3.355 \(\int \frac{x^3 (A+B x)}{\sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=104 \[ \frac{3 a^2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}-\frac{a \sqrt{a+c x^2} (16 A+9 B x)}{24 c^2}+\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c} \]

[Out]

(A*x^2*Sqrt[a + c*x^2])/(3*c) + (B*x^3*Sqrt[a + c*x^2])/(4*c) - (a*(16*A + 9*B*x)*Sqrt[a + c*x^2])/(24*c^2) +
(3*a^2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

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Rubi [A]  time = 0.0687031, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {833, 780, 217, 206} \[ \frac{3 a^2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}-\frac{a \sqrt{a+c x^2} (16 A+9 B x)}{24 c^2}+\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x^2*Sqrt[a + c*x^2])/(3*c) + (B*x^3*Sqrt[a + c*x^2])/(4*c) - (a*(16*A + 9*B*x)*Sqrt[a + c*x^2])/(24*c^2) +
(3*a^2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

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 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)}{\sqrt{a+c x^2}} \, dx &=\frac{B x^3 \sqrt{a+c x^2}}{4 c}+\frac{\int \frac{x^2 (-3 a B+4 A c x)}{\sqrt{a+c x^2}} \, dx}{4 c}\\ &=\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c}+\frac{\int \frac{x (-8 a A c-9 a B c x)}{\sqrt{a+c x^2}} \, dx}{12 c^2}\\ &=\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c}-\frac{a (16 A+9 B x) \sqrt{a+c x^2}}{24 c^2}+\frac{\left (3 a^2 B\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 c^2}\\ &=\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c}-\frac{a (16 A+9 B x) \sqrt{a+c x^2}}{24 c^2}+\frac{\left (3 a^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 c^2}\\ &=\frac{A x^2 \sqrt{a+c x^2}}{3 c}+\frac{B x^3 \sqrt{a+c x^2}}{4 c}-\frac{a (16 A+9 B x) \sqrt{a+c x^2}}{24 c^2}+\frac{3 a^2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0446965, size = 76, normalized size = 0.73 \[ \frac{9 a^2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+\sqrt{c} \sqrt{a+c x^2} \left (-16 a A-9 a B x+8 A c x^2+6 B c x^3\right )}{24 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + c*x^2]*(-16*a*A - 9*a*B*x + 8*A*c*x^2 + 6*B*c*x^3) + 9*a^2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*
x^2]])/(24*c^(5/2))

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Maple [A]  time = 0.007, size = 96, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}B}{4\,c}\sqrt{c{x}^{2}+a}}-{\frac{3\,aBx}{8\,{c}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{3\,B{a}^{2}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{A{x}^{2}}{3\,c}\sqrt{c{x}^{2}+a}}-{\frac{2\,aA}{3\,{c}^{2}}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68334, size = 385, normalized size = 3.7 \begin{align*} \left [\frac{9 \, B a^{2} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (6 \, B c^{2} x^{3} + 8 \, A c^{2} x^{2} - 9 \, B a c x - 16 \, A a c\right )} \sqrt{c x^{2} + a}}{48 \, c^{3}}, -\frac{9 \, B a^{2} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (6 \, B c^{2} x^{3} + 8 \, A c^{2} x^{2} - 9 \, B a c x - 16 \, A a c\right )} \sqrt{c x^{2} + a}}{24 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(9*B*a^2*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*B*c^2*x^3 + 8*A*c^2*x^2 - 9*B*a*
c*x - 16*A*a*c)*sqrt(c*x^2 + a))/c^3, -1/24*(9*B*a^2*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*B*c^2*x^
3 + 8*A*c^2*x^2 - 9*B*a*c*x - 16*A*a*c)*sqrt(c*x^2 + a))/c^3]

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Sympy [A]  time = 6.14437, size = 150, normalized size = 1.44 \begin{align*} A \left (\begin{cases} - \frac{2 a \sqrt{a + c x^{2}}}{3 c^{2}} + \frac{x^{2} \sqrt{a + c x^{2}}}{3 c} & \text{for}\: c \neq 0 \\\frac{x^{4}}{4 \sqrt{a}} & \text{otherwise} \end{cases}\right ) - \frac{3 B a^{\frac{3}{2}} x}{8 c^{2} \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{B \sqrt{a} x^{3}}{8 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 B a^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{8 c^{\frac{5}{2}}} + \frac{B x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.16965, size = 100, normalized size = 0.96 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left (\frac{3 \, B x}{c} + \frac{4 \, A}{c}\right )} x - \frac{9 \, B a}{c^{2}}\right )} x - \frac{16 \, A a}{c^{2}}\right )} - \frac{3 \, B a^{2} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{8 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + a)*((2*(3*B*x/c + 4*A/c)*x - 9*B*a/c^2)*x - 16*A*a/c^2) - 3/8*B*a^2*log(abs(-sqrt(c)*x + sqr
t(c*x^2 + a)))/c^(5/2)