\(\int \frac {\sqrt {x}}{a+b \sec (c+d \sqrt {x})} \, dx\) [62]

Optimal result

Integrand size = 22, antiderivative size = 393 \[ \int \frac {\sqrt {x}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{3/2}}{3 a}+\frac {2 i b x \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {4 i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \] Output:

2/3*x^(3/2)/a+2*I*b*x*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*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+4*b*x^(1/2)*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^2-4*b*x^(1/2)*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+4*I*b*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^3-4*I*b 
*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
 

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {x}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {-a^2+b^2} d^3 x^{3/2}+3 i b d^2 x \log \left (1-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-3 i b d^2 x \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 b d \sqrt {x} \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 b d \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 i b \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )\right )}{3 a \sqrt {-a^2+b^2} d^3} \] Input:

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

Output:

(2*(Sqrt[-a^2 + b^2]*d^3*x^(3/2) + (3*I)*b*d^2*x*Log[1 - (a*E^(I*(c + d*Sq 
rt[x])))/(-b + Sqrt[-a^2 + b^2])] - (3*I)*b*d^2*x*Log[1 + (a*E^(I*(c + d*S 
qrt[x])))/(b + Sqrt[-a^2 + b^2])] + 6*b*d*Sqrt[x]*PolyLog[2, (a*E^(I*(c + 
d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - 6*b*d*Sqrt[x]*PolyLog[2, -((a*E^(I 
*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))] + (6*I)*b*PolyLog[3, (a*E^(I*( 
c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - (6*I)*b*PolyLog[3, -((a*E^(I*( 
c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))]))/(3*a*Sqrt[-a^2 + b^2]*d^3)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2*(x^(3/2)/(3*a) + (I*b*x*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*Log[1 + (a*E^(I*(c + d*Sqrt[x]) 
))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (2*b*Sqrt[x]*PolyLog[ 
2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^ 
2]*d^2) - (2*b*Sqrt[x]*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a 
^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + ((2*I)*b*PolyLog[3, -((a*E^(I*(c 
+ d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - ((2*I) 
*b*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqr 
t[-a^2 + b^2]*d^3))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^(1/2)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {x}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {2 \sqrt {x}\, x +6 \left (\int \frac {\sqrt {x}\, \tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) a b +6 \left (\int \frac {\sqrt {x}\, \tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) b^{2}}{3 a +3 b} \] Input:

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

Output:

(2*(sqrt(x)*x + 3*int((sqrt(x)*tan((sqrt(x)*d + c)/2)**2)/(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 + 3*int((sqrt(x)*tan((sqrt(x)*d + c)/2)**2)/(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))/( 
3*(a + b))