3.1.50 \(\int \frac {\sqrt {a+b x^2}}{(c+d x^2)^2} \, dx\) [50]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {386, 385, 214} \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 214

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

Rule 385

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 386

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 \text {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*}

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Mathematica [A]
time = 0.31, size = 99, normalized size = 1.21 \begin {gather*} \frac {x \sqrt {a+b x^2}}{2 c^2+2 c d x^2}-\frac {a \tan ^{-1}\left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{2 c^{3/2} \sqrt {-b c+a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1944\) vs. \(2(66)=132\).
time = 0.06, size = 1945, normalized size = 23.72

method	result	size
default	\(\text {Expression too large to display}\)	\(1945\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (66) = 132\).
time = 1.03, 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)]

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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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (66) = 132\).
time = 1.66, 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)

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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)

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