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

Optimal. Leaf size=81 \[ \frac {\sqrt {a+c x^2} (4 A+3 B x)}{2 c^2}-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}-\frac {3 a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 81, 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, 780, 217, 206} \begin {gather*} \frac {\sqrt {a+c x^2} (4 A+3 B x)}{2 c^2}-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}-\frac {3 a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^2*(A + B*x))/(c*Sqrt[a + c*x^2])) + ((4*A + 3*B*x)*Sqrt[a + c*x^2])/(2*c^2) - (3*a*B*ArcTanh[(Sqrt[c]*x)/
Sqrt[a + c*x^2]])/(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 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 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 )^{3/2}} \, dx &=-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}+\frac {\int \frac {x (2 a A+3 a B x)}{\sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}+\frac {(4 A+3 B x) \sqrt {a+c x^2}}{2 c^2}-\frac {(3 a B) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 c^2}\\ &=-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}+\frac {(4 A+3 B x) \sqrt {a+c x^2}}{2 c^2}-\frac {(3 a B) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 c^2}\\ &=-\frac {x^2 (A+B x)}{c \sqrt {a+c x^2}}+\frac {(4 A+3 B x) \sqrt {a+c x^2}}{2 c^2}-\frac {3 a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 72, normalized size = 0.89 \begin {gather*} \frac {a (4 A+3 B x)+c x^2 (2 A+B x)}{2 c^2 \sqrt {a+c x^2}}-\frac {3 a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.39, size = 74, normalized size = 0.91 \begin {gather*} \frac {4 a A+3 a B x+2 A c x^2+B c x^3}{2 c^2 \sqrt {a+c x^2}}+\frac {3 a B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 197, normalized size = 2.43 \begin {gather*} \left [\frac {3 \, {\left (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 (B c^{2} x^{3} + 2 \, A c^{2} x^{2} + 3 \, B a c x + 4 \, A a c\right )} \sqrt {c x^{2} + a}}{4 \, {\left (c^{4} x^{2} + a c^{3}\right )}}, \frac {3 \, {\left (B a c x^{2} + B a^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (B c^{2} x^{3} + 2 \, A c^{2} x^{2} + 3 \, B a c x + 4 \, A a c\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{4} x^{2} + a 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)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.21, size = 70, normalized size = 0.86 \begin {gather*} \frac {{\left ({\left (\frac {B x}{c} + \frac {2 \, A}{c}\right )} x + \frac {3 \, B a}{c^{2}}\right )} x + \frac {4 \, A a}{c^{2}}}{2 \, \sqrt {c x^{2} + a}} + \frac {3 \, B a \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, 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)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.10, size = 93, normalized size = 1.15 \begin {gather*} \frac {B \,x^{3}}{2 \sqrt {c \,x^{2}+a}\, c}+\frac {A \,x^{2}}{\sqrt {c \,x^{2}+a}\, c}+\frac {3 B a x}{2 \sqrt {c \,x^{2}+a}\, c^{2}}-\frac {3 B a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {5}{2}}}+\frac {2 A a}{\sqrt {c \,x^{2}+a}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.67, size = 85, normalized size = 1.05 \begin {gather*} \frac {B x^{3}}{2 \, \sqrt {c x^{2} + a} c} + \frac {A x^{2}}{\sqrt {c x^{2} + a} c} + \frac {3 \, B a x}{2 \, \sqrt {c x^{2} + a} c^{2}} - \frac {3 \, B a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} + \frac {2 \, A a}{\sqrt {c x^{2} + a} 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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 10.93, size = 117, normalized size = 1.44 \begin {gather*} A \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 ) + B \left (\frac {3 \sqrt {a} x}{2 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} c \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)**(3/2),x)

[Out]

A*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)) +
B*(3*sqrt(a)*x/(2*c**2*sqrt(1 + c*x**2/a)) - 3*a*asinh(sqrt(c)*x/sqrt(a))/(2*c**(5/2)) + x**3/(2*sqrt(a)*c*sqr
t(1 + c*x**2/a)))

________________________________________________________________________________________