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

Optimal. Leaf size=66 \[ -\frac {x (A+B x)}{c \sqrt {a+c x^2}}+\frac {A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}+\frac {2 B \sqrt {a+c x^2}}{c^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {819, 641, 217, 206} \begin {gather*} -\frac {x (A+B x)}{c \sqrt {a+c x^2}}+\frac {A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}+\frac {2 B \sqrt {a+c x^2}}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 67, normalized size = 1.02 \begin {gather*} \frac {A \sqrt {c} \sqrt {a+c x^2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+2 a B+c x (B x-A)}{c^2 \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*B + c*x*(-A + B*x) + A*Sqrt[c]*Sqrt[a + c*x^2]*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(c^2*Sqrt[a + c*x^2]
)

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IntegrateAlgebraic [A]  time = 0.36, size = 61, normalized size = 0.92 \begin {gather*} \frac {2 a B-A c x+B c x^2}{c^2 \sqrt {a+c x^2}}-\frac {A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 164, normalized size = 2.48 \begin {gather*} \left [\frac {{\left (A c x^{2} + A a\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (B c x^{2} - A c x + 2 \, B a\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}}, -\frac {{\left (A c x^{2} + A a\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (B c x^{2} - A c x + 2 \, B a\right )} \sqrt {c x^{2} + a}}{c^{3} x^{2} + a c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((A*c*x^2 + A*a)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(B*c*x^2 - A*c*x + 2*B*a)*sq
rt(c*x^2 + a))/(c^3*x^2 + a*c^2), -((A*c*x^2 + A*a)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (B*c*x^2 - A
*c*x + 2*B*a)*sqrt(c*x^2 + a))/(c^3*x^2 + a*c^2)]

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giac [A]  time = 0.21, size = 58, normalized size = 0.88 \begin {gather*} \frac {{\left (\frac {B x}{c} - \frac {A}{c}\right )} x + \frac {2 \, B a}{c^{2}}}{\sqrt {c x^{2} + a}} - \frac {A \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*x/c - A/c)*x + 2*B*a/c^2)/sqrt(c*x^2 + a) - A*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)

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maple [A]  time = 0.05, size = 72, normalized size = 1.09 \begin {gather*} \frac {B \,x^{2}}{\sqrt {c \,x^{2}+a}\, c}-\frac {A x}{\sqrt {c \,x^{2}+a}\, c}+\frac {A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}+\frac {2 B a}{\sqrt {c \,x^{2}+a}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x^2/c/(c*x^2+a)^(1/2)+2*B*a/c^2/(c*x^2+a)^(1/2)-A*x/c/(c*x^2+a)^(1/2)+A/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2)
)

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maxima [A]  time = 0.53, size = 64, normalized size = 0.97 \begin {gather*} \frac {B x^{2}}{\sqrt {c x^{2} + a} c} - \frac {A x}{\sqrt {c x^{2} + a} c} + \frac {A \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, B a}{\sqrt {c x^{2} + a} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x^2/(sqrt(c*x^2 + a)*c) - A*x/(sqrt(c*x^2 + a)*c) + A*arcsinh(c*x/sqrt(a*c))/c^(3/2) + 2*B*a/(sqrt(c*x^2 + a
)*c^2)

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mupad [B]  time = 1.47, size = 61, normalized size = 0.92 \begin {gather*} \frac {A\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{c^{3/2}}-\frac {A\,x}{c\,\sqrt {c\,x^2+a}}+\frac {B\,\left (c\,x^2+2\,a\right )}{c^2\,\sqrt {c\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*log(c^(1/2)*x + (a + c*x^2)^(1/2)))/c^(3/2) - (A*x)/(c*(a + c*x^2)^(1/2)) + (B*(2*a + c*x^2))/(c^2*(a + c*x
^2)^(1/2))

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sympy [A]  time = 8.87, size = 83, normalized size = 1.26 \begin {gather*} A \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{c^{\frac {3}{2}}} - \frac {x}{\sqrt {a} c \sqrt {1 + \frac {c x^{2}}{a}}}\right ) + B \left (\begin {cases} \frac {2 a}{c^{2} \sqrt {a + c x^{2}}} + \frac {x^{2}}{c \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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