x} \operatorname {PolyLog}\left (4,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )+120 b d \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )-120 i b \operatorname {PolyLog}\left (5,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )+120 i b \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )\right )}{5 a \sqrt {-a^2+b^2} d^5} \] Input:

Integrate[x^(3/2)/(a + b*Sec[c + d*Sqrt[x]]),x]
 

Output:

(2*(Sqrt[-a^2 + b^2]*d^5*x^(5/2) + (5*I)*b*d^4*x^2*Log[1 - (a*E^(I*(c + d* 
Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - (5*I)*b*d^4*x^2*Log[1 + (a*E^(I*(c + 
 d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])] + 20*b*d^3*x^(3/2)*PolyLog[2, (a*E^( 
I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - 20*b*d^3*x^(3/2)*PolyLog[2, 
 -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))] + (60*I)*b*d^2*x*Pol 
yLog[3, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - (60*I)*b*d^2* 
x*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))] - 120*b* 
d*Sqrt[x]*PolyLog[4, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] + 
120*b*d*Sqrt[x]*PolyLog[4, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^ 
2]))] - (120*I)*b*PolyLog[5, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b 
^2])] + (120*I)*b*PolyLog[5, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + 
b^2]))]))/(5*a*Sqrt[-a^2 + b^2]*d^5)
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 655, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4692, 3042, 4679, 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.

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

\(\Big \downarrow \) 4692

\(\displaystyle 2 \int \frac {x^2}{a+b \sec \left (c+d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x^2}{a+b \csc \left (c+d \sqrt {x}+\frac {\pi }{2}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (\frac {x^2}{a}-\frac {b x^2}{a \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {24 i b \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}+\frac {24 i b \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}-\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {12 i b x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {12 i b x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^{5/2}}{5 a}\right )\)

Input:

Int[x^(3/2)/(a + b*Sec[c + d*Sqrt[x]]),x]
 

Output:

2*(x^(5/2)/(5*a) + (I*b*x^2*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a 
^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (I*b*x^2*Log[1 + (a*E^(I*(c + d*Sqrt 
[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (4*b*x^(3/2)*Poly 
Log[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 
+ b^2]*d^2) - (4*b*x^(3/2)*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqr 
t[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + ((12*I)*b*x*PolyLog[3, -((a*E 
^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - 
 ((12*I)*b*x*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]) 
)])/(a*Sqrt[-a^2 + b^2]*d^3) - (24*b*Sqrt[x]*PolyLog[4, -((a*E^(I*(c + d*S 
qrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^4) + (24*b*Sqrt[ 
x]*PolyLog[4, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqr 
t[-a^2 + b^2]*d^4) - ((24*I)*b*PolyLog[5, -((a*E^(I*(c + d*Sqrt[x])))/(b - 
 Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^5) + ((24*I)*b*PolyLog[5, -((a 
*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^5) 
)
 

Defintions of rubi rules used

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 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4692

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

Maple [F]

\[\int \frac {x^{\frac {3}{2}}}{a +b \sec \left (c +d \sqrt {x}\right )}d x\]

Input:

int(x^(3/2)/(a+b*sec(c+d*x^(1/2))),x)
 

Output:

int(x^(3/2)/(a+b*sec(c+d*x^(1/2))),x)
 

Fricas [F]

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

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

Output:

integral(x^(3/2)/(b*sec(d*sqrt(x) + c) + a), x)
 

Sympy [F]

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

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

Output:

Integral(x**(3/2)/(a + b*sec(c + d*sqrt(x))), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{3/2}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

integrate(x^(3/2)/(b*sec(d*sqrt(x) + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^{3/2}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 \sqrt {x}\, x^{2}+10 \left (\int \frac {\sqrt {x}\, \tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} x}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) a b +10 \left (\int \frac {\sqrt {x}\, \tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} x}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) b^{2}}{5 a +5 b} \] Input:

int(x^(3/2)/(a+b*sec(c+d*x^(1/2))),x)
 

Output:

(2*(sqrt(x)*x**2 + 5*int((sqrt(x)*tan((sqrt(x)*d + c)/2)**2*x)/(tan((sqrt( 
x)*d + c)/2)**2*a**2 - tan((sqrt(x)*d + c)/2)**2*b**2 - a**2 - 2*a*b - b** 
2),x)*a*b + 5*int((sqrt(x)*tan((sqrt(x)*d + c)/2)**2*x)/(tan((sqrt(x)*d + 
c)/2)**2*a**2 - tan((sqrt(x)*d + c)/2)**2*b**2 - a**2 - 2*a*b - b**2),x)*b 
**2))/(5*(a + b))