Integrand size = 29, antiderivative size = 69 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {x^2}{2 b c}-\frac {d \sqrt {a+b x^2}}{b^2 c^2}-\frac {\left (a c^2-d^2\right ) \log \left (d+c \sqrt {a+b x^2}\right )}{b^2 c^3} \] Output:
1/2*x^2/b/c-d*(b*x^2+a)^(1/2)/b^2/c^2-(a*c^2-d^2)*ln(d+c*(b*x^2+a)^(1/2))/ b^2/c^3
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {c \left (a c+b c x^2-2 d \sqrt {a+b x^2}\right )+\left (-2 a c^2+2 d^2\right ) \log \left (d+c \sqrt {a+b x^2}\right )}{2 b^2 c^3} \] Input:
Integrate[x^3/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]
Output:
(c*(a*c + b*c*x^2 - 2*d*Sqrt[a + b*x^2]) + (-2*a*c^2 + 2*d^2)*Log[d + c*Sq rt[a + b*x^2]])/(2*b^2*c^3)
Time = 0.68 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2586, 7267, 25, 476, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3}{d \sqrt {a+b x^2}+a c+b c x^2} \, dx\) |
\(\Big \downarrow \) 2586 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{b c x^2+a c+d \sqrt {b x^2+a}}dx^2\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {\int -\frac {a-x^4}{\sqrt {b x^2+a} c+d}d\sqrt {b x^2+a}}{b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {a-x^4}{\sqrt {b x^2+a} c+d}d\sqrt {b x^2+a}}{b^2}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle -\frac {\int \left (\frac {d}{c^2}-\frac {\sqrt {b x^2+a}}{c}+\frac {a c^2-d^2}{c^2 \left (\sqrt {b x^2+a} c+d\right )}\right )d\sqrt {b x^2+a}}{b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {d \sqrt {a+b x^2}}{c^2}-\frac {\left (a c^2-d^2\right ) \log \left (c \sqrt {a+b x^2}+d\right )}{c^3}+\frac {x^4}{2 c}}{b^2}\) |
Input:
Int[x^3/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]
Output:
(x^4/(2*c) - (d*Sqrt[a + b*x^2])/c^2 - ((a*c^2 - d^2)*Log[d + c*Sqrt[a + b *x^2]])/c^3)/b^2
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[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]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(1698\) vs. \(2(63)=126\).
Time = 0.03 (sec) , antiderivative size = 1699, normalized size of antiderivative = 24.62
Input:
int(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x,method=_RETURNVERBOSE)
Output:
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*(((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2) *(x-(-a*b)^(1/2)/b))^(1/2)+(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*b+(-a*b)^(1 /2))/b^(1/2)+((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^(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*(((x+(-a*b)^(1/2)/b)^2*b-2*(- a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2)-(-a*b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*b -(-a*b)^(1/2))/b^(1/2)+((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1 /2)/b))^(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*(((x-(-(a*c^2 -d^2)*b*c^2)^(1/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2 -d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2)+1/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)* ln((1/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)+(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)* b)/b^(1/2)+((x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d^2)*b *c^2)^(1/2)*(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/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-( -(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+2*(d^2/c^2)^(1/2)*((x-(-(a*c^2-d^2)*b*c^2 )^(1/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(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...
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (63) = 126\).
Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.33 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {2 \, b c^{2} x^{2} - 4 \, \sqrt {b x^{2} + a} c d - 2 \, {\left (a c^{2} - d^{2}\right )} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - {\left (a c^{2} - d^{2}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + {\left (a c^{2} - d^{2}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right )}{4 \, b^{2} c^{3}} \] Input:
integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="fricas")
Output:
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)
Time = 1.92 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\begin {cases} \frac {2 \left (\frac {a + b x^{2}}{4 c} - \frac {d \sqrt {a + b x^{2}}}{2 c^{2}} - \frac {\left (a c^{2} - d^{2}\right ) \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{2}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{2 c^{2}}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{2 \cdot \left (2 \sqrt {a} d + 2 a c\right )} & \text {otherwise} \end {cases} \] Input:
integrate(x**3/(a*c+b*c*x**2+d*(b*x**2+a)**(1/2)),x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {\frac {{\left (b x^{2} + a\right )} c - 2 \, \sqrt {b x^{2} + a} d}{c^{2}} - \frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left (\sqrt {b x^{2} + a} c + d\right )}{c^{3}}}{2 \, b^{2}} \] Input:
integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="maxima")
Output:
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
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=-\frac {\frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{b c^{3}} - \frac {{\left (b x^{2} + a\right )} b c - 2 \, \sqrt {b x^{2} + a} b d}{b^{2} c^{2}}}{2 \, b} \] Input:
integrate(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x, algorithm="giac")
Output:
-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
Time = 23.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.78 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {x^2}{2\,b\,c}-\frac {d\,\sqrt {b\,x^2+a}}{b^2\,c^2}+\frac {\mathrm {atanh}\left (\frac {c\,\left (a\,c^2-d^2\right )\,\sqrt {b\,x^2+a}}{d^3-a\,c^2\,d}\right )\,\left (a\,c^2-d^2\right )}{b^2\,c^3}-\frac {\ln \left (b\,c^2\,x^2+a\,c^2-d^2\right )\,\left (a\,c^2-d^2\right )}{2\,b^2\,c^3} \] Input:
int(x^3/(a*c + d*(a + b*x^2)^(1/2) + b*c*x^2),x)
Output:
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)
Time = 0.16 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.57 \[ \int \frac {x^3}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, c d -2 \,\mathrm {log}\left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, a c x +\sqrt {b \,x^{2}+a}\, a d +\sqrt {b}\, a d x +a^{2} c +a b c \,x^{2}}{\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x}\right ) a \,c^{2}+2 \,\mathrm {log}\left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, a c x +\sqrt {b \,x^{2}+a}\, a d +\sqrt {b}\, a d x +a^{2} c +a b c \,x^{2}}{\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x}\right ) d^{2}+a \,c^{2}+b \,c^{2} x^{2}}{2 b^{2} c^{3}} \] Input:
int(x^3/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x)
Output:
( - 2*sqrt(a + b*x**2)*c*d - 2*log((sqrt(b)*sqrt(a + b*x**2)*a*c*x + sqrt( a + b*x**2)*a*d + sqrt(b)*a*d*x + a**2*c + a*b*c*x**2)/(sqrt(a)*sqrt(a + b *x**2) + sqrt(b)*sqrt(a)*x))*a*c**2 + 2*log((sqrt(b)*sqrt(a + b*x**2)*a*c* x + sqrt(a + b*x**2)*a*d + sqrt(b)*a*d*x + a**2*c + a*b*c*x**2)/(sqrt(a)*s qrt(a + b*x**2) + sqrt(b)*sqrt(a)*x))*d**2 + a*c**2 + b*c**2*x**2)/(2*b**2 *c**3)