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

Optimal. Leaf size=86 \[ \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 \sqrt {b} d^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \begin {gather*} \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 \sqrt {b} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.29, size = 116, normalized size = 1.35 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right )-(b c-a d)^{3/2} \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 )}{2 b d^{3/2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.50, size = 117, normalized size = 1.36 \begin {gather*} \frac {(a d-b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{2 \sqrt {b} d^{3/2}}+\frac {\sqrt {c+d x^2} (a d-b c)}{2 d \sqrt {a+b x^2} \left (d-\frac {b \left (c+d x^2\right )}{a+b x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(b*c) + a*d)*Sqrt[c + d*x^2])/(2*d*Sqrt[a + b*x^2]*(d - (b*(c + d*x^2))/(a + b*x^2))) + ((-(b*c) + a*d)*Arc
Tanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/(2*Sqrt[b]*d^(3/2))

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fricas [A]  time = 1.62, size = 259, normalized size = 3.01 \begin {gather*} \left [\frac {4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b d - {\left (b c - a d\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 )}{8 \, b d^{2}}, \frac {2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b d + {\left (b c - a d\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 )}{4 \, b d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.55, size = 106, normalized size = 1.23 \begin {gather*} \frac {b {\left (\frac {{\left (b c - a d\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} + \frac {\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a}}{b d}\right )}}{2 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 198, normalized size = 2.30 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (a d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b c \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}\right )}{4 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}\, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^(1/2)/(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 [B]  time = 5.00, size = 280, normalized size = 3.26 \begin {gather*} \frac {\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (a\,d+b\,c\right )}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}+\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (c\,b^2+a\,d\,b\right )}{d^3\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}-\frac {4\,\sqrt {a}\,b\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}+\frac {b^2}{d^2}-\frac {2\,b\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )}{\sqrt {b}\,d^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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