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

Optimal. Leaf size=136 \[ \frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}+\frac {a+b \sinh ^{-1}(c x)}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {b \sqrt {c^2 d x^2+d} \tan ^{-1}(c x)}{c^4 d^2 \sqrt {c^2 x^2+1}}-\frac {b x \sqrt {c^2 d x^2+d}}{c^3 d^2 \sqrt {c^2 x^2+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5751, 5717, 8, 321, 203} \[ \frac {2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {b x \sqrt {c^2 x^2+1}}{c^3 d \sqrt {c^2 d x^2+d}}-\frac {b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{c^4 d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 203

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

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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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]

Rule 5751

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p
+ 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 143, normalized size = 1.05 \[ \frac {\sqrt {c^2 d x^2+d} \left (a \sqrt {c^2 x^2+1} \left (c^2 x^2+2\right )-b \left (c^3 x^3+c x\right )+b \sqrt {c^2 x^2+1} \left (c^2 x^2+2\right ) \sinh ^{-1}(c x)\right )}{c^4 d^2 \left (c^2 x^2+1\right )^{3/2}}-\frac {b \sqrt {d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{c^4 d^2 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 166, normalized size = 1.22 \[ \frac {{\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 2 \, {\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a c^{2} x^{2} - \sqrt {c^{2} x^{2} + 1} b c x + 2 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} + c^{4} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 + 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 2*(
b*c^2*x^2 + 2*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(a*c^2*x^2 - sqrt(c^2*x^2 + 1)*b*c*x + 2
*a)*sqrt(c^2*d*x^2 + d))/(c^6*d^2*x^2 + c^4*d^2)

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [C]  time = 0.22, size = 260, normalized size = 1.91 \[ \frac {a \,x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2 a}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}-\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x^2/c^2/d/(c^2*d*x^2+d)^(1/2)+2*a/d/c^4/(c^2*d*x^2+d)^(1/2)+b*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcs
inh(c*x)*x^2-b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)*x+2*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)
*arcsinh(c*x)+I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^2*ln(c*x+(c^2*x^2+1)^(1/2)-I)-I*b*(d*(c^2*x^2+
1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^2*ln(c*x+(c^2*x^2+1)^(1/2)+I)

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 119, normalized size = 0.88 \[ -b c {\left (\frac {x}{c^{4} d^{\frac {3}{2}}} + \frac {\arctan \left (c x\right )}{c^{5} d^{\frac {3}{2}}}\right )} + b {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*c*(x/(c^4*d^(3/2)) + arctan(c*x)/(c^5*d^(3/2))) + b*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 +
d)*c^4*d))*arcsinh(c*x) + a*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 + d)*c^4*d))

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________