3.3.2 \(\int (d+c^2 d x^2) (a+b \sinh ^{-1}(c x))^2 \, dx\) [202]

Optimal. Leaf size=125 \[ \frac {14}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3-\frac {4 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \]

[Out]

14/9*b^2*d*x+2/27*b^2*c^2*d*x^3-2/9*b*d*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/c+2/3*d*x*(a+b*arcsinh(c*x))^2+1/
3*d*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2-4/3*b*d*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5786, 5772, 5798, 8} \begin {gather*} \frac {1}{3} d x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}-\frac {4 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac {2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {2}{27} b^2 c^2 d x^3+\frac {14}{9} b^2 d x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 5772

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

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 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} (2 d) \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{3} (2 b c d) \int x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{9} \left (2 b^2 d\right ) \int \left (1+c^2 x^2\right ) \, dx-\frac {1}{3} (4 b c d) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=\frac {2}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3-\frac {4 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} \left (4 b^2 d\right ) \int 1 \, dx\\ &=\frac {14}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3-\frac {4 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 135, normalized size = 1.08 \begin {gather*} \frac {d \left (9 a^2 c x \left (3+c^2 x^2\right )-6 a b \sqrt {1+c^2 x^2} \left (7+c^2 x^2\right )+2 b^2 c x \left (21+c^2 x^2\right )-6 b \left (-3 a c x \left (3+c^2 x^2\right )+b \sqrt {1+c^2 x^2} \left (7+c^2 x^2\right )\right ) \sinh ^{-1}(c x)+9 b^2 c x \left (3+c^2 x^2\right ) \sinh ^{-1}(c x)^2\right )}{27 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(9*a^2*c*x*(3 + c^2*x^2) - 6*a*b*Sqrt[1 + c^2*x^2]*(7 + c^2*x^2) + 2*b^2*c*x*(21 + c^2*x^2) - 6*b*(-3*a*c*x
*(3 + c^2*x^2) + b*Sqrt[1 + c^2*x^2]*(7 + c^2*x^2))*ArcSinh[c*x] + 9*b^2*c*x*(3 + c^2*x^2)*ArcSinh[c*x]^2))/(2
7*c)

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (c^{2} d \,x^{2}+d \right ) \left (a +b \arcsinh \left (c x \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (109) = 218\).
time = 0.29, size = 230, normalized size = 1.84 \begin {gather*} \frac {1}{3} \, b^{2} c^{2} d x^{3} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{3} \, a^{2} c^{2} d x^{3} + \frac {2}{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 )} a b c^{2} d - \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {c^{2} x^{3} - 6 \, x}{c^{2}}\right )} b^{2} c^{2} d + b^{2} d x \operatorname {arsinh}\left (c x\right )^{2} + 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} \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))^2,x, algorithm="maxima")

[Out]

1/3*b^2*c^2*d*x^3*arcsinh(c*x)^2 + 1/3*a^2*c^2*d*x^3 + 2/9*(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))*a*b*c^2*d - 2/27*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4)*arcsin
h(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*c^2*d + b^2*d*x*arcsinh(c*x)^2 + 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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 178, normalized size = 1.42 \begin {gather*} \frac {{\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{3} d x^{3} + 3 \, {\left (9 \, a^{2} + 14 \, b^{2}\right )} c d x + 9 \, {\left (b^{2} c^{3} d x^{3} + 3 \, b^{2} c d x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 6 \, {\left (3 \, a b c^{3} d x^{3} + 9 \, a b c d x - {\left (b^{2} c^{2} d x^{2} + 7 \, b^{2} d\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 6 \, {\left (a b c^{2} d x^{2} + 7 \, a b d\right )} \sqrt {c^{2} x^{2} + 1}}{27 \, 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))^2,x, algorithm="fricas")

[Out]

1/27*((9*a^2 + 2*b^2)*c^3*d*x^3 + 3*(9*a^2 + 14*b^2)*c*d*x + 9*(b^2*c^3*d*x^3 + 3*b^2*c*d*x)*log(c*x + sqrt(c^
2*x^2 + 1))^2 + 6*(3*a*b*c^3*d*x^3 + 9*a*b*c*d*x - (b^2*c^2*d*x^2 + 7*b^2*d)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt
(c^2*x^2 + 1)) - 6*(a*b*c^2*d*x^2 + 7*a*b*d)*sqrt(c^2*x^2 + 1))/c

________________________________________________________________________________________

Sympy [A]
time = 0.24, size = 224, normalized size = 1.79 \begin {gather*} \begin {cases} \frac {a^{2} c^{2} d x^{3}}{3} + a^{2} d x + \frac {2 a b c^{2} d x^{3} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {2 a b c d x^{2} \sqrt {c^{2} x^{2} + 1}}{9} + 2 a b d x \operatorname {asinh}{\left (c x \right )} - \frac {14 a b d \sqrt {c^{2} x^{2} + 1}}{9 c} + \frac {b^{2} c^{2} d x^{3} \operatorname {asinh}^{2}{\left (c x \right )}}{3} + \frac {2 b^{2} c^{2} d x^{3}}{27} - \frac {2 b^{2} c d x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{9} + b^{2} d x \operatorname {asinh}^{2}{\left (c x \right )} + \frac {14 b^{2} d x}{9} - \frac {14 b^{2} d \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{9 c} & \text {for}\: c \neq 0 \\a^{2} 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))**2,x)

[Out]

Piecewise((a**2*c**2*d*x**3/3 + a**2*d*x + 2*a*b*c**2*d*x**3*asinh(c*x)/3 - 2*a*b*c*d*x**2*sqrt(c**2*x**2 + 1)
/9 + 2*a*b*d*x*asinh(c*x) - 14*a*b*d*sqrt(c**2*x**2 + 1)/(9*c) + b**2*c**2*d*x**3*asinh(c*x)**2/3 + 2*b**2*c**
2*d*x**3/27 - 2*b**2*c*d*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/9 + b**2*d*x*asinh(c*x)**2 + 14*b**2*d*x/9 - 14*b
**2*d*sqrt(c**2*x**2 + 1)*asinh(c*x)/(9*c), Ne(c, 0)), (a**2*d*x, True))

________________________________________________________________________________________

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))^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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________