3.158 \(\int \frac {(c e+d e x)^2}{a+b \sinh ^{-1}(c+d x)} \, dx\)

Optimal. Leaf size=141 \[ -\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{4 b d}+\frac {e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{4 b d}+\frac {e^2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{4 b d}-\frac {e^2 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{4 b d} \]

[Out]

-1/4*e^2*Chi((a+b*arcsinh(d*x+c))/b)*cosh(a/b)/b/d+1/4*e^2*Chi(3*(a+b*arcsinh(d*x+c))/b)*cosh(3*a/b)/b/d+1/4*e
^2*Shi((a+b*arcsinh(d*x+c))/b)*sinh(a/b)/b/d-1/4*e^2*Shi(3*(a+b*arcsinh(d*x+c))/b)*sinh(3*a/b)/b/d

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Rubi [A]  time = 0.29, antiderivative size = 137, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5865, 12, 5669, 5448, 3303, 3298, 3301} \[ -\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{4 b d}-\frac {e^2 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2/(a + b*ArcSinh[c + d*x]),x]

[Out]

-(e^2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]])/(4*b*d) + (e^2*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*Ar
cSinh[c + d*x]])/(4*b*d) + (e^2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(4*b*d) - (e^2*Sinh[(3*a)/b]*S
inhIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(4*b*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^2}{a+b \sinh ^{-1}(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^2 x^2}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int \frac {x^2}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int \left (-\frac {\cosh (x)}{4 (a+b x)}+\frac {\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=-\frac {e^2 \operatorname {Subst}\left (\int \frac {\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}+\frac {e^2 \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}\\ &=-\frac {\left (e^2 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}+\frac {\left (e^2 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}+\frac {\left (e^2 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}-\frac {\left (e^2 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}\\ &=-\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{4 b d}-\frac {e^2 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 102, normalized size = 0.72 \[ \frac {e^2 \left (-\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^2/(a + b*ArcSinh[c + d*x]),x]

[Out]

(e^2*(-(Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]]) + Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c + d*x]
)] + Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] - Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])]))
/(4*b*d)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{b \operatorname {arsinh}\left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

integral((d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2)/(b*arcsinh(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^2/(b*arcsinh(d*x + c) + a), x)

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maple [A]  time = 0.16, size = 130, normalized size = 0.92 \[ \frac {-\frac {e^{2} {\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \arcsinh \left (d x +c \right )+\frac {3 a}{b}\right )}{8 b}+\frac {e^{2} {\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{8 b}+\frac {e^{2} {\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{8 b}-\frac {e^{2} {\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \arcsinh \left (d x +c \right )-\frac {3 a}{b}\right )}{8 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c)),x)

[Out]

1/d*(-1/8*e^2/b*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+3*a/b)+1/8*e^2/b*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)+1/8*e^2/b*
exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b)-1/8*e^2/b*exp(-3*a/b)*Ei(1,-3*arcsinh(d*x+c)-3*a/b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c)),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^2/(b*arcsinh(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^2/(a + b*asinh(c + d*x)),x)

[Out]

int((c*e + d*e*x)^2/(a + b*asinh(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**2/(a+b*asinh(d*x+c)),x)

[Out]

e**2*(Integral(c**2/(a + b*asinh(c + d*x)), x) + Integral(d**2*x**2/(a + b*asinh(c + d*x)), x) + Integral(2*c*
d*x/(a + b*asinh(c + d*x)), x))

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