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

Optimal. Leaf size=48 \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {A+B x}{c \sqrt {a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {778, 217, 206} \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{3/2}}-\frac {A+B x}{c \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + B*x)/(c*Sqrt[a + c*x^2])) + (B*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 778

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

Rubi steps

\begin {align*} \int \frac {x (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac {A+B x}{c \sqrt {a+c x^2}}+\frac {B \int \frac {1}{\sqrt {a+c x^2}} \, dx}{c}\\ &=-\frac {A+B x}{c \sqrt {a+c x^2}}+\frac {B \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{c}\\ &=-\frac {A+B x}{c \sqrt {a+c x^2}}+\frac {B \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 = 0.05, size = 64, normalized size = 1.33 \[ \frac {\sqrt {a} B \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )-\sqrt {c} (A+B x)}{c^{3/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[c]*(A + B*x)) + Sqrt[a]*B*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(c^(3/2)*Sqrt[a + c*x^2])

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fricas [A]  time = 0.75, size = 147, normalized size = 3.06 \[ \left [\frac {{\left (B c x^{2} + B a\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (B c x + A c\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{3} x^{2} + a c^{2}\right )}}, -\frac {{\left (B c x^{2} + B a\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (B c x + A c\right )} \sqrt {c x^{2} + a}}{c^{3} x^{2} + a c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 48, normalized size = 1.00 \[ -\frac {\frac {B x}{c} + \frac {A}{c}}{\sqrt {c x^{2} + a}} - \frac {B \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 54, normalized size = 1.12 \[ -\frac {B x}{\sqrt {c \,x^{2}+a}\, c}+\frac {B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}-\frac {A}{\sqrt {c \,x^{2}+a}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 46, normalized size = 0.96 \[ -\frac {B x}{\sqrt {c x^{2} + a} c} + \frac {B \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {A}{\sqrt {c x^{2} + a} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.22, size = 53, normalized size = 1.10 \[ \frac {B\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{c^{3/2}}-\frac {A}{c\,\sqrt {c\,x^2+a}}-\frac {B\,x}{c\,\sqrt {c\,x^2+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.03, size = 66, normalized size = 1.38 \[ A \left (\begin {cases} - \frac {1}{c \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{2}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{c^{\frac {3}{2}}} - \frac {x}{\sqrt {a} c \sqrt {1 + \frac {c x^{2}}{a}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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