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

Optimal. Leaf size=264 \[ \frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {b d^4 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e}-\frac {b d e^2 x \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{12 c^2}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (9 c^2 d^2+e^2\right )}{6 c^4}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (2 c^2 d^2+e^2\right ) \sin ^{-1}(c x)}{2 c^3} \]

[Out]

1/4*(e*x+d)^4*(a+b*arcsech(c*x))/e+1/2*b*d*(2*c^2*d^2+e^2)*arcsin(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^3-1/4
*b*d^4*arctanh((-c^2*x^2+1)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/e-1/6*b*e*(9*c^2*d^2+e^2)*(1/(c*x+1))^(1/2)
*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^4-1/2*b*d*e^2*x*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2-1/1
2*b*e^3*x^2*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2

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Rubi [A]  time = 0.36, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6288, 1809, 844, 216, 266, 63, 208} \[ \frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (9 c^2 d^2+e^2\right )}{6 c^4}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (2 c^2 d^2+e^2\right ) \sin ^{-1}(c x)}{2 c^3}-\frac {b d^4 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e}-\frac {b d e^2 x \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{12 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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 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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6288

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

Rubi steps

\begin {align*} \int (d+e x)^3 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(d+e x)^4}{x \sqrt {1-c^2 x^2}} \, dx}{4 e}\\ &=-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-3 c^2 d^4-12 c^2 d^3 e x-2 e^2 \left (9 c^2 d^2+e^2\right ) x^2-12 c^2 d e^3 x^3}{x \sqrt {1-c^2 x^2}} \, dx}{12 c^2 e}\\ &=-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {6 c^4 d^4+12 c^2 d e \left (2 c^2 d^2+e^2\right ) x+4 c^2 e^2 \left (9 c^2 d^2+e^2\right ) x^2}{x \sqrt {1-c^2 x^2}} \, dx}{24 c^4 e}\\ &=-\frac {b e \left (9 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^4}-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-6 c^6 d^4-12 c^4 d e \left (2 c^2 d^2+e^2\right ) x}{x \sqrt {1-c^2 x^2}} \, dx}{24 c^6 e}\\ &=-\frac {b e \left (9 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^4}-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x \sqrt {1-c^2 x^2}} \, dx}{4 e}+\frac {\left (b d \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{2 c^2}\\ &=-\frac {b e \left (9 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^4}-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {b d \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{2 c^3}+\frac {\left (b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{8 e}\\ &=-\frac {b e \left (9 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^4}-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {b d \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{2 c^3}-\frac {\left (b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{4 c^2 e}\\ &=-\frac {b e \left (9 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^4}-\frac {b d e^2 x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{2 c^2}-\frac {b e^3 x^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{12 c^2}+\frac {(d+e x)^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {b d \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{2 c^3}-\frac {b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 190, normalized size = 0.72 \[ \frac {1}{4} \left (4 a d^3 x+6 a d^2 e x^2+4 a d e^2 x^3+a e^3 x^4-\frac {b e \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (c^2 \left (18 d^2+6 d e x+e^2 x^2\right )+2 e^2\right )}{3 c^4}+\frac {2 i b d \left (2 c^2 d^2+e^2\right ) \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{c^3}+b x \text {sech}^{-1}(c x) \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.76, size = 358, normalized size = 1.36 \[ \frac {3 \, a c^{3} e^{3} x^{4} + 12 \, a c^{3} d e^{2} x^{3} + 18 \, a c^{3} d^{2} e x^{2} + 12 \, a c^{3} d^{3} x - 12 \, {\left (2 \, b c^{2} d^{3} + b d e^{2}\right )} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 3 \, {\left (4 \, b c^{3} d^{3} + 6 \, b c^{3} d^{2} e + 4 \, b c^{3} d e^{2} + b c^{3} e^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 3 \, {\left (b c^{3} e^{3} x^{4} + 4 \, b c^{3} d e^{2} x^{3} + 6 \, b c^{3} d^{2} e x^{2} + 4 \, b c^{3} d^{3} x - 4 \, b c^{3} d^{3} - 6 \, b c^{3} d^{2} e - 4 \, b c^{3} d e^{2} - b c^{3} e^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c^{2} e^{3} x^{3} + 6 \, b c^{2} d e^{2} x^{2} + 2 \, {\left (9 \, b c^{2} d^{2} e + b e^{3}\right )} x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{12 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arcsech(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 - 12*(2*b*c^2*d^3 + b*d*e^2)*
arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(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)) - 1)/x) + 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{3} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 283, normalized size = 1.07 \[ \frac {\frac {\left (c x e +c d \right )^{4} a}{4 c^{3} e}+\frac {b \left (\frac {e^{3} \mathrm {arcsech}\left (c x \right ) c^{4} x^{4}}{4}+e^{2} \mathrm {arcsech}\left (c x \right ) c^{4} x^{3} d +\frac {3 e \,\mathrm {arcsech}\left (c x \right ) c^{4} x^{2} d^{2}}{2}+\mathrm {arcsech}\left (c x \right ) c^{4} x \,d^{3}+\frac {\mathrm {arcsech}\left (c x \right ) c^{4} d^{4}}{4 e}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (-3 c^{4} d^{4} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+12 c^{3} d^{3} e \arcsin \left (c x \right )-c^{2} x^{2} e^{4} \sqrt {-c^{2} x^{2}+1}-6 c^{2} d \,e^{3} x \sqrt {-c^{2} x^{2}+1}-18 c^{2} d^{2} e^{2} \sqrt {-c^{2} x^{2}+1}+6 c d \,e^{3} \arcsin \left (c x \right )-2 e^{4} \sqrt {-c^{2} x^{2}+1}\right )}{12 e \sqrt {-c^{2} x^{2}+1}}\right )}{c^{3}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arcsech(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*arcsech(c*x)*c^4*x^4+e^2*arcsech(c*x)*c^4*x^3*d+3/2*e*arcsech(c*
x)*c^4*x^2*d^2+arcsech(c*x)*c^4*x*d^3+1/4/e*arcsech(c*x)*c^4*d^4+1/12/e*(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)
^(1/2)*(-3*c^4*d^4*arctanh(1/(-c^2*x^2+1)^(1/2))+12*c^3*d^3*e*arcsin(c*x)-c^2*x^2*e^4*(-c^2*x^2+1)^(1/2)-6*c^2
*d*e^3*x*(-c^2*x^2+1)^(1/2)-18*c^2*d^2*e^2*(-c^2*x^2+1)^(1/2)+6*c*d*e^3*arcsin(c*x)-2*e^4*(-c^2*x^2+1)^(1/2))/
(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.42, size = 221, normalized size = 0.84 \[ \frac {1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac {3}{2} \, a d^{2} e x^{2} + \frac {3}{2} \, {\left (x^{2} \operatorname {arsech}\left (c x\right ) - \frac {x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c}\right )} b d^{2} e + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {arsech}\left (c x\right ) - \frac {\frac {\sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b d e^{2} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arsech}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{3}}\right )} b e^{3} + a d^{3} x + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b d^{3}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arcsech(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*arcsech(c*x) - x*sqrt(1/(c^2*x^2) - 1)/c)*b*d^2*e + 1
/2*(2*x^3*arcsech(c*x) - (sqrt(1/(c^2*x^2) - 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + arctan(sqrt(1/(c^2*x^2) - 1))/
c^2)/c)*b*d*e^2 + 1/12*(3*x^4*arcsech(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 + (c*x*arcsech(c*x) - arctan(sqrt(1/(c^2*x^2) - 1)))*b*d^3/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))*(d + e*x)^3,x)

[Out]

int((a + b*acosh(1/(c*x)))*(d + e*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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