\(\int \frac {x^3}{a+b \sec (c+d x^2)} \, dx\) [18]

Optimal result

Integrand size = 18, antiderivative size = 261 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^4}{4 a}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2} \] Output:

1/4*x^4/a+1/2*I*b*x^2*ln(1+a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a/(-a^ 
2+b^2)^(1/2)/d-1/2*I*b*x^2*ln(1+a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)))/a 
/(-a^2+b^2)^(1/2)/d+1/2*b*polylog(2,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2 
)))/a/(-a^2+b^2)^(1/2)/d^2-1/2*b*polylog(2,-a*exp(I*(d*x^2+c))/(b+(-a^2+b^ 
2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(845\) vs. \(2(261)=522\).

Time = 1.03 (sec) , antiderivative size = 845, normalized size of antiderivative = 3.24 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

((b + a*Cos[c + d*x^2])*(x^4 - (2*b*(2*(c + d*x^2)*ArcTanh[((a + b)*Cot[(c 
 + d*x^2)/2])/Sqrt[a^2 - b^2]] - 2*(c + ArcCos[-(b/a)])*ArcTanh[((a - b)*T 
an[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] - (2*I)*ArcTanh[((a 
+ b)*Cot[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + (2*I)*ArcTanh[((a - b)*Tan[(c 
+ d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[Sqrt[a^2 - b^2]/(Sqrt[2]*Sqrt[a]*E^((I/ 
2)*(c + d*x^2))*Sqrt[b + a*Cos[c + d*x^2]])] + (ArcCos[-(b/a)] + (2*I)*(Ar 
cTanh[((a + b)*Cot[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] - ArcTanh[((a - b)*Tan 
[(c + d*x^2)/2])/Sqrt[a^2 - b^2]]))*Log[(Sqrt[a^2 - b^2]*E^((I/2)*(c + d*x 
^2)))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cos[c + d*x^2]])] - (ArcCos[-(b/a)] - (2 
*I)*ArcTanh[((a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(a 
 - b - I*Sqrt[a^2 - b^2])*(1 + I*Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 
 - b^2]*Tan[(c + d*x^2)/2]))] - (ArcCos[-(b/a)] + (2*I)*ArcTanh[((a - b)*T 
an[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*((-I)*a + I*b + Sqrt[a^2 
 - b^2])*(I + Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d* 
x^2)/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^ 
2]*Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))] 
- PolyLog[2, ((b + I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*Tan[(c + d* 
x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))])))/(Sqrt[a^2 - 
 b^2]*d^2))*Sec[c + d*x^2])/(4*a*(a + b*Sec[c + d*x^2]))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.

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

\(\Big \downarrow \) 4692

\(\displaystyle \frac {1}{2} \int \frac {x^2}{a+b \sec \left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {x^2}{a+b \csc \left (d x^2+c+\frac {\pi }{2}\right )}dx^2\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {1}{2} \int \left (\frac {x^2}{a}-\frac {b x^2}{a \left (b+a \cos \left (d x^2+c\right )\right )}\right )dx^2\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1060 vs. \(2 (221) = 442\).

Time = 0.21 (sec) , antiderivative size = 1060, normalized size of antiderivative = 4.06 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((a^2 - b^2)*d^2*x^4 - I*a*b*c*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^ 
2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) + I*a*b* 
c*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 2 
*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - I*a*b*c*sqrt(-(a^2 - b^2)/a^2)*log(-2*a 
*cos(d*x^2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) 
 + I*a*b*c*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^ 
2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - a*b*sqrt(-(a^2 - b^2)/a^2)*di 
log(-(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) + I*a*sin( 
d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + a*b*sqrt(-(a^2 - b^2)/a^2 
)*dilog(-(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a* 
sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - a*b*sqrt(-(a^2 - b^2) 
/a^2)*dilog(-(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) - 
I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + a*b*sqrt(-(a^2 - 
b^2)/a^2)*dilog(-(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c 
) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - (I*a*b*d*x^2 
+ I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(d*x^2 + c) + I*b*sin(d*x^2 + 
c) + (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a 
) - (-I*a*b*d*x^2 - I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(d*x^2 + c) 
+ I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 
- b^2)/a^2) + a)/a) - (-I*a*b*d*x^2 - I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*l...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/4*(x^4 - 8*a*b*integrate((a*x^3*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + 2*b* 
x^3*cos(d*x^2 + c)^2 + a*x^3*sin(2*d*x^2 + 2*c)*sin(d*x^2 + c) + 2*b*x^3*s 
in(d*x^2 + c)^2 + a*x^3*cos(d*x^2 + c))/(a^3*cos(2*d*x^2 + 2*c)^2 + 4*a*b^ 
2*cos(d*x^2 + c)^2 + a^3*sin(2*d*x^2 + 2*c)^2 + 4*a^2*b*sin(2*d*x^2 + 2*c) 
*sin(d*x^2 + c) + 4*a*b^2*sin(d*x^2 + c)^2 + 4*a^2*b*cos(d*x^2 + c) + a^3 
+ 2*(2*a^2*b*cos(d*x^2 + c) + a^3)*cos(2*d*x^2 + 2*c)), x))/a
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {8 \left (\int \frac {\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} x^{3}}{\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) a b +8 \left (\int \frac {\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} x^{3}}{\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b^{2}-a^{2}-2 a b -b^{2}}d x \right ) b^{2}+x^{4}}{4 a +4 b} \] Input:

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

Output:

(8*int((tan((c + d*x**2)/2)**2*x**3)/(tan((c + d*x**2)/2)**2*a**2 - tan((c 
 + d*x**2)/2)**2*b**2 - a**2 - 2*a*b - b**2),x)*a*b + 8*int((tan((c + d*x* 
*2)/2)**2*x**3)/(tan((c + d*x**2)/2)**2*a**2 - tan((c + d*x**2)/2)**2*b**2 
 - a**2 - 2*a*b - b**2),x)*b**2 + x**4)/(4*(a + b))