3.981 \(\int \frac{x^3}{(a+b x^2)^{3/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=83 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{b^{3/2} \sqrt{d}}+\frac{a \sqrt{c+d x^2}}{b \sqrt{a+b x^2} (b c-a d)} \]

[Out]

(a*Sqrt[c + d*x^2])/(b*(b*c - a*d)*Sqrt[a + b*x^2]) + ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^
2])]/(b^(3/2)*Sqrt[d])

________________________________________________________________________________________

Rubi [A]  time = 0.0885643, antiderivative size = 83, normalized size of antiderivative = 1., 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 = {446, 78, 63, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{b^{3/2} \sqrt{d}}+\frac{a \sqrt{c+d x^2}}{b \sqrt{a+b x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[c + d*x^2])/(b*(b*c - a*d)*Sqrt[a + b*x^2]) + ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^
2])]/(b^(3/2)*Sqrt[d])

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.494791, size = 118, normalized size = 1.42 \[ \frac{\frac{a b \left (c+d x^2\right )}{\sqrt{a+b x^2} (b c-a d)}+\frac{\sqrt{b c-a d} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{d}}}{b^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 320, normalized size = 3.9 \begin{align*}{\frac{1}{2\, \left ( ad-bc \right ) b} \left ( \ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ){x}^{2}abd-\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ){x}^{2}{b}^{2}c+\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ){a}^{2}d-\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) abc-2\,\sqrt{bd}\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }a \right ) \sqrt{d{x}^{2}+c}{\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x)

[Out]

1/2*(ln(1/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a*b*d-ln(1/
2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^2*c+ln(1/2*(2*d*x^2
*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*d-ln(1/2*(2*d*x^2*b+2*(b*d*x^4+
a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c-2*(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)*a
)/b*(d*x^2+c)^(1/2)/(b*x^2+a)^(1/2)/(b*d)^(1/2)/(a*d-b*c)/((b*x^2+a)*(d*x^2+c))^(1/2)

________________________________________________________________________________________

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/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.46672, size = 790, normalized size = 9.52 \begin{align*} \left [\frac{4 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} a b d +{\left (a b c - a^{2} d +{\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \,{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{b d}\right )}{4 \,{\left (a b^{3} c d - a^{2} b^{2} d^{2} +{\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}, \frac{2 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} a b d -{\left (a b c - a^{2} d +{\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-b d}}{2 \,{\left (b^{2} d^{2} x^{4} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right )}{2 \,{\left (a b^{3} c d - a^{2} b^{2} d^{2} +{\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a*b*d + (a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*sqrt(b*d)*log(8*b^2*d^2*
x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sq
rt(d*x^2 + c)*sqrt(b*d)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2), 1/2*(2*sqrt(b*x^2 + a)*sqrt(d
*x^2 + c)*a*b*d - (a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x
^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)))/(a*b^3*c*d - a^2*b^2*d^
2 + (b^4*c*d - a*b^3*d^2)*x^2)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b x^{2}\right )^{\frac{3}{2}} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.22551, size = 177, normalized size = 2.13 \begin{align*} \frac{\frac{4 \, \sqrt{b d} a b}{b^{2} c - a b d -{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}} - \frac{\sqrt{b d} \log \left ({\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}{d}}{2 \, b{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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