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

Optimal. Leaf size=100 \[ \frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac{b e^4 \left ((c+d x)^2+1\right )^{5/2}}{25 d}+\frac{2 b e^4 \left ((c+d x)^2+1\right )^{3/2}}{15 d}-\frac{b e^4 \sqrt{(c+d x)^2+1}}{5 d} \]

[Out]

-(b*e^4*Sqrt[1 + (c + d*x)^2])/(5*d) + (2*b*e^4*(1 + (c + d*x)^2)^(3/2))/(15*d) - (b*e^4*(1 + (c + d*x)^2)^(5/
2))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x]))/(5*d)

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Rubi [A]  time = 0.0789751, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {5865, 12, 5661, 266, 43} \[ \frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac{b e^4 \left ((c+d x)^2+1\right )^{5/2}}{25 d}+\frac{2 b e^4 \left ((c+d x)^2+1\right )^{3/2}}{15 d}-\frac{b e^4 \sqrt{(c+d x)^2+1}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSinh[c + d*x]),x]

[Out]

-(b*e^4*Sqrt[1 + (c + d*x)^2])/(5*d) + (2*b*e^4*(1 + (c + d*x)^2)^(3/2))/(15*d) - (b*e^4*(1 + (c + d*x)^2)^(5/
2))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x]))/(5*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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (c e+d e x)^4 \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^4 x^4 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int x^4 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (b e^4\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{5 d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (b e^4\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x}} \, dx,x,(c+d x)^2\right )}{10 d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (b e^4\right ) \operatorname{Subst}\left (\int \left (\frac{1}{\sqrt{1+x}}-2 \sqrt{1+x}+(1+x)^{3/2}\right ) \, dx,x,(c+d x)^2\right )}{10 d}\\ &=-\frac{b e^4 \sqrt{1+(c+d x)^2}}{5 d}+\frac{2 b e^4 \left (1+(c+d x)^2\right )^{3/2}}{15 d}-\frac{b e^4 \left (1+(c+d x)^2\right )^{5/2}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0977003, size = 71, normalized size = 0.71 \[ \frac{e^4 \left (\frac{1}{5} (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )-\frac{1}{75} b \sqrt{(c+d x)^2+1} \left (-10 (c+d x)^2+3 \left ((c+d x)^2+1\right )^2+5\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSinh[c + d*x]),x]

[Out]

(e^4*(-(b*Sqrt[1 + (c + d*x)^2]*(5 - 10*(c + d*x)^2 + 3*(1 + (c + d*x)^2)^2))/75 + ((c + d*x)^5*(a + b*ArcSinh
[c + d*x]))/5))/d

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Maple [A]  time = 0.01, size = 93, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{5}{e}^{4}a}{5}}+{e}^{4}b \left ({\frac{ \left ( dx+c \right ) ^{5}{\it Arcsinh} \left ( dx+c \right ) }{5}}-{\frac{ \left ( dx+c \right ) ^{4}}{25}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{4\, \left ( dx+c \right ) ^{2}}{75}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{8}{75}\sqrt{1+ \left ( dx+c \right ) ^{2}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c)),x)

[Out]

1/d*(1/5*(d*x+c)^5*e^4*a+e^4*b*(1/5*(d*x+c)^5*arcsinh(d*x+c)-1/25*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2
*(1+(d*x+c)^2)^(1/2)-8/75*(1+(d*x+c)^2)^(1/2)))

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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)^4*(a+b*arcsinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.43287, size = 608, normalized size = 6.08 \begin{align*} \frac{15 \, a d^{5} e^{4} x^{5} + 75 \, a c d^{4} e^{4} x^{4} + 150 \, a c^{2} d^{3} e^{4} x^{3} + 150 \, a c^{3} d^{2} e^{4} x^{2} + 75 \, a c^{4} d e^{4} x + 15 \,{\left (b d^{5} e^{4} x^{5} + 5 \, b c d^{4} e^{4} x^{4} + 10 \, b c^{2} d^{3} e^{4} x^{3} + 10 \, b c^{3} d^{2} e^{4} x^{2} + 5 \, b c^{4} d e^{4} x + b c^{5} e^{4}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (3 \, b d^{4} e^{4} x^{4} + 12 \, b c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, b c^{2} - 2 \, b\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, b c^{3} - 2 \, b c\right )} d e^{4} x +{\left (3 \, b c^{4} - 4 \, b c^{2} + 8 \, b\right )} e^{4}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{75 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

1/75*(15*a*d^5*e^4*x^5 + 75*a*c*d^4*e^4*x^4 + 150*a*c^2*d^3*e^4*x^3 + 150*a*c^3*d^2*e^4*x^2 + 75*a*c^4*d*e^4*x
+ 15*(b*d^5*e^4*x^5 + 5*b*c*d^4*e^4*x^4 + 10*b*c^2*d^3*e^4*x^3 + 10*b*c^3*d^2*e^4*x^2 + 5*b*c^4*d*e^4*x + b*c
^5*e^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (3*b*d^4*e^4*x^4 + 12*b*c*d^3*e^4*x^3 + 2*(9*b*c^2
- 2*b)*d^2*e^4*x^2 + 4*(3*b*c^3 - 2*b*c)*d*e^4*x + (3*b*c^4 - 4*b*c^2 + 8*b)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2
+ 1))/d

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Sympy [A]  time = 4.66256, size = 527, normalized size = 5.27 \begin{align*} \begin{cases} a c^{4} e^{4} x + 2 a c^{3} d e^{4} x^{2} + 2 a c^{2} d^{2} e^{4} x^{3} + a c d^{3} e^{4} x^{4} + \frac{a d^{4} e^{4} x^{5}}{5} + \frac{b c^{5} e^{4} \operatorname{asinh}{\left (c + d x \right )}}{5 d} + b c^{4} e^{4} x \operatorname{asinh}{\left (c + d x \right )} - \frac{b c^{4} e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25 d} + 2 b c^{3} d e^{4} x^{2} \operatorname{asinh}{\left (c + d x \right )} - \frac{4 b c^{3} e^{4} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + 2 b c^{2} d^{2} e^{4} x^{3} \operatorname{asinh}{\left (c + d x \right )} - \frac{6 b c^{2} d e^{4} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac{4 b c^{2} e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75 d} + b c d^{3} e^{4} x^{4} \operatorname{asinh}{\left (c + d x \right )} - \frac{4 b c d^{2} e^{4} x^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac{8 b c e^{4} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75} + \frac{b d^{4} e^{4} x^{5} \operatorname{asinh}{\left (c + d x \right )}}{5} - \frac{b d^{3} e^{4} x^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac{4 b d e^{4} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75} - \frac{8 b e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75 d} & \text{for}\: d \neq 0 \\c^{4} e^{4} x \left (a + b \operatorname{asinh}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4*(a+b*asinh(d*x+c)),x)

[Out]

Piecewise((a*c**4*e**4*x + 2*a*c**3*d*e**4*x**2 + 2*a*c**2*d**2*e**4*x**3 + a*c*d**3*e**4*x**4 + a*d**4*e**4*x
**5/5 + b*c**5*e**4*asinh(c + d*x)/(5*d) + b*c**4*e**4*x*asinh(c + d*x) - b*c**4*e**4*sqrt(c**2 + 2*c*d*x + d*
*2*x**2 + 1)/(25*d) + 2*b*c**3*d*e**4*x**2*asinh(c + d*x) - 4*b*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 +
1)/25 + 2*b*c**2*d**2*e**4*x**3*asinh(c + d*x) - 6*b*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25
+ 4*b*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(75*d) + b*c*d**3*e**4*x**4*asinh(c + d*x) - 4*b*c*d**2*e
**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 8*b*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/75 + b*d*
*4*e**4*x**5*asinh(c + d*x)/5 - b*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 4*b*d*e**4*x**2*sqr
t(c**2 + 2*c*d*x + d**2*x**2 + 1)/75 - 8*b*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(75*d), Ne(d, 0)), (c**4*
e**4*x*(a + b*asinh(c)), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError