Integrand size = 16, antiderivative size = 92 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {a x^4}{4}-\frac {i b x^2 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{2 d^2} \] Output:
1/4*a*x^4-I*b*x^2*arctan(exp(I*(d*x^2+c)))/d+1/2*I*b*polylog(2,-I*exp(I*(d *x^2+c)))/d^2-1/2*I*b*polylog(2,I*exp(I*(d*x^2+c)))/d^2
Time = 0.02 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {a x^4}{4}-\frac {i b x^2 \arctan \left (e^{i c+i d x^2}\right )}{d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{2 d^2} \] Input:
Integrate[x^3*(a + b*Sec[c + d*x^2]),x]
Output:
(a*x^4)/4 - (I*b*x^2*ArcTan[E^(I*c + I*d*x^2)])/d + ((I/2)*b*PolyLog[2, (- I)*E^(I*(c + d*x^2))])/d^2 - ((I/2)*b*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 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.
\(\displaystyle \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a x^3+b x^3 \sec \left (c+d x^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x^4}{4}-\frac {i b x^2 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b \operatorname {PolyLog}\left (2,i e^{i \left (d x^2+c\right )}\right )}{2 d^2}\) |
Input:
Int[x^3*(a + b*Sec[c + d*x^2]),x]
Output:
(a*x^4)/4 - (I*b*x^2*ArcTan[E^(I*(c + d*x^2))])/d + ((I/2)*b*PolyLog[2, (- I)*E^(I*(c + d*x^2))])/d^2 - ((I/2)*b*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
\[\int x^{3} \left (a +b \sec \left (d \,x^{2}+c \right )\right )d x\]
Input:
int(x^3*(a+b*sec(d*x^2+c)),x)
Output:
int(x^3*(a+b*sec(d*x^2+c)),x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (68) = 136\).
Time = 0.10 (sec) , antiderivative size = 346, normalized size of antiderivative = 3.76 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\frac {a d^{2} x^{4} - b c \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - b c \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right )}{4 \, d^{2}} \] Input:
integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="fricas")
Output:
1/4*(a*d^2*x^4 - b*c*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) + b*c*log( cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) - b*c*log(-cos(d*x^2 + c) + I*sin(d *x^2 + c) + I) + b*c*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) - I*b*dil og(I*cos(d*x^2 + c) + sin(d*x^2 + c)) - I*b*dilog(I*cos(d*x^2 + c) - sin(d *x^2 + c)) + I*b*dilog(-I*cos(d*x^2 + c) + sin(d*x^2 + c)) + I*b*dilog(-I* cos(d*x^2 + c) - sin(d*x^2 + c)) + (b*d*x^2 + b*c)*log(I*cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - (b*d*x^2 + b*c)*log(I*cos(d*x^2 + c) - sin(d*x^2 + c ) + 1) + (b*d*x^2 + b*c)*log(-I*cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - (b* d*x^2 + b*c)*log(-I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1))/d^2
\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int x^{3} \left (a + b \sec {\left (c + d x^{2} \right )}\right )\, dx \] Input:
integrate(x**3*(a+b*sec(d*x**2+c)),x)
Output:
Integral(x**3*(a + b*sec(c + d*x**2)), x)
\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{3} \,d x } \] Input:
integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="maxima")
Output:
1/4*a*x^4 + 2*b*integrate((x^3*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + x^3*sin (2*d*x^2 + 2*c)*sin(d*x^2 + c) + x^3*cos(d*x^2 + c))/(cos(2*d*x^2 + 2*c)^2 + sin(2*d*x^2 + 2*c)^2 + 2*cos(2*d*x^2 + 2*c) + 1), x)
\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{3} \,d x } \] Input:
integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="giac")
Output:
integrate((b*sec(d*x^2 + c) + a)*x^3, x)
Timed out. \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=\int x^3\,\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right ) \,d x \] Input:
int(x^3*(a + b/cos(c + d*x^2)),x)
Output:
int(x^3*(a + b/cos(c + d*x^2)), x)
\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx=-2 \left (\int \frac {\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} x^{3}}{\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}-1}d x \right ) b +\frac {a \,x^{4}}{4}+\frac {b \,x^{4}}{4} \] 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 - 1),x)*b + a*x**4 + b*x**4)/4