3.3.60 \(\int x \sqrt {d+c^2 d x^2} (a+b \sinh ^{-1}(c x))^2 \, dx\) [260]

Optimal. Leaf size=180 \[ \frac {4 b^2 \sqrt {d+c^2 d x^2}}{9 c^2}+\frac {2 b^2 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{27 c^2}-\frac {2 b x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {1+c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d} \]

[Out]

1/3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2/c^2/d+4/9*b^2*(c^2*d*x^2+d)^(1/2)/c^2+2/27*b^2*(c^2*x^2+1)*(c^2*d
*x^2+d)^(1/2)/c^2-2/3*b*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-2/9*b*c*x^3*(a+b*arcsinh(
c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5798, 5784, 455, 45} \begin {gather*} -\frac {2 b x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {c^2 x^2+1}}+\frac {\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 b c x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {c^2 x^2+1}}+\frac {2 b^2 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}}{27 c^2}+\frac {4 b^2 \sqrt {c^2 d x^2+d}}{9 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*Sqrt[d + c^2*d*x^2])/(9*c^2) + (2*b^2*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/(27*c^2) - (2*b*x*Sqrt[d + c^2
*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c*Sqrt[1 + c^2*x^2]) - (2*b*c*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(
9*Sqrt[1 + c^2*x^2]) + ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/(3*c^2*d)

Rule 45

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])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 5784

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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]

Rubi steps

\begin {align*} \int x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {\left (2 b \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{3 c \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {1+c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\left (2 b^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (1+\frac {c^2 x^2}{3}\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {1+c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\left (b^2 \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \frac {1+\frac {c^2 x}{3}}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {1+c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\left (b^2 \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \left (\frac {2}{3 \sqrt {1+c^2 x}}+\frac {1}{3} \sqrt {1+c^2 x}\right ) \, dx,x,x^2\right )}{3 \sqrt {1+c^2 x^2}}\\ &=\frac {4 b^2 \sqrt {d+c^2 d x^2}}{9 c^2}+\frac {2 b^2 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{27 c^2}-\frac {2 b x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {1+c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 166, normalized size = 0.92 \begin {gather*} \frac {\sqrt {d+c^2 d x^2} \left (-6 a b c x \sqrt {1+c^2 x^2} \left (3+c^2 x^2\right )+9 \left (a+a c^2 x^2\right )^2+2 b^2 \left (7+8 c^2 x^2+c^4 x^4\right )+6 b \left (3 a \left (1+c^2 x^2\right )^2-b c x \sqrt {1+c^2 x^2} \left (3+c^2 x^2\right )\right ) \sinh ^{-1}(c x)+9 \left (b+b c^2 x^2\right )^2 \sinh ^{-1}(c x)^2\right )}{27 c^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(-6*a*b*c*x*Sqrt[1 + c^2*x^2]*(3 + c^2*x^2) + 9*(a + a*c^2*x^2)^2 + 2*b^2*(7 + 8*c^2*x^2
+ c^4*x^4) + 6*b*(3*a*(1 + c^2*x^2)^2 - b*c*x*Sqrt[1 + c^2*x^2]*(3 + c^2*x^2))*ArcSinh[c*x] + 9*(b + b*c^2*x^2
)^2*ArcSinh[c*x]^2))/(27*c^2*(1 + c^2*x^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(656\) vs. \(2(156)=312\).
time = 0.98, size = 657, normalized size = 3.65

method	result	size
default	\(\frac {a^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3 c^{2} d}+b^{2} \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}+4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}+3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (9 \arcsinh \left (c x \right )^{2}-6 \arcsinh \left (c x \right )+2\right )}{216 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (\arcsinh \left (c x \right )^{2}-2 \arcsinh \left (c x \right )+2\right )}{8 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (\arcsinh \left (c x \right )^{2}+2 \arcsinh \left (c x \right )+2\right )}{8 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}-3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (9 \arcsinh \left (c x \right )^{2}+6 \arcsinh \left (c x \right )+2\right )}{216 c^{2} \left (c^{2} x^{2}+1\right )}\right )+2 a b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}+4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}+3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (-1+3 \arcsinh \left (c x \right )\right )}{72 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (\arcsinh \left (c x \right )-1\right )}{8 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (1+\arcsinh \left (c x \right )\right )}{8 c^{2} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+5 c^{2} x^{2}-3 \sqrt {c^{2} x^{2}+1}\, c x +1\right ) \left (1+3 \arcsinh \left (c x \right )\right )}{72 c^{2} \left (c^{2} x^{2}+1\right )}\right )\)	\(657\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*a^2/c^2/d*(c^2*d*x^2+d)^(3/2)+b^2*(1/216*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^
2*x^2+3*(c^2*x^2+1)^(1/2)*c*x+1)*(9*arcsinh(c*x)^2-6*arcsinh(c*x)+2)/c^2/(c^2*x^2+1)+1/8*(d*(c^2*x^2+1))^(1/2)
*(c^2*x^2+(c^2*x^2+1)^(1/2)*c*x+1)*(arcsinh(c*x)^2-2*arcsinh(c*x)+2)/c^2/(c^2*x^2+1)+1/8*(d*(c^2*x^2+1))^(1/2)
*(c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*(arcsinh(c*x)^2+2*arcsinh(c*x)+2)/c^2/(c^2*x^2+1)+1/216*(d*(c^2*x^2+1))^(1/
2)*(4*c^4*x^4-4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^2-3*(c^2*x^2+1)^(1/2)*c*x+1)*(9*arcsinh(c*x)^2+6*arcsinh(c*x
)+2)/c^2/(c^2*x^2+1))+2*a*b*(1/72*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^2+3*(c^
2*x^2+1)^(1/2)*c*x+1)*(-1+3*arcsinh(c*x))/c^2/(c^2*x^2+1)+1/8*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+(c^2*x^2+1)^(1/2)
*c*x+1)*(arcsinh(c*x)-1)/c^2/(c^2*x^2+1)+1/8*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*(1+arcsin
h(c*x))/c^2/(c^2*x^2+1)+1/72*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*(c^2*x^2+1)^(1/2)*x^3*c^3+5*c^2*x^2-3*(c^2*x^2
+1)^(1/2)*c*x+1)*(1+3*arcsinh(c*x))/c^2/(c^2*x^2+1))

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Maxima [A]
time = 0.29, size = 183, normalized size = 1.02 \begin {gather*} \frac {2}{27} \, b^{2} {\left (\frac {\sqrt {c^{2} x^{2} + 1} d^{\frac {3}{2}} x^{2} + \frac {7 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {3}{2}}}{c^{2}}}{d} - \frac {3 \, {\left (c^{2} d^{\frac {3}{2}} x^{3} + 3 \, d^{\frac {3}{2}} x\right )} \operatorname {arsinh}\left (c x\right )}{c d}\right )} + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} b^{2} \operatorname {arsinh}\left (c x\right )^{2}}{3 \, c^{2} d} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} a b \operatorname {arsinh}\left (c x\right )}{3 \, c^{2} d} - \frac {2 \, {\left (c^{2} d^{\frac {3}{2}} x^{3} + 3 \, d^{\frac {3}{2}} x\right )} a b}{9 \, c d} + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} a^{2}}{3 \, c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/27*b^2*((sqrt(c^2*x^2 + 1)*d^(3/2)*x^2 + 7*sqrt(c^2*x^2 + 1)*d^(3/2)/c^2)/d - 3*(c^2*d^(3/2)*x^3 + 3*d^(3/2)
*x)*arcsinh(c*x)/(c*d)) + 1/3*(c^2*d*x^2 + d)^(3/2)*b^2*arcsinh(c*x)^2/(c^2*d) + 2/3*(c^2*d*x^2 + d)^(3/2)*a*b
*arcsinh(c*x)/(c^2*d) - 2/9*(c^2*d^(3/2)*x^3 + 3*d^(3/2)*x)*a*b/(c*d) + 1/3*(c^2*d*x^2 + d)^(3/2)*a^2/(c^2*d)

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Fricas [A]
time = 0.42, size = 249, normalized size = 1.38 \begin {gather*} \frac {9 \, {\left (b^{2} c^{4} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 6 \, {\left (3 \, a b c^{4} x^{4} + 6 \, a b c^{2} x^{2} + 3 \, a b - {\left (b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left ({\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{4} x^{4} + 2 \, {\left (9 \, a^{2} + 8 \, b^{2}\right )} c^{2} x^{2} + 9 \, a^{2} + 14 \, b^{2} - 6 \, {\left (a b c^{3} x^{3} + 3 \, a b c x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{27 \, {\left (c^{4} x^{2} + c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

1/27*(9*(b^2*c^4*x^4 + 2*b^2*c^2*x^2 + b^2)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 6*(3*a*b*c^4*
x^4 + 6*a*b*c^2*x^2 + 3*a*b - (b^2*c^3*x^3 + 3*b^2*c*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(
c^2*x^2 + 1)) + ((9*a^2 + 2*b^2)*c^4*x^4 + 2*(9*a^2 + 8*b^2)*c^2*x^2 + 9*a^2 + 14*b^2 - 6*(a*b*c^3*x^3 + 3*a*b
*c*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d))/(c^4*x^2 + c^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {d \left (c^{2} x^{2} + 1\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**2*(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x*sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d\,c^2\,x^2+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(1/2),x)

[Out]

int(x*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(1/2), x)

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