\(\int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx\) [545]

Optimal result

Integrand size = 29, antiderivative size = 69 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {x^2}{2 b c}-\frac {d \sqrt {a+b x^2}}{b^2 c^2}-\frac {\left (a c^2-d^2\right ) \log \left (d+c \sqrt {a+b x^2}\right )}{b^2 c^3} \]

[Out]

1/2*x^2/b/c-(a*c^2-d^2)*ln(d+c*(b*x^2+a)^(1/2))/b^2/c^3-d*(b*x^2+a)^(1/2)/b^2/c^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2186, 711} \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=-\frac {d \sqrt {a+b x^2}}{b^2 c^2}-\frac {\left (a c^2-d^2\right ) \log \left (c \sqrt {a+b x^2}+d\right )}{b^2 c^3}+\frac {x^2}{2 b c} \]

[In]

Int[x^3/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]

[Out]

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

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2186

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/n, Subst[Int
[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c
- a*d, 0] && IntegerQ[(m + 1)/n]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{a c+b c x+d \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {-a+x^2}{d+c x} \, dx,x,\sqrt {a+b x^2}\right )}{b^2} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {d}{c^2}+\frac {x}{c}+\frac {-a c^2+d^2}{c^2 (d+c x)}\right ) \, dx,x,\sqrt {a+b x^2}\right )}{b^2} \\ & = \frac {x^2}{2 b c}-\frac {d \sqrt {a+b x^2}}{b^2 c^2}-\frac {\left (a c^2-d^2\right ) \log \left (d+c \sqrt {a+b x^2}\right )}{b^2 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {c \left (a c+b c x^2-2 d \sqrt {a+b x^2}\right )+\left (-2 a c^2+2 d^2\right ) \log \left (d+c \sqrt {a+b x^2}\right )}{2 b^2 c^3} \]

[In]

Integrate[x^3/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1698\) vs. \(2(63)=126\).

Time = 0.08 (sec) , antiderivative size = 1699, normalized size of antiderivative = 24.62

method	result	size
default	\(\text {Expression too large to display}\)	\(1699\)

[In]

int(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

d*(-1/2*c^2*a/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/(-(-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/b*(
(b*(x-1/b*(-a*b)^(1/2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2)+(-a*b)^(1/2)*ln(((x-1/b*(-a*b)^(1/2))*b+(
-a*b)^(1/2))/b^(1/2)+(b*(x-1/b*(-a*b)^(1/2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))-1/2*c^2*a/
((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/(-(-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/b*((b*(x+1/b*(-a*
b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2)-(-a*b)^(1/2)*ln(((x+1/b*(-a*b)^(1/2))*b-(-a*b)^(1/2))/b
^(1/2)+(b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))+1/2*(a*c^2-d^2)/((-a*b)^
(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/(-(-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/b*((b*(x-(-(a*c^2-d^2)*b*
c^2)^(1/2)/b/c^2)^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2)+(-(a*
c^2-d^2)*b*c^2)^(1/2)/c^2*ln(((-(a*c^2-d^2)*b*c^2)^(1/2)/c^2+(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/b^(1/2)+(
b*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)
+d^2/c^2)^(1/2))/b^(1/2)-d^2/c^2/(d^2/c^2)^(1/2)*ln((2*d^2/c^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c^2-d^
2)*b*c^2)^(1/2)/b/c^2)+2*(d^2/c^2)^(1/2)*(b*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2
)/c^2*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2))/(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)))+1/2*(a*c^2-d
^2)/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/(-(-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/b*((b*(x+(-(a
*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)
^(1/2)-(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*ln((-(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2+(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)
*b)/b^(1/2)+(b*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)
^(1/2)/b/c^2)+d^2/c^2)^(1/2))/b^(1/2)-d^2/c^2/(d^2/c^2)^(1/2)*ln((2*d^2/c^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(
x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+2*(d^2/c^2)^(1/2)*(b*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2
)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2))/(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)))
)-a*c*(1/2*(a*c^2-d^2)/b^2/d^2/c^2*ln(b*c^2*x^2+a*c^2-d^2)-1/2*a/b^2/d^2*ln(b*x^2+a))-b*c*(-1/2/b^2/c^2*x^2+1/
2*(-a^2*c^4+2*a*c^2*d^2-d^4)/d^2/c^4/b^3*ln(b*c^2*x^2+a*c^2-d^2)+1/2/b^3*a^2/d^2*ln(b*x^2+a))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (63) = 126\).

Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.33 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {2 \, b c^{2} x^{2} - 4 \, \sqrt {b x^{2} + a} c d - 2 \, {\left (a c^{2} - d^{2}\right )} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - {\left (a c^{2} - d^{2}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + {\left (a c^{2} - d^{2}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right )}{4 \, b^{2} c^{3}} \]

[In]

integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="fricas")

[Out]

1/4*(2*b*c^2*x^2 - 4*sqrt(b*x^2 + a)*c*d - 2*(a*c^2 - d^2)*log(b*c^2*x^2 + a*c^2 - d^2) - (a*c^2 - d^2)*log(-(
b*c^2*x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) + (a*c^2 - d^2)*log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 +
a)*c*d + d^2)/x^2))/(b^2*c^3)

Sympy [A] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\begin {cases} \frac {2 \left (\frac {a + b x^{2}}{4 c} - \frac {d \sqrt {a + b x^{2}}}{2 c^{2}} - \frac {\left (a c^{2} - d^{2}\right ) \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{2}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{2 c^{2}}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{2 \cdot \left (2 \sqrt {a} d + 2 a c\right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3/(a*c+b*c*x**2+d*(b*x**2+a)**(1/2)),x)

[Out]

Piecewise((2*((a + b*x**2)/(4*c) - d*sqrt(a + b*x**2)/(2*c**2) - (a*c**2 - d**2)*Piecewise((sqrt(a + b*x**2)/d
, Eq(c, 0)), (log(c*sqrt(a + b*x**2) + d)/c, True))/(2*c**2))/b**2, Ne(b, 0)), (x**4/(2*(2*sqrt(a)*d + 2*a*c))
, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {\frac {{\left (b x^{2} + a\right )} c - 2 \, \sqrt {b x^{2} + a} d}{c^{2}} - \frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left (\sqrt {b x^{2} + a} c + d\right )}{c^{3}}}{2 \, b^{2}} \]

[In]

integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="maxima")

[Out]

1/2*(((b*x^2 + a)*c - 2*sqrt(b*x^2 + a)*d)/c^2 - 2*(a*c^2 - d^2)*log(sqrt(b*x^2 + a)*c + d)/c^3)/b^2

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=-\frac {\frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{b c^{3}} - \frac {{\left (b x^{2} + a\right )} b c - 2 \, \sqrt {b x^{2} + a} b d}{b^{2} c^{2}}}{2 \, b} \]

[In]

integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="giac")

[Out]

-1/2*(2*(a*c^2 - d^2)*log(abs(sqrt(b*x^2 + a)*c + d))/(b*c^3) - ((b*x^2 + a)*b*c - 2*sqrt(b*x^2 + a)*b*d)/(b^2
*c^2))/b

Mupad [B] (verification not implemented)

Time = 17.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.78 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {x^2}{2\,b\,c}-\frac {d\,\sqrt {b\,x^2+a}}{b^2\,c^2}+\frac {\mathrm {atanh}\left (\frac {c\,\left (a\,c^2-d^2\right )\,\sqrt {b\,x^2+a}}{d^3-a\,c^2\,d}\right )\,\left (a\,c^2-d^2\right )}{b^2\,c^3}-\frac {\ln \left (b\,c^2\,x^2+a\,c^2-d^2\right )\,\left (a\,c^2-d^2\right )}{2\,b^2\,c^3} \]

[In]

int(x^3/(a*c + d*(a + b*x^2)^(1/2) + b*c*x^2),x)

[Out]

x^2/(2*b*c) - (d*(a + b*x^2)^(1/2))/(b^2*c^2) + (atanh((c*(a*c^2 - d^2)*(a + b*x^2)^(1/2))/(d^3 - a*c^2*d))*(a
*c^2 - d^2))/(b^2*c^3) - (log(a*c^2 - d^2 + b*c^2*x^2)*(a*c^2 - d^2))/(2*b^2*c^3)