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

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

[Out]

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

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Rubi [A]  time = 0.0468163, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {833, 780, 195, 217, 206} \[ -\frac{a^2 A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2}}-\frac{\left (a+c x^2\right )^{3/2} (8 a B-15 A c x)}{60 c^2}-\frac{a A x \sqrt{a+c x^2}}{8 c}+\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x)*Sqrt[a + c*x^2],x]

[Out]

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

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])

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 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 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 x^2 (A+B x) \sqrt{a+c x^2} \, dx &=\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{\int x (-2 a B+5 A c x) \sqrt{a+c x^2} \, dx}{5 c}\\ &=\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c}-\frac{(8 a B-15 A c x) \left (a+c x^2\right )^{3/2}}{60 c^2}-\frac{(a A) \int \sqrt{a+c x^2} \, dx}{4 c}\\ &=-\frac{a A x \sqrt{a+c x^2}}{8 c}+\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c}-\frac{(8 a B-15 A c x) \left (a+c x^2\right )^{3/2}}{60 c^2}-\frac{\left (a^2 A\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 c}\\ &=-\frac{a A x \sqrt{a+c x^2}}{8 c}+\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c}-\frac{(8 a B-15 A c x) \left (a+c x^2\right )^{3/2}}{60 c^2}-\frac{\left (a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 c}\\ &=-\frac{a A x \sqrt{a+c x^2}}{8 c}+\frac{B x^2 \left (a+c x^2\right )^{3/2}}{5 c}-\frac{(8 a B-15 A c x) \left (a+c x^2\right )^{3/2}}{60 c^2}-\frac{a^2 A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.17218, size = 93, normalized size = 0.89 \[ \frac{\sqrt{a+c x^2} \left (-\frac{15 a^{3/2} A \sqrt{c} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}-16 a^2 B+a c x (15 A+8 B x)+6 c^2 x^3 (5 A+4 B x)\right )}{120 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 94, normalized size = 0.9 \begin{align*}{\frac{B{x}^{2}}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{2\,aB}{15\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ax}{4\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{aAx}{8\,c}\sqrt{c{x}^{2}+a}}-{\frac{A{a}^{2}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*x^2*(c*x^2+a)^(3/2)/c-2/15*B*a/c^2*(c*x^2+a)^(3/2)+1/4*A*x*(c*x^2+a)^(3/2)/c-1/8*a*A*x*(c*x^2+a)^(1/2)/c
-1/8*A*a^2/c^(3/2)*ln(x*c^(1/2)+(c*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^2*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72529, size = 435, normalized size = 4.18 \begin{align*} \left [\frac{15 \, A a^{2} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (24 \, B c^{2} x^{4} + 30 \, A c^{2} x^{3} + 8 \, B a c x^{2} + 15 \, A a c x - 16 \, B a^{2}\right )} \sqrt{c x^{2} + a}}{240 \, c^{2}}, \frac{15 \, A a^{2} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) +{\left (24 \, B c^{2} x^{4} + 30 \, A c^{2} x^{3} + 8 \, B a c x^{2} + 15 \, A a c x - 16 \, B a^{2}\right )} \sqrt{c x^{2} + a}}{120 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.39678, size = 165, normalized size = 1.59 \begin{align*} \frac{A a^{\frac{3}{2}} x}{8 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 A \sqrt{a} x^{3}}{8 \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{A a^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{8 c^{\frac{3}{2}}} + \frac{A c x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} + B \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + c x^{2}}}{15 c^{2}} + \frac{a x^{2} \sqrt{a + c x^{2}}}{15 c} + \frac{x^{4} \sqrt{a + c x^{2}}}{5} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13145, size = 109, normalized size = 1.05 \begin{align*} \frac{A a^{2} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{8 \, c^{\frac{3}{2}}} + \frac{1}{120} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left (3 \,{\left (4 \, B x + 5 \, A\right )} x + \frac{4 \, B a}{c}\right )} x + \frac{15 \, A a}{c}\right )} x - \frac{16 \, B a^{2}}{c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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