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

Optimal. Leaf size=85 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{2 a \sqrt {b c-a d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^4}}{\sqrt {c}}\right )}{2 a \sqrt {c}} \]

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Rubi [A]  time = 0.07, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 86, 63, 208} \begin {gather*} \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{2 a \sqrt {b c-a d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^4}}{\sqrt {c}}\right )}{2 a \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 86

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

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]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 81, normalized size = 0.95 \begin {gather*} \frac {\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^4}}{\sqrt {c}}\right )}{\sqrt {c}}}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 95, normalized size = 1.12 \begin {gather*} \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4} \sqrt {a d-b c}}{b c-a d}\right )}{2 a \sqrt {a d-b c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^4}}{\sqrt {c}}\right )}{2 a \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 431, normalized size = 5.07 \begin {gather*} \left [\frac {c \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 ) + \sqrt {c} \log \left (\frac {d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {c} + 2 \, c}{x^{4}}\right )}{4 \, a c}, \frac {2 \, c \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 ) + \sqrt {c} \log \left (\frac {d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {c} + 2 \, c}{x^{4}}\right )}{4 \, a c}, \frac {c \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 ) + 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{4} + c} \sqrt {-c}}{c}\right )}{4 \, a c}, \frac {c \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 ) + \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{4} + c} \sqrt {-c}}{c}\right )}{2 \, a c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 71, normalized size = 0.84 \begin {gather*} -\frac {b \arctan \left (\frac {\sqrt {d x^{4} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a} + \frac {\arctan \left (\frac {\sqrt {d x^{4} + c}}{\sqrt {-c}}\right )}{2 \, a \sqrt {-c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.29, size = 347, normalized size = 4.08 \begin {gather*} \frac {\ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-\frac {a d -b c}{b}}\, a}+\frac {\ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {\ln \left (\frac {2 c +2 \sqrt {d \,x^{4}+c}\, \sqrt {c}}{x^{2}}\right )}{2 a \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{4} + a\right )} \sqrt {d x^{4} + c} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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mupad [B]  time = 5.03, size = 652, normalized size = 7.67 \begin {gather*} -\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^4+c}}{\sqrt {c}}\right )}{2\,a\,\sqrt {c}}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (b^3\,d^2\,\sqrt {d\,x^4+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^4+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{4\,\left (a^2\,d-a\,b\,c\right )}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (b^3\,d^2\,\sqrt {d\,x^4+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^4+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{4\,\left (a^2\,d-a\,b\,c\right )}}{\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (b^3\,d^2\,\sqrt {d\,x^4+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^4+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (b^3\,d^2\,\sqrt {d\,x^4+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^4+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{4\,\left (a^2\,d-a\,b\,c\right )}}\right )\,\sqrt {b^2\,c-a\,b\,d}\,1{}\mathrm {i}}{2\,\left (a^2\,d-a\,b\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- atanh((c + d*x^4)^(1/2)/c^(1/2))/(2*a*c^(1/2)) - (atan((((b^2*c - a*b*d)^(1/2)*(b^3*d^2*(c + d*x^4)^(1/2) -
((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^4)^(1/2)*(b^2*c - a*b*d)^
(1/2))/(4*(a^2*d - a*b*c))))/(4*(a^2*d - a*b*c)))*1i)/(4*(a^2*d - a*b*c)) + ((b^2*c - a*b*d)^(1/2)*(b^3*d^2*(c
+ d*x^4)^(1/2) + ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^4)^(1/2
)*(b^2*c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(4*(a^2*d - a*b*c)))*1i)/(4*(a^2*d - a*b*c)))/(((b^2*c - a*b*d)
^(1/2)*(b^3*d^2*(c + d*x^4)^(1/2) - ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2
)*(c + d*x^4)^(1/2)*(b^2*c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(4*(a^2*d - a*b*c))))/(4*(a^2*d - a*b*c)) - (
(b^2*c - a*b*d)^(1/2)*(b^3*d^2*(c + d*x^4)^(1/2) + ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 1
6*a^2*b^3*c*d^2)*(c + d*x^4)^(1/2)*(b^2*c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(4*(a^2*d - a*b*c))))/(4*(a^2*
d - a*b*c))))*(b^2*c - a*b*d)^(1/2)*1i)/(2*(a^2*d - a*b*c))

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sympy [A]  time = 20.21, size = 66, normalized size = 0.78 \begin {gather*} - \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{4}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a \sqrt {\frac {a d - b c}{b}}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{4}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-atan(sqrt(c + d*x**4)/sqrt((a*d - b*c)/b))/(2*a*sqrt((a*d - b*c)/b)) + atan(sqrt(c + d*x**4)/sqrt(-c))/(2*a*s
qrt(-c))

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