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

Optimal. Leaf size=125 \[ -\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}{8 d^2}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{8 \sqrt{b} d^{5/2}}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d} \]

[Out]

(-3*(b*c - a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*d^2) + ((a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(4*d) + (3*(b*c
- a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(8*Sqrt[b]*d^(5/2))

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Rubi [A]  time = 0.104095, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {444, 50, 63, 217, 206} \[ -\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}{8 d^2}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{8 \sqrt{b} d^{5/2}}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*d^2) + ((a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(4*d) + (3*(b*c
- a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(8*Sqrt[b]*d^(5/2))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \left (a+b x^2\right )^{3/2}}{\sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d}-\frac{(3 (b c-a d)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{8 d}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 d^2}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d}+\frac{\left (3 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{16 d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 d^2}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d}+\frac{\left (3 (b c-a d)^2\right ) \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 )}{8 b d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 d^2}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d}+\frac{\left (3 (b c-a d)^2\right ) \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 )}{8 b d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 d^2}+\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{8 \sqrt{b} d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.361895, size = 131, normalized size = 1.05 \[ \frac{\sqrt{d} \sqrt{a+b x^2} \left (c+d x^2\right ) \left (5 a d-3 b c+2 b d x^2\right )+\frac{3 (b c-a d)^{5/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{b}}{8 d^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(-3*b*c + 5*a*d + 2*b*d*x^2) + (3*(b*c - a*d)^(5/2)*Sqrt[(b*(c + d*x^2))/
(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/b)/(8*d^(5/2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.013, size = 337, normalized size = 2.7 \begin{align*}{\frac{1}{16\,{d}^{2}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 4\,\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}bd+3\,{d}^{2}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}-6\,\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) cabd+3\,{b}^{2}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{2}+10\,\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}ad-6\,\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}bc \right ){\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*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x)

[Out]

1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*b*d+3*d^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))*a^2-6*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*a*b*d+3*b^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))*c^2+10*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*a*d-6*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*c)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/d^2/(b*d
)^(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*(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.04739, size = 751, normalized size = 6.01 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{32 \, b d^{3}}, -\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{16 \, b d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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)*sqrt(d*x^2 + c)*sqrt(b*d)) + 4*(2*b^2*d^2*x^2 -
3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3), -1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqr
t(-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)) - 2*(2*b^2*d^2*x^2 - 3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3
)]

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

[Out]

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

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Giac [A]  time = 1.23552, size = 201, normalized size = 1.61 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}{\left (\frac{2 \,{\left (b x^{2} + a\right )}}{b d} - \frac{3 \,{\left (b c d - a d^{2}\right )}}{b d^{3}}\right )} - \frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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^{2}}\right )} b}{8 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)/(b*d) - 3*(b*c*d - a*d^2)/(b*d^3)) -
3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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^2))*b/abs(b)