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

Optimal. Leaf size=104 \[ -\frac {a^2 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}-\frac {\left (a+b x^2\right )^{3/2} (8 a B-15 A b x)}{60 b^2}-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {833, 780, 195, 217, 206} \begin {gather*} -\frac {a^2 A \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} (8 a B-15 A b x)}{60 b^2}-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

-(a*A*x*Sqrt[a + b*x^2])/(8*b) + (B*x^2*(a + b*x^2)^(3/2))/(5*b) - ((8*a*B - 15*A*b*x)*(a + b*x^2)^(3/2))/(60*
b^2) - (a^2*A*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]

Rule 833

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

Rubi steps

\begin {align*} \int x^2 (A+B x) \sqrt {a+b x^2} \, dx &=\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {\int x (-2 a B+5 A b x) \sqrt {a+b x^2} \, dx}{5 b}\\ &=\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {(a A) \int \sqrt {a+b x^2} \, dx}{4 b}\\ &=-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {\left (a^2 A\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b}\\ &=-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {\left (a^2 A\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 A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {a^2 A \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 = 0.19, size = 93, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-\frac {15 a^{3/2} A \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}-16 a^2 B+a b x (15 A+8 B x)+6 b^2 x^3 (5 A+4 B x)\right )}{120 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-16*a^2*B + 6*b^2*x^3*(5*A + 4*B*x) + a*b*x*(15*A + 8*B*x) - (15*a^(3/2)*A*Sqrt[b]*ArcSinh[(
Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(120*b^2)

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IntegrateAlgebraic [A]  time = 0.23, size = 92, normalized size = 0.88 \begin {gather*} \frac {a^2 A \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 b^{3/2}}+\frac {\sqrt {a+b x^2} \left (-16 a^2 B+15 a A b x+8 a b B x^2+30 A b^2 x^3+24 b^2 B x^4\right )}{120 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-16*a^2*B + 15*a*A*b*x + 8*a*b*B*x^2 + 30*A*b^2*x^3 + 24*b^2*B*x^4))/(120*b^2) + (a^2*A*Log[
-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(3/2))

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fricas [A]  time = 1.00, size = 175, normalized size = 1.68 \begin {gather*} \left [\frac {15 \, A a^{2} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, B b^{2} x^{4} + 30 \, A b^{2} x^{3} + 8 \, B a b x^{2} + 15 \, A a b x - 16 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}, \frac {15 \, A a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, B b^{2} x^{4} + 30 \, A b^{2} x^{3} + 8 \, B a b x^{2} + 15 \, A a b x - 16 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(15*A*a^2*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(24*B*b^2*x^4 + 30*A*b^2*x^3 + 8*
B*a*b*x^2 + 15*A*a*b*x - 16*B*a^2)*sqrt(b*x^2 + a))/b^2, 1/120*(15*A*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2
+ a)) + (24*B*b^2*x^4 + 30*A*b^2*x^3 + 8*B*a*b*x^2 + 15*A*a*b*x - 16*B*a^2)*sqrt(b*x^2 + a))/b^2]

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giac [A]  time = 0.47, size = 81, normalized size = 0.78 \begin {gather*} \frac {A a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} + \frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, B x + 5 \, A\right )} x + \frac {4 \, B a}{b}\right )} x + \frac {15 \, A a}{b}\right )} x - \frac {16 \, B a^{2}}{b^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*A*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/120*sqrt(b*x^2 + a)*((2*(3*(4*B*x + 5*A)*x + 4*B*
a/b)*x + 15*A*a/b)*x - 16*B*a^2/b^2)

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maple [A]  time = 0.01, size = 94, normalized size = 0.90 \begin {gather*} -\frac {A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, A a x}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,x^{2}}{5 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A x}{4 b}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B a}{15 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*x^2*(b*x^2+a)^(3/2)/b-2/15*B*a/b^2*(b*x^2+a)^(3/2)+1/4*A*x*(b*x^2+a)^(3/2)/b-1/8*a*A*x*(b*x^2+a)^(1/2)/b
-1/8*A*a^2/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.36, size = 86, normalized size = 0.83 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x}{4 \, b} - \frac {\sqrt {b x^{2} + a} A a x}{8 \, b} - \frac {A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a}{15 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b*x^2 + a)^(3/2)*B*x^2/b + 1/4*(b*x^2 + a)^(3/2)*A*x/b - 1/8*sqrt(b*x^2 + a)*A*a*x/b - 1/8*A*a^2*arcsinh(
b*x/sqrt(a*b))/b^(3/2) - 2/15*(b*x^2 + a)^(3/2)*B*a/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 10.14, size = 165, normalized size = 1.59 \begin {gather*} \frac {A a^{\frac {3}{2}} x}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A \sqrt {a} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {A b x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*x/(8*b*sqrt(1 + b*x**2/a)) + 3*A*sqrt(a)*x**3/(8*sqrt(1 + b*x**2/a)) - A*a**2*asinh(sqrt(b)*x/sqrt(
a))/(8*b**(3/2)) + A*b*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + B*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) +
a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True))

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