\(\int x^3 (a+b \text {sech}(c+d \sqrt {x})) \, dx\) [39]

Optimal result

Integrand size = 18, antiderivative size = 426 \[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {4 b x^{7/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 i b x^3 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 i b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 i b x^2 \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 i b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 i b x \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 i b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \operatorname {PolyLog}\left (8,-i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \operatorname {PolyLog}\left (8,i e^{c+d \sqrt {x}}\right )}{d^8} \] Output:

1/4*a*x^4+4*b*x^(7/2)*arctan(exp(c+d*x^(1/2)))/d+420*I*b*x^2*polylog(4,I*e 
xp(c+d*x^(1/2)))/d^4+10080*I*b*polylog(8,I*exp(c+d*x^(1/2)))/d^8-1680*I*b* 
x^(3/2)*polylog(5,I*exp(c+d*x^(1/2)))/d^5+10080*I*b*x^(1/2)*polylog(7,-I*e 
xp(c+d*x^(1/2)))/d^7-5040*I*b*x*polylog(6,-I*exp(c+d*x^(1/2)))/d^6-10080*I 
*b*x^(1/2)*polylog(7,I*exp(c+d*x^(1/2)))/d^7+14*I*b*x^3*polylog(2,I*exp(c+ 
d*x^(1/2)))/d^2-10080*I*b*polylog(8,-I*exp(c+d*x^(1/2)))/d^8+84*I*b*x^(5/2 
)*polylog(3,-I*exp(c+d*x^(1/2)))/d^3+5040*I*b*x*polylog(6,I*exp(c+d*x^(1/2 
)))/d^6-420*I*b*x^2*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+1680*I*b*x^(3/2)*po 
lylog(5,-I*exp(c+d*x^(1/2)))/d^5-84*I*b*x^(5/2)*polylog(3,I*exp(c+d*x^(1/2 
)))/d^3-14*I*b*x^3*polylog(2,-I*exp(c+d*x^(1/2)))/d^2
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 415, normalized size of antiderivative = 0.97 \[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {2 i b \left (d^7 x^{7/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-d^7 x^{7/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-7 d^6 x^3 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+7 d^6 x^3 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+42 d^5 x^{5/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-42 d^5 x^{5/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-210 d^4 x^2 \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+210 d^4 x^2 \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+840 d^3 x^{3/2} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-840 d^3 x^{3/2} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )-2520 d^2 x \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )+2520 d^2 x \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )+5040 d \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{c+d \sqrt {x}}\right )-5040 d \sqrt {x} \operatorname {PolyLog}\left (7,i e^{c+d \sqrt {x}}\right )-5040 \operatorname {PolyLog}\left (8,-i e^{c+d \sqrt {x}}\right )+5040 \operatorname {PolyLog}\left (8,i e^{c+d \sqrt {x}}\right )\right )}{d^8} \] Input:

Integrate[x^3*(a + b*Sech[c + d*Sqrt[x]]),x]
 

Output:

(a*x^4)/4 + ((2*I)*b*(d^7*x^(7/2)*Log[1 - I*E^(c + d*Sqrt[x])] - d^7*x^(7/ 
2)*Log[1 + I*E^(c + d*Sqrt[x])] - 7*d^6*x^3*PolyLog[2, (-I)*E^(c + d*Sqrt[ 
x])] + 7*d^6*x^3*PolyLog[2, I*E^(c + d*Sqrt[x])] + 42*d^5*x^(5/2)*PolyLog[ 
3, (-I)*E^(c + d*Sqrt[x])] - 42*d^5*x^(5/2)*PolyLog[3, I*E^(c + d*Sqrt[x]) 
] - 210*d^4*x^2*PolyLog[4, (-I)*E^(c + d*Sqrt[x])] + 210*d^4*x^2*PolyLog[4 
, I*E^(c + d*Sqrt[x])] + 840*d^3*x^(3/2)*PolyLog[5, (-I)*E^(c + d*Sqrt[x]) 
] - 840*d^3*x^(3/2)*PolyLog[5, I*E^(c + d*Sqrt[x])] - 2520*d^2*x*PolyLog[6 
, (-I)*E^(c + d*Sqrt[x])] + 2520*d^2*x*PolyLog[6, I*E^(c + d*Sqrt[x])] + 5 
040*d*Sqrt[x]*PolyLog[7, (-I)*E^(c + d*Sqrt[x])] - 5040*d*Sqrt[x]*PolyLog[ 
7, I*E^(c + d*Sqrt[x])] - 5040*PolyLog[8, (-I)*E^(c + d*Sqrt[x])] + 5040*P 
olyLog[8, I*E^(c + d*Sqrt[x])]))/d^8
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 426, 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.

\(\displaystyle \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (a x^3+b x^3 \text {sech}\left (c+d \sqrt {x}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^4}{4}+\frac {4 b x^{7/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10080 i b \operatorname {PolyLog}\left (8,-i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \operatorname {PolyLog}\left (8,i e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 i b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 i b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{c+d \sqrt {x}}\right )}{d^7}-\frac {5040 i b x \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 i b x \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {1680 i b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 i b x^{3/2} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {420 i b x^2 \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 i b x^2 \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {84 i b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 i b x^{5/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {14 i b x^3 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 i b x^3 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}\)

Input:

Int[x^3*(a + b*Sech[c + d*Sqrt[x]]),x]
 

Output:

(a*x^4)/4 + (4*b*x^(7/2)*ArcTan[E^(c + d*Sqrt[x])])/d - ((14*I)*b*x^3*Poly 
Log[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((14*I)*b*x^3*PolyLog[2, I*E^(c + d* 
Sqrt[x])])/d^2 + ((84*I)*b*x^(5/2)*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 
 - ((84*I)*b*x^(5/2)*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 - ((420*I)*b*x^2 
*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((420*I)*b*x^2*PolyLog[4, I*E^( 
c + d*Sqrt[x])])/d^4 + ((1680*I)*b*x^(3/2)*PolyLog[5, (-I)*E^(c + d*Sqrt[x 
])])/d^5 - ((1680*I)*b*x^(3/2)*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 - ((50 
40*I)*b*x*PolyLog[6, (-I)*E^(c + d*Sqrt[x])])/d^6 + ((5040*I)*b*x*PolyLog[ 
6, I*E^(c + d*Sqrt[x])])/d^6 + ((10080*I)*b*Sqrt[x]*PolyLog[7, (-I)*E^(c + 
 d*Sqrt[x])])/d^7 - ((10080*I)*b*Sqrt[x]*PolyLog[7, I*E^(c + d*Sqrt[x])])/ 
d^7 - ((10080*I)*b*PolyLog[8, (-I)*E^(c + d*Sqrt[x])])/d^8 + ((10080*I)*b* 
PolyLog[8, I*E^(c + d*Sqrt[x])])/d^8
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

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]]
 

Maple [F]

\[\int x^{3} \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )d x\]

Input:

int(x^3*(a+b*sech(c+d*x^(1/2))),x)
 

Output:

int(x^3*(a+b*sech(c+d*x^(1/2))),x)
 

Fricas [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*sech(c+d*x^(1/2))),x, algorithm="fricas")
 

Output:

integral(b*x^3*sech(d*sqrt(x) + c) + a*x^3, x)
 

Sympy [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^{3} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )\, dx \] Input:

integrate(x**3*(a+b*sech(c+d*x**(1/2))),x)
 

Output:

Integral(x**3*(a + b*sech(c + d*sqrt(x))), x)
 

Maxima [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*sech(c+d*x^(1/2))),x, algorithm="maxima")
 

Output:

1/4*a*x^4 + 2*b*integrate(x^3*e^(d*sqrt(x) + c)/(e^(2*d*sqrt(x) + 2*c) + 1 
), x)
 

Giac [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*sech(c+d*x^(1/2))),x, algorithm="giac")
 

Output:

integrate((b*sech(d*sqrt(x) + c) + a)*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^3\,\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \] Input:

int(x^3*(a + b/cosh(c + d*x^(1/2))),x)
 

Output:

int(x^3*(a + b/cosh(c + d*x^(1/2))), x)
 

Reduce [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\left (\int \mathrm {sech}\left (\sqrt {x}\, d +c \right ) x^{3}d x \right ) b +\frac {a \,x^{4}}{4} \] Input:

int(x^3*(a+b*sech(c+d*x^(1/2))),x)
 

Output:

(4*int(sech(sqrt(x)*d + c)*x**3,x)*b + a*x**4)/4