3.5.6 \(\int \frac {(c+a^2 c x^2)^3}{\sinh ^{-1}(a x)^2} \, dx\) [406]

Optimal. Leaf size=94 \[ -\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \]

[Out]

-c^3*(a^2*x^2+1)^(7/2)/a/arcsinh(a*x)+35/64*c^3*Shi(arcsinh(a*x))/a+63/64*c^3*Shi(3*arcsinh(a*x))/a+35/64*c^3*
Shi(5*arcsinh(a*x))/a+7/64*c^3*Shi(7*arcsinh(a*x))/a

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Rubi [A]
time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5790, 5819, 5556, 3379} \begin {gather*} -\frac {c^3 \left (a^2 x^2+1\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^3/ArcSinh[a*x]^2,x]

[Out]

-((c^3*(1 + a^2*x^2)^(7/2))/(a*ArcSinh[a*x])) + (35*c^3*SinhIntegral[ArcSinh[a*x]])/(64*a) + (63*c^3*SinhInteg
ral[3*ArcSinh[a*x]])/(64*a) + (35*c^3*SinhIntegral[5*ArcSinh[a*x]])/(64*a) + (7*c^3*SinhIntegral[7*ArcSinh[a*x
]])/(64*a)

Rule 3379

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 5556

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 5790

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[Simp[Sqrt[1 + c^2*
x^2]*(d + e*x^2)^p]*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[c*((2*p + 1)/(b*(n + 1)))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b
, c, d, e, p}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
0]

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^3}{\sinh ^{-1}(a x)^2} \, dx &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\left (7 a c^3\right ) \int \frac {x \left (1+a^2 x^2\right )^{5/2}}{\sinh ^{-1}(a x)} \, dx\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \frac {\cosh ^6(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \left (\frac {5 \sinh (x)}{64 x}+\frac {9 \sinh (3 x)}{64 x}+\frac {5 \sinh (5 x)}{64 x}+\frac {\sinh (7 x)}{64 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \frac {\sinh (7 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int \frac {\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (63 c^3\right ) \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 82, normalized size = 0.87 \begin {gather*} \frac {c^3 \left (-64 \left (1+a^2 x^2\right )^{7/2}+35 \sinh ^{-1}(a x) \text {Shi}\left (\sinh ^{-1}(a x)\right )+63 \sinh ^{-1}(a x) \text {Shi}\left (3 \sinh ^{-1}(a x)\right )+35 \sinh ^{-1}(a x) \text {Shi}\left (5 \sinh ^{-1}(a x)\right )+7 \sinh ^{-1}(a x) \text {Shi}\left (7 \sinh ^{-1}(a x)\right )\right )}{64 a \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + a^2*c*x^2)^3/ArcSinh[a*x]^2,x]

[Out]

(c^3*(-64*(1 + a^2*x^2)^(7/2) + 35*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] + 63*ArcSinh[a*x]*SinhIntegral[3*Ar
cSinh[a*x]] + 35*ArcSinh[a*x]*SinhIntegral[5*ArcSinh[a*x]] + 7*ArcSinh[a*x]*SinhIntegral[7*ArcSinh[a*x]]))/(64
*a*ArcSinh[a*x])

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Maple [A]
time = 3.36, size = 106, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {c^{3} \left (7 \hyperbolicSineIntegral \left (7 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+63 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-21 \cosh \left (3 \arcsinh \left (a x \right )\right )-7 \cosh \left (5 \arcsinh \left (a x \right )\right )-\cosh \left (7 \arcsinh \left (a x \right )\right )-35 \sqrt {a^{2} x^{2}+1}\right )}{64 a \arcsinh \left (a x \right )}\)	\(106\)
default	\(\frac {c^{3} \left (7 \hyperbolicSineIntegral \left (7 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+63 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-21 \cosh \left (3 \arcsinh \left (a x \right )\right )-7 \cosh \left (5 \arcsinh \left (a x \right )\right )-\cosh \left (7 \arcsinh \left (a x \right )\right )-35 \sqrt {a^{2} x^{2}+1}\right )}{64 a \arcsinh \left (a x \right )}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^3/arcsinh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/64/a*c^3*(7*Shi(7*arcsinh(a*x))*arcsinh(a*x)+63*Shi(3*arcsinh(a*x))*arcsinh(a*x)+35*Shi(5*arcsinh(a*x))*arcs
inh(a*x)+35*Shi(arcsinh(a*x))*arcsinh(a*x)-21*cosh(3*arcsinh(a*x))-7*cosh(5*arcsinh(a*x))-cosh(7*arcsinh(a*x))
-35*(a^2*x^2+1)^(1/2))/arcsinh(a*x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^9*c^3*x^9 + 4*a^7*c^3*x^7 + 6*a^5*c^3*x^5 + 4*a^3*c^3*x^3 + a*c^3*x + (a^8*c^3*x^8 + 4*a^6*c^3*x^6 + 6*a^4
*c^3*x^4 + 4*a^2*c^3*x^2 + c^3)*sqrt(a^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a^2
*x^2 + 1))) + integrate((7*a^10*c^3*x^10 + 29*a^8*c^3*x^8 + 46*a^6*c^3*x^6 + 34*a^4*c^3*x^4 + 11*a^2*c^3*x^2 +
c^3 + (7*a^8*c^3*x^8 + 20*a^6*c^3*x^6 + 18*a^4*c^3*x^4 + 4*a^2*c^3*x^2 - c^3)*(a^2*x^2 + 1) + 7*(2*a^9*c^3*x^
9 + 7*a^7*c^3*x^7 + 9*a^5*c^3*x^5 + 5*a^3*c^3*x^3 + a*c^3*x)*sqrt(a^2*x^2 + 1))/((a^4*x^4 + (a^2*x^2 + 1)*a^2*
x^2 + 2*a^2*x^2 + 2*(a^3*x^3 + a*x)*sqrt(a^2*x^2 + 1) + 1)*log(a*x + sqrt(a^2*x^2 + 1))), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)/arcsinh(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{3} \left (\int \frac {3 a^{2} x^{2}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {3 a^{4} x^{4}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {a^{6} x^{6}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**3/asinh(a*x)**2,x)

[Out]

c**3*(Integral(3*a**2*x**2/asinh(a*x)**2, x) + Integral(3*a**4*x**4/asinh(a*x)**2, x) + Integral(a**6*x**6/asi
nh(a*x)**2, x) + Integral(asinh(a*x)**(-2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

integrate((a^2*c*x^2 + c)^3/arcsinh(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,a^2\,x^2+c\right )}^3}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + a^2*c*x^2)^3/asinh(a*x)^2,x)

[Out]

int((c + a^2*c*x^2)^3/asinh(a*x)^2, x)

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