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

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

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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 c^2 \sqrt {a+c x^2}}-\frac {x^2 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (2*A + 3*B*x)/(3*c^2*Sqrt[a + c*x^2]) + (B*ArcTanh[(Sqrt[c]*x)/Sqrt
[a + c*x^2]])/c^(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+c x^2\right )^{5/2}} \, dx &=-\frac {x^2 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {\int \frac {x (2 a A+3 a B x)}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac {x^2 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {2 A+3 B x}{3 c^2 \sqrt {a+c x^2}}+\frac {B \int \frac {1}{\sqrt {a+c x^2}} \, dx}{c^2}\\ &=-\frac {x^2 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {2 A+3 B x}{3 c^2 \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^2}\\ &=-\frac {x^2 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {2 A+3 B x}{3 c^2 \sqrt {a+c x^2}}+\frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 69, normalized size = 0.87 \begin {gather*} \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}-\frac {a (2 A+3 B x)+c x^2 (3 A+4 B x)}{3 c^2 \left (a+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*(2*A + 3*B*x) + c*x^2*(3*A + 4*B*x))/(c^2*(a + c*x^2)^(3/2)) + (B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]]
)/c^(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 c x^2-4 B c x^3}{3 c^2 \left (a+c x^2\right )^{3/2}}-\frac {B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 239, normalized size = 3.03 \begin {gather*} \left [\frac {3 \, {\left (B c^{2} x^{4} + 2 \, B a c x^{2} + B a^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (4 \, B c^{2} x^{3} + 3 \, A c^{2} x^{2} + 3 \, B a c x + 2 \, A a c\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{5} x^{4} + 2 \, a c^{4} x^{2} + a^{2} c^{3}\right )}}, -\frac {3 \, {\left (B c^{2} x^{4} + 2 \, B a c x^{2} + B a^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (4 \, B c^{2} x^{3} + 3 \, A c^{2} x^{2} + 3 \, B a c x + 2 \, A a c\right )} \sqrt {c x^{2} + a}}{3 \, {\left (c^{5} x^{4} + 2 \, a c^{4} x^{2} + a^{2} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 70, normalized size = 0.89 \begin {gather*} -\frac {{\left ({\left (\frac {4 \, B x}{c} + \frac {3 \, A}{c}\right )} x + \frac {3 \, B a}{c^{2}}\right )} x + \frac {2 \, A a}{c^{2}}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}}} - \frac {B \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 91, normalized size = 1.15 \begin {gather*} -\frac {B \,x^{3}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c}-\frac {A \,x^{2}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c}-\frac {2 A a}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{2}}-\frac {B x}{\sqrt {c \,x^{2}+a}\, c^{2}}+\frac {B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 102, normalized size = 1.29 \begin {gather*} -\frac {1}{3} \, B x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {2 \, a}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}}\right )} - \frac {A x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} - \frac {B x}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {B \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {5}{2}}} - \frac {2 \, A a}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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sympy [A]  time = 15.98, size = 400, normalized size = 5.06 \begin {gather*} A \left (\begin {cases} - \frac {2 a}{3 a c^{2} \sqrt {a + c x^{2}} + 3 c^{3} x^{2} \sqrt {a + c x^{2}}} - \frac {3 c x^{2}}{3 a c^{2} \sqrt {a + c x^{2}} + 3 c^{3} x^{2} \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {3 a^{\frac {39}{2}} c^{11} \sqrt {1 + \frac {c x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} c^{12} x^{2} \sqrt {1 + \frac {c x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a^{19} c^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {4 a^{18} c^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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