\(\int \frac {\sqrt {x}}{(a+b \sec (c+d \sqrt {x}))^2} \, dx\) [67]

Optimal result

Integrand size = 22, antiderivative size = 1125 \[ \int \frac {\sqrt {x}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Output:

-8*I*b*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b 
^2)^(1/2)/d^3+2/3*x^(3/2)/a^2+4*b^2*x^(1/2)*ln(1+a*exp(I*(c+d*x^(1/2)))/(b 
-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^2+4*b^2*x^(1/2)*ln(1+a*exp(I*(c+d*x^( 
1/2)))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^2+4*I*b*x*ln(1+a*exp(I*(c+d* 
x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d-4*I*b^2*polylog(2,- 
a*exp(I*(c+d*x^(1/2)))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^3+4*I*b^3*po 
lylog(3,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2) 
/d^3-2*I*b^3*x*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2 
+b^2)^(3/2)/d-4*I*b*x*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^ 
2/(-a^2+b^2)^(1/2)/d-4*I*b^2*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b-I*(a^2-b 
^2)^(1/2)))/a^2/(a^2-b^2)/d^3-4*b^3*x^(1/2)*polylog(2,-a*exp(I*(c+d*x^(1/2 
)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2+8*b*x^(1/2)*polylog(2,- 
a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2+4*b^ 
3*x^(1/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^ 
2+b^2)^(3/2)/d^2-8*b*x^(1/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^ 
2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2-4*I*b^3*polylog(3,-a*exp(I*(c+d*x^(1/2 
)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^3-2*I*b^2*x/a^2/(a^2-b^2) 
/d+8*I*b*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2 
+b^2)^(1/2)/d^3+2*I*b^3*x*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)) 
)/a^2/(-a^2+b^2)^(3/2)/d+2*b^2*x*sin(c+d*x^(1/2))/a/(a^2-b^2)/d/(b+a*co...
 

Mathematica [A] (verified)

Time = 6.03 (sec) , antiderivative size = 1210, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {x}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(b + a*Cos[c + d*Sqrt[x]])*Sec[c + d*Sqrt[x]]^2*(x^(3/2)*(b + a*Cos[c + 
 d*Sqrt[x]]) + (3*b*(b + a*Cos[c + d*Sqrt[x]])*(((-2*I)*b*d^2*E^((2*I)*c)* 
x)/(1 + E^((2*I)*c)) + (2*b*d*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*Sqrt[x]*Log[1 
 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)] 
)] + (2*I)*a^2*d^2*E^(I*c)*x*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c 
) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - I*b^2*d^2*E^(I*c)*x*Log[1 + (a*E^(I 
*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*b*d 
*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*Sqrt[x]*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])) 
)/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (2*I)*a^2*d^2*E^(I*c)*x* 
Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I 
)*c)])] + I*b^2*d^2*E^(I*c)*x*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I* 
c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*((-I)*b*Sqrt[(-a^2 + b^2)*E^((2* 
I)*c)] + 2*a^2*d*E^(I*c)*Sqrt[x] - b^2*d*E^(I*c)*Sqrt[x])*PolyLog[2, -((a* 
E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 
2*((-I)*b*Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - 2*a^2*d*E^(I*c)*Sqrt[x] + b^2*d 
*E^(I*c)*Sqrt[x])*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sq 
rt[(-a^2 + b^2)*E^((2*I)*c)]))] + (4*I)*a^2*E^(I*c)*PolyLog[3, -((a*E^(I*( 
2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - (2*I)* 
b^2*E^(I*c)*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a 
^2 + b^2)*E^((2*I)*c)]))] - (4*I)*a^2*E^(I*c)*PolyLog[3, -((a*E^(I*(2*c...
 

Rubi [A] (verified)

Time = 2.29 (sec) , antiderivative size = 1126, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2*(((-I)*b^2*x)/(a^2*(a^2 - b^2)*d) + x^(3/2)/(3*a^2) + (2*b^2*Sqrt[x]*Log 
[1 + (a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2])])/(a^2*(a^2 - b^2)* 
d^2) + (2*b^2*Sqrt[x]*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 - 
b^2])])/(a^2*(a^2 - b^2)*d^2) - (I*b^3*x*Log[1 + (a*E^(I*(c + d*Sqrt[x]))) 
/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x*Log[1 + 
(a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d 
) + (I*b^3*x*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a 
^2*(-a^2 + b^2)^(3/2)*d) - ((2*I)*b*x*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b 
 + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) - ((2*I)*b^2*PolyLog[2, -( 
(a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^3) 
 - ((2*I)*b^2*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 - b^2 
]))])/(a^2*(a^2 - b^2)*d^3) - (2*b^3*Sqrt[x]*PolyLog[2, -((a*E^(I*(c + d*S 
qrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (4*b*Sq 
rt[x]*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a^ 
2*Sqrt[-a^2 + b^2]*d^2) + (2*b^3*Sqrt[x]*PolyLog[2, -((a*E^(I*(c + d*Sqrt[ 
x])))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) - (4*b*Sqrt[x 
]*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a^2*Sq 
rt[-a^2 + b^2]*d^2) - ((2*I)*b^3*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b 
 - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^3) + ((4*I)*b*PolyLog[3, 
 -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 +...
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {x}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {too large to display} \] Input:

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

Output:

(2*( - 24*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*a - tan((sqrt( 
x)*d + c)/2)*b)/sqrt( - a**2 + b**2))*cos(sqrt(x)*d + c)*a**2*b + 24*sqrt( 
 - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*a - tan((sqrt(x)*d + c)/2)*b) 
/sqrt( - a**2 + b**2))*cos(sqrt(x)*d + c)*a*b**2 - 24*sqrt( - a**2 + b**2) 
*atan((tan((sqrt(x)*d + c)/2)*a - tan((sqrt(x)*d + c)/2)*b)/sqrt( - a**2 + 
 b**2))*a*b**2 + 24*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*a - 
tan((sqrt(x)*d + c)/2)*b)/sqrt( - a**2 + b**2))*b**3 + 2*sqrt(x)*cos(sqrt( 
x)*d + c)*a**4*d**3*x + sqrt(x)*cos(sqrt(x)*d + c)*a**3*b*d**3*x - sqrt(x) 
*cos(sqrt(x)*d + c)*a**2*b**2*d**3*x - 24*cos(sqrt(x)*d + c)*int(sqrt(x)/( 
2*tan((sqrt(x)*d + c)/2)**4*a**5 - 7*tan((sqrt(x)*d + c)/2)**4*a**4*b + 8* 
tan((sqrt(x)*d + c)/2)**4*a**3*b**2 - 2*tan((sqrt(x)*d + c)/2)**4*a**2*b** 
3 - 2*tan((sqrt(x)*d + c)/2)**4*a*b**4 + tan((sqrt(x)*d + c)/2)**4*b**5 - 
4*tan((sqrt(x)*d + c)/2)**2*a**5 + 6*tan((sqrt(x)*d + c)/2)**2*a**4*b + 4* 
tan((sqrt(x)*d + c)/2)**2*a**3*b**2 - 8*tan((sqrt(x)*d + c)/2)**2*a**2*b** 
3 + 2*tan((sqrt(x)*d + c)/2)**2*b**5 + 2*a**5 + a**4*b - 4*a**3*b**2 - 2*a 
**2*b**3 + 2*a*b**4 + b**5),x)*a**8*b*d**3 + 12*cos(sqrt(x)*d + c)*int(sqr 
t(x)/(2*tan((sqrt(x)*d + c)/2)**4*a**5 - 7*tan((sqrt(x)*d + c)/2)**4*a**4* 
b + 8*tan((sqrt(x)*d + c)/2)**4*a**3*b**2 - 2*tan((sqrt(x)*d + c)/2)**4*a* 
*2*b**3 - 2*tan((sqrt(x)*d + c)/2)**4*a*b**4 + tan((sqrt(x)*d + c)/2)**4*b 
**5 - 4*tan((sqrt(x)*d + c)/2)**2*a**5 + 6*tan((sqrt(x)*d + c)/2)**2*a*...