Integrand size = 18, antiderivative size = 476 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \operatorname {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \operatorname {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {10080 i b \operatorname {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \operatorname {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8} \] Output:
1/4*a*x^4-5040*I*b*x*polylog(6,I*exp(I*(c+d*x^(1/2))))/d^6+5040*I*b*x*poly log(6,-I*exp(I*(c+d*x^(1/2))))/d^6+10080*I*b*polylog(8,I*exp(I*(c+d*x^(1/2 ))))/d^8-84*b*x^(5/2)*polylog(3,-I*exp(I*(c+d*x^(1/2))))/d^3+84*b*x^(5/2)* polylog(3,I*exp(I*(c+d*x^(1/2))))/d^3+420*I*b*x^2*polylog(4,I*exp(I*(c+d*x ^(1/2))))/d^4-4*I*b*x^(7/2)*arctan(exp(I*(c+d*x^(1/2))))/d+1680*b*x^(3/2)* polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-1680*b*x^(3/2)*polylog(5,I*exp(I*(c +d*x^(1/2))))/d^5-420*I*b*x^2*polylog(4,-I*exp(I*(c+d*x^(1/2))))/d^4-14*I* b*x^3*polylog(2,I*exp(I*(c+d*x^(1/2))))/d^2-10080*b*x^(1/2)*polylog(7,-I*e xp(I*(c+d*x^(1/2))))/d^7+10080*b*x^(1/2)*polylog(7,I*exp(I*(c+d*x^(1/2)))) /d^7+14*I*b*x^3*polylog(2,-I*exp(I*(c+d*x^(1/2))))/d^2-10080*I*b*polylog(8 ,-I*exp(I*(c+d*x^(1/2))))/d^8
Time = 0.11 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.01 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \arctan \left (e^{i c+i d \sqrt {x}}\right )}{d}+\frac {14 i b x^3 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \operatorname {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \operatorname {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {10080 i b \operatorname {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \operatorname {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8} \] Input:
Integrate[x^3*(a + b*Sec[c + d*Sqrt[x]]),x]
Output:
(a*x^4)/4 - ((4*I)*b*x^(7/2)*ArcTan[E^(I*c + I*d*Sqrt[x])])/d + ((14*I)*b* x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((14*I)*b*x^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, (-I)*E^(I*(c + d *Sqrt[x]))])/d^3 + (84*b*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((420*I)*b* x^2*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/ d^6 - ((5040*I)*b*x*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 - (10080*b*Sq rt[x]*PolyLog[7, (-I)*E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyL og[7, I*E^(I*(c + d*Sqrt[x]))])/d^7 - ((10080*I)*b*PolyLog[8, (-I)*E^(I*(c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, I*E^(I*(c + d*Sqrt[x]))])/d ^8
Time = 0.78 (sec) , antiderivative size = 476, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a x^3+b x^3 \sec \left (c+d \sqrt {x}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x^4}{4}-\frac {4 i b x^{7/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {10080 i b \operatorname {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \operatorname {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {5040 i b x \operatorname {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \operatorname {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {420 i b x^2 \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 i b x^3 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}\) |
Input:
Int[x^3*(a + b*Sec[c + d*Sqrt[x]]),x]
Output:
(a*x^4)/4 - ((4*I)*b*x^(7/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((14*I)*b* x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((14*I)*b*x^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, (-I)*E^(I*(c + d *Sqrt[x]))])/d^3 + (84*b*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((420*I)*b* x^2*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/ d^6 - ((5040*I)*b*x*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 - (10080*b*Sq rt[x]*PolyLog[7, (-I)*E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyL og[7, I*E^(I*(c + d*Sqrt[x]))])/d^7 - ((10080*I)*b*PolyLog[8, (-I)*E^(I*(c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, I*E^(I*(c + d*Sqrt[x]))])/d ^8
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]]
\[\int x^{3} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )d x\]
Input:
int(x^3*(a+b*sec(c+d*x^(1/2))),x)
Output:
int(x^3*(a+b*sec(c+d*x^(1/2))),x)
\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:
integrate(x^3*(a+b*sec(c+d*x^(1/2))),x, algorithm="fricas")
Output:
integral(b*x^3*sec(d*sqrt(x) + c) + a*x^3, x)
\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\int x^{3} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )\, dx \] Input:
integrate(x**3*(a+b*sec(c+d*x**(1/2))),x)
Output:
Integral(x**3*(a + b*sec(c + d*sqrt(x))), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1512 vs. \(2 (352) = 704\).
Time = 0.22 (sec) , antiderivative size = 1512, normalized size of antiderivative = 3.18 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate(x^3*(a+b*sec(c+d*x^(1/2))),x, algorithm="maxima")
Output:
1/4*((d*sqrt(x) + c)^8*a - 8*(d*sqrt(x) + c)^7*a*c + 28*(d*sqrt(x) + c)^6* a*c^2 - 56*(d*sqrt(x) + c)^5*a*c^3 + 70*(d*sqrt(x) + c)^4*a*c^4 - 56*(d*sq rt(x) + c)^3*a*c^5 + 28*(d*sqrt(x) + c)^2*a*c^6 - 8*(d*sqrt(x) + c)*a*c^7 - 8*b*c^7*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) - 8*(I*(d*sqrt(x) + c)^7*b - 7*I*(d*sqrt(x) + c)^6*b*c + 21*I*(d*sqrt(x) + c)^5*b*c^2 - 35*I* (d*sqrt(x) + c)^4*b*c^3 + 35*I*(d*sqrt(x) + c)^3*b*c^4 - 21*I*(d*sqrt(x) + c)^2*b*c^5 + 7*I*(d*sqrt(x) + c)*b*c^6)*arctan2(cos(d*sqrt(x) + c), sin(d *sqrt(x) + c) + 1) - 8*(I*(d*sqrt(x) + c)^7*b - 7*I*(d*sqrt(x) + c)^6*b*c + 21*I*(d*sqrt(x) + c)^5*b*c^2 - 35*I*(d*sqrt(x) + c)^4*b*c^3 + 35*I*(d*sq rt(x) + c)^3*b*c^4 - 21*I*(d*sqrt(x) + c)^2*b*c^5 + 7*I*(d*sqrt(x) + c)*b* c^6)*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) - 56*(I*(d*sqrt( x) + c)^6*b - 6*I*(d*sqrt(x) + c)^5*b*c + 15*I*(d*sqrt(x) + c)^4*b*c^2 - 2 0*I*(d*sqrt(x) + c)^3*b*c^3 + 15*I*(d*sqrt(x) + c)^2*b*c^4 - 6*I*(d*sqrt(x ) + c)*b*c^5 + I*b*c^6)*dilog(I*e^(I*d*sqrt(x) + I*c)) - 56*(-I*(d*sqrt(x) + c)^6*b + 6*I*(d*sqrt(x) + c)^5*b*c - 15*I*(d*sqrt(x) + c)^4*b*c^2 + 20* I*(d*sqrt(x) + c)^3*b*c^3 - 15*I*(d*sqrt(x) + c)^2*b*c^4 + 6*I*(d*sqrt(x) + c)*b*c^5 - I*b*c^6)*dilog(-I*e^(I*d*sqrt(x) + I*c)) + 4*((d*sqrt(x) + c) ^7*b - 7*(d*sqrt(x) + c)^6*b*c + 21*(d*sqrt(x) + c)^5*b*c^2 - 35*(d*sqrt(x ) + c)^4*b*c^3 + 35*(d*sqrt(x) + c)^3*b*c^4 - 21*(d*sqrt(x) + c)^2*b*c^5 + 7*(d*sqrt(x) + c)*b*c^6)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)...
\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:
integrate(x^3*(a+b*sec(c+d*x^(1/2))),x, algorithm="giac")
Output:
integrate((b*sec(d*sqrt(x) + c) + a)*x^3, x)
Timed out. \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\int x^3\,\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right ) \,d x \] Input:
int(x^3*(a + b/cos(c + d*x^(1/2))),x)
Output:
int(x^3*(a + b/cos(c + d*x^(1/2))), x)
\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=-2 \left (\int \frac {\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} x^{3}}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2}-1}d x \right ) b +\frac {a \,x^{4}}{4}+\frac {b \,x^{4}}{4} \] Input:
int(x^3*(a+b*sec(c+d*x^(1/2))),x)
Output:
( - 8*int((tan((sqrt(x)*d + c)/2)**2*x**3)/(tan((sqrt(x)*d + c)/2)**2 - 1) ,x)*b + a*x**4 + b*x**4)/4