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

Optimal. Leaf size=75 \[ -\frac {2 b d \sqrt {1+c^2 x^2}}{3 c}-\frac {b d \left (1+c^2 x^2\right )^{3/2}}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5784, 12, 455, 45} \begin {gather*} \frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )+d x \left (a+b \sinh ^{-1}(c x)\right )-\frac {b d \left (c^2 x^2+1\right )^{3/2}}{9 c}-\frac {2 b d \sqrt {c^2 x^2+1}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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]

Rubi steps

\begin {align*} \int \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac {d x \left (3+c^2 x^2\right )}{3 \sqrt {1+c^2 x^2}} \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{3} (b c d) \int \frac {x \left (3+c^2 x^2\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{6} (b c d) \text {Subst}\left (\int \frac {3+c^2 x}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{6} (b c d) \text {Subst}\left (\int \left (\frac {2}{\sqrt {1+c^2 x}}+\sqrt {1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {2 b d \sqrt {1+c^2 x^2}}{3 c}-\frac {b d \left (1+c^2 x^2\right )^{3/2}}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 86, normalized size = 1.15 \begin {gather*} a d x+\frac {1}{3} a c^2 d x^3-\frac {7 b d \sqrt {1+c^2 x^2}}{9 c}-\frac {1}{9} b c d x^2 \sqrt {1+c^2 x^2}+b d x \sinh ^{-1}(c x)+\frac {1}{3} b c^2 d x^3 \sinh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.48, size = 76, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {d a \left (\frac {1}{3} c^{3} x^{3}+c x \right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+\arcsinh \left (c x \right ) c x -\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {7 \sqrt {c^{2} x^{2}+1}}{9}\right )}{c}\)	\(76\)
default	\(\frac {d a \left (\frac {1}{3} c^{3} x^{3}+c x \right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+\arcsinh \left (c x \right ) c x -\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {7 \sqrt {c^{2} x^{2}+1}}{9}\right )}{c}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 97, normalized size = 1.29 \begin {gather*} \frac {1}{3} \, a c^{2} d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{2} d + a d x + \frac {{\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b d}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 83, normalized size = 1.11 \begin {gather*} \frac {3 \, a c^{3} d x^{3} + 9 \, a c d x + 3 \, {\left (b c^{3} d x^{3} + 3 \, b c d x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c^{2} d x^{2} + 7 \, b d\right )} \sqrt {c^{2} x^{2} + 1}}{9 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.16, size = 90, normalized size = 1.20 \begin {gather*} \begin {cases} \frac {a c^{2} d x^{3}}{3} + a d x + \frac {b c^{2} d x^{3} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {b c d x^{2} \sqrt {c^{2} x^{2} + 1}}{9} + b d x \operatorname {asinh}{\left (c x \right )} - \frac {7 b d \sqrt {c^{2} x^{2} + 1}}{9 c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**3/3 + a*d*x + b*c**2*d*x**3*asinh(c*x)/3 - b*c*d*x**2*sqrt(c**2*x**2 + 1)/9 + b*d*x*asi
nh(c*x) - 7*b*d*sqrt(c**2*x**2 + 1)/(9*c), Ne(c, 0)), (a*d*x, True))

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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)*(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 = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right ) \,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),x)

[Out]

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

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