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

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

[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)

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Rubi [A]  time = 0.170185, antiderivative size = 176, normalized size of antiderivative = 1., 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 = {5801, 743, 833, 780, 215} \[ \frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 e}-\frac{b \sqrt{c^2 x^2+1} \left (e x \left (26 c^2 d^2-9 e^2\right )+4 d \left (19 c^2 d^2-16 e^2\right )\right )}{96 c^3}-\frac{b \left (-24 c^2 d^2 e^2+8 c^4 d^4+3 e^4\right ) \sinh ^{-1}(c x)}{32 c^4 e}-\frac{b \sqrt{c^2 x^2+1} (d+e x)^3}{16 c}-\frac{7 b d \sqrt{c^2 x^2+1} (d+e x)^2}{48 c} \]

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 5801

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]

Rule 743

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 833

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 780

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 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]

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*}

Mathematica [A]  time = 0.137749, size = 166, normalized size = 0.94 \[ \frac{24 a c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )-b c \sqrt{c^2 x^2+1} \left (c^2 \left (72 d^2 e x+96 d^3+32 d e^2 x^2+6 e^3 x^3\right )-e^2 (64 d+9 e x)\right )+3 b \sinh ^{-1}(c x) \left (8 c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+24 c^2 d^2 e-3 e^3\right )}{96 c^4} \]

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 = 0.008, size = 259, normalized size = 1.5 \begin{align*}{\frac{1}{c} \left ({\frac{ \left ( cex+cd \right ) ^{4}a}{4\,{c}^{3}e}}+{\frac{b}{{c}^{3}} \left ({\frac{{e}^{3}{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{4}}+{e}^{2}{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{3}d+{\frac{3\,e{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{2}{d}^{2}}{2}}+{\it Arcsinh} \left ( cx \right ){c}^{4}x{d}^{3}+{\frac{{c}^{4}{d}^{4}{\it Arcsinh} \left ( cx \right ) }{4\,e}}-{\frac{1}{4\,e} \left ({e}^{4} \left ({\frac{{c}^{3}{x}^{3}}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{3\,cx}{8}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{3\,{\it Arcsinh} \left ( cx \right ) }{8}} \right ) +4\,cd{e}^{3} \left ( 1/3\,{c}^{2}{x}^{2}\sqrt{{c}^{2}{x}^{2}+1}-2/3\,\sqrt{{c}^{2}{x}^{2}+1} \right ) +6\,{c}^{2}{d}^{2}{e}^{2} \left ( 1/2\,cx\sqrt{{c}^{2}{x}^{2}+1}-1/2\,{\it Arcsinh} \left ( cx \right ) \right ) +4\,{c}^{3}{d}^{3}e\sqrt{{c}^{2}{x}^{2}+1}+{c}^{4}{d}^{4}{\it Arcsinh} \left ( cx \right ) \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.04887, size = 343, normalized size = 1.95 \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}{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 (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b d^{2} e + \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 (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b e^{3} + a d^{3} x + \frac{{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b d^{3}}{c} \end{align*}

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

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Fricas [A]  time = 2.4354, size = 468, normalized size = 2.66 \begin{align*} \frac{24 \, a c^{4} e^{3} x^{4} + 96 \, a c^{4} d e^{2} x^{3} + 144 \, a c^{4} d^{2} e x^{2} + 96 \, a c^{4} d^{3} x + 3 \,{\left (8 \, b c^{4} e^{3} x^{4} + 32 \, b c^{4} d e^{2} x^{3} + 48 \, b c^{4} d^{2} e x^{2} + 32 \, b c^{4} d^{3} x + 24 \, b c^{2} d^{2} e - 3 \, b e^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (6 \, b c^{3} e^{3} x^{3} + 32 \, b c^{3} d e^{2} x^{2} + 96 \, b c^{3} d^{3} - 64 \, b c d e^{2} + 9 \,{\left (8 \, b c^{3} d^{2} e - b c e^{3}\right )} x\right )} \sqrt{c^{2} x^{2} + 1}}{96 \, c^{4}} \end{align*}

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

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Sympy [A]  time = 2.12706, size = 316, normalized size = 1.8 \begin{align*} \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{align*}

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 [A]  time = 1.66671, size = 400, normalized size = 2.27 \begin{align*}{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{c}\right )} b d^{3} + a d^{3} x + \frac{1}{32} \,{\left (8 \, a x^{4} +{\left (8 \, x^{4} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1} x{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{3}{c^{4}}\right )} - \frac{3 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{4}{\left | c \right |}}\right )} c\right )} b\right )} e^{3} + \frac{1}{3} \,{\left (3 \, a d x^{3} +{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} b d\right )} e^{2} + \frac{3}{4} \,{\left (2 \, a d^{2} x^{2} +{\left (2 \, x^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} + \frac{\log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{2}{\left | c \right |}}\right )}\right )} b d^{2}\right )} e \end{align*}

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]

(x*log(c*x + sqrt(c^2*x^2 + 1)) - sqrt(c^2*x^2 + 1)/c)*b*d^3 + a*d^3*x + 1/32*(8*a*x^4 + (8*x^4*log(c*x + sqrt
(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*x*(2*x^2/c^2 - 3/c^4) - 3*log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^4*abs
(c)))*c)*b)*e^3 + 1/3*(3*a*d*x^3 + (3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)^(3/2) - 3*sqrt(c^2*x^2
+ 1))/c^3)*b*d)*e^2 + 3/4*(2*a*d^2*x^2 + (2*x^2*log(c*x + sqrt(c^2*x^2 + 1)) - c*(sqrt(c^2*x^2 + 1)*x/c^2 + l
og(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^2*abs(c))))*b*d^2)*e