3.14 \(\int (d+e x) (a+b \sinh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=140 \[ -\frac {2 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c}-\frac {b e x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac {e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 e}+2 b^2 d x+\frac {1}{4} b^2 e x^2 \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5801, 5821, 5675, 5717, 8, 5758, 30} \[ -\frac {2 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c}-\frac {b e x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac {e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 e}+2 b^2 d x+\frac {1}{4} b^2 e x^2 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSinh[c*x])^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -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 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 5821

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

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 142, normalized size = 1.01 \[ \frac {c \left (2 a^2 c x (2 d+e x)-2 a b \sqrt {c^2 x^2+1} (4 d+e x)+b^2 c x (8 d+e x)\right )+2 b \sinh ^{-1}(c x) \left (a \left (4 c^2 d x+2 c^2 e x^2+e\right )-b c \sqrt {c^2 x^2+1} (4 d+e x)\right )+b^2 \sinh ^{-1}(c x)^2 \left (4 c^2 d x+2 c^2 e x^2+e\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(a + b*ArcSinh[c*x])^2,x]

[Out]

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

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fricas [A]  time = 0.63, size = 183, normalized size = 1.31 \[ \frac {{\left (2 \, a^{2} + b^{2}\right )} c^{2} e x^{2} + 4 \, {\left (a^{2} + 2 \, b^{2}\right )} c^{2} d x + {\left (2 \, b^{2} c^{2} e x^{2} + 4 \, b^{2} c^{2} d x + b^{2} e\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 2 \, {\left (2 \, a b c^{2} e x^{2} + 4 \, a b c^{2} d x + a b e - {\left (b^{2} c e x + 4 \, b^{2} c d\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (a b c e x + 4 \, a b c d\right )} \sqrt {c^{2} x^{2} + 1}}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.07, size = 193, normalized size = 1.38 \[ \frac {\frac {a^{2} \left (\frac {1}{2} c^{2} x^{2} e +c^{2} d x \right )}{c}+\frac {b^{2} \left (\frac {e \left (2 \arcsinh \left (c x \right )^{2} c^{2} x^{2}-2 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c x +\arcsinh \left (c x \right )^{2}+c^{2} x^{2}+1\right )}{4}+c d \left (\arcsinh \left (c x \right )^{2} c x -2 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}+2 c x \right )\right )}{c}+\frac {2 a b \left (\frac {\arcsinh \left (c x \right ) c^{2} x^{2} e}{2}+\arcsinh \left (c x \right ) c^{2} x d -\frac {e \left (\frac {c x \sqrt {c^{2} x^{2}+1}}{2}-\frac {\arcsinh \left (c x \right )}{2}\right )}{2}-c d \sqrt {c^{2} x^{2}+1}\right )}{c}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(a+b*arcsinh(c*x))^2,x)

[Out]

1/c*(a^2/c*(1/2*c^2*x^2*e+c^2*d*x)+b^2/c*(1/4*e*(2*arcsinh(c*x)^2*c^2*x^2-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c*x
+arcsinh(c*x)^2+c^2*x^2+1)+c*d*(arcsinh(c*x)^2*c*x-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2*c*x))+2*a*b/c*(1/2*arcsi
nh(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 = 219, normalized size = 1.56 \[ \frac {1}{2} \, b^{2} e x^{2} \operatorname {arsinh}\left (c x\right )^{2} + b^{2} d x \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{2} \, a^{2} e x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )}\right )} a b e + \frac {1}{4} \, {\left (c^{2} {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{4}}\right )} - 2 \, c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )} \operatorname {arsinh}\left (c x\right )\right )} b^{2} e + 2 \, b^{2} d {\left (x - \frac {\sqrt {c^{2} x^{2} + 1} \operatorname {arsinh}\left (c x\right )}{c}\right )} + a^{2} d x + \frac {2 \, {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} a b d}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2*(d + e*x),x)

[Out]

int((a + b*asinh(c*x))^2*(d + e*x), x)

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sympy [A]  time = 0.85, size = 233, normalized size = 1.66 \[ \begin {cases} a^{2} d x + \frac {a^{2} e x^{2}}{2} + 2 a b d x \operatorname {asinh}{\left (c x \right )} + a b e x^{2} \operatorname {asinh}{\left (c x \right )} - \frac {2 a b d \sqrt {c^{2} x^{2} + 1}}{c} - \frac {a b e x \sqrt {c^{2} x^{2} + 1}}{2 c} + \frac {a b e \operatorname {asinh}{\left (c x \right )}}{2 c^{2}} + b^{2} d x \operatorname {asinh}^{2}{\left (c x \right )} + 2 b^{2} d x + \frac {b^{2} e x^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{2} + \frac {b^{2} e x^{2}}{4} - \frac {2 b^{2} d \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{c} - \frac {b^{2} e x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{2 c} + \frac {b^{2} e \operatorname {asinh}^{2}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*asinh(c*x))**2,x)

[Out]

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

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