3.3.68 \(\int x (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))^2 \, dx\) [268]

Optimal. Leaf size=267 \[ \frac {16 b^2 d \sqrt {d+c^2 d x^2}}{75 c^2}+\frac {8 b^2 d \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{225 c^2}+\frac {2 b^2 d \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}}{125 c^2}-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d} \]

[Out]

1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/c^2/d+16/75*b^2*d*(c^2*d*x^2+d)^(1/2)/c^2+8/225*b^2*d*(c^2*x^2+1)
*(c^2*d*x^2+d)^(1/2)/c^2+2/125*b^2*d*(c^2*x^2+1)^2*(c^2*d*x^2+d)^(1/2)/c^2-2/5*b*d*x*(a+b*arcsinh(c*x))*(c^2*d
*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-4/15*b*c*d*x^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2/25
*b*c^3*d*x^5*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5798, 200, 5784, 12, 1261, 712} \begin {gather*} -\frac {2 b d x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {c^2 x^2+1}}+\frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {4 b c d x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {c^2 x^2+1}}-\frac {2 b c^3 d x^5 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {c^2 x^2+1}}+\frac {2 b^2 d \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d}}{125 c^2}+\frac {16 b^2 d \sqrt {c^2 d x^2+d}}{75 c^2}+\frac {8 b^2 d \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}}{225 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*d*Sqrt[d + c^2*d*x^2])/(75*c^2) + (8*b^2*d*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/(225*c^2) + (2*b^2*d*(1
+ c^2*x^2)^2*Sqrt[d + c^2*d*x^2])/(125*c^2) - (2*b*d*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(5*c*Sqrt[1 +
c^2*x^2]) - (4*b*c*d*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(15*Sqrt[1 + c^2*x^2]) - (2*b*c^3*d*x^5*Sq
rt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(25*Sqrt[1 + c^2*x^2]) + ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2
)/(5*c^2*d)

Rule 12

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

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

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 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {\left (2 b d \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{5 c \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {\left (2 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (15+10 c^2 x^2+3 c^4 x^4\right )}{15 \sqrt {1+c^2 x^2}} \, dx}{5 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {\left (2 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (15+10 c^2 x^2+3 c^4 x^4\right )}{\sqrt {1+c^2 x^2}} \, dx}{75 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \frac {15+10 c^2 x+3 c^4 x^2}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{75 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \left (\frac {8}{\sqrt {1+c^2 x}}+4 \sqrt {1+c^2 x}+3 \left (1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )}{75 \sqrt {1+c^2 x^2}}\\ &=\frac {16 b^2 d \sqrt {d+c^2 d x^2}}{75 c^2}+\frac {8 b^2 d \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{225 c^2}+\frac {2 b^2 d \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}}{125 c^2}-\frac {2 b d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{5 c^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 198, normalized size = 0.74 \begin {gather*} \frac {d \sqrt {d+c^2 d x^2} \left (225 a^2 \left (1+c^2 x^2\right )^3-30 a b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )+2 b^2 \left (149+187 c^2 x^2+47 c^4 x^4+9 c^6 x^6\right )+30 b \left (15 a \left (1+c^2 x^2\right )^3-b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )\right ) \sinh ^{-1}(c x)+225 b^2 \left (1+c^2 x^2\right )^3 \sinh ^{-1}(c x)^2\right )}{1125 c^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d + c^2*d*x^2]*(225*a^2*(1 + c^2*x^2)^3 - 30*a*b*c*x*Sqrt[1 + c^2*x^2]*(15 + 10*c^2*x^2 + 3*c^4*x^4) +
2*b^2*(149 + 187*c^2*x^2 + 47*c^4*x^4 + 9*c^6*x^6) + 30*b*(15*a*(1 + c^2*x^2)^3 - b*c*x*Sqrt[1 + c^2*x^2]*(15
+ 10*c^2*x^2 + 3*c^4*x^4))*ArcSinh[c*x] + 225*b^2*(1 + c^2*x^2)^3*ArcSinh[c*x]^2))/(1125*c^2*(1 + c^2*x^2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1148\) vs. \(2(233)=466\).
time = 0.95, size = 1149, normalized size = 4.30

method	result	size
default	\(\text {Expression too large to display}\)	\(1149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/5*a^2/c^2/d*(c^2*d*x^2+d)^(5/2)+b^2*(1/4000*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6+16*(c^2*x^2+1)^(1/2)*x^5*c^5+2
8*c^4*x^4+20*(c^2*x^2+1)^(1/2)*x^3*c^3+13*c^2*x^2+5*(c^2*x^2+1)^(1/2)*c*x+1)*(25*arcsinh(c*x)^2-10*arcsinh(c*x
)+2)*d/c^2/(c^2*x^2+1)+1/288*(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)*d/c^2/(c^2*x^2+1)+1/16*(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)*d/c^2/(c^2*x^2+1)+1/16*(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)*d/c^2/(c^2*x^2+1)+1/288*(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)*d/c
^2/(c^2*x^2+1)+1/4000*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6-16*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4-20*(c^2*x^2+1)
^(1/2)*x^3*c^3+13*c^2*x^2-5*(c^2*x^2+1)^(1/2)*c*x+1)*(25*arcsinh(c*x)^2+10*arcsinh(c*x)+2)*d/c^2/(c^2*x^2+1))+
2*a*b*(1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6+16*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4+20*(c^2*x^2+1)^(1/2)*x^
3*c^3+13*c^2*x^2+5*(c^2*x^2+1)^(1/2)*c*x+1)*(-1+5*arcsinh(c*x))*d/c^2/(c^2*x^2+1)+1/96*(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))*d/c^2/(c^2*x^2+
1)+1/16*(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)*d/c^2/(c^2*x^2+1)+1/16*(d*(c^
2*x^2+1))^(1/2)*(c^2*x^2-(c^2*x^2+1)^(1/2)*c*x+1)*(1+arcsinh(c*x))*d/c^2/(c^2*x^2+1)+1/96*(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))*d/c^2/(c^2*x^
2+1)+1/800*(d*(c^2*x^2+1))^(1/2)*(16*x^6*c^6-16*(c^2*x^2+1)^(1/2)*x^5*c^5+28*c^4*x^4-20*(c^2*x^2+1)^(1/2)*x^3*
c^3+13*c^2*x^2-5*(c^2*x^2+1)^(1/2)*c*x+1)*(1+5*arcsinh(c*x))*d/c^2/(c^2*x^2+1))

________________________________________________________________________________________

Maxima [A]
time = 0.33, size = 230, normalized size = 0.86 \begin {gather*} \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b^{2} \operatorname {arsinh}\left (c x\right )^{2}}{5 \, c^{2} d} + \frac {2}{1125} \, b^{2} {\left (\frac {9 \, \sqrt {c^{2} x^{2} + 1} c^{2} d^{\frac {5}{2}} x^{4} + 38 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}} x^{2} + \frac {149 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}}}{c^{2}}}{d} - \frac {15 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} \operatorname {arsinh}\left (c x\right )}{c d}\right )} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a b \operatorname {arsinh}\left (c x\right )}{5 \, c^{2} d} + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a^{2}}{5 \, c^{2} d} - \frac {2 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} a b}{75 \, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c^2*d*x^2 + d)^(5/2)*b^2*arcsinh(c*x)^2/(c^2*d) + 2/1125*b^2*((9*sqrt(c^2*x^2 + 1)*c^2*d^(5/2)*x^4 + 38*s
qrt(c^2*x^2 + 1)*d^(5/2)*x^2 + 149*sqrt(c^2*x^2 + 1)*d^(5/2)/c^2)/d - 15*(3*c^4*d^(5/2)*x^5 + 10*c^2*d^(5/2)*x
^3 + 15*d^(5/2)*x)*arcsinh(c*x)/(c*d)) + 2/5*(c^2*d*x^2 + d)^(5/2)*a*b*arcsinh(c*x)/(c^2*d) + 1/5*(c^2*d*x^2 +
d)^(5/2)*a^2/(c^2*d) - 2/75*(3*c^4*d^(5/2)*x^5 + 10*c^2*d^(5/2)*x^3 + 15*d^(5/2)*x)*a*b/(c*d)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 332, normalized size = 1.24 \begin {gather*} \frac {225 \, {\left (b^{2} c^{6} d x^{6} + 3 \, b^{2} c^{4} d x^{4} + 3 \, b^{2} c^{2} d x^{2} + b^{2} d\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 30 \, {\left (15 \, a b c^{6} d x^{6} + 45 \, a b c^{4} d x^{4} + 45 \, a b c^{2} d x^{2} + 15 \, a b d - {\left (3 \, b^{2} c^{5} d x^{5} + 10 \, b^{2} c^{3} d x^{3} + 15 \, b^{2} c d 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 (9 \, {\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{6} d x^{6} + {\left (675 \, a^{2} + 94 \, b^{2}\right )} c^{4} d x^{4} + {\left (675 \, a^{2} + 374 \, b^{2}\right )} c^{2} d x^{2} + {\left (225 \, a^{2} + 298 \, b^{2}\right )} d - 30 \, {\left (3 \, a b c^{5} d x^{5} + 10 \, a b c^{3} d x^{3} + 15 \, a b c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{1125 \, {\left (c^{4} x^{2} + c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1125*(225*(b^2*c^6*d*x^6 + 3*b^2*c^4*d*x^4 + 3*b^2*c^2*d*x^2 + b^2*d)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2
*x^2 + 1))^2 + 30*(15*a*b*c^6*d*x^6 + 45*a*b*c^4*d*x^4 + 45*a*b*c^2*d*x^2 + 15*a*b*d - (3*b^2*c^5*d*x^5 + 10*b
^2*c^3*d*x^3 + 15*b^2*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (9*(25*a^2
+ 2*b^2)*c^6*d*x^6 + (675*a^2 + 94*b^2)*c^4*d*x^4 + (675*a^2 + 374*b^2)*c^2*d*x^2 + (225*a^2 + 298*b^2)*d - 30
*(3*a*b*c^5*d*x^5 + 10*a*b*c^3*d*x^3 + 15*a*b*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d))/(c^4*x^2 + c^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \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*(c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))**2,x)

[Out]

Integral(x*(d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))**2, x)

________________________________________________________________________________________

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*(c^2*d*x^2+d)^(3/2)*(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 = -0.00 \begin {gather*} \int x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,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)^(3/2),x)

[Out]

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

________________________________________________________________________________________