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

Optimal. Leaf size=119 \[ \frac {a^2 x^4}{4}-\frac {i a b \text {Li}_2\left (-i e^{d x^2+c}\right )}{d^2}+\frac {i a b \text {Li}_2\left (i e^{d x^2+c}\right )}{d^2}+\frac {2 a b x^2 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 \log \left (\cosh \left (c+d x^2\right )\right )}{2 d^2}+\frac {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5436, 4190, 4180, 2279, 2391, 4184, 3475} \[ -\frac {i a b \text {PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac {i a b \text {PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac {a^2 x^4}{4}+\frac {2 a b x^2 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac {b^2 \log \left (\cosh \left (c+d x^2\right )\right )}{2 d^2}+\frac {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sech[c + d*x^2])^2,x]

[Out]

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4190

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 5436

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(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[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int x^3 \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b \text {sech}(c+d x))^2 \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (a^2 x+2 a b x \text {sech}(c+d x)+b^2 x \text {sech}^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac {a^2 x^4}{4}+(a b) \operatorname {Subst}\left (\int x \text {sech}(c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int x \text {sech}^2(c+d x) \, dx,x,x^2\right )\\ &=\frac {a^2 x^4}{4}+\frac {2 a b x^2 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}+\frac {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d}-\frac {(i a b) \operatorname {Subst}\left (\int \log \left (1-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}+\frac {(i a b) \operatorname {Subst}\left (\int \log \left (1+i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}-\frac {b^2 \operatorname {Subst}\left (\int \tanh (c+d x) \, dx,x,x^2\right )}{2 d}\\ &=\frac {a^2 x^4}{4}+\frac {2 a b x^2 \tan ^{-1}\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 {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d}-\frac {(i a b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^2}+\frac {(i a b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^2}\\ &=\frac {a^2 x^4}{4}+\frac {2 a b x^2 \tan ^{-1}\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 \text {Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac {i a b \text {Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac {b^2 x^2 \tanh \left (c+d x^2\right )}{2 d}\\ \end {align*}

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Mathematica [B]  time = 3.20, size = 273, normalized size = 2.29 \[ \frac {\cosh \left (c+d x^2\right ) \left (a+b \text {sech}\left (c+d x^2\right )\right )^2 \left (d x^2 \cosh \left (c+d x^2\right ) \left (a^2 d x^2+2 b^2 \tanh (c)\right )+4 a b \cosh \left (c+d x^2\right ) \left (\frac {\text {csch}(c) \left (\text {Li}_2\left (-e^{-d x^2-\tanh ^{-1}(\coth (c))}\right )-\text {Li}_2\left (e^{-d x^2-\tanh ^{-1}(\coth (c))}\right )+\left (\tanh ^{-1}(\coth (c))+d x^2\right ) \left (\log \left (1-e^{-\tanh ^{-1}(\coth (c))-d x^2}\right )-\log \left (e^{-\tanh ^{-1}(\coth (c))-d x^2}+1\right )\right )\right )}{\sqrt {-\text {csch}^2(c)}}-2 \tanh ^{-1}(\coth (c)) \tan ^{-1}\left (\cosh (c) \tanh \left (\frac {d x^2}{2}\right )+\sinh (c)\right )\right )-2 b^2 d x^2 \tanh (c) \cosh \left (c+d x^2\right )+2 b^2 d x^2 \text {sech}(c) \sinh \left (d x^2\right )-2 b^2 \cosh \left (c+d x^2\right ) \left (\log \left (\cosh \left (c+d x^2\right )\right )-d x^2 \tanh (c)\right )\right )}{4 d^2 \left (a \cosh \left (c+d x^2\right )+b\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cosh[c + d*x^2]*(a + b*Sech[c + d*x^2])^2*(4*a*b*Cosh[c + d*x^2]*(-2*ArcTan[Sinh[c] + Cosh[c]*Tanh[(d*x^2)/2]
]*ArcTanh[Coth[c]] + (Csch[c]*((d*x^2 + ArcTanh[Coth[c]])*(Log[1 - E^(-(d*x^2) - ArcTanh[Coth[c]])] - Log[1 +
E^(-(d*x^2) - ArcTanh[Coth[c]])]) + PolyLog[2, -E^(-(d*x^2) - ArcTanh[Coth[c]])] - PolyLog[2, E^(-(d*x^2) - Ar
cTanh[Coth[c]])]))/Sqrt[-Csch[c]^2]) + 2*b^2*d*x^2*Sech[c]*Sinh[d*x^2] - 2*b^2*d*x^2*Cosh[c + d*x^2]*Tanh[c] +
d*x^2*Cosh[c + d*x^2]*(a^2*d*x^2 + 2*b^2*Tanh[c]) - 2*b^2*Cosh[c + d*x^2]*(Log[Cosh[c + d*x^2]] - d*x^2*Tanh[
c])))/(4*d^2*(b + a*Cosh[c + d*x^2])^2)

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fricas [B]  time = 0.48, size = 782, normalized size = 6.57 \[ \frac {a^{2} d^{2} x^{4} + 4 \, b^{2} c + {\left (a^{2} d^{2} x^{4} + 4 \, b^{2} d x^{2} + 4 \, b^{2} c\right )} \cosh \left (d x^{2} + c\right )^{2} + 2 \, {\left (a^{2} d^{2} x^{4} + 4 \, b^{2} d x^{2} + 4 \, b^{2} c\right )} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) + {\left (a^{2} d^{2} x^{4} + 4 \, b^{2} d x^{2} + 4 \, b^{2} c\right )} \sinh \left (d x^{2} + c\right )^{2} + {\left (4 i \, a b \cosh \left (d x^{2} + c\right )^{2} + 8 i \, a b \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) + 4 i \, a b \sinh \left (d x^{2} + c\right )^{2} + 4 i \, a b\right )} {\rm Li}_2\left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right )\right ) + {\left (-4 i \, a b \cosh \left (d x^{2} + c\right )^{2} - 8 i \, a b \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) - 4 i \, a b \sinh \left (d x^{2} + c\right )^{2} - 4 i \, a b\right )} {\rm Li}_2\left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right )\right ) + {\left (-4 i \, a b c - 2 \, {\left (2 i \, a b c + b^{2}\right )} \cosh \left (d x^{2} + c\right )^{2} - 4 \, {\left (2 i \, a b c + b^{2}\right )} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) - 2 \, {\left (2 i \, a b c + b^{2}\right )} \sinh \left (d x^{2} + c\right )^{2} - 2 \, b^{2}\right )} \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) + i\right ) + {\left (4 i \, a b c - 2 \, {\left (-2 i \, a b c + b^{2}\right )} \cosh \left (d x^{2} + c\right )^{2} - 4 \, {\left (-2 i \, a b c + b^{2}\right )} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) - 2 \, {\left (-2 i \, a b c + b^{2}\right )} \sinh \left (d x^{2} + c\right )^{2} - 2 \, b^{2}\right )} \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) - i\right ) + {\left (-4 i \, a b d x^{2} - 4 i \, a b c + {\left (-4 i \, a b d x^{2} - 4 i \, a b c\right )} \cosh \left (d x^{2} + c\right )^{2} + {\left (-8 i \, a b d x^{2} - 8 i \, a b c\right )} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) + {\left (-4 i \, a b d x^{2} - 4 i \, a b c\right )} \sinh \left (d x^{2} + c\right )^{2}\right )} \log \left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right ) + 1\right ) + {\left (4 i \, a b d x^{2} + 4 i \, a b c + {\left (4 i \, a b d x^{2} + 4 i \, a b c\right )} \cosh \left (d x^{2} + c\right )^{2} + {\left (8 i \, a b d x^{2} + 8 i \, a b c\right )} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) + {\left (4 i \, a b d x^{2} + 4 i \, a b c\right )} \sinh \left (d x^{2} + c\right )^{2}\right )} \log \left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right ) + 1\right )}{4 \, {\left (d^{2} \cosh \left (d x^{2} + c\right )^{2} + 2 \, d^{2} \cosh \left (d x^{2} + c\right ) \sinh \left (d x^{2} + c\right ) + d^{2} \sinh \left (d x^{2} + c\right )^{2} + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 + 8*I*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) + 4*I*a*b*sinh(d*x^2 + c)^2 + 4*I*a*b)*dil
og(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c)) + (-4*I*a*b*cosh(d*x^2 + c)^2 - 8*I*a*b*cosh(d*x^2 + c)*sinh(d*x^2 +
c) - 4*I*a*b*sinh(d*x^2 + c)^2 - 4*I*a*b)*dilog(-I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c)) + (-4*I*a*b*c - 2*(2*
I*a*b*c + b^2)*cosh(d*x^2 + c)^2 - 4*(2*I*a*b*c + b^2)*cosh(d*x^2 + c)*sinh(d*x^2 + c) - 2*(2*I*a*b*c + b^2)*s
inh(d*x^2 + c)^2 - 2*b^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) + I) + (4*I*a*b*c - 2*(-2*I*a*b*c + b^2)*cosh(
d*x^2 + c)^2 - 4*(-2*I*a*b*c + b^2)*cosh(d*x^2 + c)*sinh(d*x^2 + c) - 2*(-2*I*a*b*c + b^2)*sinh(d*x^2 + c)^2 -
2*b^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) - I) + (-4*I*a*b*d*x^2 - 4*I*a*b*c + (-4*I*a*b*d*x^2 - 4*I*a*b*c
)*cosh(d*x^2 + c)^2 + (-8*I*a*b*d*x^2 - 8*I*a*b*c)*cosh(d*x^2 + c)*sinh(d*x^2 + c) + (-4*I*a*b*d*x^2 - 4*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 + 4*I*a*b*c + (4*I*a*b*
d*x^2 + 4*I*a*b*c)*cosh(d*x^2 + c)^2 + (8*I*a*b*d*x^2 + 8*I*a*b*c)*cosh(d*x^2 + c)*sinh(d*x^2 + c) + (4*I*a*b*
d*x^2 + 4*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {sech}\left (d x^{2} + c\right ) + a\right )}^{2} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \,\mathrm {sech}\left (d \,x^{2}+c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a^{2} x^{4} + \frac {1}{2} \, {\left (\frac {2 \, x^{2} e^{\left (2 \, d x^{2} + 2 \, c\right )}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} + d} - \frac {\log \left ({\left (e^{\left (2 \, d x^{2} + 2 \, c\right )} + 1\right )} e^{\left (-2 \, c\right )}\right )}{d^{2}}\right )} b^{2} + 4 \, a b \int \frac {x^{3} e^{\left (d x^{2} + c\right )}}{e^{\left (2 \, d x^{2} + 2 \, c\right )} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\left (a+\frac {b}{\mathrm {cosh}\left (d\,x^2+c\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/cosh(c + d*x^2))^2,x)

[Out]

int(x^3*(a + b/cosh(c + d*x^2))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \operatorname {sech}{\left (c + d x^{2} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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