\(\int x (A+B x) \sqrt {a+b x^2} \, dx\) [3]

Optimal result

Integrand size = 18, antiderivative size = 80 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=-\frac {a B x \sqrt {a+b x^2}}{8 b}+\frac {(4 A+3 B x) \left (a+b x^2\right )^{3/2}}{12 b}-\frac {a^2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {794, 201, 223, 212} \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=-\frac {a^2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {\left (a+b x^2\right )^{3/2} (4 A+3 B x)}{12 b}-\frac {a B x \sqrt {a+b x^2}}{8 b} \]

[In]

Int[x*(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

-1/8*(a*B*x*Sqrt[a + b*x^2])/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 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 794

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*} \text {integral}& = \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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (8 a A+3 a B x+8 A b x^2+6 b B x^3\right )}{24 b}+\frac {a^2 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\left (6 b B \,x^{3}+8 A b \,x^{2}+3 B a x +8 A a \right ) \sqrt {b \,x^{2}+a}}{24 b}-\frac {B \,a^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}\)	\(65\)
default	\(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}\)	\(76\)

[In]

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

[Out]

1/24*(6*B*b*x^3+8*A*b*x^2+3*B*a*x+8*A*a)/b*(b*x^2+a)^(1/2)-1/8*B*a^2/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.96 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\left [\frac {3 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 3 \, B a b x + 8 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}, \frac {3 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 3 \, B a b x + 8 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b^{2}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.34 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\begin {cases} - \frac {B a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x^{2}} \left (\frac {A a}{3 b} + \frac {A x^{2}}{3} + \frac {B a x}{8 b} + \frac {B x^{3}}{4}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-B*a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**
2), True))/(8*b) + sqrt(a + b*x**2)*(A*a/(3*b) + A*x**2/3 + B*a*x/(8*b) + B*x**3/4), Ne(b, 0)), (sqrt(a)*(A*x*
*2/2 + B*x**3/3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\frac {B a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, B x + 4 \, A\right )} x + \frac {3 \, B a}{b}\right )} x + \frac {8 \, A a}{b}\right )} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x (A+B x) \sqrt {a+b x^2} \, dx=\int x\,\sqrt {b\,x^2+a}\,\left (A+B\,x\right ) \,d x \]

[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)