3.6.29 \(\int \frac {1}{x^3 (a+b x^4) \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=80 \[ -\frac {b \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^4}}{2 a c x^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {465, 480, 12, 377, 205} \begin {gather*} -\frac {b \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^4}}{2 a c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 377

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

Rule 465

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

Rule 480

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

Rubi steps

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

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Mathematica [C]  time = 4.83, size = 179, normalized size = 2.24 \begin {gather*} -\frac {\left (\frac {d x^4}{c}+1\right ) \left (\frac {4 x^4 \left (c+d x^4\right ) (b c-a d) \, _2F_1\left (2,2;\frac {5}{2};\frac {(b c-a d) x^4}{c \left (b x^4+a\right )}\right )}{3 c^2 \left (a+b x^4\right )}+\frac {\left (c+2 d x^4\right ) \sin ^{-1}\left (\sqrt {\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}}\right )}{c \sqrt {\frac {a x^4 \left (c+d x^4\right ) (b c-a d)}{c^2 \left (a+b x^4\right )^2}}}\right )}{2 x^2 \left (a+b x^4\right ) \sqrt {c+d x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^3*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

-1/2*((1 + (d*x^4)/c)*(((c + 2*d*x^4)*ArcSin[Sqrt[((b*c - a*d)*x^4)/(c*(a + b*x^4))]])/(c*Sqrt[(a*(b*c - a*d)*
x^4*(c + d*x^4))/(c^2*(a + b*x^4)^2)]) + (4*(b*c - a*d)*x^4*(c + d*x^4)*Hypergeometric2F1[2, 2, 5/2, ((b*c - a
*d)*x^4)/(c*(a + b*x^4))])/(3*c^2*(a + b*x^4))))/(x^2*(a + b*x^4)*Sqrt[c + d*x^4])

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IntegrateAlgebraic [A]  time = 0.48, size = 142, normalized size = 1.78 \begin {gather*} \frac {b \sqrt {b c-a d} \tan ^{-1}\left (\frac {b \sqrt {d} x^4}{\sqrt {a} \sqrt {b c-a d}}+\frac {b x^2 \sqrt {c+d x^4}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{2 a^{3/2} (a d-b c)}-\frac {\sqrt {c+d x^4}}{2 a c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*(a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

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fricas [B]  time = 0.53, size = 332, normalized size = 4.15 \begin {gather*} \left [-\frac {\sqrt {-a b c + a^{2} d} b c x^{2} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) + 4 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )}}{8 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}, -\frac {\sqrt {a b c - a^{2} d} b c x^{2} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} + {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )}}{4 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(-a*b*c + a^2*d)*b*c*x^2*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4
+ a^2*c^2 + 4*((b*c - 2*a*d)*x^6 - a*c*x^2)*sqrt(d*x^4 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^8 + 2*a*b*x^4 + a^2)
) + 4*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2), -1/4*(sqrt(a*b*c - a^2*d)*b*c*x^2*arctan(1
/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^6 + (a*b*c^2 - a^2*c*d
)*x^2)) + 2*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2)]

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giac [A]  time = 0.22, size = 116, normalized size = 1.45 \begin {gather*} \frac {1}{2} \, d^{\frac {3}{2}} {\left (\frac {b \arctan \left (\frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a d} + \frac {2}{{\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} - c\right )} a d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.30, size = 350, normalized size = 4.38 \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 {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, a}-\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 {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {\sqrt {d \,x^{4}+c}}{2 a c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x)

[Out]

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

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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^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,\left (b\,x^4+a\right )\,\sqrt {d\,x^4+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^4)*(c + d*x^4)^(1/2)),x)

[Out]

int(1/(x^3*(a + b*x^4)*(c + d*x^4)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \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**3/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

Integral(1/(x**3*(a + b*x**4)*sqrt(c + d*x**4)), x)

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