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

Optimal. Leaf size=217 \[ \frac{\left (c^2 d x^2+d\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4 d^2}-\frac{\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d}-\frac{b c^3 d x^7 \sqrt{c^2 d x^2+d}}{49 \sqrt{c^2 x^2+1}}-\frac{8 b c d x^5 \sqrt{c^2 d x^2+d}}{175 \sqrt{c^2 x^2+1}}-\frac{b d x^3 \sqrt{c^2 d x^2+d}}{105 c \sqrt{c^2 x^2+1}}+\frac{2 b d x \sqrt{c^2 d x^2+d}}{35 c^3 \sqrt{c^2 x^2+1}} \]

[Out]

(2*b*d*x*Sqrt[d + c^2*d*x^2])/(35*c^3*Sqrt[1 + c^2*x^2]) - (b*d*x^3*Sqrt[d + c^2*d*x^2])/(105*c*Sqrt[1 + c^2*x
^2]) - (8*b*c*d*x^5*Sqrt[d + c^2*d*x^2])/(175*Sqrt[1 + c^2*x^2]) - (b*c^3*d*x^7*Sqrt[d + c^2*d*x^2])/(49*Sqrt[
1 + c^2*x^2]) - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4*d) + ((d + c^2*d*x^2)^(7/2)*(a + b*ArcSinh
[c*x]))/(7*c^4*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.17704, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {266, 43, 5734, 12, 373} \[ \frac{\left (c^2 d x^2+d\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4 d^2}-\frac{\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 d}-\frac{b c^3 d x^7 \sqrt{c^2 d x^2+d}}{49 \sqrt{c^2 x^2+1}}-\frac{8 b c d x^5 \sqrt{c^2 d x^2+d}}{175 \sqrt{c^2 x^2+1}}-\frac{b d x^3 \sqrt{c^2 d x^2+d}}{105 c \sqrt{c^2 x^2+1}}+\frac{2 b d x \sqrt{c^2 d x^2+d}}{35 c^3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d*x*Sqrt[d + c^2*d*x^2])/(35*c^3*Sqrt[1 + c^2*x^2]) - (b*d*x^3*Sqrt[d + c^2*d*x^2])/(105*c*Sqrt[1 + c^2*x
^2]) - (8*b*c*d*x^5*Sqrt[d + c^2*d*x^2])/(175*Sqrt[1 + c^2*x^2]) - (b*c^3*d*x^7*Sqrt[d + c^2*d*x^2])/(49*Sqrt[
1 + c^2*x^2]) - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4*d) + ((d + c^2*d*x^2)^(7/2)*(a + b*ArcSinh
[c*x]))/(7*c^4*d^2)

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 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 5734

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], Int[x^m*(d + e*x^2)^p, x], x] - Dist[(b*c*d^(p - 1/2)*Sqrt[d
+ e*x^2])/Sqrt[1 + c^2*x^2], 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 + 1/2, 0] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 12

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

Rule 373

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

Rubi steps

\begin{align*} \int x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=-\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int \frac{\left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right )}{35 c^4} \, dx}{\sqrt{1+c^2 x^2}}+\left (a+b \sinh ^{-1}(c x)\right ) \int x^3 \left (d+c^2 d x^2\right )^{3/2} \, dx\\ &=-\frac{\left (b d \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right ) \, dx}{35 c^3 \sqrt{1+c^2 x^2}}+\frac{1}{2} \left (a+b \sinh ^{-1}(c x)\right ) \operatorname{Subst}\left (\int x \left (d+c^2 d x\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac{\left (b d \sqrt{d+c^2 d x^2}\right ) \int \left (-2+c^2 x^2+8 c^4 x^4+5 c^6 x^6\right ) \, dx}{35 c^3 \sqrt{1+c^2 x^2}}+\frac{1}{2} \left (a+b \sinh ^{-1}(c x)\right ) \operatorname{Subst}\left (\int \left (-\frac{\left (d+c^2 d x\right )^{3/2}}{c^2}+\frac{\left (d+c^2 d x\right )^{5/2}}{c^2 d}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b d x \sqrt{d+c^2 d x^2}}{35 c^3 \sqrt{1+c^2 x^2}}-\frac{b d x^3 \sqrt{d+c^2 d x^2}}{105 c \sqrt{1+c^2 x^2}}-\frac{8 b c d x^5 \sqrt{d+c^2 d x^2}}{175 \sqrt{1+c^2 x^2}}-\frac{b c^3 d x^7 \sqrt{d+c^2 d x^2}}{49 \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 )}{5 c^4 d}+\frac{\left (d+c^2 d x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.164445, size = 130, normalized size = 0.6 \[ \frac{d \sqrt{c^2 d x^2+d} \left (105 a \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3-b c x \left (75 c^6 x^6+168 c^4 x^4+35 c^2 x^2-210\right ) \sqrt{c^2 x^2+1}+105 b \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3 \sinh ^{-1}(c x)\right )}{3675 c^4 \left (c^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d + c^2*d*x^2]*(105*a*(1 + c^2*x^2)^3*(-2 + 5*c^2*x^2) - b*c*x*Sqrt[1 + c^2*x^2]*(-210 + 35*c^2*x^2 +
168*c^4*x^4 + 75*c^6*x^6) + 105*b*(1 + c^2*x^2)^3*(-2 + 5*c^2*x^2)*ArcSinh[c*x]))/(3675*c^4*(1 + c^2*x^2))

________________________________________________________________________________________

Maple [B]  time = 0.211, size = 872, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(1/7*x^2*(c^2*d*x^2+d)^(5/2)/c^2/d-2/35/d/c^4*(c^2*d*x^2+d)^(5/2))+b*(1/6272*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x
^8+64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6+112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4+56*c^3*x^3*(c^2*x^2+1)^(
1/2)+25*c^2*x^2+7*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+7*arcsinh(c*x))*d/c^4/(c^2*x^2+1)+1/3200*(d*(c^2*x^2+1))^(1/2)*
(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^
(1/2)+1)*(-1+5*arcsinh(c*x))*d/c^4/(c^2*x^2+1)-1/384*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1
/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+3*arcsinh(c*x))*d/c^4/(c^2*x^2+1)-3/128*(d*(c^2*x^2+1))^(1/2)*(c^
2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(-1+arcsinh(c*x))*d/c^4/(c^2*x^2+1)-3/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c
^2*x^2+1)^(1/2)+1)*(1+arcsinh(c*x))*d/c^4/(c^2*x^2+1)-1/384*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^
2+1)^(1/2)+5*c^2*x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(1+3*arcsinh(c*x))*d/c^4/(c^2*x^2+1)+1/3200*(d*(c^2*x^2+1))^(1
/2)*(16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2
+1)^(1/2)+1)*(1+5*arcsinh(c*x))*d/c^4/(c^2*x^2+1)+1/6272*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x^8-64*c^7*x^7*(c^2*x^2
+1)^(1/2)+144*c^6*x^6-112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4-56*c^3*x^3*(c^2*x^2+1)^(1/2)+25*c^2*x^2-7*c*x*
(c^2*x^2+1)^(1/2)+1)*(1+7*arcsinh(c*x))*d/c^4/(c^2*x^2+1))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.51518, size = 460, normalized size = 2.12 \begin{align*} \frac{105 \,{\left (5 \, b c^{8} d x^{8} + 13 \, b c^{6} d x^{6} + 9 \, b c^{4} d x^{4} - b c^{2} d x^{2} - 2 \, b d\right )} \sqrt{c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (525 \, a c^{8} d x^{8} + 1365 \, a c^{6} d x^{6} + 945 \, a c^{4} d x^{4} - 105 \, a c^{2} d x^{2} - 210 \, a d -{\left (75 \, b c^{7} d x^{7} + 168 \, b c^{5} d x^{5} + 35 \, b c^{3} d x^{3} - 210 \, b c d x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{c^{2} d x^{2} + d}}{3675 \,{\left (c^{6} x^{2} + c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3675*(105*(5*b*c^8*d*x^8 + 13*b*c^6*d*x^6 + 9*b*c^4*d*x^4 - b*c^2*d*x^2 - 2*b*d)*sqrt(c^2*d*x^2 + d)*log(c*x
+ sqrt(c^2*x^2 + 1)) + (525*a*c^8*d*x^8 + 1365*a*c^6*d*x^6 + 945*a*c^4*d*x^4 - 105*a*c^2*d*x^2 - 210*a*d - (7
5*b*c^7*d*x^7 + 168*b*c^5*d*x^5 + 35*b*c^3*d*x^3 - 210*b*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d))/(c^6*x
^2 + c^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError