3.4.45 \(\int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx\) [345]

Optimal. Leaf size=148 \[ -\frac {(3 b+4 a c) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 a^2 d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{4 a d^2}+\frac {b (3 b+4 a c) \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} d^2} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1985, 1981, 1980, 393, 205, 214} \begin {gather*} \frac {b (4 a c+3 b) \tanh ^{-1}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} d^2}-\frac {(4 a c+3 b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{8 a^2 d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + b/(c + d*x^2)],x]

[Out]

-1/8*((3*b + 4*a*c)*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(a^2*d^2) + ((c + d*x^2)^2*Sqrt[(b + a*
c + a*d*x^2)/(c + d*x^2)])/(4*a*d^2) + (b*(3*b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]]
)/(8*a^(5/2)*d^2)

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 214

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

Rule 393

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 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

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

Rule 1985

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u*((b + a*c + a*d*x^n)/(c + d*x^n))^p
, x] /; FreeQ[{a, b, c, d, n, p}, x]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^3 \sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^3 \sqrt {c+d x^2}}{\sqrt {b+a c+a d x^2}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {\sqrt {b+a \left (c+d x^2\right )} \text {Subst}\left (\int \frac {x \sqrt {c+d x}}{\sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {\left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a d^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left ((3 b+4 a c) \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{\sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{8 a d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(3 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (b (3 b+4 a c) \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x} \sqrt {b+a c+a d x}} \, dx,x,x^2\right )}{16 a^2 d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(3 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (b (3 b+4 a c) \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+a x^2}} \, dx,x,\sqrt {c+d x^2}\right )}{8 a^2 d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(3 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (b (3 b+4 a c) \sqrt {b+a \left (c+d x^2\right )}\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}}\right )}{8 a^2 d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(3 b+4 a c) \left (b+a \left (c+d x^2\right )\right )}{8 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (c+d x^2\right ) \left (b+a \left (c+d x^2\right )\right )}{4 a d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {b (3 b+4 a c) \sqrt {b+a \left (c+d x^2\right )} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}}\right )}{8 a^{5/2} d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 108, normalized size = 0.73 \begin {gather*} \frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (-3 b-2 a c+2 a d x^2\right )+b (3 b+4 a c) \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-3*b - 2*a*c + 2*a*d*x^2) + b*(3*b + 4*a*c)*ArcTan
h[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(8*a^(5/2)*d^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(132)=264\).
time = 0.06, size = 354, normalized size = 2.39

method	result	size
risch	\(-\frac {\left (-2 a d \,x^{2}+2 a c +3 b \right ) \left (a d \,x^{2}+a c +b \right )}{8 d^{2} a^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}+\frac {\left (\frac {b \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) c}{4 d a \sqrt {a \,d^{2}}}+\frac {3 b^{2} \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{16 d \,a^{2} \sqrt {a \,d^{2}}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(267\)
default	\(-\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}-4 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a c -3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d +6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, b \right )}{16 d^{2} \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} \sqrt {a \,d^{2}}}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 223, normalized size = 1.51 \begin {gather*} -\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{4} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a^{3} d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} a^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} - \frac {{\left (4 \, a c + 3 \, b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, a^{\frac {5}{2}} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*a*b*c + 3*b^2)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c + 5*a*b^2)*sqrt((a*d*x^2 + a*c +
b)/(d*x^2 + c)))/(a^4*d^2 - 2*(a*d*x^2 + a*c + b)*a^3*d^2/(d*x^2 + c) + (a*d*x^2 + a*c + b)^2*a^2*d^2/(d*x^2 +
c)^2) - 1/16*(4*a*c + 3*b)*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(sqrt(a) + sqrt((a*d*x^2
+ a*c + b)/(d*x^2 + c))))/(a^(5/2)*d^2)

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Fricas [A]
time = 0.38, size = 333, normalized size = 2.25 \begin {gather*} \left [\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} - 3 \, a b d x^{2} - 2 \, a^{2} c^{2} - 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a^{3} d^{2}}, -\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x^{4} - 3 \, a b d x^{2} - 2 \, a^{2} c^{2} - 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a^{3} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt((a*c + a*d*x**2 + b)/(c + d*x**2)), x)

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Giac [A]
time = 4.46, size = 167, normalized size = 1.13 \begin {gather*} \frac {2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, x^{2}}{a d} - \frac {2 \, a c d + 3 \, b d}{a^{2} d^{3}}\right )} - \frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \log \left ({\left | -2 \, a c d - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} - b d \right |}\right )}{a^{\frac {5}{2}} d {\left | d \right |}}}{16 \, \mathrm {sgn}\left (d x^{2} + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*x^2/(a*d) - (2*a*c*d + 3*b*d)/(a^2*d^3)) - (4
*a*b*c + 3*b^2)*log(abs(-2*a*c*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))
*sqrt(a)*abs(d) - b*d))/(a^(5/2)*d*abs(d)))/sgn(d*x^2 + c)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b/(c + d*x^2))^(1/2),x)

[Out]

int(x^3/(a + b/(c + d*x^2))^(1/2), x)

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