3.2.39 \(\int (d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [139]

Optimal. Leaf size=254 \[ -\frac {25 b c d^2 x^2 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {5 b c^3 d^2 x^4 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {d+c^2 d x^2}}{36 c}+\frac {5}{16} d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {1+c^2 x^2}} \]

[Out]

5/24*d*x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))+1/6*x*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))-1/36*b*d^2*(c^2*x
^2+1)^(5/2)*(c^2*d*x^2+d)^(1/2)/c+5/16*d^2*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)-25/96*b*c*d^2*x^2*(c^2*d*x
^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-5/96*b*c^3*d^2*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+5/32*d^2*(a+b*arcsinh(c
*x))^2*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5786, 5785, 5783, 30, 14, 267} \begin {gather*} \frac {5}{16} d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {c^2 x^2+1}}+\frac {1}{6} x \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {25 b c d^2 x^2 \sqrt {c^2 d x^2+d}}{96 \sqrt {c^2 x^2+1}}-\frac {b d^2 \left (c^2 x^2+1\right )^{5/2} \sqrt {c^2 d x^2+d}}{36 c}-\frac {5 b c^3 d^2 x^4 \sqrt {c^2 d x^2+d}}{96 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(-25*b*c*d^2*x^2*Sqrt[d + c^2*d*x^2])/(96*Sqrt[1 + c^2*x^2]) - (5*b*c^3*d^2*x^4*Sqrt[d + c^2*d*x^2])/(96*Sqrt[
1 + c^2*x^2]) - (b*d^2*(1 + c^2*x^2)^(5/2)*Sqrt[d + c^2*d*x^2])/(36*c) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*A
rcSinh[c*x]))/16 + (5*d*x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (x*(d + c^2*d*x^2)^(5/2)*(a + b*Arc
Sinh[c*x]))/6 + (5*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c*Sqrt[1 + c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5785

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

Rule 5786

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

Rubi steps

\begin {align*} \int \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} (5 d) \int \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right )^2 \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=-\frac {b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {d+c^2 d x^2}}{36 c}+\frac {5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} \left (5 d^2\right ) \int \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{24 \sqrt {1+c^2 x^2}}\\ &=-\frac {b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {d+c^2 d x^2}}{36 c}+\frac {5}{16} d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (5 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{16 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{24 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{16 \sqrt {1+c^2 x^2}}\\ &=-\frac {25 b c d^2 x^2 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {5 b c^3 d^2 x^4 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {d+c^2 d x^2}}{36 c}+\frac {5}{16} d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 317, normalized size = 1.25 \begin {gather*} \frac {d^2 \left (1584 a c x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}+1248 a c^3 x^3 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}+384 a c^5 x^5 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}+360 b \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)^2-270 b \sqrt {d+c^2 d x^2} \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b \sqrt {d+c^2 d x^2} \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \sqrt {d+c^2 d x^2} \cosh \left (6 \sinh ^{-1}(c x)\right )+720 a \sqrt {d} \sqrt {1+c^2 x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+12 b \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x) \left (45 \sinh \left (2 \sinh ^{-1}(c x)\right )+9 \sinh \left (4 \sinh ^{-1}(c x)\right )+\sinh \left (6 \sinh ^{-1}(c x)\right )\right )\right )}{2304 c \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(d^2*(1584*a*c*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 1248*a*c^3*x^3*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]
+ 384*a*c^5*x^5*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 360*b*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]^2 - 270*b*Sqrt[
d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] - 27*b*Sqrt[d + c^2*d*x^2]*Cosh[4*ArcSinh[c*x]] - 2*b*Sqrt[d + c^2*d*x^2]*
Cosh[6*ArcSinh[c*x]] + 720*a*Sqrt[d]*Sqrt[1 + c^2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 12*b*Sqrt[d
+ c^2*d*x^2]*ArcSinh[c*x]*(45*Sinh[2*ArcSinh[c*x]] + 9*Sinh[4*ArcSinh[c*x]] + Sinh[6*ArcSinh[c*x]])))/(2304*c*
Sqrt[1 + c^2*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(800\) vs. \(2(218)=436\).
time = 1.46, size = 801, normalized size = 3.15

method	result	size
default	\(\frac {x \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} a}{6}+\frac {5 a d x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{24}+\frac {5 a \,d^{2} x \sqrt {c^{2} d \,x^{2}+d}}{16}+\frac {5 a \,d^{3} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{16 \sqrt {c^{2} d}}+b \left (\frac {5 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2} d^{2}}{32 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (32 c^{7} x^{7}+32 c^{6} x^{6} \sqrt {c^{2} x^{2}+1}+64 c^{5} x^{5}+48 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+38 c^{3} x^{3}+18 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+6 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+6 \arcsinh \left (c x \right )\right ) d^{2}}{2304 c \left (c^{2} x^{2}+1\right )}+\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}+8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}+8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+4 \arcsinh \left (c x \right )\right ) d^{2}}{512 c \left (c^{2} x^{2}+1\right )}+\frac {15 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \arcsinh \left (c x \right )\right ) d^{2}}{256 c \left (c^{2} x^{2}+1\right )}+\frac {15 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \arcsinh \left (c x \right )\right ) d^{2}}{256 c \left (c^{2} x^{2}+1\right )}+\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}-8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}-8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+4 \arcsinh \left (c x \right )\right ) d^{2}}{512 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (32 c^{7} x^{7}-32 c^{6} x^{6} \sqrt {c^{2} x^{2}+1}+64 c^{5} x^{5}-48 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+38 c^{3} x^{3}-18 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+6 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+6 \arcsinh \left (c x \right )\right ) d^{2}}{2304 c \left (c^{2} x^{2}+1\right )}\right )\)	\(801\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/6*x*(c^2*d*x^2+d)^(5/2)*a+5/24*a*d*x*(c^2*d*x^2+d)^(3/2)+5/16*a*d^2*x*(c^2*d*x^2+d)^(1/2)+5/16*a*d^3*ln(x*c^
2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+b*(5/32*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh
(c*x)^2*d^2+1/2304*(d*(c^2*x^2+1))^(1/2)*(32*c^7*x^7+32*c^6*x^6*(c^2*x^2+1)^(1/2)+64*c^5*x^5+48*c^4*x^4*(c^2*x
^2+1)^(1/2)+38*c^3*x^3+18*c^2*x^2*(c^2*x^2+1)^(1/2)+6*c*x+(c^2*x^2+1)^(1/2))*(-1+6*arcsinh(c*x))*d^2/c/(c^2*x^
2+1)+3/512*(d*(c^2*x^2+1))^(1/2)*(8*c^5*x^5+8*c^4*x^4*(c^2*x^2+1)^(1/2)+12*c^3*x^3+8*c^2*x^2*(c^2*x^2+1)^(1/2)
+4*c*x+(c^2*x^2+1)^(1/2))*(-1+4*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+15/256*(d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3+2*c^2*
x^2*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2+1)^(1/2))*(-1+2*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+15/256*(d*(c^2*x^2+1))^(1
/2)*(2*c^3*x^3-2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x-(c^2*x^2+1)^(1/2))*(1+2*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+3/512
*(d*(c^2*x^2+1))^(1/2)*(8*c^5*x^5-8*c^4*x^4*(c^2*x^2+1)^(1/2)+12*c^3*x^3-8*c^2*x^2*(c^2*x^2+1)^(1/2)+4*c*x-(c^
2*x^2+1)^(1/2))*(1+4*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+1/2304*(d*(c^2*x^2+1))^(1/2)*(32*c^7*x^7-32*c^6*x^6*(c^2*
x^2+1)^(1/2)+64*c^5*x^5-48*c^4*x^4*(c^2*x^2+1)^(1/2)+38*c^3*x^3-18*c^2*x^2*(c^2*x^2+1)^(1/2)+6*c*x-(c^2*x^2+1)
^(1/2))*(1+6*arcsinh(c*x))*d^2/c/(c^2*x^2+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x)),x)

[Out]

Integral((d*(c**2*x**2 + 1))**(5/2)*(a + b*asinh(c*x)), 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((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),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 = -0.00 \begin {gather*} \int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2),x)

[Out]

int((a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2), x)

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