\(\int \frac {x^3}{(a+b \sec (c+d x^2))^2} \, dx\) [25]

Optimal result

Integrand size = 18, antiderivative size = 596 \[ \int \frac {x^3}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\frac {x^4}{4 a^2}-\frac {i b^3 x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}+\frac {i b^3 x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}+\frac {b^2 \log \left (b+a \cos \left (c+d x^2\right )\right )}{2 a^2 \left (a^2-b^2\right ) d^2}-\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d^2}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}+\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d^2}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}+\frac {b^2 x^2 \sin \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (b+a \cos \left (c+d x^2\right )\right )} \] Output:

1/4*x^4/a^2-1/2*I*b^3*x^2*ln(1+a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a^ 
2/(-a^2+b^2)^(3/2)/d+I*b*x^2*ln(1+a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2))) 
/a^2/(-a^2+b^2)^(1/2)/d+1/2*I*b^3*x^2*ln(1+a*exp(I*(d*x^2+c))/(b+(-a^2+b^2 
)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d-I*b*x^2*ln(1+a*exp(I*(d*x^2+c))/(b+(-a^2+ 
b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d+1/2*b^2*ln(b+a*cos(d*x^2+c))/a^2/(a^2- 
b^2)/d^2-1/2*b^3*polylog(2,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a^2/( 
-a^2+b^2)^(3/2)/d^2+b*polylog(2,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/ 
a^2/(-a^2+b^2)^(1/2)/d^2+1/2*b^3*polylog(2,-a*exp(I*(d*x^2+c))/(b+(-a^2+b^ 
2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2-b*polylog(2,-a*exp(I*(d*x^2+c))/(b+(-a 
^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2+1/2*b^2*x^2*sin(d*x^2+c)/a/(a^2-b 
^2)/d/(b+a*cos(d*x^2+c))
 

Mathematica [A] (warning: unable to verify)

Time = 9.87 (sec) , antiderivative size = 1118, normalized size of antiderivative = 1.88 \[ \int \frac {x^3}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((-c + d*x^2)*(c + d*x^2)*(b + a*Cos[c + d*x^2])^2*Sec[c + d*x^2]^2)/(4*a^ 
2*d^2*(a + b*Sec[c + d*x^2])^2) + ((b + a*Cos[c + d*x^2])*Sec[c + d*x^2]^2 
*(b^2*c*Sin[c + d*x^2] - b^2*(c + d*x^2)*Sin[c + d*x^2]))/(2*a*(-a + b)*(a 
 + b)*d^2*(a + b*Sec[c + d*x^2])^2) + (b*Cos[(c + d*x^2)/2]^2*(b + a*Cos[c 
 + d*x^2])*(2*(2*a^2 - b^2)*c*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x^2)/2])/Sqr 
t[a + b]] - Sqrt[a - b]*b*Sqrt[a + b]*Log[Sec[(c + d*x^2)/2]^2] + Sqrt[a - 
 b]*b*Sqrt[a + b]*Log[(b + a*Cos[c + d*x^2])*Sec[(c + d*x^2)/2]^2] + I*(2* 
a^2 - b^2)*(Log[1 - I*Tan[(c + d*x^2)/2]]*Log[(Sqrt[a + b] - Sqrt[a - b]*T 
an[(c + d*x^2)/2])/(I*Sqrt[a - b] + Sqrt[a + b])] + PolyLog[2, (Sqrt[a - b 
]*(1 - I*Tan[(c + d*x^2)/2]))/(Sqrt[a - b] - I*Sqrt[a + b])]) - I*(2*a^2 - 
 b^2)*(Log[1 - I*Tan[(c + d*x^2)/2]]*Log[(I*(Sqrt[a + b] + Sqrt[a - b]*Tan 
[(c + d*x^2)/2]))/(Sqrt[a - b] + I*Sqrt[a + b])] + PolyLog[2, (Sqrt[a - b] 
*(1 - I*Tan[(c + d*x^2)/2]))/(Sqrt[a - b] + I*Sqrt[a + b])]) + I*(2*a^2 - 
b^2)*(Log[1 + I*Tan[(c + d*x^2)/2]]*Log[(Sqrt[a + b] + Sqrt[a - b]*Tan[(c 
+ d*x^2)/2])/(I*Sqrt[a - b] + Sqrt[a + b])] + PolyLog[2, (Sqrt[a - b]*(1 + 
 I*Tan[(c + d*x^2)/2]))/(Sqrt[a - b] - I*Sqrt[a + b])]) - I*(2*a^2 - b^2)* 
(Log[1 + I*Tan[(c + d*x^2)/2]]*Log[(I*(Sqrt[a + b] - Sqrt[a - b]*Tan[(c + 
d*x^2)/2]))/(Sqrt[a - b] + I*Sqrt[a + b])] + PolyLog[2, (Sqrt[a - b]*(1 + 
I*Tan[(c + d*x^2)/2]))/(Sqrt[a - b] + I*Sqrt[a + b])]))*Sec[c + d*x^2]^2*( 
(2*a^2 - b^2)*d*x^2 + a*b*Sin[c + d*x^2])*(Sqrt[a + b] - Sqrt[a - b]*Ta...
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 586, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(x^4/(2*a^2) - (I*b^3*x^2*Log[1 + (a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b 
^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^2*Log[1 + (a*E^(I*(c + d*x^ 
2)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) + (I*b^3*x^2*Log[1 
+ (a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d 
) - ((2*I)*b*x^2*Log[1 + (a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a 
^2*Sqrt[-a^2 + b^2]*d) + (b^2*Log[b + a*Cos[c + d*x^2]])/(a^2*(a^2 - b^2)* 
d^2) - (b^3*PolyLog[2, -((a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^2]))])/( 
a^2*(-a^2 + b^2)^(3/2)*d^2) + (2*b*PolyLog[2, -((a*E^(I*(c + d*x^2)))/(b - 
 Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (b^3*PolyLog[2, -((a*E^ 
(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) - 
(2*b*PolyLog[2, -((a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a^2*Sqr 
t[-a^2 + b^2]*d^2) + (b^2*x^2*Sin[c + d*x^2])/(a*(a^2 - b^2)*d*(b + a*Cos[ 
c + d*x^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]

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

Input:

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

Output:

int(x^3/(a+b*sec(d*x^2+c))^2,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. 1928 vs. \(2 (522) = 1044\).

Time = 0.23 (sec) , antiderivative size = 1928, normalized size of antiderivative = 3.23 \[ \int \frac {x^3}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((a^5 - 2*a^3*b^2 + a*b^4)*d^2*x^4*cos(d*x^2 + c) + (a^4*b - 2*a^2*b^3 
 + b^5)*d^2*x^4 + 2*(a^3*b^2 - a*b^4)*d*x^2*sin(d*x^2 + c) - (2*a^3*b^2 - 
a*b^4 + (2*a^4*b - a^2*b^3)*cos(d*x^2 + c))*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) + (2*a^3*b^2 - a*b^4 + (2*a^4*b 
- a^2*b^3)*cos(d*x^2 + c))*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) - (2*a^3*b^2 - a*b^4 + (2*a^4*b - a^2*b^3)*cos(d* 
x^2 + c))*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) + (2*a^3*b^2 - a*b^4 + (2*a^4*b - a^2*b^3)*cos(d*x^2 + c))*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*(2* 
a^3*b^2 - a*b^4)*d*x^2 - I*(2*a^3*b^2 - a*b^4)*c + (-I*(2*a^4*b - a^2*b^3) 
*d*x^2 - I*(2*a^4*b - a^2*b^3)*c)*cos(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2)*l 
og((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*(2*a^3*b^2 - a*b^4)*d*x^2 + 
I*(2*a^3*b^2 - a*b^4)*c + (I*(2*a^4*b - a^2*b^3)*d*x^2 + I*(2*a^4*b - a^2* 
b^3)*c)*cos(d*x^2 + 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 - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^3/(a+b*sec(d*x^2+c))^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*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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