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

Optimal. Leaf size=68 \[ \frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}-\frac{b e \sqrt{(c+d x)^2+1} (c+d x)}{4 d}+\frac{b e \sinh ^{-1}(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0387322, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {5865, 12, 5661, 321, 215} \[ \frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}-\frac{b e \sqrt{(c+d x)^2+1} (c+d x)}{4 d}+\frac{b e \sinh ^{-1}(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0619275, size = 57, normalized size = 0.84 \[ \frac{e \left (2 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )-b \sqrt{(c+d x)^2+1} (c+d x)+b \sinh ^{-1}(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 62, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{2}ea}{2}}+eb \left ({\frac{{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}}{2}}-{\frac{dx+c}{4}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{{\it Arcsinh} \left ( dx+c \right ) }{4}} \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)),x)

[Out]

1/d*(1/2*(d*x+c)^2*e*a+e*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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.40071, size = 257, normalized size = 3.78 \begin{align*} \frac{2 \, a d^{2} e x^{2} + 4 \, a c d e x +{\left (2 \, b d^{2} e x^{2} + 4 \, b c d e x +{\left (2 \, b c^{2} + b\right )} e\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (b d e x + b c e\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{4 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.437205, size = 148, normalized size = 2.18 \begin{align*} \begin{cases} a c e x + \frac{a d e x^{2}}{2} + \frac{b c^{2} e \operatorname{asinh}{\left (c + d x \right )}}{2 d} + b c e x \operatorname{asinh}{\left (c + d x \right )} - \frac{b c e \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{4 d} + \frac{b d e x^{2} \operatorname{asinh}{\left (c + d x \right )}}{2} - \frac{b e x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{4} + \frac{b e \operatorname{asinh}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c e 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)*(a+b*asinh(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.96356, size = 331, normalized size = 4.87 \begin{align*} \frac{1}{4} \,{\left (2 \, a d x^{2} - 4 \,{\left (d{\left (\frac{c \log \left ({\left | -c d -{\left (x{\left | d \right |} - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\left | d \right |} \right |}\right )}{d{\left | d \right |}} + \frac{\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d^{2}}\right )} - x \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )} b c +{\left (2 \, x^{2} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}{\left (\frac{x}{d^{2}} - \frac{3 \, c}{d^{3}}\right )} - \frac{{\left (2 \, c^{2} - 1\right )} \log \left ({\left | -c d -{\left (x{\left | d \right |} - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\left | d \right |} \right |}\right )}{d^{2}{\left | d \right |}}\right )} d\right )} b d + 4 \, a c x\right )} e \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)),x, algorithm="giac")

[Out]

1/4*(2*a*d*x^2 - 4*(d*(c*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d*abs(d)) + s
qrt(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2) - x*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))*b*c + (2*x^2*log(d
*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x/d^2 - 3*c/d^3) - (2*c^2 -
1)*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d^2*abs(d)))*d)*b*d + 4*a*c*x)*e