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

Optimal. Leaf size=161 \[ \frac{3 b^2 e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{4 d}-\frac{3 b e (c+d x) \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b^3 e (c+d x) \sqrt{(c+d x)^2+1}}{8 d}+\frac{3 b^3 e \sinh ^{-1}(c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.212607, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5865, 12, 5661, 5758, 5675, 321, 215} \[ \frac{3 b^2 e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{4 d}-\frac{3 b e (c+d x) \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b^3 e (c+d x) \sqrt{(c+d x)^2+1}}{8 d}+\frac{3 b^3 e \sinh ^{-1}(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 12

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

Rule 5661

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

Rule 5758

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

Rule 5675

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 (c e+d e x) \left (a+b \sinh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e x \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e \operatorname{Subst}\left (\int x \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}-\frac{(3 b e) \operatorname{Subst}\left (\int \frac{x^2 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac{3 b e (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}+\frac{(3 b e) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{4 d}+\frac{\left (3 b^2 e\right ) \operatorname{Subst}\left (\int x \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}\\ &=\frac{3 b^2 e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{4 d}-\frac{3 b e (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}-\frac{\left (3 b^3 e\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac{3 b^3 e (c+d x) \sqrt{1+(c+d x)^2}}{8 d}+\frac{3 b^2 e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{4 d}-\frac{3 b e (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}+\frac{\left (3 b^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{8 d}\\ &=-\frac{3 b^3 e (c+d x) \sqrt{1+(c+d x)^2}}{8 d}+\frac{3 b^3 e \sinh ^{-1}(c+d x)}{8 d}+\frac{3 b^2 e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{4 d}-\frac{3 b e (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.214428, size = 200, normalized size = 1.24 \[ \frac{e \left (2 a \left (2 a^2+3 b^2\right ) (c+d x)^2-3 b \left (2 a^2+b^2\right ) (c+d x) \sqrt{(c+d x)^2+1}+3 b \left (2 a^2+b^2\right ) \sinh ^{-1}(c+d x)-6 b (c+d x) \sinh ^{-1}(c+d x) \left (-2 a^2 (c+d x)+2 a b \sqrt{(c+d x)^2+1}-b^2 (c+d x)\right )+6 b^2 \sinh ^{-1}(c+d x)^2 \left (2 a (c+d x)^2+a-b \sqrt{(c+d x)^2+1} (c+d x)\right )+2 b^3 \left (2 (c+d x)^2+1\right ) \sinh ^{-1}(c+d x)^3\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*a*(2*a^2 + 3*b^2)*(c + d*x)^2 - 3*b*(2*a^2 + b^2)*(c + d*x)*Sqrt[1 + (c + d*x)^2] + 3*b*(2*a^2 + b^2)*Ar
cSinh[c + d*x] - 6*b*(c + d*x)*(-2*a^2*(c + d*x) - b^2*(c + d*x) + 2*a*b*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*
x] + 6*b^2*(a + 2*a*(c + d*x)^2 - b*(c + d*x)*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 + 2*b^3*(1 + 2*(c + d*
x)^2)*ArcSinh[c + d*x]^3))/(8*d)

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Maple [A]  time = 0.032, size = 243, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{2}e{a}^{3}}{2}}+e{b}^{3} \left ({\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2}}-{\frac{3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) }{4}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{ \left ( 3+3\, \left ( dx+c \right ) ^{2} \right ){\it Arcsinh} \left ( dx+c \right ) }{4}}-{\frac{3\,dx+3\,c}{8}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{3\,{\it Arcsinh} \left ( dx+c \right ) }{8}} \right ) +3\,ea{b}^{2} \left ( 1/2\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) -1/2\,{\it Arcsinh} \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \left ( dx+c \right ) -1/4\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}+1/4\, \left ( dx+c \right ) ^{2}+1/4 \right ) +3\,e{a}^{2}b \left ( 1/2\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}-1/4\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+1/4\,{\it Arcsinh} \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.77998, size = 891, normalized size = 5.53 \begin{align*} \frac{2 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} d^{2} e x^{2} + 4 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} c d e x + 2 \,{\left (2 \, b^{3} d^{2} e x^{2} + 4 \, b^{3} c d e x +{\left (2 \, b^{3} c^{2} + b^{3}\right )} e\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + 6 \,{\left (2 \, a b^{2} d^{2} e x^{2} + 4 \, a b^{2} c d e x +{\left (2 \, a b^{2} c^{2} + a b^{2}\right )} e -{\left (b^{3} d e x + b^{3} c e\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 3 \,{\left (2 \,{\left (2 \, a^{2} b + b^{3}\right )} d^{2} e x^{2} + 4 \,{\left (2 \, a^{2} b + b^{3}\right )} c d e x +{\left (2 \, a^{2} b + b^{3} + 2 \,{\left (2 \, a^{2} b + b^{3}\right )} c^{2}\right )} e - 4 \,{\left (a b^{2} d e x + a b^{2} c e\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \,{\left ({\left (2 \, a^{2} b + b^{3}\right )} d e x +{\left (2 \, a^{2} b + b^{3}\right )} c e\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.80014, size = 685, normalized size = 4.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*c*e*x + a**3*d*e*x**2/2 + 3*a**2*b*c**2*e*asinh(c + d*x)/(2*d) + 3*a**2*b*c*e*x*asinh(c + d*x)
- 3*a**2*b*c*e*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(4*d) + 3*a**2*b*d*e*x**2*asinh(c + d*x)/2 - 3*a**2*b*e*x
*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/4 + 3*a**2*b*e*asinh(c + d*x)/(4*d) + 3*a*b**2*c**2*e*asinh(c + d*x)**2/
(2*d) + 3*a*b**2*c*e*x*asinh(c + d*x)**2 + 3*a*b**2*c*e*x/2 - 3*a*b**2*c*e*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1
)*asinh(c + d*x)/(2*d) + 3*a*b**2*d*e*x**2*asinh(c + d*x)**2/2 + 3*a*b**2*d*e*x**2/4 - 3*a*b**2*e*x*sqrt(c**2
+ 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/2 + 3*a*b**2*e*asinh(c + d*x)**2/(4*d) + b**3*c**2*e*asinh(c + d*x)*
*3/(2*d) + 3*b**3*c**2*e*asinh(c + d*x)/(4*d) + b**3*c*e*x*asinh(c + d*x)**3 + 3*b**3*c*e*x*asinh(c + d*x)/2 -
3*b**3*c*e*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/(4*d) - 3*b**3*c*e*sqrt(c**2 + 2*c*d*x + d*
*2*x**2 + 1)/(8*d) + b**3*d*e*x**2*asinh(c + d*x)**3/2 + 3*b**3*d*e*x**2*asinh(c + d*x)/4 - 3*b**3*e*x*sqrt(c*
*2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/4 - 3*b**3*e*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/8 + b**3*e
*asinh(c + d*x)**3/(4*d) + 3*b**3*e*asinh(c + d*x)/(8*d), Ne(d, 0)), (c*e*x*(a + b*asinh(c))**3, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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