3.1.4 \(\int (d+e x)^3 (a+b \sinh ^{-1}(c x)) \, dx\) [4]

Optimal. Leaf size=176 \[ -\frac {7 b d (d+e x)^2 \sqrt {1+c^2 x^2}}{48 c}-\frac {b (d+e x)^3 \sqrt {1+c^2 x^2}}{16 c}-\frac {b \left (4 d \left (19 c^2 d^2-16 e^2\right )+e \left (26 c^2 d^2-9 e^2\right ) x\right ) \sqrt {1+c^2 x^2}}{96 c^3}-\frac {b \left (8 c^4 d^4-24 c^2 d^2 e^2+3 e^4\right ) \sinh ^{-1}(c x)}{32 c^4 e}+\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e} \]

[Out]

-1/32*b*(8*c^4*d^4-24*c^2*d^2*e^2+3*e^4)*arcsinh(c*x)/c^4/e+1/4*(e*x+d)^4*(a+b*arcsinh(c*x))/e-7/48*b*d*(e*x+d
)^2*(c^2*x^2+1)^(1/2)/c-1/16*b*(e*x+d)^3*(c^2*x^2+1)^(1/2)/c-1/96*b*(4*d*(19*c^2*d^2-16*e^2)+e*(26*c^2*d^2-9*e
^2)*x)*(c^2*x^2+1)^(1/2)/c^3

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Rubi [A]
time = 0.12, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5828, 757, 847, 794, 221} \begin {gather*} \frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac {b \sqrt {c^2 x^2+1} (d+e x)^3}{16 c}-\frac {7 b d \sqrt {c^2 x^2+1} (d+e x)^2}{48 c}-\frac {b \left (8 c^4 d^4-24 c^2 d^2 e^2+3 e^4\right ) \sinh ^{-1}(c x)}{32 c^4 e}-\frac {b \sqrt {c^2 x^2+1} \left (e x \left (26 c^2 d^2-9 e^2\right )+4 d \left (19 c^2 d^2-16 e^2\right )\right )}{96 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*b*d*(d + e*x)^2*Sqrt[1 + c^2*x^2])/(48*c) - (b*(d + e*x)^3*Sqrt[1 + c^2*x^2])/(16*c) - (b*(4*d*(19*c^2*d^2
- 16*e^2) + e*(26*c^2*d^2 - 9*e^2)*x)*Sqrt[1 + c^2*x^2])/(96*c^3) - (b*(8*c^4*d^4 - 24*c^2*d^2*e^2 + 3*e^4)*A
rcSinh[c*x])/(32*c^4*e) + ((d + e*x)^4*(a + b*ArcSinh[c*x]))/(4*e)

Rule 221

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 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

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

Rule 5828

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

Rubi steps

\begin {align*} \int (d+e x)^3 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac {(b c) \int \frac {(d+e x)^4}{\sqrt {1+c^2 x^2}} \, dx}{4 e}\\ &=-\frac {b (d+e x)^3 \sqrt {1+c^2 x^2}}{16 c}+\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac {b \int \frac {(d+e x)^2 \left (4 c^2 d^2-3 e^2+7 c^2 d e x\right )}{\sqrt {1+c^2 x^2}} \, dx}{16 c e}\\ &=-\frac {7 b d (d+e x)^2 \sqrt {1+c^2 x^2}}{48 c}-\frac {b (d+e x)^3 \sqrt {1+c^2 x^2}}{16 c}+\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac {b \int \frac {(d+e x) \left (c^2 d \left (12 c^2 d^2-23 e^2\right )+c^2 e \left (26 c^2 d^2-9 e^2\right ) x\right )}{\sqrt {1+c^2 x^2}} \, dx}{48 c^3 e}\\ &=-\frac {7 b d (d+e x)^2 \sqrt {1+c^2 x^2}}{48 c}-\frac {b (d+e x)^3 \sqrt {1+c^2 x^2}}{16 c}-\frac {b \left (4 d \left (19 c^2 d^2-16 e^2\right )+e \left (26 c^2 d^2-9 e^2\right ) x\right ) \sqrt {1+c^2 x^2}}{96 c^3}+\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac {\left (b \left (8 c^4 d^4-24 c^2 d^2 e^2+3 e^4\right )\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{32 c^3 e}\\ &=-\frac {7 b d (d+e x)^2 \sqrt {1+c^2 x^2}}{48 c}-\frac {b (d+e x)^3 \sqrt {1+c^2 x^2}}{16 c}-\frac {b \left (4 d \left (19 c^2 d^2-16 e^2\right )+e \left (26 c^2 d^2-9 e^2\right ) x\right ) \sqrt {1+c^2 x^2}}{96 c^3}-\frac {b \left (8 c^4 d^4-24 c^2 d^2 e^2+3 e^4\right ) \sinh ^{-1}(c x)}{32 c^4 e}+\frac {(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 166, normalized size = 0.94 \begin {gather*} \frac {24 a c^4 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )-b c \sqrt {1+c^2 x^2} \left (-e^2 (64 d+9 e x)+c^2 \left (96 d^3+72 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )\right )+3 b \left (24 c^2 d^2 e-3 e^3+8 c^4 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right ) \sinh ^{-1}(c x)}{96 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) - b*c*Sqrt[1 + c^2*x^2]*(-(e^2*(64*d + 9*e*x)) + c^2*(
96*d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) + 3*b*(24*c^2*d^2*e - 3*e^3 + 8*c^4*x*(4*d^3 + 6*d^2*e*x + 4*
d*e^2*x^2 + e^3*x^3))*ArcSinh[c*x])/(96*c^4)

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Maple [A]
time = 0.68, size = 259, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {\frac {\left (c e x +c d \right )^{4} a}{4 c^{3} e}+\frac {b \left (\frac {\arcsinh \left (c x \right ) c^{4} d^{4}}{4 e}+\arcsinh \left (c x \right ) c^{4} d^{3} x +\frac {3 e \arcsinh \left (c x \right ) c^{4} d^{2} x^{2}}{2}+e^{2} \arcsinh \left (c x \right ) c^{4} d \,x^{3}+\frac {e^{3} \arcsinh \left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{4} d^{4} \arcsinh \left (c x \right )+4 d^{3} c^{3} e \sqrt {c^{2} x^{2}+1}+6 d^{2} c^{2} e^{2} \left (\frac {\sqrt {c^{2} x^{2}+1}\, c x}{2}-\frac {\arcsinh \left (c x \right )}{2}\right )+4 d c \,e^{3} \left (\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {c^{2} x^{2}+1}}{3}\right )+e^{4} \left (\frac {\sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}}{4}-\frac {3 \sqrt {c^{2} x^{2}+1}\, c x}{8}+\frac {3 \arcsinh \left (c x \right )}{8}\right )}{4 e}\right )}{c^{3}}}{c}\)	\(259\)
default	\(\frac {\frac {\left (c e x +c d \right )^{4} a}{4 c^{3} e}+\frac {b \left (\frac {\arcsinh \left (c x \right ) c^{4} d^{4}}{4 e}+\arcsinh \left (c x \right ) c^{4} d^{3} x +\frac {3 e \arcsinh \left (c x \right ) c^{4} d^{2} x^{2}}{2}+e^{2} \arcsinh \left (c x \right ) c^{4} d \,x^{3}+\frac {e^{3} \arcsinh \left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{4} d^{4} \arcsinh \left (c x \right )+4 d^{3} c^{3} e \sqrt {c^{2} x^{2}+1}+6 d^{2} c^{2} e^{2} \left (\frac {\sqrt {c^{2} x^{2}+1}\, c x}{2}-\frac {\arcsinh \left (c x \right )}{2}\right )+4 d c \,e^{3} \left (\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {c^{2} x^{2}+1}}{3}\right )+e^{4} \left (\frac {\sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}}{4}-\frac {3 \sqrt {c^{2} x^{2}+1}\, c x}{8}+\frac {3 \arcsinh \left (c x \right )}{8}\right )}{4 e}\right )}{c^{3}}}{c}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 228, normalized size = 1.30 \begin {gather*} \frac {1}{4} \, a x^{4} e^{3} + a d x^{3} e^{2} + \frac {3}{2} \, a d^{2} x^{2} e + a d^{3} x + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )}\right )} b d^{2} e + \frac {{\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b d^{3}}{c} + \frac {1}{3} \, {\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 d e^{2} + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} b e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4*e^3 + a*d*x^3*e^2 + 3/2*a*d^2*x^2*e + a*d^3*x + 3/4*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x/c^2
- arcsinh(c*x)/c^3))*b*d^2*e + (c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*d^3/c + 1/3*(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*d*e^2 + 1/32*(8*x^4*arcsinh(c*x) - (2*sqrt(c^2*x^2 + 1
)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c)*b*e^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (158) = 316\).
time = 0.38, size = 525, normalized size = 2.98 \begin {gather*} \frac {24 \, a c^{4} x^{4} \cosh \left (1\right )^{3} + 24 \, a c^{4} x^{4} \sinh \left (1\right )^{3} + 96 \, a c^{4} d x^{3} \cosh \left (1\right )^{2} + 144 \, a c^{4} d^{2} x^{2} \cosh \left (1\right ) + 96 \, a c^{4} d^{3} x + 24 \, {\left (3 \, a c^{4} x^{4} \cosh \left (1\right ) + 4 \, a c^{4} d x^{3}\right )} \sinh \left (1\right )^{2} + 3 \, {\left (32 \, b c^{4} d x^{3} \cosh \left (1\right )^{2} + 32 \, b c^{4} d^{3} x + {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \cosh \left (1\right )^{3} + {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \sinh \left (1\right )^{3} + {\left (32 \, b c^{4} d x^{3} + 3 \, {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \cosh \left (1\right )\right )} \sinh \left (1\right )^{2} + 24 \, {\left (2 \, b c^{4} d^{2} x^{2} + b c^{2} d^{2}\right )} \cosh \left (1\right ) + {\left (64 \, b c^{4} d x^{3} \cosh \left (1\right ) + 48 \, b c^{4} d^{2} x^{2} + 24 \, b c^{2} d^{2} + 3 \, {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \cosh \left (1\right )^{2}\right )} \sinh \left (1\right )\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 24 \, {\left (3 \, a c^{4} x^{4} \cosh \left (1\right )^{2} + 8 \, a c^{4} d x^{3} \cosh \left (1\right ) + 6 \, a c^{4} d^{2} x^{2}\right )} \sinh \left (1\right ) - {\left (72 \, b c^{3} d^{2} x \cosh \left (1\right ) + 96 \, b c^{3} d^{3} + 3 \, {\left (2 \, b c^{3} x^{3} - 3 \, b c x\right )} \cosh \left (1\right )^{3} + 3 \, {\left (2 \, b c^{3} x^{3} - 3 \, b c x\right )} \sinh \left (1\right )^{3} + 32 \, {\left (b c^{3} d x^{2} - 2 \, b c d\right )} \cosh \left (1\right )^{2} + {\left (32 \, b c^{3} d x^{2} - 64 \, b c d + 9 \, {\left (2 \, b c^{3} x^{3} - 3 \, b c x\right )} \cosh \left (1\right )\right )} \sinh \left (1\right )^{2} + {\left (72 \, b c^{3} d^{2} x + 9 \, {\left (2 \, b c^{3} x^{3} - 3 \, b c x\right )} \cosh \left (1\right )^{2} + 64 \, {\left (b c^{3} d x^{2} - 2 \, b c d\right )} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \sqrt {c^{2} x^{2} + 1}}{96 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*a*c^4*x^4*cosh(1)^3 + 24*a*c^4*x^4*sinh(1)^3 + 96*a*c^4*d*x^3*cosh(1)^2 + 144*a*c^4*d^2*x^2*cosh(1) +
96*a*c^4*d^3*x + 24*(3*a*c^4*x^4*cosh(1) + 4*a*c^4*d*x^3)*sinh(1)^2 + 3*(32*b*c^4*d*x^3*cosh(1)^2 + 32*b*c^4*
d^3*x + (8*b*c^4*x^4 - 3*b)*cosh(1)^3 + (8*b*c^4*x^4 - 3*b)*sinh(1)^3 + (32*b*c^4*d*x^3 + 3*(8*b*c^4*x^4 - 3*b
)*cosh(1))*sinh(1)^2 + 24*(2*b*c^4*d^2*x^2 + b*c^2*d^2)*cosh(1) + (64*b*c^4*d*x^3*cosh(1) + 48*b*c^4*d^2*x^2 +
24*b*c^2*d^2 + 3*(8*b*c^4*x^4 - 3*b)*cosh(1)^2)*sinh(1))*log(c*x + sqrt(c^2*x^2 + 1)) + 24*(3*a*c^4*x^4*cosh(
1)^2 + 8*a*c^4*d*x^3*cosh(1) + 6*a*c^4*d^2*x^2)*sinh(1) - (72*b*c^3*d^2*x*cosh(1) + 96*b*c^3*d^3 + 3*(2*b*c^3*
x^3 - 3*b*c*x)*cosh(1)^3 + 3*(2*b*c^3*x^3 - 3*b*c*x)*sinh(1)^3 + 32*(b*c^3*d*x^2 - 2*b*c*d)*cosh(1)^2 + (32*b*
c^3*d*x^2 - 64*b*c*d + 9*(2*b*c^3*x^3 - 3*b*c*x)*cosh(1))*sinh(1)^2 + (72*b*c^3*d^2*x + 9*(2*b*c^3*x^3 - 3*b*c
*x)*cosh(1)^2 + 64*(b*c^3*d*x^2 - 2*b*c*d)*cosh(1))*sinh(1))*sqrt(c^2*x^2 + 1))/c^4

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Sympy [A]
time = 0.28, size = 316, normalized size = 1.80 \begin {gather*} \begin {cases} a d^{3} x + \frac {3 a d^{2} e x^{2}}{2} + a d e^{2} x^{3} + \frac {a e^{3} x^{4}}{4} + b d^{3} x \operatorname {asinh}{\left (c x \right )} + \frac {3 b d^{2} e x^{2} \operatorname {asinh}{\left (c x \right )}}{2} + b d e^{2} x^{3} \operatorname {asinh}{\left (c x \right )} + \frac {b e^{3} x^{4} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {b d^{3} \sqrt {c^{2} x^{2} + 1}}{c} - \frac {3 b d^{2} e x \sqrt {c^{2} x^{2} + 1}}{4 c} - \frac {b d e^{2} x^{2} \sqrt {c^{2} x^{2} + 1}}{3 c} - \frac {b e^{3} x^{3} \sqrt {c^{2} x^{2} + 1}}{16 c} + \frac {3 b d^{2} e \operatorname {asinh}{\left (c x \right )}}{4 c^{2}} + \frac {2 b d e^{2} \sqrt {c^{2} x^{2} + 1}}{3 c^{3}} + \frac {3 b e^{3} x \sqrt {c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b e^{3} \operatorname {asinh}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\a \left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**3*x + 3*a*d**2*e*x**2/2 + a*d*e**2*x**3 + a*e**3*x**4/4 + b*d**3*x*asinh(c*x) + 3*b*d**2*e*x**
2*asinh(c*x)/2 + b*d*e**2*x**3*asinh(c*x) + b*e**3*x**4*asinh(c*x)/4 - b*d**3*sqrt(c**2*x**2 + 1)/c - 3*b*d**2
*e*x*sqrt(c**2*x**2 + 1)/(4*c) - b*d*e**2*x**2*sqrt(c**2*x**2 + 1)/(3*c) - b*e**3*x**3*sqrt(c**2*x**2 + 1)/(16
*c) + 3*b*d**2*e*asinh(c*x)/(4*c**2) + 2*b*d*e**2*sqrt(c**2*x**2 + 1)/(3*c**3) + 3*b*e**3*x*sqrt(c**2*x**2 + 1
)/(32*c**3) - 3*b*e**3*asinh(c*x)/(32*c**4), Ne(c, 0)), (a*(d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4
/4), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst 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 \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d+e\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))*(d + e*x)^3,x)

[Out]

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

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