\(\int \frac {x^2}{a+b \sec (c+d \sqrt {x})} \, dx\) [42]

Optimal result

Integrand size = 20, antiderivative size = 781 \[ \int \frac {x^2}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {x^3}{3 a}+\frac {2 i b x^{5/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x^{5/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5}+\frac {240 b \operatorname {PolyLog}\left (6,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^6}-\frac {240 b \operatorname {PolyLog}\left (6,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^6} \] Output:

1/3*x^3/a+2*I*b*x^(5/2)*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^(5/2)*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+10*b*x^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^2-10*b*x^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+40*I*b*x^(3/2) 
*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^3-40*I*b*x^(3/2)*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^3-120*b*x*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+120*b*x*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^4-240*I*b*x^(1/2)*polylo 
g(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+2 
40*I*b*x^(1/2)*polylog(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+240*b*polylog(6,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^ 
(1/2)))/a/(-a^2+b^2)^(1/2)/d^6-240*b*polylog(6,-a*exp(I*(c+d*x^(1/2)))/(b+ 
(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^6
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 608, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx=\frac {\sqrt {-a^2+b^2} d^6 x^3+6 i b d^5 x^{5/2} \log \left (1-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 i b d^5 x^{5/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+30 b d^4 x^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-30 b d^4 x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+120 i b d^3 x^{3/2} \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-120 i b d^3 x^{3/2} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )-360 b d^2 x \operatorname {PolyLog}\left (4,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )+360 b d^2 x \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )-720 i b d \sqrt {x} \operatorname {PolyLog}\left (5,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )+720 i b d \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )+720 b \operatorname {PolyLog}\left (6,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-b+\sqrt {-a^2+b^2}}\right )-720 b \operatorname {PolyLog}\left (6,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{3 a \sqrt {-a^2+b^2} d^6} \] Input:

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

Output:

(Sqrt[-a^2 + b^2]*d^6*x^3 + (6*I)*b*d^5*x^(5/2)*Log[1 - (a*E^(I*(c + d*Sqr 
t[x])))/(-b + Sqrt[-a^2 + b^2])] - (6*I)*b*d^5*x^(5/2)*Log[1 + (a*E^(I*(c 
+ d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])] + 30*b*d^4*x^2*PolyLog[2, (a*E^(I*( 
c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - 30*b*d^4*x^2*PolyLog[2, -((a*E 
^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))] + (120*I)*b*d^3*x^(3/2)*Pol 
yLog[3, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] - (120*I)*b*d^3 
*x^(3/2)*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))] - 
 360*b*d^2*x*PolyLog[4, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt[-a^2 + b^2])] 
 + 360*b*d^2*x*PolyLog[4, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2 
]))] - (720*I)*b*d*Sqrt[x]*PolyLog[5, (a*E^(I*(c + d*Sqrt[x])))/(-b + Sqrt 
[-a^2 + b^2])] + (720*I)*b*d*Sqrt[x]*PolyLog[5, -((a*E^(I*(c + d*Sqrt[x])) 
)/(b + Sqrt[-a^2 + b^2]))] + 720*b*PolyLog[6, (a*E^(I*(c + d*Sqrt[x])))/(- 
b + Sqrt[-a^2 + b^2])] - 720*b*PolyLog[6, -((a*E^(I*(c + d*Sqrt[x])))/(b + 
 Sqrt[-a^2 + b^2]))])/(3*a*Sqrt[-a^2 + b^2]*d^6)
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 783, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.

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

\(\Big \downarrow \) 4692

\(\displaystyle 2 \int \frac {x^{5/2}}{a+b \sec \left (c+d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (\frac {x^{5/2}}{a}-\frac {b x^{5/2}}{a \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {120 b \operatorname {PolyLog}\left (6,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^6 \sqrt {b^2-a^2}}-\frac {120 b \operatorname {PolyLog}\left (6,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^6 \sqrt {b^2-a^2}}-\frac {120 i b \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}+\frac {120 i b \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}-\frac {60 b x \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {60 b x \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {20 i b x^{3/2} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {20 i b x^{3/2} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {5 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {5 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^{5/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^{5/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^3}{6 a}\right )\)

Input:

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

Output:

2*(x^3/(6*a) + (I*b*x^(5/2)*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^(5/2)*Log[1 + (a*E^(I*(c + d* 
Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (5*b*x^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) - (5*b*x^2*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a 
^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + ((20*I)*b*x^(3/2)*PolyLog[3, -((a 
*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) 
 - ((20*I)*b*x^(3/2)*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 
 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (60*b*x*PolyLog[4, -((a*E^(I*(c + d 
*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^4) + (60*b*x*P 
olyLog[4, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a 
^2 + b^2]*d^4) - ((120*I)*b*Sqrt[x]*PolyLog[5, -((a*E^(I*(c + d*Sqrt[x]))) 
/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^5) + ((120*I)*b*Sqrt[x]*P 
olyLog[5, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a 
^2 + b^2]*d^5) + (120*b*PolyLog[6, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[- 
a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^6) - (120*b*PolyLog[6, -((a*E^(I*(c + 
 d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^6))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(6*int((tan((sqrt(x)*d + c)/2)**2*x**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 + 6*int((tan( 
(sqrt(x)*d + c)/2)**2*x**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 + x**3)/(3*(a + b))