3.809 \(\int \frac{1}{x^5 (a+b x^4) \sqrt{c+d x^4}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{2 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{4 a^2 c^{3/2}}-\frac{\sqrt{c+d x^4}}{4 a c x^4} \]

[Out]

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

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Rubi [A]  time = 0.119991, antiderivative size = 117, 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} \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{2 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{4 a^2 c^{3/2}}-\frac{\sqrt{c+d x^4}}{4 a c x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

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

Mathematica [A]  time = 0.115368, size = 151, normalized size = 1.29 \[ \frac{b^{3/2} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^4}}{\sqrt{b c-a d}}\right )}{2 a^2 (a d-b c)}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{2 a^2 \sqrt{c}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^4}}{\sqrt{c}}\right )}{4 a c^{3/2}}-\frac{\sqrt{c+d x^4}}{4 a c x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

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Maple [B]  time = 0.016, size = 402, normalized size = 3.4 \begin{align*} -{\frac{b}{4\,{a}^{2}}\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{-{\frac{ad-bc}{b}}}}}}-{\frac{b}{4\,{a}^{2}}\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{-{\frac{ad-bc}{b}}}}}}-{\frac{1}{4\,ac{x}^{4}}\sqrt{d{x}^{4}+c}}+{\frac{d}{4\,a}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{4}+c} \right ) } \right ){c}^{-{\frac{3}{2}}}}+{\frac{b}{2\,{a}^{2}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{4}+c} \right ) } \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(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^5), x)

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Fricas [A]  time = 1.70135, size = 1261, normalized size = 10.78 \begin{align*} \left [\frac{2 \, b c^{2} x^{4} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{4} + 2 \, b c - a d - 2 \, \sqrt{d x^{4} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a 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^{2} x^{4}}, -\frac{4 \, b c^{2} x^{4} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{4} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a 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^{2} x^{4}}, \frac{b c^{2} x^{4} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{4} + 2 \, b c - a d - 2 \, \sqrt{d x^{4} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{4} + a}\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^{2} x^{4}}, -\frac{2 \, b c^{2} x^{4} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{4} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a 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^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{5} \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**5/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

Integral(1/(x**5*(a + b*x**4)*sqrt(c + d*x**4)), x)

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Giac [A]  time = 1.09855, size = 159, normalized size = 1.36 \begin{align*} \frac{1}{4} \, d^{2}{\left (\frac{2 \, b^{2} \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} c d^{2}} - \frac{\sqrt{d x^{4} + c}}{a c d^{2} x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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