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

Optimal. Leaf size=91 \[ \frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^3}}{\sqrt {a} \sqrt {c+d x^3}}\right )}{3 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {446, 96, 93, 208} \begin {gather*} \frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^3}}{\sqrt {a} \sqrt {c+d x^3}}\right )}{3 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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^4 \sqrt {a+b x^3} \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3}-\frac {(b c+a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^3\right )}{6 a c}\\ &=-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3}-\frac {(b c+a d) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^3}}{\sqrt {c+d x^3}}\right )}{3 a c}\\ &=-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3}+\frac {(b c+a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^3}}{\sqrt {a} \sqrt {c+d x^3}}\right )}{3 a^{3/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 91, normalized size = 1.00 \begin {gather*} \frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^3}}{\sqrt {a} \sqrt {c+d x^3}}\right )}{3 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.65, size = 119, normalized size = 1.31 \begin {gather*} \frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^3}}{\sqrt {a} \sqrt {c+d x^3}}\right )}{3 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x^3} (a d-b c)}{3 a c \sqrt {c+d x^3} \left (a-\frac {c \left (a+b x^3\right )}{c+d x^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 278, normalized size = 3.05 \begin {gather*} \left [\frac {\sqrt {a c} {\left (b c + a d\right )} x^{3} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{6} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{3} + 4 \, {\left ({\left (b c + a d\right )} x^{3} + 2 \, a c\right )} \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} \sqrt {a c}}{x^{6}}\right ) - 4 \, \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} a c}{12 \, a^{2} c^{2} x^{3}}, -\frac {\sqrt {-a c} {\left (b c + a d\right )} x^{3} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{3} + 2 \, a c\right )} \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{6} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{3}\right )}}\right ) + 2 \, \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(sqrt(a*c)*(b*c + a*d)*x^3*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^6 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*
x^3 + 4*((b*c + a*d)*x^3 + 2*a*c)*sqrt(b*x^3 + a)*sqrt(d*x^3 + c)*sqrt(a*c))/x^6) - 4*sqrt(b*x^3 + a)*sqrt(d*x
^3 + c)*a*c)/(a^2*c^2*x^3), -1/6*(sqrt(-a*c)*(b*c + a*d)*x^3*arctan(1/2*((b*c + a*d)*x^3 + 2*a*c)*sqrt(b*x^3 +
a)*sqrt(d*x^3 + c)*sqrt(-a*c)/(a*b*c*d*x^6 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^3)) + 2*sqrt(b*x^3 + a)*sqrt(d*x
^3 + c)*a*c)/(a^2*c^2*x^3)]

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giac [B]  time = 0.24, size = 413, normalized size = 4.54 \begin {gather*} \frac {\sqrt {b d} b^{4} d {\left (\frac {{\left (b c + a d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a b^{3} c d} - \frac {2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2} - {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{2} b c - {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{2} a d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{3} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}\right )}^{4}\right )} a b^{2} c d}\right )}}{3 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \,x^{3}+a}\, \sqrt {d \,x^{3}+c}\, x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 10.77, size = 481, normalized size = 5.29 \begin {gather*} \frac {\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,\left (\frac {c\,b^2}{12}+\frac {a\,d\,b}{12}\right )}{a^{3/2}\,c^{3/2}\,d\,\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}-\frac {b^2}{12\,a\,c\,d}+\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^2\,\left (\frac {a^2\,d^2}{12}-\frac {a\,b\,c\,d}{4}+\frac {b^2\,c^2}{12}\right )}{a^2\,c^2\,d\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3}{{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3}+\frac {b\,\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}{d\,\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}-\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^2\,\left (a\,d+b\,c\right )}{\sqrt {a}\,\sqrt {c}\,d\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^2}}+\frac {\ln \left (\frac {\sqrt {b\,x^3+a}-\sqrt {a}}{\sqrt {d\,x^3+c}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}+a^{3/2}\,\sqrt {c}\,d\right )}{6\,a^2\,c^2}-\frac {\ln \left (\frac {\left (\sqrt {c}\,\sqrt {b\,x^3+a}-\sqrt {a}\,\sqrt {d\,x^3+c}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}{\sqrt {d\,x^3+c}-\sqrt {c}}\right )}{\sqrt {d\,x^3+c}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}+a^{3/2}\,\sqrt {c}\,d\right )}{6\,a^2\,c^2}-\frac {d\,\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}{12\,a\,c\,\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x^3)^(1/2) - a^(1/2))*((b^2*c)/12 + (a*b*d)/12))/(a^(3/2)*c^(3/2)*d*((c + d*x^3)^(1/2) - c^(1/2))) -
b^2/(12*a*c*d) + (((a + b*x^3)^(1/2) - a^(1/2))^2*((a^2*d^2)/12 + (b^2*c^2)/12 - (a*b*c*d)/4))/(a^2*c^2*d*((c
+ d*x^3)^(1/2) - c^(1/2))^2))/(((a + b*x^3)^(1/2) - a^(1/2))^3/((c + d*x^3)^(1/2) - c^(1/2))^3 + (b*((a + b*x
^3)^(1/2) - a^(1/2)))/(d*((c + d*x^3)^(1/2) - c^(1/2))) - (((a + b*x^3)^(1/2) - a^(1/2))^2*(a*d + b*c))/(a^(1/
2)*c^(1/2)*d*((c + d*x^3)^(1/2) - c^(1/2))^2)) + (log(((a + b*x^3)^(1/2) - a^(1/2))/((c + d*x^3)^(1/2) - c^(1/
2)))*(a^(1/2)*b*c^(3/2) + a^(3/2)*c^(1/2)*d))/(6*a^2*c^2) - (log(((c^(1/2)*(a + b*x^3)^(1/2) - a^(1/2)*(c + d*
x^3)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x^3)^(1/2) - a^(1/2)))/((c + d*x^3)^(1/2) - c^(1/2))))/((c + d*x^3
)^(1/2) - c^(1/2)))*(a^(1/2)*b*c^(3/2) + a^(3/2)*c^(1/2)*d))/(6*a^2*c^2) - (d*((a + b*x^3)^(1/2) - a^(1/2)))/(
12*a*c*((c + d*x^3)^(1/2) - c^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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