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

Optimal. Leaf size=125 \[ \frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 \sqrt {b} d^{5/2}}-\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {444, 50, 63, 217, 206} \[ -\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 d^2}+\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 \sqrt {b} d^{5/2}}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 131, normalized size = 1.05 \[ \frac {\sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right ) \left (5 a d-3 b c+2 b d x^2\right )+\frac {3 (b c-a d)^{5/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{b}}{8 d^{5/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 334, normalized size = 2.67 \[ \left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, b d^{3}}, -\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, b d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(b*d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*
c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) + 4*(2*b^2*d^2*x^2 -
3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3), -1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqr
t(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d +
(b^2*c*d + a*b*d^2)*x^2)) - 2*(2*b^2*d^2*x^2 - 3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3
)]

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giac [A]  time = 0.52, size = 149, normalized size = 1.19 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (\frac {2 \, {\left (b x^{2} + a\right )}}{b d} - \frac {3 \, {\left (b c d - a d^{2}\right )}}{b d^{3}}\right )} - \frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} b}{8 \, {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 337, normalized size = 2.70 \[ \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 a^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-6 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}+10 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a d -6 \sqrt {b d}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, b c \right )}{16 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*d*x^2+3*a^2*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))-6*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))+3*b^2*c^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))+10*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*a*d-6*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*c)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/d^2/(b*d
)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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