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

Optimal. Leaf size=82 \[ \frac {a \tanh ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} \sqrt {b c-a d}}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {378, 377, 208} \begin {gather*} \frac {a \tanh ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} \sqrt {b c-a d}}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/(c + d*x^2)^2,x]

[Out]

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

Rule 208

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

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 378

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.23, size = 165, normalized size = 2.01 \begin {gather*} \frac {x \sqrt {a+b x^2} \left (\sqrt {x^2 \left (\frac {d}{c}-\frac {b}{a}\right )} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}+\sqrt {\frac {d x^2}{c}+1} \sin ^{-1}\left (\frac {\sqrt {x^2 \left (\frac {d}{c}-\frac {b}{a}\right )}}{\sqrt {\frac {d x^2}{c}+1}}\right )\right )}{2 c^2 \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \sqrt {x^2 \left (\frac {d}{c}-\frac {b}{a}\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(Sqrt[(-(b/a) + d/c)*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))] + Sqrt[1 + (d*x^2)/c]*ArcSi
n[Sqrt[(-(b/a) + d/c)*x^2]/Sqrt[1 + (d*x^2)/c]]))/(2*c^2*Sqrt[(-(b/a) + d/c)*x^2]*Sqrt[1 + (b*x^2)/a]*Sqrt[1 +
(d*x^2)/c])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.39, size = 145, normalized size = 1.77 \begin {gather*} \frac {a \sqrt {a d-b c} \tan ^{-1}\left (\frac {\sqrt {b} d x^2}{\sqrt {c} \sqrt {a d-b c}}-\frac {d x \sqrt {a+b x^2}}{\sqrt {c} \sqrt {a d-b c}}+\frac {\sqrt {b} \sqrt {c}}{\sqrt {a d-b c}}\right )}{2 c^{3/2} (b c-a d)}+\frac {x \sqrt {a+b x^2}}{2 c \left (c+d x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.22, size = 369, normalized size = 4.50 \begin {gather*} \left [\frac {4 \, {\left (b c^{2} - a c d\right )} \sqrt {b x^{2} + a} x + {\left (a d x^{2} + a c\right )} \sqrt {b c^{2} - a c d} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt {b c^{2} - a c d} \sqrt {b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{8 \, {\left (b c^{4} - a c^{3} d + {\left (b c^{3} d - a c^{2} d^{2}\right )} x^{2}\right )}}, \frac {2 \, {\left (b c^{2} - a c d\right )} \sqrt {b x^{2} + a} x - {\left (a d x^{2} + a c\right )} \sqrt {-b c^{2} + a c d} \arctan \left (\frac {\sqrt {-b c^{2} + a c d} {\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a}}{2 \, {\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right )}{4 \, {\left (b c^{4} - a c^{3} d + {\left (b c^{3} d - a c^{2} d^{2}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.64, size = 217, normalized size = 2.65 \begin {gather*} -\frac {a \sqrt {b} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{2 \, \sqrt {-b^{2} c^{2} + a b c d} c} + \frac {2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} c - {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d + a^{2} \sqrt {b} d}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )} c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 2521, normalized size = 30.74

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________