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

Optimal. Leaf size=103 \[ \frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac {b^2 e (c+d x)^2}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5865, 12, 5661, 5758, 5675, 30} \[ \frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d}-\frac {b e \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac {b^2 e (c+d x)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 30

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

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 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 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 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
]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 120, normalized size = 1.17 \[ \frac {e \left (\left (2 a^2+b^2\right ) (c+d x)^2-2 a b \sqrt {(c+d x)^2+1} (c+d x)+2 b (c+d x) \sinh ^{-1}(c+d x) \left (2 a (c+d x)-b \sqrt {(c+d x)^2+1}\right )+2 a b \sinh ^{-1}(c+d x)+b^2 \left (2 (c+d x)^2+1\right ) \sinh ^{-1}(c+d x)^2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 230, normalized size = 2.23 \[ \frac {{\left (2 \, a^{2} + b^{2}\right )} d^{2} e x^{2} + 2 \, {\left (2 \, a^{2} + b^{2}\right )} c d e x + {\left (2 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} c d e x + {\left (2 \, b^{2} c^{2} + b^{2}\right )} e\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 2 \, {\left (2 \, a b d^{2} e x^{2} + 4 \, a b c d e x + {\left (2 \, a b c^{2} + a b\right )} e - {\left (b^{2} d e x + b^{2} c e\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 2 \, {\left (a b d e x + a b c e\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 135, normalized size = 1.31 \[ \frac {\frac {\left (d x +c \right )^{2} e \,a^{2}}{2}+e \,b^{2} \left (\frac {\arcsinh \left (d x +c \right )^{2} \left (1+\left (d x +c \right )^{2}\right )}{2}-\frac {\arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right )}{2}-\frac {\arcsinh \left (d x +c \right )^{2}}{4}+\frac {\left (d x +c \right )^{2}}{4}+\frac {1}{4}\right )+2 e a b \left (\frac {\left (d x +c \right )^{2} \arcsinh \left (d x +c \right )}{2}-\frac {\left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{4}+\frac {\arcsinh \left (d x +c \right )}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} d e x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (d x + c\right ) - d {\left (\frac {3 \, c^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c}{d^{3}}\right )}\right )} a b d e + a^{2} c e x + \frac {2 \, {\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} a b c e}{d} + \frac {1}{2} \, {\left (b^{2} d e x^{2} + 2 \, b^{2} c e x\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - \int \frac {{\left (b^{2} d^{4} e x^{4} + 4 \, b^{2} c d^{3} e x^{3} + {\left (5 \, c^{2} d^{2} e + d^{2} e\right )} b^{2} x^{2} + 2 \, {\left (c^{3} d e + c d e\right )} b^{2} x + {\left (b^{2} d^{3} e x^{3} + 3 \, b^{2} c d^{2} e x^{2} + 2 \, b^{2} c^{2} d e x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*d*e*x^2 + 1/2*(2*x^2*arcsinh(d*x + c) - d*(3*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)
*d^2))/d^3 + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x/d^2 - (c^2 + 1)*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(
c^2 + 1)*d^2))/d^3 - 3*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c/d^3))*a*b*d*e + a^2*c*e*x + 2*((d*x + c)*arcsinh(d*
x + c) - sqrt((d*x + c)^2 + 1))*a*b*c*e/d + 1/2*(b^2*d*e*x^2 + 2*b^2*c*e*x)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d
*x + c^2 + 1))^2 - integrate((b^2*d^4*e*x^4 + 4*b^2*c*d^3*e*x^3 + (5*c^2*d^2*e + d^2*e)*b^2*x^2 + 2*(c^3*d*e +
c*d*e)*b^2*x + (b^2*d^3*e*x^3 + 3*b^2*c*d^2*e*x^2 + 2*b^2*c^2*d*e*x)*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))/(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*
x + c^2 + 1)^(3/2) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.86, size = 335, normalized size = 3.25 \[ \begin {cases} a^{2} c e x + \frac {a^{2} d e x^{2}}{2} + \frac {a b c^{2} e \operatorname {asinh}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname {asinh}{\left (c + d x \right )} - \frac {a b c e \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{2 d} + a b d e x^{2} \operatorname {asinh}{\left (c + d x \right )} - \frac {a b e x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{2} + \frac {a b e \operatorname {asinh}{\left (c + d x \right )}}{2 d} + \frac {b^{2} c^{2} e \operatorname {asinh}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname {asinh}^{2}{\left (c + d x \right )} + \frac {b^{2} c e x}{2} - \frac {b^{2} c e \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{2 d} + \frac {b^{2} d e x^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{2} + \frac {b^{2} d e x^{2}}{4} - \frac {b^{2} e x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{2} + \frac {b^{2} e \operatorname {asinh}^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\c e x \left (a + b \operatorname {asinh}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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