3.1.36 \(\int \frac {x^3 (A+B x)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac {2 A+3 B x}{3 b^2 \sqrt {a+b x^2}}-\frac {x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}+\frac {B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 79, 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, 778, 217, 206} \begin {gather*} -\frac {2 A+3 B x}{3 b^2 \sqrt {a+b x^2}}-\frac {x^2 (A+B x)}{3 b \left (a+b x^2\right )^{3/2}}+\frac {B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]

[Out]

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

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]

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

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Mathematica [A]  time = 0.12, size = 69, normalized size = 0.87 \begin {gather*} \frac {B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}}-\frac {a (2 A+3 B x)+b x^2 (3 A+4 B x)}{3 b^2 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]

[Out]

-1/3*(a*(2*A + 3*B*x) + b*x^2*(3*A + 4*B*x))/(b^2*(a + b*x^2)^(3/2)) + (B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]
)/b^(5/2)

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IntegrateAlgebraic [A]  time = 0.51, size = 72, normalized size = 0.91 \begin {gather*} \frac {-2 a A-3 a B x-3 A b x^2-4 b B x^3}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]

[Out]

(-2*a*A - 3*a*B*x - 3*A*b*x^2 - 4*b*B*x^3)/(3*b^2*(a + b*x^2)^(3/2)) - (B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])
/b^(5/2)

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fricas [A]  time = 0.82, size = 239, normalized size = 3.03 \begin {gather*} \left [\frac {3 \, {\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(B*b^2*x^4 + 2*B*a*b*x^2 + B*a^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(4*B*b^2
*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x + 2*A*a*b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), -1/3*(3*(B*b^2*x
^4 + 2*B*a*b*x^2 + B*a^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (4*B*b^2*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x
+ 2*A*a*b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]

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giac [A]  time = 0.48, size = 70, normalized size = 0.89 \begin {gather*} -\frac {{\left ({\left (\frac {4 \, B x}{b} + \frac {3 \, A}{b}\right )} x + \frac {3 \, B a}{b^{2}}\right )} x + \frac {2 \, A a}{b^{2}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {B \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 91, normalized size = 1.15 \begin {gather*} -\frac {B \,x^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b}-\frac {A \,x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} b}-\frac {2 A a}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}-\frac {B x}{\sqrt {b \,x^{2}+a}\, b^{2}}+\frac {B \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(b*x^2+a)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.41, size = 102, normalized size = 1.29 \begin {gather*} -\frac {1}{3} \, B x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {A x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {B x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {B \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, A a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (A+B\,x\right )}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x))/(a + b*x^2)^(5/2),x)

[Out]

int((x^3*(A + B*x))/(a + b*x^2)^(5/2), x)

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sympy [A]  time = 18.32, size = 400, normalized size = 5.06 \begin {gather*} A \left (\begin {cases} - \frac {2 a}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {3 b x^{2}}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {3 a^{\frac {39}{2}} b^{11} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} b^{12} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a^{19} b^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {4 a^{18} b^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(b*x**2+a)**(5/2),x)

[Out]

A*Piecewise((-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)) - 3*b*x**2/(3*a*b**2*sqrt(a + b*x
**2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(5/2)), True)) + B*(3*a**(39/2)*b**11*sqrt(1 + b*
x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1
+ b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt
(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2)*s
qrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(2
7/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)))

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