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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, 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 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 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 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]

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])

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*}

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Mathematica [A]  time = 0.16, size = 130, normalized size = 0.60 \[ \frac {d \sqrt {c^2 d x^2+d} \left (105 a \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3+105 b \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3 \sinh ^{-1}(c x)-b c x \left (75 c^6 x^6+168 c^4 x^4+35 c^2 x^2-210\right ) \sqrt {c^2 x^2+1}\right )}{3675 c^4 \left (c^2 x^2+1\right )} \]

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))

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fricas [A]  time = 0.67, size = 199, normalized size = 0.92 \[ \frac {105 \, {\left (5 \, b c^{8} d x^{8} + 13 \, b c^{6} d x^{6} + 9 \, b c^{4} d x^{4} - b c^{2} d x^{2} - 2 \, b d\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (525 \, a c^{8} d x^{8} + 1365 \, a c^{6} d x^{6} + 945 \, a c^{4} d x^{4} - 105 \, a c^{2} d x^{2} - 210 \, a d - {\left (75 \, b c^{7} d x^{7} + 168 \, b c^{5} d x^{5} + 35 \, b c^{3} d x^{3} - 210 \, b c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{3675 \, {\left (c^{6} x^{2} + c^{4}\right )}} \]

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)

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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)^(3/2)*(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 [B]  time = 0.27, size = 872, normalized size = 4.02 \[ a \left (\frac {x^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{7 c^{2} d}-\frac {2 \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{35 d \,c^{4}}\right )+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (64 c^{8} x^{8}+64 c^{7} x^{7} \sqrt {c^{2} x^{2}+1}+144 c^{6} x^{6}+112 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}+104 c^{4} x^{4}+56 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+25 c^{2} x^{2}+7 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (-1+7 \arcsinh \left (c x \right )\right ) d}{6272 c^{4} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 c^{6} x^{6}+16 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}+28 c^{4} x^{4}+20 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+13 c^{2} x^{2}+5 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (-1+5 \arcsinh \left (c x \right )\right ) d}{3200 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}+4 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+5 c^{2} x^{2}+3 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (-1+3 \arcsinh \left (c x \right )\right ) d}{384 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (-1+\arcsinh \left (c x \right )\right ) d}{128 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (1+\arcsinh \left (c x \right )\right ) d}{128 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 c^{4} x^{4}-4 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+5 c^{2} x^{2}-3 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (1+3 \arcsinh \left (c x \right )\right ) d}{384 c^{4} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (16 c^{6} x^{6}-16 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}+28 c^{4} x^{4}-20 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+13 c^{2} x^{2}-5 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (1+5 \arcsinh \left (c x \right )\right ) d}{3200 c^{4} \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (64 c^{8} x^{8}-64 c^{7} x^{7} \sqrt {c^{2} x^{2}+1}+144 c^{6} x^{6}-112 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}+104 c^{4} x^{4}-56 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+25 c^{2} x^{2}-7 c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (1+7 \arcsinh \left (c x \right )\right ) d}{6272 c^{4} \left (c^{2} x^{2}+1\right )}\right ) \]

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))

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maxima [A]  time = 0.39, size = 145, normalized size = 0.67 \[ \frac {1}{35} \, {\left (\frac {5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{35} \, {\left (\frac {5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} a - \frac {{\left (75 \, c^{6} d^{\frac {3}{2}} x^{7} + 168 \, c^{4} d^{\frac {3}{2}} x^{5} + 35 \, c^{2} d^{\frac {3}{2}} x^{3} - 210 \, d^{\frac {3}{2}} x\right )} b}{3675 \, c^{3}} \]

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]

1/35*(5*(c^2*d*x^2 + d)^(5/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(5/2)/(c^4*d))*b*arcsinh(c*x) + 1/35*(5*(c^2*d*x
^2 + d)^(5/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(5/2)/(c^4*d))*a - 1/3675*(75*c^6*d^(3/2)*x^7 + 168*c^4*d^(3/2)*
x^5 + 35*c^2*d^(3/2)*x^3 - 210*d^(3/2)*x)*b/c^3

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

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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )\, dx \]

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]

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

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