3.44 \(\int (d+e x)^3 (a+b \text{csch}^{-1}(c x)) \, dx\)

Optimal. Leaf size=167 \[ \frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b e x \sqrt{\frac{1}{c^2 x^2}+1} \left (9 c^2 d^2-e^2\right )}{6 c^3}+\frac{b d \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{2 c^3}+\frac{b d e^2 x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{2 c}+\frac{b e^3 x^3 \sqrt{\frac{1}{c^2 x^2}+1}}{12 c}-\frac{b d^4 \text{csch}^{-1}(c x)}{4 e} \]

[Out]

(b*e*(9*c^2*d^2 - e^2)*Sqrt[1 + 1/(c^2*x^2)]*x)/(6*c^3) + (b*d*e^2*Sqrt[1 + 1/(c^2*x^2)]*x^2)/(2*c) + (b*e^3*S
qrt[1 + 1/(c^2*x^2)]*x^3)/(12*c) - (b*d^4*ArcCsch[c*x])/(4*e) + ((d + e*x)^4*(a + b*ArcCsch[c*x]))/(4*e) + (b*
d*(2*c^2*d^2 - e^2)*ArcTanh[Sqrt[1 + 1/(c^2*x^2)]])/(2*c^3)

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Rubi [A]  time = 0.38363, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {6290, 1568, 1475, 1807, 844, 215, 266, 63, 208} \[ \frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b e x \sqrt{\frac{1}{c^2 x^2}+1} \left (9 c^2 d^2-e^2\right )}{6 c^3}+\frac{b d \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{2 c^3}+\frac{b d e^2 x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{2 c}+\frac{b e^3 x^3 \sqrt{\frac{1}{c^2 x^2}+1}}{12 c}-\frac{b d^4 \text{csch}^{-1}(c x)}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*ArcCsch[c*x]),x]

[Out]

(b*e*(9*c^2*d^2 - e^2)*Sqrt[1 + 1/(c^2*x^2)]*x)/(6*c^3) + (b*d*e^2*Sqrt[1 + 1/(c^2*x^2)]*x^2)/(2*c) + (b*e^3*S
qrt[1 + 1/(c^2*x^2)]*x^3)/(12*c) - (b*d^4*ArcCsch[c*x])/(4*e) + ((d + e*x)^4*(a + b*ArcCsch[c*x]))/(4*e) + (b*
d*(2*c^2*d^2 - e^2)*ArcTanh[Sqrt[1 + 1/(c^2*x^2)]])/(2*c^3)

Rule 6290

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a +
b*ArcCsch[c*x]))/(e*(m + 1)), x] + Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 + 1/(c^2*x^2)]), x]
, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rule 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 266

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b \text{csch}^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b \int \frac{(d+e x)^4}{\sqrt{1+\frac{1}{c^2 x^2}} x^2} \, dx}{4 c e}\\ &=\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b \int \frac{\left (e+\frac{d}{x}\right )^4 x^2}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{4 c e}\\ &=\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{b \operatorname{Subst}\left (\int \frac{(e+d x)^4}{x^4 \sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{4 c e}\\ &=\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b \operatorname{Subst}\left (\int \frac{-12 d e^3-2 e^2 \left (9 d^2-\frac{e^2}{c^2}\right ) x-12 d^3 e x^2-3 d^4 x^3}{x^3 \sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{12 c e}\\ &=\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{b \operatorname{Subst}\left (\int \frac{4 e^2 \left (9 d^2-\frac{e^2}{c^2}\right )+12 d e \left (2 d^2-\frac{e^2}{c^2}\right ) x+6 d^4 x^2}{x^2 \sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{24 c e}\\ &=\frac{b e \left (9 c^2 d^2-e^2\right ) \sqrt{1+\frac{1}{c^2 x^2}} x}{6 c^3}+\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b \operatorname{Subst}\left (\int \frac{-12 d e \left (2 d^2-\frac{e^2}{c^2}\right )-6 d^4 x}{x \sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{24 c e}\\ &=\frac{b e \left (9 c^2 d^2-e^2\right ) \sqrt{1+\frac{1}{c^2 x^2}} x}{6 c^3}+\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{\left (b d^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{4 c e}-\frac{\left (b d \left (2 d^2-\frac{e^2}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c}\\ &=\frac{b e \left (9 c^2 d^2-e^2\right ) \sqrt{1+\frac{1}{c^2 x^2}} x}{6 c^3}+\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}-\frac{b d^4 \text{csch}^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{\left (b d \left (2 c^2 d^2-e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 c^3}\\ &=\frac{b e \left (9 c^2 d^2-e^2\right ) \sqrt{1+\frac{1}{c^2 x^2}} x}{6 c^3}+\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}-\frac{b d^4 \text{csch}^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{\left (b d \left (2 c^2 d^2-e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-c^2+c^2 x^2} \, dx,x,\sqrt{1+\frac{1}{c^2 x^2}}\right )}{2 c}\\ &=\frac{b e \left (9 c^2 d^2-e^2\right ) \sqrt{1+\frac{1}{c^2 x^2}} x}{6 c^3}+\frac{b d e^2 \sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c}+\frac{b e^3 \sqrt{1+\frac{1}{c^2 x^2}} x^3}{12 c}-\frac{b d^4 \text{csch}^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{b d \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\sqrt{1+\frac{1}{c^2 x^2}}\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.279445, size = 165, normalized size = 0.99 \[ \frac{3 a c^3 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+b e x \sqrt{\frac{1}{c^2 x^2}+1} \left (c^2 \left (18 d^2+6 d e x+e^2 x^2\right )-2 e^2\right )+6 b d \left (2 c^2 d^2-e^2\right ) \log \left (x \left (\sqrt{\frac{1}{c^2 x^2}+1}+1\right )\right )+3 b c^3 x \text{csch}^{-1}(c x) \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a + b*ArcCsch[c*x]),x]

[Out]

(3*a*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b*e*Sqrt[1 + 1/(c^2*x^2)]*x*(-2*e^2 + c^2*(18*d^2 + 6
*d*e*x + e^2*x^2)) + 3*b*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcCsch[c*x] + 6*b*d*(2*c^2*d^2 - e
^2)*Log[(1 + Sqrt[1 + 1/(c^2*x^2)])*x])/(12*c^3)

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Maple [A]  time = 0.178, size = 269, normalized size = 1.6 \begin{align*}{\frac{1}{c} \left ({\frac{ \left ( cxe+cd \right ) ^{4}a}{4\,{c}^{3}e}}+{\frac{b}{{c}^{3}} \left ({\frac{{e}^{3}{\rm arccsch} \left (cx\right ){c}^{4}{x}^{4}}{4}}+{e}^{2}{\rm arccsch} \left (cx\right ){c}^{4}{x}^{3}d+{\frac{3\,e{\rm arccsch} \left (cx\right ){c}^{4}{x}^{2}{d}^{2}}{2}}+{\rm arccsch} \left (cx\right ){c}^{4}x{d}^{3}+{\frac{{\rm arccsch} \left (cx\right ){c}^{4}{d}^{4}}{4\,e}}+{\frac{1}{12\,cxe}\sqrt{{c}^{2}{x}^{2}+1} \left ( -3\,{c}^{4}{d}^{4}{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) +12\,{c}^{3}{d}^{3}e{\it Arcsinh} \left ( cx \right ) +{e}^{4}{c}^{2}{x}^{2}\sqrt{{c}^{2}{x}^{2}+1}+6\,{c}^{2}d{e}^{3}x\sqrt{{c}^{2}{x}^{2}+1}+18\,{c}^{2}{d}^{2}{e}^{2}\sqrt{{c}^{2}{x}^{2}+1}-6\,cd{e}^{3}{\it Arcsinh} \left ( cx \right ) -2\,{e}^{4}\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arccsch(c*x)),x)

[Out]

1/c*(1/4*(c*e*x+c*d)^4*a/c^3/e+b/c^3*(1/4*e^3*arccsch(c*x)*c^4*x^4+e^2*arccsch(c*x)*c^4*x^3*d+3/2*e*arccsch(c*
x)*c^4*x^2*d^2+arccsch(c*x)*c^4*x*d^3+1/4/e*arccsch(c*x)*c^4*d^4+1/12/e*(c^2*x^2+1)^(1/2)*(-3*c^4*d^4*arctanh(
1/(c^2*x^2+1)^(1/2))+12*c^3*d^3*e*arcsinh(c*x)+e^4*c^2*x^2*(c^2*x^2+1)^(1/2)+6*c^2*d*e^3*x*(c^2*x^2+1)^(1/2)+1
8*c^2*d^2*e^2*(c^2*x^2+1)^(1/2)-6*c*d*e^3*arcsinh(c*x)-2*e^4*(c^2*x^2+1)^(1/2))/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/
x))

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Maxima [A]  time = 1.01557, size = 352, normalized size = 2.11 \begin{align*} \frac{1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac{3}{2} \, a d^{2} e x^{2} + \frac{3}{2} \,{\left (x^{2} \operatorname{arcsch}\left (c x\right ) + \frac{x \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c}\right )} b d^{2} e + \frac{1}{4} \,{\left (4 \, x^{3} \operatorname{arcsch}\left (c x\right ) + \frac{\frac{2 \, \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} + 1\right )} - c^{2}} - \frac{\log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} + \frac{\log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b d e^{2} + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arcsch}\left (c x\right ) + \frac{c^{2} x^{3}{\left (\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 3 \, x \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e^{3} + a d^{3} x + \frac{{\left (2 \, c x \operatorname{arcsch}\left (c x\right ) + \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + 1\right ) - \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} - 1\right )\right )} b d^{3}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e^3*x^4 + a*d*e^2*x^3 + 3/2*a*d^2*e*x^2 + 3/2*(x^2*arccsch(c*x) + x*sqrt(1/(c^2*x^2) + 1)/c)*b*d^2*e + 1
/4*(4*x^3*arccsch(c*x) + (2*sqrt(1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) + 1) - c^2) - log(sqrt(1/(c^2*x^2) + 1) +
1)/c^2 + log(sqrt(1/(c^2*x^2) + 1) - 1)/c^2)/c)*b*d*e^2 + 1/12*(3*x^4*arccsch(c*x) + (c^2*x^3*(1/(c^2*x^2) + 1
)^(3/2) - 3*x*sqrt(1/(c^2*x^2) + 1))/c^3)*b*e^3 + a*d^3*x + 1/2*(2*c*x*arccsch(c*x) + log(sqrt(1/(c^2*x^2) + 1
) + 1) - log(sqrt(1/(c^2*x^2) + 1) - 1))*b*d^3/c

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Fricas [B]  time = 3.64093, size = 898, normalized size = 5.38 \begin{align*} \frac{3 \, a c^{3} e^{3} x^{4} + 12 \, a c^{3} d e^{2} x^{3} + 18 \, a c^{3} d^{2} e x^{2} + 12 \, a c^{3} d^{3} x + 3 \,{\left (4 \, b c^{3} d^{3} + 6 \, b c^{3} d^{2} e + 4 \, b c^{3} d e^{2} + b c^{3} e^{3}\right )} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - 6 \,{\left (2 \, b c^{2} d^{3} - b d e^{2}\right )} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right ) - 3 \,{\left (4 \, b c^{3} d^{3} + 6 \, b c^{3} d^{2} e + 4 \, b c^{3} d e^{2} + b c^{3} e^{3}\right )} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) + 3 \,{\left (b c^{3} e^{3} x^{4} + 4 \, b c^{3} d e^{2} x^{3} + 6 \, b c^{3} d^{2} e x^{2} + 4 \, b c^{3} d^{3} x - 4 \, b c^{3} d^{3} - 6 \, b c^{3} d^{2} e - 4 \, b c^{3} d e^{2} - b c^{3} e^{3}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) +{\left (b c^{2} e^{3} x^{3} + 6 \, b c^{2} d e^{2} x^{2} + 2 \,{\left (9 \, b c^{2} d^{2} e - b e^{3}\right )} x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{12 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^3*e^3*x^4 + 12*a*c^3*d*e^2*x^3 + 18*a*c^3*d^2*e*x^2 + 12*a*c^3*d^3*x + 3*(4*b*c^3*d^3 + 6*b*c^3*d^
2*e + 4*b*c^3*d*e^2 + b*c^3*e^3)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x + 1) - 6*(2*b*c^2*d^3 - b*d*e^2)*
log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x) - 3*(4*b*c^3*d^3 + 6*b*c^3*d^2*e + 4*b*c^3*d*e^2 + b*c^3*e^3)*log
(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) + 3*(b*c^3*e^3*x^4 + 4*b*c^3*d*e^2*x^3 + 6*b*c^3*d^2*e*x^2 + 4*b
*c^3*d^3*x - 4*b*c^3*d^3 - 6*b*c^3*d^2*e - 4*b*c^3*d*e^2 - b*c^3*e^3)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) +
1)/(c*x)) + (b*c^2*e^3*x^3 + 6*b*c^2*d*e^2*x^2 + 2*(9*b*c^2*d^2*e - b*e^3)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/
c^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{acsch}{\left (c x \right )}\right ) \left (d + e x\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*acsch(c*x)),x)

[Out]

Integral((a + b*acsch(c*x))*(d + e*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{3}{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^3*(b*arccsch(c*x) + a), x)