3.3 \(\int x (A+B x) \sqrt {a+b x^2} \, dx\)

Optimal. Leaf size=80 \[ -\frac {a^2 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {\left (a+b x^2\right )^{3/2} (4 A+3 B x)}{12 b}-\frac {a B x \sqrt {a+b x^2}}{8 b} \]

[Out]

1/12*(3*B*x+4*A)*(b*x^2+a)^(3/2)/b-1/8*a^2*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)-1/8*a*B*x*(b*x^2+a)^(1
/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {780, 195, 217, 206} \[ -\frac {a^2 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {\left (a+b x^2\right )^{3/2} (4 A+3 B x)}{12 b}-\frac {a B x \sqrt {a+b x^2}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

-(a*B*x*Sqrt[a + b*x^2])/(8*b) + ((4*A + 3*B*x)*(a + b*x^2)^(3/2))/(12*b) - (a^2*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a
+ b*x^2]])/(8*b^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

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 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]

Rubi steps

\begin {align*} \int x (A+B x) \sqrt {a+b x^2} \, dx &=\frac {(4 A+3 B x) \left (a+b x^2\right )^{3/2}}{12 b}-\frac {(a B) \int \sqrt {a+b x^2} \, dx}{4 b}\\ &=-\frac {a B x \sqrt {a+b x^2}}{8 b}+\frac {(4 A+3 B x) \left (a+b x^2\right )^{3/2}}{12 b}-\frac {\left (a^2 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b}\\ &=-\frac {a B x \sqrt {a+b x^2}}{8 b}+\frac {(4 A+3 B x) \left (a+b x^2\right )^{3/2}}{12 b}-\frac {\left (a^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b}\\ &=-\frac {a B x \sqrt {a+b x^2}}{8 b}+\frac {(4 A+3 B x) \left (a+b x^2\right )^{3/2}}{12 b}-\frac {a^2 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 86, normalized size = 1.08 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {b} \left (8 a A+3 a B x+8 A b x^2+6 b B x^3\right )-\frac {3 a^{3/2} B \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{24 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*(8*a*A + 3*a*B*x + 8*A*b*x^2 + 6*b*B*x^3) - (3*a^(3/2)*B*ArcSinh[(Sqrt[b]*x)/Sqrt[a]
])/Sqrt[1 + (b*x^2)/a]))/(24*b^(3/2))

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fricas [A]  time = 0.58, size = 157, normalized size = 1.96 \[ \left [\frac {3 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 3 \, B a b x + 8 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}, \frac {3 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 3 \, B a b x + 8 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*B*a^2*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*B*b^2*x^3 + 8*A*b^2*x^2 + 3*B*a*
b*x + 8*A*a*b)*sqrt(b*x^2 + a))/b^2, 1/24*(3*B*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*B*b^2*x^3
+ 8*A*b^2*x^2 + 3*B*a*b*x + 8*A*a*b)*sqrt(b*x^2 + a))/b^2]

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giac [A]  time = 0.44, size = 68, normalized size = 0.85 \[ \frac {B a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, B x + 4 \, A\right )} x + \frac {3 \, B a}{b}\right )} x + \frac {8 \, A a}{b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/8*B*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/24*sqrt(b*x^2 + a)*((2*(3*B*x + 4*A)*x + 3*B*a/b)
*x + 8*A*a/b)

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maple [A]  time = 0.00, size = 75, normalized size = 0.94 \[ -\frac {B \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, B a x}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B x}{4 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)*(b*x^2+a)^(1/2),x)

[Out]

1/4*B*x*(b*x^2+a)^(3/2)/b-1/8*a*B*x*(b*x^2+a)^(1/2)/b-1/8*B*a^2/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/3*A*(b
*x^2+a)^(3/2)/b

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maxima [A]  time = 1.32, size = 67, normalized size = 0.84 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/4*(b*x^2 + a)^(3/2)*B*x/b - 1/8*sqrt(b*x^2 + a)*B*a*x/b - 1/8*B*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 1/3*(b*
x^2 + a)^(3/2)*A/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\sqrt {b\,x^2+a}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2)^(1/2)*(A + B*x),x)

[Out]

int(x*(a + b*x^2)^(1/2)*(A + B*x), x)

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sympy [A]  time = 14.41, size = 124, normalized size = 1.55 \[ A \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + \frac {B a^{\frac {3}{2}} x}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {B b x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x**2+a)**(1/2),x)

[Out]

A*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + B*a**(3/2)*x/(8*b*sqrt(1 + b*x**2
/a)) + 3*B*sqrt(a)*x**3/(8*sqrt(1 + b*x**2/a)) - B*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*b**(3/2)) + B*b*x**5/(4*sq
rt(a)*sqrt(1 + b*x**2/a))

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