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

Optimal. Leaf size=88 \[ \frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 b d}-\frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0934393, antiderivative size = 88, 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, 80, 63, 217, 206} \[ \frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 b d}-\frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

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

Mathematica [A]  time = 0.166655, size = 123, normalized size = 1.4 \[ \frac{b \sqrt{d} \sqrt{a+b x^2} \left (c+d x^2\right )-\sqrt{b c-a d} (a d+b c) \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{2 b^2 d^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.016, size = 200, normalized size = 2.3 \begin{align*} -{\frac{1}{4\,bd} \left ( a\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 ) d+b\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 ) c-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd} \right ) \sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c}{\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.17643, size = 595, normalized size = 6.76 \begin{align*} \left [\frac{4 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} b d +{\left (b c + a d\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 )}{8 \, b^{2} d^{2}}, \frac{2 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} b d +{\left (b c + a d\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 )}{4 \, b^{2} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b*d + (b*c + a*d)*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)*sqrt(d*x^2 + c)*sqrt(b*d)))/(b
^2*d^2), 1/4*(2*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b*d + (b*c + a*d)*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)))/(b^2*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{a + b x^{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)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.22905, size = 140, normalized size = 1.59 \begin{align*} \frac{\frac{{\left (b c + a d\right )} \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 |}\right )}{\sqrt{b d} d} + \frac{\sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}}{b d}}{2 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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