3.29 \(\int \frac{x^3 (A+B x)}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=81 \[ \frac{\sqrt{a+b x^2} (4 A+3 B x)}{2 b^2}-\frac{x^2 (A+B x)}{b \sqrt{a+b x^2}}-\frac{3 a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}} \]

[Out]

-((x^2*(A + B*x))/(b*Sqrt[a + b*x^2])) + ((4*A + 3*B*x)*Sqrt[a + b*x^2])/(2*b^2) - (3*a*B*ArcTanh[(Sqrt[b]*x)/
Sqrt[a + b*x^2]])/(2*b^(5/2))

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Rubi [A]  time = 0.0434866, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {819, 780, 217, 206} \[ \frac{\sqrt{a+b x^2} (4 A+3 B x)}{2 b^2}-\frac{x^2 (A+B x)}{b \sqrt{a+b x^2}}-\frac{3 a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a + b*x^2)^(3/2),x]

[Out]

-((x^2*(A + B*x))/(b*Sqrt[a + b*x^2])) + ((4*A + 3*B*x)*Sqrt[a + b*x^2])/(2*b^2) - (3*a*B*ArcTanh[(Sqrt[b]*x)/
Sqrt[a + b*x^2]])/(2*b^(5/2))

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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

Mathematica [A]  time = 0.0522584, size = 72, normalized size = 0.89 \[ \frac{a (4 A+3 B x)+b x^2 (2 A+B x)}{2 b^2 \sqrt{a+b x^2}}-\frac{3 a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(a + b*x^2)^(3/2),x]

[Out]

(b*x^2*(2*A + B*x) + a*(4*A + 3*B*x))/(2*b^2*Sqrt[a + b*x^2]) - (3*a*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(
2*b^(5/2))

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Maple [A]  time = 0.008, size = 93, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}B}{2\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,Bax}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,Ba}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{A{x}^{2}}{b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+2\,{\frac{Aa}{{b}^{2}\sqrt{b{x}^{2}+a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(b*x^2+a)^(3/2),x)

[Out]

1/2*B*x^3/b/(b*x^2+a)^(1/2)+3/2*B/b^2*a*x/(b*x^2+a)^(1/2)-3/2*B/b^(5/2)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+A*x^2/
b/(b*x^2+a)^(1/2)+2*A*a/b^2/(b*x^2+a)^(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+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59985, size = 448, normalized size = 5.53 \begin{align*} \left [\frac{3 \,{\left (B a b x^{2} + B a^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (B b^{2} x^{3} + 2 \, A b^{2} x^{2} + 3 \, B a b x + 4 \, A a b\right )} \sqrt{b x^{2} + a}}{4 \,{\left (b^{4} x^{2} + a b^{3}\right )}}, \frac{3 \,{\left (B a b x^{2} + B a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (B b^{2} x^{3} + 2 \, A b^{2} x^{2} + 3 \, B a b x + 4 \, A a b\right )} \sqrt{b x^{2} + a}}{2 \,{\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(3*(B*a*b*x^2 + B*a^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(B*b^2*x^3 + 2*A*b^2*x
^2 + 3*B*a*b*x + 4*A*a*b)*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3), 1/2*(3*(B*a*b*x^2 + B*a^2)*sqrt(-b)*arctan(sqrt(
-b)*x/sqrt(b*x^2 + a)) + (B*b^2*x^3 + 2*A*b^2*x^2 + 3*B*a*b*x + 4*A*a*b)*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3)]

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Sympy [A]  time = 7.67201, size = 117, normalized size = 1.44 \begin{align*} A \left (\begin{cases} \frac{2 a}{b^{2} \sqrt{a + b x^{2}}} + \frac{x^{2}}{b \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{3}{2}}} & \text{otherwise} \end{cases}\right ) + B \left (\frac{3 \sqrt{a} x}{2 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{5}{2}}} + \frac{x^{3}}{2 \sqrt{a} b \sqrt{1 + \frac{b x^{2}}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(b*x**2+a)**(3/2),x)

[Out]

A*Piecewise((2*a/(b**2*sqrt(a + b*x**2)) + x**2/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(3/2)), True)) +
B*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqr
t(1 + b*x**2/a)))

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Giac [A]  time = 1.21729, size = 95, normalized size = 1.17 \begin{align*} \frac{{\left ({\left (\frac{B x}{b} + \frac{2 \, A}{b}\right )} x + \frac{3 \, B a}{b^{2}}\right )} x + \frac{4 \, A a}{b^{2}}}{2 \, \sqrt{b x^{2} + a}} + \frac{3 \, B a \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/2*(((B*x/b + 2*A/b)*x + 3*B*a/b^2)*x + 4*A*a/b^2)/sqrt(b*x^2 + a) + 3/2*B*a*log(abs(-sqrt(b)*x + sqrt(b*x^2
+ a)))/b^(5/2)