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

Optimal. Leaf size=81 \[ \frac {\sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b}+\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \]

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Rubi [A]  time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {402, 217, 206, 377, 205} \begin {gather*} \frac {\sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b}+\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(a + b*x^2),x]

[Out]

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

Rule 205

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

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 377

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

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 84, normalized size = 1.04 \begin {gather*} \frac {\sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b}+\frac {\sqrt {d} \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 138, normalized size = 1.70 \begin {gather*} -\frac {\sqrt {b c-a d} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{\sqrt {a} b}-\frac {\sqrt {d} \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 596, normalized size = 7.36 \begin {gather*} \left [\frac {2 \, \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b}, -\frac {4 \, \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b}, \frac {\sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{2 \, b}, -\frac {2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right )}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.01, size = 948, normalized size = 11.70 \begin {gather*} \frac {a d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {a d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {c \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {c \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {\sqrt {d}\, \ln \left (\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) d -\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b}+\frac {\sqrt {d}\, \ln \left (\frac {\left (x -\frac {\sqrt {-a b}}{b}\right ) d +\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{2 \sqrt {-a b}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{2 \sqrt {-a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \left \{\begin {array}{cl} \frac {\sqrt {-d}\,\mathrm {asin}\left (x\,\sqrt {-\frac {d}{c}}\right )}{a} & \text {\ if\ \ }\left (\left (c+a\,d=0\wedge b=-1\right )\vee a\,d=b\,c\right )\wedge d<0\\ \frac {\sqrt {d}\,\ln \left (2\,\sqrt {d}\,x+2\,\sqrt {d\,x^2+c}\right )}{b}+\frac {\mathrm {atan}\left (\frac {x\,\sqrt {b\,c-a\,d}}{\sqrt {a}\,\sqrt {d\,x^2+c}}\right )\,\sqrt {b\,c-a\,d}}{\sqrt {a}\,b} & \text {\ if\ \ }c\neq 0\wedge \left (\left (\left (c+a\,d\neq 0\vee b\neq -1\right )\wedge a\,d\neq b\,c\right )\vee \neg d<0\right )\\ \int \frac {\sqrt {d\,x^2+c}}{b\,x^2+a} \,d x & \text {\ if\ \ }\left (\left (\left (\left (c+a\,d=0\wedge b=-1\right )\vee a\,d=b\,c\right )\wedge d<0\right )\vee c=0\right )\wedge \left (\left (\left (c+a\,d\neq 0\vee b\neq -1\right )\wedge a\,d\neq b\,c\right )\vee \neg d<0\right ) \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(1/2)/(a + b*x^2),x)

[Out]

piecewise((c + a*d == 0 & b == -1 | a*d == b*c) & d < 0, ((-d)^(1/2)*asin(x*(-d/c)^(1/2)))/a, c ~= 0 & ((c + a
*d ~= 0 | b ~= -1) & a*d ~= b*c | ~d < 0), (d^(1/2)*log(2*d^(1/2)*x + 2*(c + d*x^2)^(1/2)))/b + (atan((x*(- a*
d + b*c)^(1/2))/(a^(1/2)*(c + d*x^2)^(1/2)))*(- a*d + b*c)^(1/2))/(a^(1/2)*b), ((c + a*d == 0 & b == -1 | a*d
== b*c) & d < 0 | c == 0) & ((c + a*d ~= 0 | b ~= -1) & a*d ~= b*c | ~d < 0), int((c + d*x^2)^(1/2)/(a + b*x^2
), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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