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

Optimal. Leaf size=149 \[ \frac{x \sqrt{c+d x^2} (3 b c-4 a d)}{8 a^2 \left (a+b x^2\right ) (b c-a d)}+\frac{c (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}+\frac{b x \left (c+d x^2\right )^{3/2}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0794741, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {382, 378, 377, 205} \[ \frac{x \sqrt{c+d x^2} (3 b c-4 a d)}{8 a^2 \left (a+b x^2\right ) (b c-a d)}+\frac{c (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}+\frac{b x \left (c+d x^2\right )^{3/2}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(a + b*x^2)^3,x]

[Out]

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

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 5.17001, size = 130, normalized size = 0.87 \[ \frac{\frac{\sqrt{a} x \sqrt{c+d x^2} \left (-4 a^2 d+a b \left (5 c-2 d x^2\right )+3 b^2 c x^2\right )}{\left (a+b x^2\right )^2 (b c-a d)}+\frac{c (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{3/2}}}{8 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 5177, normalized size = 34.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{2} + c}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.89617, size = 1439, normalized size = 9.66 \begin{align*} \left [-\frac{{\left (3 \, a^{2} b c^{2} - 4 \, a^{3} c d +{\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{4} + 2 \,{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\right )} \sqrt{-a b c + a^{2} d} \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 ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \,{\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{3} +{\left (5 \, a^{2} b^{2} c^{2} - 9 \, a^{3} b c d + 4 \, a^{4} d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{32 \,{\left (a^{5} b^{2} c^{2} - 2 \, a^{6} b c d + a^{7} d^{2} +{\left (a^{3} b^{4} c^{2} - 2 \, a^{4} b^{3} c d + a^{5} b^{2} d^{2}\right )} x^{4} + 2 \,{\left (a^{4} b^{3} c^{2} - 2 \, a^{5} b^{2} c d + a^{6} b d^{2}\right )} x^{2}\right )}}, \frac{{\left (3 \, a^{2} b c^{2} - 4 \, a^{3} c d +{\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{4} + 2 \,{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \,{\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{3} +{\left (5 \, a^{2} b^{2} c^{2} - 9 \, a^{3} b c d + 4 \, a^{4} d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{16 \,{\left (a^{5} b^{2} c^{2} - 2 \, a^{6} b c d + a^{7} d^{2} +{\left (a^{3} b^{4} c^{2} - 2 \, a^{4} b^{3} c d + a^{5} b^{2} d^{2}\right )} x^{4} + 2 \,{\left (a^{4} b^{3} c^{2} - 2 \, a^{5} b^{2} c d + a^{6} b d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((3*a^2*b*c^2 - 4*a^3*c*d + (3*b^3*c^2 - 4*a*b^2*c*d)*x^4 + 2*(3*a*b^2*c^2 - 4*a^2*b*c*d)*x^2)*sqrt(-a*
b*c + a^2*d)*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*((b*c -
2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((3*a*b^3*c^2 - 5*a
^2*b^2*c*d + 2*a^3*b*d^2)*x^3 + (5*a^2*b^2*c^2 - 9*a^3*b*c*d + 4*a^4*d^2)*x)*sqrt(d*x^2 + c))/(a^5*b^2*c^2 - 2
*a^6*b*c*d + a^7*d^2 + (a^3*b^4*c^2 - 2*a^4*b^3*c*d + a^5*b^2*d^2)*x^4 + 2*(a^4*b^3*c^2 - 2*a^5*b^2*c*d + a^6*
b*d^2)*x^2), 1/16*((3*a^2*b*c^2 - 4*a^3*c*d + (3*b^3*c^2 - 4*a*b^2*c*d)*x^4 + 2*(3*a*b^2*c^2 - 4*a^2*b*c*d)*x^
2)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^
2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*((3*a*b^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^3 + (5*a^2*b^2*c^2 - 9
*a^3*b*c*d + 4*a^4*d^2)*x)*sqrt(d*x^2 + c))/(a^5*b^2*c^2 - 2*a^6*b*c*d + a^7*d^2 + (a^3*b^4*c^2 - 2*a^4*b^3*c*
d + a^5*b^2*d^2)*x^4 + 2*(a^4*b^3*c^2 - 2*a^5*b^2*c*d + a^6*b*d^2)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{2}}}{\left (a + b x^{2}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.77492, size = 657, normalized size = 4.41 \begin{align*} -\frac{{\left (3 \, b c^{2} \sqrt{d} - 4 \, a c d^{\frac{3}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{8 \,{\left (a^{2} b c - a^{3} d\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b^{3} c^{2} \sqrt{d} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b^{2} c d^{\frac{3}{2}} - 9 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{3} c^{3} \sqrt{d} + 30 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b^{2} c^{2} d^{\frac{3}{2}} - 40 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} b c d^{\frac{5}{2}} + 16 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{3} d^{\frac{7}{2}} + 9 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c^{4} \sqrt{d} - 28 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} c^{3} d^{\frac{3}{2}} + 16 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} b c^{2} d^{\frac{5}{2}} - 3 \, b^{3} c^{5} \sqrt{d} + 2 \, a b^{2} c^{4} d^{\frac{3}{2}}}{4 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}^{2}{\left (a^{2} b^{2} c - a^{3} b d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*b*c^2*sqrt(d) - 4*a*c*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c
*d - a^2*d^2))/((a^2*b*c - a^3*d)*sqrt(a*b*c*d - a^2*d^2)) - 1/4*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^3*c^2*sq
rt(d) - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b^2*c*d^(3/2) - 9*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^3*c^3*sqrt(d)
+ 30*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b^2*c^2*d^(3/2) - 40*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*b*c*d^(5/2) +
16*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^3*d^(7/2) + 9*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^4*sqrt(d) - 28*(sqrt(
d)*x - sqrt(d*x^2 + c))^2*a*b^2*c^3*d^(3/2) + 16*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b*c^2*d^(5/2) - 3*b^3*c^5
*sqrt(d) + 2*a*b^2*c^4*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c +
4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)^2*(a^2*b^2*c - a^3*b*d))