3.712 \(\int \frac{1}{x^4 (a+b x^2) \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^2} (2 a d+3 b c)}{3 a^2 c^2 x}-\frac{\sqrt{c+d x^2}}{3 a c x^3} \]

[Out]

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

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Rubi [A]  time = 0.122726, antiderivative size = 110, normalized size of antiderivative = 1., 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 = {480, 583, 12, 377, 205} \[ \frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^2} (2 a d+3 b c)}{3 a^2 c^2 x}-\frac{\sqrt{c+d x^2}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

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

Rule 480

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 5.10724, size = 96, normalized size = 0.87 \[ \frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^2} \left (-a c+2 a d x^2+3 b c x^2\right )}{3 a^2 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

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

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Maple [B]  time = 0.012, size = 379, normalized size = 3.5 \begin{align*} -{\frac{1}{3\,ac{x}^{3}}\sqrt{d{x}^{2}+c}}+{\frac{2\,d}{3\,a{c}^{2}x}\sqrt{d{x}^{2}+c}}+{\frac{{b}^{2}}{2\,{a}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{{b}^{2}}{2\,{a}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}+{\frac{b}{{a}^{2}cx}\sqrt{d{x}^{2}+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )} \sqrt{d x^{2} + c} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*sqrt(d*x^2 + c)*x^4), x)

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Fricas [B]  time = 1.99592, size = 849, normalized size = 7.72 \begin{align*} \left [-\frac{3 \, \sqrt{-a b c + a^{2} d} b^{2} c^{2} x^{3} \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 (a^{2} b c^{2} - a^{3} c d -{\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{12 \,{\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{3}}, \frac{3 \, \sqrt{a b c - a^{2} d} b^{2} c^{2} x^{3} \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 (a^{2} b c^{2} - a^{3} c d -{\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{6 \,{\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/12*(3*sqrt(-a*b*c + a^2*d)*b^2*c^2*x^3*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*(a^2*b*c^2 - a^3*c*d - (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x^2)*sqrt(d*x^2 + c))/((a^3*b*c^3 - a^
4*c^2*d)*x^3), 1/6*(3*sqrt(a*b*c - a^2*d)*b^2*c^2*x^3*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*(a^2*b*c^2 - a^3*c*d - (3*a*b^2*c^2 -
a^2*b*c*d - 2*a^3*d^2)*x^2)*sqrt(d*x^2 + c))/((a^3*b*c^3 - a^4*c^2*d)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b x^{2}\right ) \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/(x**4*(a + b*x**2)*sqrt(c + d*x**2)), x)

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Giac [B]  time = 1.89253, size = 263, normalized size = 2.39 \begin{align*} -\frac{1}{3} \, d^{\frac{5}{2}}{\left (\frac{3 \, b^{2} \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 )}{\sqrt{a b c d - a^{2} d^{2}} a^{2} d^{2}} + \frac{2 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + 3 \, b c^{2} + 2 \, a c d\right )}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2} d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/3*d^(5/2)*(3*b^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))/(sqr
t(a*b*c*d - a^2*d^2)*a^2*d^2) + 2*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c
- 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + 3*b*c^2 + 2*a*c*d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^2*d^2
))