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

Optimal. Leaf size=132 \[ \frac{\sqrt{b} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{4 a^2 (b c-a d)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{2 a^2 \sqrt{c}}+\frac{b \sqrt{c+d x^4}}{4 a \left (a+b x^4\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.141509, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \[ \frac{\sqrt{b} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{4 a^2 (b c-a d)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{2 a^2 \sqrt{c}}+\frac{b \sqrt{c+d x^4}}{4 a \left (a+b x^4\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.237257, size = 123, normalized size = 0.93 \[ \frac{\frac{a b \sqrt{c+d x^4}}{\left (a+b x^4\right ) (b c-a d)}+\frac{\sqrt{b} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{\sqrt{c}}}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 880, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70359, size = 1831, normalized size = 13.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.37324, size = 207, normalized size = 1.57 \begin{align*} -\frac{1}{4} \, d^{2}{\left (\frac{{\left (2 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac{\sqrt{d x^{4} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{\sqrt{d x^{4} + c} b}{{\left (a b c d - a^{2} d^{2}\right )}{\left ({\left (d x^{4} + c\right )} b - b c + a d\right )}} - \frac{2 \, \arctan \left (\frac{\sqrt{d x^{4} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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