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

Optimal. Leaf size=97 \[ \frac {(d+e x)^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 e}-\frac {b \left (2 d^2-\frac {e^2}{c^2}\right ) \sinh ^{-1}(c x)}{4 e}-\frac {b \sqrt {c^2 x^2+1} (d+e x)}{4 c}-\frac {3 b d \sqrt {c^2 x^2+1}}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5801, 743, 641, 215} \[ \frac {(d+e x)^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 e}-\frac {b \left (2 d^2-\frac {e^2}{c^2}\right ) \sinh ^{-1}(c x)}{4 e}-\frac {b \sqrt {c^2 x^2+1} (d+e x)}{4 c}-\frac {3 b d \sqrt {c^2 x^2+1}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSinh[c*x]),x]

[Out]

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

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]

Rule 641

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 91, normalized size = 0.94 \[ a d x+\frac {1}{2} a e x^2-\frac {b d \sqrt {c^2 x^2+1}}{c}-\frac {b e x \sqrt {c^2 x^2+1}}{4 c}+\frac {b e \sinh ^{-1}(c x)}{4 c^2}+b d x \sinh ^{-1}(c x)+\frac {1}{2} b e x^2 \sinh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 87, normalized size = 0.90 \[ \frac {2 \, a c^{2} e x^{2} + 4 \, a c^{2} d x + {\left (2 \, b c^{2} e x^{2} + 4 \, b c^{2} d x + b e\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c e x + 4 \, b c d\right )} \sqrt {c^{2} x^{2} + 1}}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.82, size = 124, normalized size = 1.28 \[ {\left (x \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - \frac {\sqrt {c^{2} x^{2} + 1}}{c}\right )} b d + a d x + \frac {1}{4} \, {\left (2 \, a x^{2} + {\left (2 \, x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} + \frac {\log \left (-x {\left | c \right |} + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} {\left | c \right |}}\right )}\right )} b\right )} e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.01, size = 96, normalized size = 0.99 \[ \frac {\frac {a \left (\frac {1}{2} c^{2} x^{2} e +c^{2} d x \right )}{c}+\frac {b \left (\frac {\arcsinh \left (c x \right ) c^{2} x^{2} e}{2}+\arcsinh \left (c x \right ) c^{2} x d -\frac {e \left (\frac {c x \sqrt {c^{2} x^{2}+1}}{2}-\frac {\arcsinh \left (c x \right )}{2}\right )}{2}-c d \sqrt {c^{2} x^{2}+1}\right )}{c}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 82, normalized size = 0.85 \[ \frac {1}{2} \, a e x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )}\right )} b e + a d x + \frac {{\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b d}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 78, normalized size = 0.80 \[ \frac {a\,x\,\left (2\,d+e\,x\right )}{2}-\frac {b\,d\,\left (\sqrt {c^2\,x^2+1}-c\,x\,\mathrm {asinh}\left (c\,x\right )\right )}{c}-\frac {b\,e\,x\,\sqrt {c^2\,x^2+1}}{4\,c}+b\,e\,x\,\mathrm {asinh}\left (c\,x\right )\,\left (\frac {x}{2}+\frac {1}{4\,c^2\,x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))*(d + e*x),x)

[Out]

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

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sympy [A]  time = 0.30, size = 99, normalized size = 1.02 \[ \begin {cases} a d x + \frac {a e x^{2}}{2} + b d x \operatorname {asinh}{\left (c x \right )} + \frac {b e x^{2} \operatorname {asinh}{\left (c x \right )}}{2} - \frac {b d \sqrt {c^{2} x^{2} + 1}}{c} - \frac {b e x \sqrt {c^{2} x^{2} + 1}}{4 c} + \frac {b e \operatorname {asinh}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\a \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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