3.813 \(\int \frac{1}{x^3 (a+b x^4) \sqrt{c+d x^4}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{b \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{3/2} \sqrt{b c-a d}}-\frac{\sqrt{c+d x^4}}{2 a c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0880034, antiderivative size = 80, 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 = {465, 480, 12, 377, 205} \[ -\frac{b \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{3/2} \sqrt{b c-a d}}-\frac{\sqrt{c+d x^4}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [C]  time = 0.685292, size = 179, normalized size = 2.24 \[ -\frac{\left (\frac{d x^4}{c}+1\right ) \left (\frac{4 x^4 \left (c+d x^4\right ) (b c-a d) \, _2F_1\left (2,2;\frac{5}{2};\frac{(b c-a d) x^4}{c \left (b x^4+a\right )}\right )}{3 c^2 \left (a+b x^4\right )}+\frac{\left (c+2 d x^4\right ) \sin ^{-1}\left (\sqrt{\frac{x^4 (b c-a d)}{c \left (a+b x^4\right )}}\right )}{c \sqrt{\frac{a x^4 \left (c+d x^4\right ) (b c-a d)}{c^2 \left (a+b x^4\right )^2}}}\right )}{2 x^2 \left (a+b x^4\right ) \sqrt{c+d x^4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^3*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

-((1 + (d*x^4)/c)*(((c + 2*d*x^4)*ArcSin[Sqrt[((b*c - a*d)*x^4)/(c*(a + b*x^4))]])/(c*Sqrt[(a*(b*c - a*d)*x^4*
(c + d*x^4))/(c^2*(a + b*x^4)^2)]) + (4*(b*c - a*d)*x^4*(c + d*x^4)*Hypergeometric2F1[2, 2, 5/2, ((b*c - a*d)*
x^4)/(c*(a + b*x^4))])/(3*c^2*(a + b*x^4))))/(2*x^2*(a + b*x^4)*Sqrt[c + d*x^4])

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Maple [B]  time = 0.015, size = 350, normalized size = 4.4 \begin{align*} -{\frac{1}{2\,ac{x}^{2}}\sqrt{d{x}^{4}+c}}+{\frac{b}{4\,a}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ({x}^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{b}{4\,a}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ({x}^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )} \sqrt{d x^{4} + c} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)*sqrt(d*x^4 + c)*x^3), x)

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Fricas [B]  time = 2.25124, size = 698, normalized size = 8.72 \begin{align*} \left [-\frac{\sqrt{-a b c + a^{2} d} b c x^{2} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) + 4 \, \sqrt{d x^{4} + c}{\left (a b c - a^{2} d\right )}}{8 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}, -\frac{\sqrt{a b c - a^{2} d} b c x^{2} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{a b c - a^{2} d}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} +{\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt{d x^{4} + c}{\left (a b c - a^{2} d\right )}}{4 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(-a*b*c + a^2*d)*b*c*x^2*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4
+ a^2*c^2 + 4*((b*c - 2*a*d)*x^6 - a*c*x^2)*sqrt(d*x^4 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^8 + 2*a*b*x^4 + a^2)
) + 4*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2), -1/4*(sqrt(a*b*c - a^2*d)*b*c*x^2*arctan(1
/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^6 + (a*b*c^2 - a^2*c*d
)*x^2)) + 2*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b x^{4}\right ) \sqrt{c + d x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

Integral(1/(x**3*(a + b*x**4)*sqrt(c + d*x**4)), x)

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Giac [A]  time = 1.09911, size = 86, normalized size = 1.08 \begin{align*} \frac{\frac{b c \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{\sqrt{a b c - a^{2} d} a} - \frac{\sqrt{d + \frac{c}{x^{4}}}}{a}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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