3.1.16 \(\int \frac {x^5}{a+b \sec (c+d x^2)} \, dx\) [16]

3.1.16.1 Optimal result

Integrand size = 18, antiderivative size = 382 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \]

output

1/6*x^6/a+1/2*I*b*x^4*ln(1+a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a/d/(- 
a^2+b^2)^(1/2)-1/2*I*b*x^4*ln(1+a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)))/a 
/d/(-a^2+b^2)^(1/2)+b*x^2*polylog(2,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2 
)))/a/d^2/(-a^2+b^2)^(1/2)-b*x^2*polylog(2,-a*exp(I*(d*x^2+c))/(b+(-a^2+b^ 
2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)+I*b*polylog(3,-a*exp(I*(d*x^2+c))/(b-(-a 
^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)-I*b*polylog(3,-a*exp(I*(d*x^2+c))/( 
b+(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)
 

3.1.16.2 Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {\sqrt {-a^2+b^2} d^3 x^6+3 i b d^2 x^4 \log \left (1-\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-3 i b d^2 x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 b d x^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 b d x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 i b \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{6 a \sqrt {-a^2+b^2} d^3} \]

input

Integrate[x^5/(a + b*Sec[c + d*x^2]),x]
 

output

(Sqrt[-a^2 + b^2]*d^3*x^6 + (3*I)*b*d^2*x^4*Log[1 - (a*E^(I*(c + d*x^2)))/ 
(-b + Sqrt[-a^2 + b^2])] - (3*I)*b*d^2*x^4*Log[1 + (a*E^(I*(c + d*x^2)))/( 
b + Sqrt[-a^2 + b^2])] + 6*b*d*x^2*PolyLog[2, (a*E^(I*(c + d*x^2)))/(-b + 
Sqrt[-a^2 + b^2])] - 6*b*d*x^2*PolyLog[2, -((a*E^(I*(c + d*x^2)))/(b + Sqr 
t[-a^2 + b^2]))] + (6*I)*b*PolyLog[3, (a*E^(I*(c + d*x^2)))/(-b + Sqrt[-a^ 
2 + b^2])] - (6*I)*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b 
^2]))])/(6*a*Sqrt[-a^2 + b^2]*d^3)
 

3.1.16.3 Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.

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

\(\Big \downarrow \) 4692

\(\displaystyle \frac {1}{2} \int \frac {x^4}{a+b \sec \left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {x^4}{a+b \csc \left (d x^2+c+\frac {\pi }{2}\right )}dx^2\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {1}{2} \int \left (\frac {x^4}{a}-\frac {b x^4}{a \left (b+a \cos \left (d x^2+c\right )\right )}\right )dx^2\)

\(\Big \downarrow \) 2009

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

input

Int[x^5/(a + b*Sec[c + d*x^2]),x]
 

output

(x^6/(3*a) + (I*b*x^4*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^4*Log[1 + (a*E^(I*(c + d*x^2)))/(b + Sq 
rt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (2*b*x^2*PolyLog[2, -((a*E^(I*( 
c + d*x^2)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) - (2*b*x^2 
*PolyLog[2, -((a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 
 + b^2]*d^2) + ((2*I)*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 
+ b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - ((2*I)*b*PolyLog[3, -((a*E^(I*(c + d 
*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3))/2
 

3.1.16.3.1 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]
 

3.1.16.4 Maple [F]

\[\int \frac {x^{5}}{a +b \sec \left (d \,x^{2}+c \right )}d x\]

input

int(x^5/(a+b*sec(d*x^2+c)),x)
 

output

int(x^5/(a+b*sec(d*x^2+c)),x)
 

3.1.16.5 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. 1457 vs. \(2 (330) = 660\).

Time = 0.44 (sec) , antiderivative size = 1457, normalized size of antiderivative = 3.81 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\text {Too large to display} \]

input

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="fricas")
 

output

1/12*(2*(a^2 - b^2)*d^3*x^6 - 6*a*b*d*x^2*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) + 6*a*b*d*x^2*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) - 6*a*b*d*x^2*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) + 6*a*b*d* 
x^2*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 
) + 3*I*a*b*c^2*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) - 3*I*a*b*c^2*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) + 3*I*a*b*c^2*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) - 3*I*a*b* 
c^2*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) - 6*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polyl 
og(3, -(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) + I*a*si 
n(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 6*I*a*b*sqrt(-(a^2 - b^2)/a^2)* 
polylog(3, -(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) + 6*I*a*b*sqrt(-(a^2 - b^...
 

3.1.16.6 Sympy [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^{5}}{a + b \sec {\left (c + d x^{2} \right )}}\, dx \]

input

integrate(x**5/(a+b*sec(d*x**2+c)),x)
 

output

Integral(x**5/(a + b*sec(c + d*x**2)), x)
 

3.1.16.7 Maxima [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]

input

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="maxima")
 

output

1/6*(x^6 - 12*a*b*integrate((a*x^5*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + 2*b 
*x^5*cos(d*x^2 + c)^2 + a*x^5*sin(2*d*x^2 + 2*c)*sin(d*x^2 + c) + 2*b*x^5* 
sin(d*x^2 + c)^2 + a*x^5*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
 

3.1.16.8 Giac [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]

input

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="giac")
 

output

integrate(x^5/(b*sec(d*x^2 + c) + a), x)
 

3.1.16.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^5}{a+\frac {b}{\cos \left (d\,x^2+c\right )}} \,d x \]

input

int(x^5/(a + b/cos(c + d*x^2)),x)
 

output

int(x^5/(a + b/cos(c + d*x^2)), x)