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

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

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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 {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b c-a d}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{3/2}}-\frac {\sqrt {c+d x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.13, size = 151, normalized size = 1.29 \begin {gather*} \frac {2 b^{3/2} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (a d-b c)}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a c^{3/2}}-\frac {\sqrt {c+d x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.23, size = 127, normalized size = 1.09 \begin {gather*} -\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 a^2 \sqrt {a d-b c}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{3/2}}-\frac {\sqrt {c+d x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 565, normalized size = 4.83 \begin {gather*} \left [\frac {2 \, b c^{2} x^{3} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) + {\left (2 \, b c + a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, -\frac {4 \, b c^{2} x^{3} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) - {\left (2 \, b c + a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, \frac {b c^{2} x^{3} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - {\left (2 \, b c + a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - \sqrt {d x^{3} + c} a c}{3 \, a^{2} c^{2} x^{3}}, -\frac {2 \, b c^{2} x^{3} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) + {\left (2 \, b c + a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + \sqrt {d x^{3} + c} a c}{3 \, 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)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 104, normalized size = 0.89 \begin {gather*} \frac {2 \, b^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c + a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c} - \frac {\sqrt {d x^{3} + c}}{3 \, a c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.25, size = 498, normalized size = 4.26 \begin {gather*} \frac {2 b \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a^{2} \sqrt {c}}+\frac {\frac {d \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}-\frac {\sqrt {d \,x^{3}+c}}{3 c \,x^{3}}}{a}-\frac {i b^{2} \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {\left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right )\right ) b}{2 \left (a d -b c \right ) d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 a^{2} d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*(d*x^3+c)^(1/2)/c/x^3+1/3*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))-1/3*I/a^2*b^2/d^2*2^(1/2)*sum(
1/(a*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*(
(x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1
/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha
*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^
(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha
^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c
*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a))+2/3/a^2*
b*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.42, size = 142, normalized size = 1.21 \begin {gather*} \frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (a\,d+2\,b\,c\right )}{6\,a^2\,c^{3/2}}-\frac {\sqrt {d\,x^3+c}}{3\,a\,c\,x^3}+\frac {b^{3/2}\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a^2\,\sqrt {a\,d-b\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2) + c^(1/2))^3)/x^6)*(a*d + 2*b*c))/(6*a^2*c^(3/2)) - (c
+ d*x^3)^(1/2)/(3*a*c*x^3) + (b^(3/2)*log((2*b*c - a*d + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i + b*d*
x^3)/(a + b*x^3))*1i)/(3*a^2*(a*d - b*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} \left (a + b x^{3}\right ) \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)/(d*x**3+c)**(1/2),x)

[Out]

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

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