3.11 \(\int x^3 (d+c^2 d x^2)^2 (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=180 \[ \frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac {73 b d^2 \sinh ^{-1}(c x)}{3072 c^4}-\frac {43 b c d^2 x^5 \sqrt {c^2 x^2+1}}{1152}-\frac {73 b d^2 x^3 \sqrt {c^2 x^2+1}}{4608 c}+\frac {73 b d^2 x \sqrt {c^2 x^2+1}}{3072 c^3}-\frac {1}{64} b c^3 d^2 x^7 \sqrt {c^2 x^2+1} \]

[Out]

-73/3072*b*d^2*arcsinh(c*x)/c^4+1/4*d^2*x^4*(a+b*arcsinh(c*x))+1/3*c^2*d^2*x^6*(a+b*arcsinh(c*x))+1/8*c^4*d^2*
x^8*(a+b*arcsinh(c*x))+73/3072*b*d^2*x*(c^2*x^2+1)^(1/2)/c^3-73/4608*b*d^2*x^3*(c^2*x^2+1)^(1/2)/c-43/1152*b*c
*d^2*x^5*(c^2*x^2+1)^(1/2)-1/64*b*c^3*d^2*x^7*(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 43, 5730, 12, 1267, 459, 321, 215} \[ \frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{64} b c^3 d^2 x^7 \sqrt {c^2 x^2+1}-\frac {43 b c d^2 x^5 \sqrt {c^2 x^2+1}}{1152}-\frac {73 b d^2 x^3 \sqrt {c^2 x^2+1}}{4608 c}+\frac {73 b d^2 x \sqrt {c^2 x^2+1}}{3072 c^3}-\frac {73 b d^2 \sinh ^{-1}(c x)}{3072 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + c^2*d*x^2)^2*(a + b*ArcSinh[c*x]),x]

[Out]

(73*b*d^2*x*Sqrt[1 + c^2*x^2])/(3072*c^3) - (73*b*d^2*x^3*Sqrt[1 + c^2*x^2])/(4608*c) - (43*b*c*d^2*x^5*Sqrt[1
+ c^2*x^2])/1152 - (b*c^3*d^2*x^7*Sqrt[1 + c^2*x^2])/64 - (73*b*d^2*ArcSinh[c*x])/(3072*c^4) + (d^2*x^4*(a +
b*ArcSinh[c*x]))/4 + (c^2*d^2*x^6*(a + b*ArcSinh[c*x]))/3 + (c^4*d^2*x^8*(a + b*ArcSinh[c*x]))/8

Rule 12

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

Rule 43

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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 266

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

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

Rule 1267

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[(c^p*(f*x)^(m + 4*p - 1)*(d + e*x^2)^(q + 1))/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1)), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(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, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^3 \left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac {d^2 x^4 \left (6+8 c^2 x^2+3 c^4 x^4\right )}{24 \sqrt {1+c^2 x^2}} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{24} \left (b c d^2\right ) \int \frac {x^4 \left (6+8 c^2 x^2+3 c^4 x^4\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {1}{64} b c^3 d^2 x^7 \sqrt {1+c^2 x^2}+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (b d^2\right ) \int \frac {x^4 \left (48 c^2+43 c^4 x^2\right )}{\sqrt {1+c^2 x^2}} \, dx}{192 c}\\ &=-\frac {43 b c d^2 x^5 \sqrt {1+c^2 x^2}}{1152}-\frac {1}{64} b c^3 d^2 x^7 \sqrt {1+c^2 x^2}+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (73 b c d^2\right ) \int \frac {x^4}{\sqrt {1+c^2 x^2}} \, dx}{1152}\\ &=-\frac {73 b d^2 x^3 \sqrt {1+c^2 x^2}}{4608 c}-\frac {43 b c d^2 x^5 \sqrt {1+c^2 x^2}}{1152}-\frac {1}{64} b c^3 d^2 x^7 \sqrt {1+c^2 x^2}+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (73 b d^2\right ) \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{1536 c}\\ &=\frac {73 b d^2 x \sqrt {1+c^2 x^2}}{3072 c^3}-\frac {73 b d^2 x^3 \sqrt {1+c^2 x^2}}{4608 c}-\frac {43 b c d^2 x^5 \sqrt {1+c^2 x^2}}{1152}-\frac {1}{64} b c^3 d^2 x^7 \sqrt {1+c^2 x^2}+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (73 b d^2\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{3072 c^3}\\ &=\frac {73 b d^2 x \sqrt {1+c^2 x^2}}{3072 c^3}-\frac {73 b d^2 x^3 \sqrt {1+c^2 x^2}}{4608 c}-\frac {43 b c d^2 x^5 \sqrt {1+c^2 x^2}}{1152}-\frac {1}{64} b c^3 d^2 x^7 \sqrt {1+c^2 x^2}-\frac {73 b d^2 \sinh ^{-1}(c x)}{3072 c^4}+\frac {1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 115, normalized size = 0.64 \[ \frac {d^2 \left (384 a c^4 x^4 \left (3 c^4 x^4+8 c^2 x^2+6\right )+3 b \left (384 c^8 x^8+1024 c^6 x^6+768 c^4 x^4-73\right ) \sinh ^{-1}(c x)-b c x \sqrt {c^2 x^2+1} \left (144 c^6 x^6+344 c^4 x^4+146 c^2 x^2-219\right )\right )}{9216 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + c^2*d*x^2)^2*(a + b*ArcSinh[c*x]),x]

[Out]

(d^2*(384*a*c^4*x^4*(6 + 8*c^2*x^2 + 3*c^4*x^4) - b*c*x*Sqrt[1 + c^2*x^2]*(-219 + 146*c^2*x^2 + 344*c^4*x^4 +
144*c^6*x^6) + 3*b*(-73 + 768*c^4*x^4 + 1024*c^6*x^6 + 384*c^8*x^8)*ArcSinh[c*x]))/(9216*c^4)

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fricas [A]  time = 0.58, size = 161, normalized size = 0.89 \[ \frac {1152 \, a c^{8} d^{2} x^{8} + 3072 \, a c^{6} d^{2} x^{6} + 2304 \, a c^{4} d^{2} x^{4} + 3 \, {\left (384 \, b c^{8} d^{2} x^{8} + 1024 \, b c^{6} d^{2} x^{6} + 768 \, b c^{4} d^{2} x^{4} - 73 \, b d^{2}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (144 \, b c^{7} d^{2} x^{7} + 344 \, b c^{5} d^{2} x^{5} + 146 \, b c^{3} d^{2} x^{3} - 219 \, b c d^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}{9216 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c^2*d*x^2+d)^2*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

1/9216*(1152*a*c^8*d^2*x^8 + 3072*a*c^6*d^2*x^6 + 2304*a*c^4*d^2*x^4 + 3*(384*b*c^8*d^2*x^8 + 1024*b*c^6*d^2*x
^6 + 768*b*c^4*d^2*x^4 - 73*b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (144*b*c^7*d^2*x^7 + 344*b*c^5*d^2*x^5 + 146
*b*c^3*d^2*x^3 - 219*b*c*d^2*x)*sqrt(c^2*x^2 + 1))/c^4

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c^2*d*x^2+d)^2*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.01, size = 156, normalized size = 0.87 \[ \frac {d^{2} a \left (\frac {1}{8} c^{8} x^{8}+\frac {1}{3} c^{6} x^{6}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (\frac {\arcsinh \left (c x \right ) c^{8} x^{8}}{8}+\frac {\arcsinh \left (c x \right ) c^{6} x^{6}}{3}+\frac {\arcsinh \left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{7} x^{7} \sqrt {c^{2} x^{2}+1}}{64}-\frac {43 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{1152}-\frac {73 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{4608}+\frac {73 c x \sqrt {c^{2} x^{2}+1}}{3072}-\frac {73 \arcsinh \left (c x \right )}{3072}\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c^2*d*x^2+d)^2*(a+b*arcsinh(c*x)),x)

[Out]

1/c^4*(d^2*a*(1/8*c^8*x^8+1/3*c^6*x^6+1/4*c^4*x^4)+d^2*b*(1/8*arcsinh(c*x)*c^8*x^8+1/3*arcsinh(c*x)*c^6*x^6+1/
4*arcsinh(c*x)*c^4*x^4-1/64*c^7*x^7*(c^2*x^2+1)^(1/2)-43/1152*c^5*x^5*(c^2*x^2+1)^(1/2)-73/4608*c^3*x^3*(c^2*x
^2+1)^(1/2)+73/3072*c*x*(c^2*x^2+1)^(1/2)-73/3072*arcsinh(c*x)))

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maxima [A]  time = 0.62, size = 292, normalized size = 1.62 \[ \frac {1}{8} \, a c^{4} d^{2} x^{8} + \frac {1}{3} \, a c^{2} d^{2} x^{6} + \frac {1}{3072} \, {\left (384 \, x^{8} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {48 \, \sqrt {c^{2} x^{2} + 1} x^{7}}{c^{2}} - \frac {56 \, \sqrt {c^{2} x^{2} + 1} x^{5}}{c^{4}} + \frac {70 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{6}} - \frac {105 \, \sqrt {c^{2} x^{2} + 1} x}{c^{8}} + \frac {105 \, \operatorname {arsinh}\left (c x\right )}{c^{9}}\right )} c\right )} b c^{4} d^{2} + \frac {1}{4} \, a d^{2} x^{4} + \frac {1}{144} \, {\left (48 \, x^{6} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {8 \, \sqrt {c^{2} x^{2} + 1} x^{5}}{c^{2}} - \frac {10 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \operatorname {arsinh}\left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d^{2} + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c^2*d*x^2+d)^2*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

1/8*a*c^4*d^2*x^8 + 1/3*a*c^2*d^2*x^6 + 1/3072*(384*x^8*arcsinh(c*x) - (48*sqrt(c^2*x^2 + 1)*x^7/c^2 - 56*sqrt
(c^2*x^2 + 1)*x^5/c^4 + 70*sqrt(c^2*x^2 + 1)*x^3/c^6 - 105*sqrt(c^2*x^2 + 1)*x/c^8 + 105*arcsinh(c*x)/c^9)*c)*
b*c^4*d^2 + 1/4*a*d^2*x^4 + 1/144*(48*x^6*arcsinh(c*x) - (8*sqrt(c^2*x^2 + 1)*x^5/c^2 - 10*sqrt(c^2*x^2 + 1)*x
^3/c^4 + 15*sqrt(c^2*x^2 + 1)*x/c^6 - 15*arcsinh(c*x)/c^7)*c)*b*c^2*d^2 + 1/32*(8*x^4*arcsinh(c*x) - (2*sqrt(c
^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c)*b*d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*asinh(c*x))*(d + c^2*d*x^2)^2,x)

[Out]

int(x^3*(a + b*asinh(c*x))*(d + c^2*d*x^2)^2, x)

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sympy [A]  time = 11.00, size = 218, normalized size = 1.21 \[ \begin {cases} \frac {a c^{4} d^{2} x^{8}}{8} + \frac {a c^{2} d^{2} x^{6}}{3} + \frac {a d^{2} x^{4}}{4} + \frac {b c^{4} d^{2} x^{8} \operatorname {asinh}{\left (c x \right )}}{8} - \frac {b c^{3} d^{2} x^{7} \sqrt {c^{2} x^{2} + 1}}{64} + \frac {b c^{2} d^{2} x^{6} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {43 b c d^{2} x^{5} \sqrt {c^{2} x^{2} + 1}}{1152} + \frac {b d^{2} x^{4} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {73 b d^{2} x^{3} \sqrt {c^{2} x^{2} + 1}}{4608 c} + \frac {73 b d^{2} x \sqrt {c^{2} x^{2} + 1}}{3072 c^{3}} - \frac {73 b d^{2} \operatorname {asinh}{\left (c x \right )}}{3072 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c**2*d*x**2+d)**2*(a+b*asinh(c*x)),x)

[Out]

Piecewise((a*c**4*d**2*x**8/8 + a*c**2*d**2*x**6/3 + a*d**2*x**4/4 + b*c**4*d**2*x**8*asinh(c*x)/8 - b*c**3*d*
*2*x**7*sqrt(c**2*x**2 + 1)/64 + b*c**2*d**2*x**6*asinh(c*x)/3 - 43*b*c*d**2*x**5*sqrt(c**2*x**2 + 1)/1152 + b
*d**2*x**4*asinh(c*x)/4 - 73*b*d**2*x**3*sqrt(c**2*x**2 + 1)/(4608*c) + 73*b*d**2*x*sqrt(c**2*x**2 + 1)/(3072*
c**3) - 73*b*d**2*asinh(c*x)/(3072*c**4), Ne(c, 0)), (a*d**2*x**4/4, True))

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