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

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

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Rubi [A]  time = 0.12, antiderivative size = 117, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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^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*}

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Mathematica [A]  time = 0.17, size = 151, normalized size = 1.29 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Sqrt[c + d*x^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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IntegrateAlgebraic [A]  time = 0.25, size = 127, normalized size = 1.09 \begin {gather*} -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4} \sqrt {a d-b c}}{b c-a d}\right )}{2 a^2 \sqrt {a d-b c}}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Sqrt[c + d*x^4]/(a*c*x^4) - (b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x^4])/(b*c - a*d)])/(2
*a^2*Sqrt[-(b*c) + a*d]) + ((2*b*c + a*d)*ArcTanh[Sqrt[c + d*x^4]/Sqrt[c]])/(4*a^2*c^(3/2))

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fricas [A]  time = 0.70, size = 565, normalized size = 4.83 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 104, normalized size = 0.89 \begin {gather*} \frac {b^{2} \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^{2}} - \frac {{\left (2 \, b c + a d\right )} \arctan \left (\frac {\sqrt {d x^{4} + c}}{\sqrt {-c}}\right )}{4 \, a^{2} \sqrt {-c} c} - \frac {\sqrt {d x^{4} + c}}{4 \, a c x^{4}} \end {gather*}

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]

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

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maple [B]  time = 0.29, size = 402, normalized size = 3.44 \begin {gather*} -\frac {b \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^{2}}-\frac {b \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^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {d \,x^{4}+c}\, \sqrt {c}}{x^{2}}\right )}{4 a \,c^{\frac {3}{2}}}+\frac {b \ln \left (\frac {2 c +2 \sqrt {d \,x^{4}+c}\, \sqrt {c}}{x^{2}}\right )}{2 a^{2} \sqrt {c}}-\frac {\sqrt {d \,x^{4}+c}}{4 a c \,x^{4}} \end {gather*}

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

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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mupad [B]  time = 5.35, size = 396, normalized size = 3.38 \begin {gather*} \frac {\ln \left (\sqrt {d\,x^4+c}\,{\left (b^4\,c-a\,b^3\,d\right )}^{3/2}+b^6\,c^2+a^2\,b^4\,d^2-2\,a\,b^5\,c\,d\right )\,\sqrt {b^4\,c-a\,b^3\,d}}{4\,a^3\,d-4\,a^2\,b\,c}-\frac {\ln \left (\sqrt {d\,x^4+c}\,{\left (b^4\,c-a\,b^3\,d\right )}^{3/2}-b^6\,c^2-a^2\,b^4\,d^2+2\,a\,b^5\,c\,d\right )\,\sqrt {b^4\,c-a\,b^3\,d}}{4\,\left (a^3\,d-a^2\,b\,c\right )}-\frac {\sqrt {d\,x^4+c}}{4\,a\,c\,x^4}-\frac {\mathrm {atan}\left (\frac {b^4\,d^4\,\sqrt {d\,x^4+c}\,3{}\mathrm {i}}{16\,\sqrt {c^3}\,\left (\frac {3\,b^4\,d^4}{16\,c}+\frac {5\,a\,b^3\,d^5}{32\,c^2}+\frac {a^2\,b^2\,d^6}{32\,c^3}\right )}+\frac {b^2\,d^6\,\sqrt {d\,x^4+c}\,1{}\mathrm {i}}{32\,\sqrt {c^3}\,\left (\frac {5\,b^3\,d^5}{32\,a}+\frac {b^2\,d^6}{32\,c}+\frac {3\,b^4\,c\,d^4}{16\,a^2}\right )}+\frac {b^3\,d^5\,\sqrt {d\,x^4+c}\,5{}\mathrm {i}}{32\,\sqrt {c^3}\,\left (\frac {3\,b^4\,d^4}{16\,a}+\frac {5\,b^3\,d^5}{32\,c}+\frac {a\,b^2\,d^6}{32\,c^2}\right )}\right )\,\left (a\,d+2\,b\,c\right )\,1{}\mathrm {i}}{4\,a^2\,\sqrt {c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((c + d*x^4)^(1/2)*(b^4*c - a*b^3*d)^(3/2) + b^6*c^2 + a^2*b^4*d^2 - 2*a*b^5*c*d)*(b^4*c - a*b^3*d)^(1/2))
/(4*a^3*d - 4*a^2*b*c) - (log((c + d*x^4)^(1/2)*(b^4*c - a*b^3*d)^(3/2) - b^6*c^2 - a^2*b^4*d^2 + 2*a*b^5*c*d)
*(b^4*c - a*b^3*d)^(1/2))/(4*(a^3*d - a^2*b*c)) - (c + d*x^4)^(1/2)/(4*a*c*x^4) - (atan((b^4*d^4*(c + d*x^4)^(
1/2)*3i)/(16*(c^3)^(1/2)*((3*b^4*d^4)/(16*c) + (5*a*b^3*d^5)/(32*c^2) + (a^2*b^2*d^6)/(32*c^3))) + (b^2*d^6*(c
+ d*x^4)^(1/2)*1i)/(32*(c^3)^(1/2)*((5*b^3*d^5)/(32*a) + (b^2*d^6)/(32*c) + (3*b^4*c*d^4)/(16*a^2))) + (b^3*d
^5*(c + d*x^4)^(1/2)*5i)/(32*(c^3)^(1/2)*((3*b^4*d^4)/(16*a) + (5*b^3*d^5)/(32*c) + (a*b^2*d^6)/(32*c^2))))*(a
*d + 2*b*c)*1i)/(4*a^2*(c^3)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \left (a + b x^{4}\right ) \sqrt {c + d x^{4}}}\, dx \end {gather*}

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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