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

Optimal. Leaf size=141 \[ -\frac {b (4 a c+b) \tanh ^{-1}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{3/2} d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 d^2}+\frac {(b-4 a c) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{8 a d^2} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 181, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 446, 80, 50, 63, 217, 206} \[ -\frac {b (4 a c+b) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {a \left (c+d x^2\right )+b}}\right )}{8 a^{3/2} d^2 \sqrt {a \left (c+d x^2\right )+b}}+\frac {\left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{4 a d^2}-\frac {(4 a c+b) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{8 a d^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + b/(c + d*x^2)],x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 97, normalized size = 0.69 \[ \frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (-2 a c+2 a d x^2+b\right )-b (4 a c+b) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{3/2} d^2} \]

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

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fricas [A]  time = 0.81, size = 325, normalized size = 2.30 \[ \left [\frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} + a b d x^{2} - 2 \, a^{2} c^{2} + a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a^{2} d^{2}}, \frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{4} + a b d x^{2} - 2 \, a^{2} c^{2} + a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a^{2} d^{2}}\right ] \]

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 + 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 + a*b*d*x^2 - 2*a^2*c^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*d^2), 1/16*((4*a*b*c + b^2)*sq
rt(-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 + a*b*d*x^2 - 2*a^2*c^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*d^2)]

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giac [A]  time = 0.44, size = 159, normalized size = 1.13 \[ \frac {1}{16} \, {\left (2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, a c d - b d}{a d^{3}}\right )} + \frac {{\left (4 \, a b c + b^{2}\right )} \log \left ({\left | -2 \, a c d - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} - b d \right |}\right )}{a^{\frac {3}{2}} d {\left | d \right |}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]

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/d - (2*a*c*d - b*d)/(a*d^3)) + (4*a*b*c +
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)*a
bs(d) - b*d))/(a^(3/2)*d*abs(d)))*sgn(d*x^2 + c)

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maple [B]  time = 0.04, size = 354, normalized size = 2.51 \[ \frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-4 a b c d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}-b^{2} d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a c +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b \right )}{16 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, a \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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maxima [A]  time = 1.75, size = 218, normalized size = 1.55 \[ -\frac {{\left (4 \, a b c - b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c + a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{3} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a^{2} d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} a d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} + \frac {{\left (4 \, a c + b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, a^{\frac {3}{2}} d^{2}} \]

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

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

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

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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