3.297 \(\int \frac {x^3}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\)

Optimal. Leaf size=169 \[ -\frac {(b c-a d) (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{5/2} d^{3/2} \sqrt {e}}-\frac {\left (c+d x^2\right ) (3 a d+b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^2 d e}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d e} \]

[Out]

-1/8*(-a*d+b*c)*(3*a*d+b*c)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b^(5/2)/d^(3/2)/e^(
1/2)-1/8*(3*a*d+b*c)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^2/d/e+1/4*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(
1/2)/b/d/e

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Rubi [A]  time = 0.13, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1960, 385, 199, 208} \[ -\frac {(b c-a d) (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{5/2} d^{3/2} \sqrt {e}}-\frac {\left (c+d x^2\right ) (3 a d+b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^2 d e}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c + 3*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(8*b^2*d*e) + (Sqrt[(e*(a + b*x^2))/(c + d*x^2)
]*(c + d*x^2)^2)/(4*b*d*e) - ((b*c - a*d)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(S
qrt[b]*Sqrt[e])])/(8*b^(5/2)*d^(3/2)*Sqrt[e])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {-a e+c x^2}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b d e}-\frac {((b c-a d) (b c+3 a d) e) \operatorname {Subst}\left (\int \frac {1}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{4 b d}\\ &=-\frac {(b c+3 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^2 d e}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b d e}-\frac {((b c-a d) (b c+3 a d)) \operatorname {Subst}\left (\int \frac {1}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b^2 d}\\ &=-\frac {(b c+3 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^2 d e}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b d e}-\frac {(b c-a d) (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{5/2} d^{3/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 172, normalized size = 1.02 \[ \frac {\sqrt {d} \left (a+b x^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \left (b \left (c+2 d x^2\right )-3 a d\right )-\sqrt {a+b x^2} \sqrt {b c-a d} (3 a d+b c) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{8 b^2 d^{3/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*(a + b*x^2)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*(-3*a*d + b*(c + 2*d*x^2)) - Sqrt[b*c - a*d]*(b*c + 3*a
*d)*Sqrt[a + b*x^2]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(8*b^2*d^(3/2)*Sqrt[(e*(a + b*x^2))/(c
+ d*x^2)]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)])

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fricas [A]  time = 0.46, size = 413, normalized size = 2.44 \[ \left [-\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d - 3 \, a b c d^{2} + 3 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, b^{3} d^{2} e}, \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {-b d e} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) + 2 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d - 3 \, a b c d^{2} + 3 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, b^{3} d^{2} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(b*d*e)*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2
*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4*(2*b*d^2*x^4 + b*c^2 + a*c*d + (3*b*c*d + a*d^2)*x^2)*sqrt(b*d*e)*sqrt((b*e*
x^2 + a*e)/(d*x^2 + c))) - 4*(2*b^2*d^3*x^4 + b^2*c^2*d - 3*a*b*c*d^2 + 3*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*e
*x^2 + a*e)/(d*x^2 + c)))/(b^3*d^2*e), 1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(-b*d*e)*arctan(1/2*(2*b*d*
x^2 + b*c + a*d)*sqrt(-b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) + 2*(2*b^2*d^3*x^4 +
b^2*c^2*d - 3*a*b*c*d^2 + 3*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^3*d^2*e)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep*d+c)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,1,0]%%%},[2,0]%%%}+%%%{%%{[-2,0]:[
1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,0,1]%%%},[0,2]%%%} / %%%{%%%{1,[0,2,0]%%%},[2,0]%%%}+%%%{%%{
[%%%{-2,[0,1,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,1,1]%%%},[0,2]%%%} Error: Bad Argume
nt Value

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maple [B]  time = 0.04, size = 342, normalized size = 2.02 \[ \frac {\left (b \,x^{2}+a \right ) \left (3 a^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b d \,x^{2}-6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c \right )}{16 \sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b*x^2+a)*(4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*d*x^2+3*d^2*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b
*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2-2*a*b*c*d*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))-ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*b^2*c^2-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*d+2*(b*d*x^4+a*d
*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c)/((b*x^2+a)/(d*x^2+c)*e)^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/b^2/d/(b*d)
^(1/2)

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maxima [A]  time = 2.23, size = 268, normalized size = 1.59 \[ \frac {1}{16} \, e {\left (\frac {2 \, {\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e\right )}}{b^{4} d e^{3} - \frac {2 \, {\left (b x^{2} + a\right )} b^{3} d^{2} e^{3}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} b^{2} d^{3} e^{3}}{{\left (d x^{2} + c\right )}^{2}}} + \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b^{2} d e}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*e*(2*((b^2*c^2*d + 2*a*b*c*d^2 - 3*a^2*d^3)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2) + (b^3*c^2 - 6*a*b^2*c*d +
5*a^2*b*d^2)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e)/(b^4*d*e^3 - 2*(b*x^2 + a)*b^3*d^2*e^3/(d*x^2 + c) + (b*x^2 +
a)^2*b^2*d^3*e^3/(d*x^2 + c)^2) + (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*log((d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - s
qrt(b*d*e))/(d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) + sqrt(b*d*e)))/(sqrt(b*d*e)*b^2*d*e))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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