\(\int x^3 (a+b \text {sech}(c+d x^2))^2 \, dx\) [11]

Optimal result

Integrand size = 18, antiderivative size = 119 \[ \int x^3 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^4}{4}+\frac {2 a b x^2 \arctan \left (e^{c+d x^2}\right )}{d}-\frac {b^2 \log \left (\cosh \left (c+d x^2\right )\right )}{2 d^2}-\frac {i a b \operatorname {PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac {i a b \operatorname {PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(324\) vs. \(2(119)=238\).

Time = 1.23 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.72 \[ \int x^3 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx=\frac {4 b^2 d e^{2 c} x^2+a^2 d^2 x^4+a^2 d^2 e^{2 c} x^4+4 i a b d x^2 \log \left (1-i e^{c+d x^2}\right )+4 i a b d e^{2 c} x^2 \log \left (1-i e^{c+d x^2}\right )-4 i a b d x^2 \log \left (1+i e^{c+d x^2}\right )-4 i a b d e^{2 c} x^2 \log \left (1+i e^{c+d x^2}\right )-2 b^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )-2 b^2 e^{2 c} \log \left (1+e^{2 \left (c+d x^2\right )}\right )-4 i a b \left (1+e^{2 c}\right ) \operatorname {PolyLog}\left (2,-i e^{c+d x^2}\right )+4 i a b \left (1+e^{2 c}\right ) \operatorname {PolyLog}\left (2,i e^{c+d x^2}\right )+2 b^2 d x^2 \text {sech}(c) \text {sech}\left (c+d x^2\right ) \sinh \left (d x^2\right )+2 b^2 d e^{2 c} x^2 \text {sech}(c) \text {sech}\left (c+d x^2\right ) \sinh \left (d x^2\right )}{4 d^2 \left (1+e^{2 c}\right )} \] Input:

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

Output:

(4*b^2*d*E^(2*c)*x^2 + a^2*d^2*x^4 + a^2*d^2*E^(2*c)*x^4 + (4*I)*a*b*d*x^2 
*Log[1 - I*E^(c + d*x^2)] + (4*I)*a*b*d*E^(2*c)*x^2*Log[1 - I*E^(c + d*x^2 
)] - (4*I)*a*b*d*x^2*Log[1 + I*E^(c + d*x^2)] - (4*I)*a*b*d*E^(2*c)*x^2*Lo 
g[1 + I*E^(c + d*x^2)] - 2*b^2*Log[1 + E^(2*(c + d*x^2))] - 2*b^2*E^(2*c)* 
Log[1 + E^(2*(c + d*x^2))] - (4*I)*a*b*(1 + E^(2*c))*PolyLog[2, (-I)*E^(c 
+ d*x^2)] + (4*I)*a*b*(1 + E^(2*c))*PolyLog[2, I*E^(c + d*x^2)] + 2*b^2*d* 
x^2*Sech[c]*Sech[c + d*x^2]*Sinh[d*x^2] + 2*b^2*d*E^(2*c)*x^2*Sech[c]*Sech 
[c + d*x^2]*Sinh[d*x^2])/(4*d^2*(1 + E^(2*c)))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5959, 3042, 4678, 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.

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

\(\Big \downarrow \) 5959

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a^2*x^4)/2 + (4*a*b*x^2*ArcTan[E^(c + d*x^2)])/d - (b^2*Log[Cosh[c + d*x 
^2]])/d^2 - ((2*I)*a*b*PolyLog[2, (-I)*E^(c + d*x^2)])/d^2 + ((2*I)*a*b*Po 
lyLog[2, I*E^(c + d*x^2)])/d^2 + (b^2*x^2*Tanh[c + d*x^2])/d)/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 4678

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

rule 5959

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[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]

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

Input:

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

Output:

int(x^3*(a+b*sech(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. 782 vs. \(2 (100) = 200\).

Time = 0.10 (sec) , antiderivative size = 782, normalized size of antiderivative = 6.57 \[ \int x^3 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*(a^2*d^2*x^4 + 4*b^2*c + (a^2*d^2*x^4 + 4*b^2*d*x^2 + 4*b^2*c)*cosh(d* 
x^2 + c)^2 + 2*(a^2*d^2*x^4 + 4*b^2*d*x^2 + 4*b^2*c)*cosh(d*x^2 + c)*sinh( 
d*x^2 + c) + (a^2*d^2*x^4 + 4*b^2*d*x^2 + 4*b^2*c)*sinh(d*x^2 + c)^2 - 4*( 
-I*a*b*cosh(d*x^2 + c)^2 - 2*I*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) - I*a*b 
*sinh(d*x^2 + c)^2 - I*a*b)*dilog(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c)) - 
 4*(I*a*b*cosh(d*x^2 + c)^2 + 2*I*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) + I* 
a*b*sinh(d*x^2 + c)^2 + I*a*b)*dilog(-I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c 
)) - 2*(2*I*a*b*c + (2*I*a*b*c + b^2)*cosh(d*x^2 + c)^2 + 2*(2*I*a*b*c + b 
^2)*cosh(d*x^2 + c)*sinh(d*x^2 + c) + (2*I*a*b*c + b^2)*sinh(d*x^2 + c)^2 
+ b^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) + I) - 2*(-2*I*a*b*c + (-2*I* 
a*b*c + b^2)*cosh(d*x^2 + c)^2 + 2*(-2*I*a*b*c + b^2)*cosh(d*x^2 + c)*sinh 
(d*x^2 + c) + (-2*I*a*b*c + b^2)*sinh(d*x^2 + c)^2 + b^2)*log(cosh(d*x^2 + 
 c) + sinh(d*x^2 + c) - I) - 4*(I*a*b*d*x^2 + I*a*b*c + (I*a*b*d*x^2 + I*a 
*b*c)*cosh(d*x^2 + c)^2 + 2*(I*a*b*d*x^2 + I*a*b*c)*cosh(d*x^2 + c)*sinh(d 
*x^2 + c) + (I*a*b*d*x^2 + I*a*b*c)*sinh(d*x^2 + c)^2)*log(I*cosh(d*x^2 + 
c) + I*sinh(d*x^2 + c) + 1) - 4*(-I*a*b*d*x^2 - I*a*b*c + (-I*a*b*d*x^2 - 
I*a*b*c)*cosh(d*x^2 + c)^2 + 2*(-I*a*b*d*x^2 - I*a*b*c)*cosh(d*x^2 + c)*si 
nh(d*x^2 + c) + (-I*a*b*d*x^2 - I*a*b*c)*sinh(d*x^2 + c)^2)*log(-I*cosh(d* 
x^2 + c) - I*sinh(d*x^2 + c) + 1))/(d^2*cosh(d*x^2 + c)^2 + 2*d^2*cosh(d*x 
^2 + c)*sinh(d*x^2 + c) + d^2*sinh(d*x^2 + c)^2 + d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^3 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx=\frac {8 e^{2 d \,x^{2}+2 c} \mathit {atan} \left (e^{d \,x^{2}+c}\right ) a b +8 \mathit {atan} \left (e^{d \,x^{2}+c}\right ) a b +32 e^{2 d \,x^{2}+3 c} \left (\int \frac {e^{d \,x^{2}} x^{3}}{e^{4 d \,x^{2}+4 c}+2 e^{2 d \,x^{2}+2 c}+1}d x \right ) a b \,d^{2}-2 e^{2 d \,x^{2}+2 c} \mathrm {log}\left (e^{2 d \,x^{2}+2 c}+1\right ) b^{2}+e^{2 d \,x^{2}+2 c} a^{2} d^{2} x^{4}+4 e^{2 d \,x^{2}+2 c} b^{2} d \,x^{2}-8 e^{d \,x^{2}+c} a b d \,x^{2}+32 e^{c} \left (\int \frac {e^{d \,x^{2}} x^{3}}{e^{4 d \,x^{2}+4 c}+2 e^{2 d \,x^{2}+2 c}+1}d x \right ) a b \,d^{2}-2 \,\mathrm {log}\left (e^{2 d \,x^{2}+2 c}+1\right ) b^{2}+a^{2} d^{2} x^{4}}{4 d^{2} \left (e^{2 d \,x^{2}+2 c}+1\right )} \] Input:

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

Output:

(8*e**(2*c + 2*d*x**2)*atan(e**(c + d*x**2))*a*b + 8*atan(e**(c + d*x**2)) 
*a*b + 32*e**(3*c + 2*d*x**2)*int((e**(d*x**2)*x**3)/(e**(4*c + 4*d*x**2) 
+ 2*e**(2*c + 2*d*x**2) + 1),x)*a*b*d**2 - 2*e**(2*c + 2*d*x**2)*log(e**(2 
*c + 2*d*x**2) + 1)*b**2 + e**(2*c + 2*d*x**2)*a**2*d**2*x**4 + 4*e**(2*c 
+ 2*d*x**2)*b**2*d*x**2 - 8*e**(c + d*x**2)*a*b*d*x**2 + 32*e**c*int((e**( 
d*x**2)*x**3)/(e**(4*c + 4*d*x**2) + 2*e**(2*c + 2*d*x**2) + 1),x)*a*b*d** 
2 - 2*log(e**(2*c + 2*d*x**2) + 1)*b**2 + a**2*d**2*x**4)/(4*d**2*(e**(2*c 
 + 2*d*x**2) + 1))