3.793 \(\int \frac{\sqrt{c+d x^4}}{x^5 (a+b x^4)} \, dx\)

Optimal. Leaf size=115 \[ \frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{4 a^2 \sqrt{c}}-\frac{\sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{2 a^2}-\frac{\sqrt{c+d x^4}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.123727, antiderivative size = 115, 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, 99, 156, 63, 208} \[ \frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{4 a^2 \sqrt{c}}-\frac{\sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{2 a^2}-\frac{\sqrt{c+d x^4}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^4]/(x^5*(a + b*x^4)),x]

[Out]

-Sqrt[c + d*x^4]/(4*a*x^4) + ((2*b*c - a*d)*ArcTanh[Sqrt[c + d*x^4]/Sqrt[c]])/(4*a^2*Sqrt[c]) - (Sqrt[b]*Sqrt[
b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^4])/Sqrt[b*c - a*d]])/(2*a^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 99

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

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

Mathematica [A]  time = 0.120966, size = 107, normalized size = 0.93 \[ \frac{\frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{\sqrt{c}}-2 \sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )-\frac{a \sqrt{c+d x^4}}{x^4}}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^4]/(x^5*(a + b*x^4)),x]

[Out]

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

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Maple [B]  time = 0.016, size = 1107, normalized size = 9.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^4+c)^(1/2)/x^5/(b*x^4+a),x)

[Out]

1/4*b/a^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)+1/4/a^2*d^(1/2)
*(-a*b)^(1/2)*ln((d*(-a*b)^(1/2)/b+(x^2-(-a*b)^(1/2)/b)*d)/d^(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))+1/4/a/(-(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))*d-1/4*b/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))*c+1/4*b/a^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)-1/4/a^2*d^(1/2)*(-a*b)^(1/2)*ln((-d*(-a*b)^(1/2)/b+(x^2+(-a*b)^(1/2)/b)*d)/d^(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))+1/4/a/(-(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))*d-1/4*b/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))*c-1/4/a/c/x^4*(d*x^4+
c)^(3/2)-1/4/a*d/c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^4+c)^(1/2))/x^2)+1/4/a*d/c*(d*x^4+c)^(1/2)-1/2/a^2*b*(d*x^4+c)
^(1/2)+1/2/a^2*b*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{\sqrt{d x^{4} + c}}{{\left (b x^{4} + a\right )} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^5/(b*x^4+a),x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^4 + c)/((b*x^4 + a)*x^5), x)

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Fricas [A]  time = 1.68249, size = 1160, normalized size = 10.09 \begin{align*} \left [\frac{2 \, \sqrt{b^{2} c - a b d} c x^{4} \log \left (\frac{b d x^{4} + 2 \, b c - a d - 2 \, \sqrt{d x^{4} + c} \sqrt{b^{2} c - a b d}}{b x^{4} + a}\right ) -{\left (2 \, b c - a d\right )} \sqrt{c} x^{4} \log \left (\frac{d x^{4} - 2 \, \sqrt{d x^{4} + c} \sqrt{c} + 2 \, c}{x^{4}}\right ) - 2 \, \sqrt{d x^{4} + c} a c}{8 \, a^{2} c x^{4}}, \frac{4 \, \sqrt{-b^{2} c + a b d} c x^{4} \arctan \left (\frac{\sqrt{d x^{4} + c} \sqrt{-b^{2} c + a b d}}{b d x^{4} + b c}\right ) -{\left (2 \, b c - a d\right )} \sqrt{c} x^{4} \log \left (\frac{d x^{4} - 2 \, \sqrt{d x^{4} + c} \sqrt{c} + 2 \, c}{x^{4}}\right ) - 2 \, \sqrt{d x^{4} + c} a c}{8 \, a^{2} c x^{4}}, -\frac{{\left (2 \, b c - a d\right )} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{d x^{4} + c} \sqrt{-c}}{c}\right ) - \sqrt{b^{2} c - a b d} c x^{4} \log \left (\frac{b d x^{4} + 2 \, b c - a d - 2 \, \sqrt{d x^{4} + c} \sqrt{b^{2} c - a b d}}{b x^{4} + a}\right ) + \sqrt{d x^{4} + c} a c}{4 \, a^{2} c x^{4}}, \frac{2 \, \sqrt{-b^{2} c + a b d} c x^{4} \arctan \left (\frac{\sqrt{d x^{4} + c} \sqrt{-b^{2} c + a b d}}{b d x^{4} + b c}\right ) -{\left (2 \, b c - a d\right )} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{d x^{4} + c} \sqrt{-c}}{c}\right ) - \sqrt{d x^{4} + c} a c}{4 \, a^{2} c x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^5/(b*x^4+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{4}}}{x^{5} \left (a + b x^{4}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**4+c)**(1/2)/x**5/(b*x**4+a),x)

[Out]

Integral(sqrt(c + d*x**4)/(x**5*(a + b*x**4)), x)

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Giac [A]  time = 1.13654, size = 163, normalized size = 1.42 \begin{align*} \frac{1}{4} \, d^{2}{\left (\frac{2 \,{\left (b^{2} c - a b d\right )} \arctan \left (\frac{\sqrt{d x^{4} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{2} d^{2}} - \frac{{\left (2 \, b c - a d\right )} \arctan \left (\frac{\sqrt{d x^{4} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} d^{2}} - \frac{\sqrt{d x^{4} + c}}{a d^{2} x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^5/(b*x^4+a),x, algorithm="giac")

[Out]

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