3.36 \(\int \frac{x^3 (A+B x)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{2 A+3 B x}{3 b^2 \sqrt{a+b x^2}}-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}+\frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.042459, antiderivative size = 79, 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, 778, 217, 206} \[ -\frac{2 A+3 B x}{3 b^2 \sqrt{a+b x^2}}-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}+\frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]

[Out]

-(x^2*(A + B*x))/(3*b*(a + b*x^2)^(3/2)) - (2*A + 3*B*x)/(3*b^2*Sqrt[a + b*x^2]) + (B*ArcTanh[(Sqrt[b]*x)/Sqrt
[a + b*x^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 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[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 )^{5/2}} \, dx &=-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}+\frac{\int \frac{x (2 a A+3 a B x)}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}-\frac{2 A+3 B x}{3 b^2 \sqrt{a+b x^2}}+\frac{B \int \frac{1}{\sqrt{a+b x^2}} \, dx}{b^2}\\ &=-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}-\frac{2 A+3 B x}{3 b^2 \sqrt{a+b x^2}}+\frac{B \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{b^2}\\ &=-\frac{x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}-\frac{2 A+3 B x}{3 b^2 \sqrt{a+b x^2}}+\frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0764119, size = 69, normalized size = 0.87 \[ \frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}}-\frac{a (2 A+3 B x)+b x^2 (3 A+4 B x)}{3 b^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.008, size = 91, normalized size = 1.2 \begin{align*} -{\frac{{x}^{3}B}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{Bx}{{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{B\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}-{\frac{A{x}^{2}}{b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,Aa}{3\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(b*x^2+a)^(5/2),x)

[Out]

-1/3*B*x^3/b/(b*x^2+a)^(3/2)-B/b^2*x/(b*x^2+a)^(1/2)+B/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-A*x^2/b/(b*x^2+a)
^(3/2)-2/3*A*a/b^2/(b*x^2+a)^(3/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7397, size = 536, normalized size = 6.78 \begin{align*} \left [\frac{3 \,{\left (B b^{2} x^{4} + 2 \, 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 (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt{b x^{2} + a}}{6 \,{\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac{3 \,{\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt{b x^{2} + a}}{3 \,{\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(B*b^2*x^4 + 2*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*(4*B*b^2
*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x + 2*A*a*b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), -1/3*(3*(B*b^2*x
^4 + 2*B*a*b*x^2 + B*a^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (4*B*b^2*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x
+ 2*A*a*b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]

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Sympy [A]  time = 12.5789, size = 400, normalized size = 5.06 \begin{align*} A \left (\begin{cases} - \frac{2 a}{3 a b^{2} \sqrt{a + b x^{2}} + 3 b^{3} x^{2} \sqrt{a + b x^{2}}} - \frac{3 b x^{2}}{3 a b^{2} \sqrt{a + b x^{2}} + 3 b^{3} x^{2} \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{5}{2}}} & \text{otherwise} \end{cases}\right ) + B \left (\frac{3 a^{\frac{39}{2}} b^{11} \sqrt{1 + \frac{b x^{2}}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a^{\frac{37}{2}} b^{12} x^{2} \sqrt{1 + \frac{b x^{2}}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a^{19} b^{\frac{23}{2}} x}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{4 a^{18} b^{\frac{25}{2}} x^{3}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \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)**(5/2),x)

[Out]

A*Piecewise((-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)) - 3*b*x**2/(3*a*b**2*sqrt(a + b*x
**2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(5/2)), True)) + B*(3*a**(39/2)*b**11*sqrt(1 + b*
x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1
+ b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt
(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2)*s
qrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(2
7/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)))

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Giac [A]  time = 1.41643, size = 95, normalized size = 1.2 \begin{align*} -\frac{{\left ({\left (\frac{4 \, B x}{b} + \frac{3 \, A}{b}\right )} x + \frac{3 \, B a}{b^{2}}\right )} x + \frac{2 \, A a}{b^{2}}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} - \frac{B \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{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)^(5/2),x, algorithm="giac")

[Out]

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