3.1.27 \(\int \frac {(d+c^2 d x^2)^3 (a+b \sinh ^{-1}(c x))}{x^4} \, dx\) [27]

Optimal. Leaf size=174 \[ -\frac {8}{3} b c^3 d^3 \sqrt {1+c^2 x^2}-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {1}{9} b c^3 d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right ) \]

[Out]

-1/9*b*c^3*d^3*(c^2*x^2+1)^(3/2)-1/3*d^3*(a+b*arcsinh(c*x))/x^3-3*c^2*d^3*(a+b*arcsinh(c*x))/x+3*c^4*d^3*x*(a+
b*arcsinh(c*x))+1/3*c^6*d^3*x^3*(a+b*arcsinh(c*x))-17/6*b*c^3*d^3*arctanh((c^2*x^2+1)^(1/2))-8/3*b*c^3*d^3*(c^
2*x^2+1)^(1/2)-1/6*b*c*d^3*(c^2*x^2+1)^(1/2)/x^2

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Rubi [A]
time = 0.17, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {276, 5803, 12, 1813, 1635, 911, 1167, 214} \begin {gather*} \frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {b c d^3 \sqrt {c^2 x^2+1}}{6 x^2}-\frac {1}{9} b c^3 d^3 \left (c^2 x^2+1\right )^{3/2}-\frac {8}{3} b c^3 d^3 \sqrt {c^2 x^2+1}-\frac {17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^3*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(-8*b*c^3*d^3*Sqrt[1 + c^2*x^2])/3 - (b*c*d^3*Sqrt[1 + c^2*x^2])/(6*x^2) - (b*c^3*d^3*(1 + c^2*x^2)^(3/2))/9 -
(d^3*(a + b*ArcSinh[c*x]))/(3*x^3) - (3*c^2*d^3*(a + b*ArcSinh[c*x]))/x + 3*c^4*d^3*x*(a + b*ArcSinh[c*x]) +
(c^6*d^3*x^3*(a + b*ArcSinh[c*x]))/3 - (17*b*c^3*d^3*ArcTanh[Sqrt[1 + c^2*x^2]])/6

Rule 12

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 276

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

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1635

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(
b*c - a*d))), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac {d^3 \left (-1-9 c^2 x^2+9 c^4 x^4+c^6 x^6\right )}{3 x^3 \sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{3} \left (b c d^3\right ) \int \frac {-1-9 c^2 x^2+9 c^4 x^4+c^6 x^6}{x^3 \sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{6} \left (b c d^3\right ) \text {Subst}\left (\int \frac {-1-9 c^2 x+9 c^4 x^2+c^6 x^3}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} \left (b c d^3\right ) \text {Subst}\left (\int \frac {\frac {17 c^2}{2}-9 c^4 x-c^6 x^2}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (b d^3\right ) \text {Subst}\left (\int \frac {\frac {33 c^2}{2}-7 c^2 x^2-c^2 x^4}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{3 c}\\ &=-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (b d^3\right ) \text {Subst}\left (\int \left (-8 c^4-c^4 x^2+\frac {17 c^2}{2 \left (-\frac {1}{c^2}+\frac {x^2}{c^2}\right )}\right ) \, dx,x,\sqrt {1+c^2 x^2}\right )}{3 c}\\ &=-\frac {8}{3} b c^3 d^3 \sqrt {1+c^2 x^2}-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {1}{9} b c^3 d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} \left (17 b c d^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )\\ &=-\frac {8}{3} b c^3 d^3 \sqrt {1+c^2 x^2}-\frac {b c d^3 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {1}{9} b c^3 d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^6 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 171, normalized size = 0.98 \begin {gather*} \frac {d^3 \left (-6 a-54 a c^2 x^2+54 a c^4 x^4+6 a c^6 x^6-3 b c x \sqrt {1+c^2 x^2}-50 b c^3 x^3 \sqrt {1+c^2 x^2}-2 b c^5 x^5 \sqrt {1+c^2 x^2}+6 b \left (-1-9 c^2 x^2+9 c^4 x^4+c^6 x^6\right ) \sinh ^{-1}(c x)+51 b c^3 x^3 \log (x)-51 b c^3 x^3 \log \left (1+\sqrt {1+c^2 x^2}\right )\right )}{18 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)^3*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(d^3*(-6*a - 54*a*c^2*x^2 + 54*a*c^4*x^4 + 6*a*c^6*x^6 - 3*b*c*x*Sqrt[1 + c^2*x^2] - 50*b*c^3*x^3*Sqrt[1 + c^2
*x^2] - 2*b*c^5*x^5*Sqrt[1 + c^2*x^2] + 6*b*(-1 - 9*c^2*x^2 + 9*c^4*x^4 + c^6*x^6)*ArcSinh[c*x] + 51*b*c^3*x^3
*Log[x] - 51*b*c^3*x^3*Log[1 + Sqrt[1 + c^2*x^2]]))/(18*x^3)

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Maple [A]
time = 0.53, size = 155, normalized size = 0.89

method	result	size
derivativedivides	\(c^{3} \left (a \,d^{3} \left (\frac {c^{3} x^{3}}{3}+3 c x -\frac {1}{3 c^{3} x^{3}}-\frac {3}{c x}\right )+d^{3} b \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+3 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \arcsinh \left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {25 \sqrt {c^{2} x^{2}+1}}{9}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {17 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(155\)
default	\(c^{3} \left (a \,d^{3} \left (\frac {c^{3} x^{3}}{3}+3 c x -\frac {1}{3 c^{3} x^{3}}-\frac {3}{c x}\right )+d^{3} b \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+3 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \arcsinh \left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {25 \sqrt {c^{2} x^{2}+1}}{9}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {17 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

c^3*(a*d^3*(1/3*c^3*x^3+3*c*x-1/3/c^3/x^3-3/c/x)+d^3*b*(1/3*arcsinh(c*x)*c^3*x^3+3*arcsinh(c*x)*c*x-1/3*arcsin
h(c*x)/c^3/x^3-3*arcsinh(c*x)/c/x-1/9*c^2*x^2*(c^2*x^2+1)^(1/2)-25/9*(c^2*x^2+1)^(1/2)-1/6/c^2/x^2*(c^2*x^2+1)
^(1/2)-17/6*arctanh(1/(c^2*x^2+1)^(1/2))))

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Maxima [A]
time = 0.26, size = 208, normalized size = 1.20 \begin {gather*} \frac {1}{3} \, a c^{6} d^{3} x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{6} d^{3} + 3 \, a c^{4} d^{3} x + 3 \, {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b c^{3} d^{3} - 3 \, {\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b c^{2} d^{3} + \frac {1}{6} \, {\left ({\left (c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac {2 \, \operatorname {arsinh}\left (c x\right )}{x^{3}}\right )} b d^{3} - \frac {3 \, a c^{2} d^{3}}{x} - \frac {a d^{3}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")

[Out]

1/3*a*c^6*d^3*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*b*c^6*d
^3 + 3*a*c^4*d^3*x + 3*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c^3*d^3 - 3*(c*arcsinh(1/(c*abs(x))) + arcsinh
(c*x)/x)*b*c^2*d^3 + 1/6*((c^2*arcsinh(1/(c*abs(x))) - sqrt(c^2*x^2 + 1)/x^2)*c - 2*arcsinh(c*x)/x^3)*b*d^3 -
3*a*c^2*d^3/x - 1/3*a*d^3/x^3

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Fricas [A]
time = 0.39, size = 289, normalized size = 1.66 \begin {gather*} \frac {6 \, a c^{6} d^{3} x^{6} + 54 \, a c^{4} d^{3} x^{4} - 51 \, b c^{3} d^{3} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) + 51 \, b c^{3} d^{3} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) - 54 \, a c^{2} d^{3} x^{2} - 6 \, {\left (b c^{6} + 9 \, b c^{4} - 9 \, b c^{2} - b\right )} d^{3} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) - 6 \, a d^{3} + 6 \, {\left (b c^{6} d^{3} x^{6} + 9 \, b c^{4} d^{3} x^{4} - 9 \, b c^{2} d^{3} x^{2} - {\left (b c^{6} + 9 \, b c^{4} - 9 \, b c^{2} - b\right )} d^{3} x^{3} - b d^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (2 \, b c^{5} d^{3} x^{5} + 50 \, b c^{3} d^{3} x^{3} + 3 \, b c d^{3} x\right )} \sqrt {c^{2} x^{2} + 1}}{18 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

1/18*(6*a*c^6*d^3*x^6 + 54*a*c^4*d^3*x^4 - 51*b*c^3*d^3*x^3*log(-c*x + sqrt(c^2*x^2 + 1) + 1) + 51*b*c^3*d^3*x
^3*log(-c*x + sqrt(c^2*x^2 + 1) - 1) - 54*a*c^2*d^3*x^2 - 6*(b*c^6 + 9*b*c^4 - 9*b*c^2 - b)*d^3*x^3*log(-c*x +
sqrt(c^2*x^2 + 1)) - 6*a*d^3 + 6*(b*c^6*d^3*x^6 + 9*b*c^4*d^3*x^4 - 9*b*c^2*d^3*x^2 - (b*c^6 + 9*b*c^4 - 9*b*
c^2 - b)*d^3*x^3 - b*d^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (2*b*c^5*d^3*x^5 + 50*b*c^3*d^3*x^3 + 3*b*c*d^3*x)*sq
rt(c^2*x^2 + 1))/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**3*(a+b*asinh(c*x))/x**4,x)

[Out]

d**3*(Integral(3*a*c**4, x) + Integral(a/x**4, x) + Integral(3*a*c**2/x**2, x) + Integral(a*c**6*x**2, x) + In
tegral(3*b*c**4*asinh(c*x), x) + Integral(b*asinh(c*x)/x**4, x) + Integral(3*b*c**2*asinh(c*x)/x**2, x) + Inte
gral(b*c**6*x**2*asinh(c*x), x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^3}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^3)/x^4,x)

[Out]

int(((a + b*asinh(c*x))*(d + c^2*d*x^2)^3)/x^4, x)

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