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

Optimal. Leaf size=115 \[ \frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}-\frac {2 \sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2}-\frac {\sqrt {c+d x^3}}{3 a x^3} \]

[Out]

1/3*(-a*d+2*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^2/c^(1/2)-2/3*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(
1/2))*b^(1/2)*(-a*d+b*c)^(1/2)/a^2-1/3*(d*x^3+c)^(1/2)/a/x^3

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Rubi [A]  time = 0.12, antiderivative size = 115, 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, 99, 156, 63, 208} \[ \frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}-\frac {2 \sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2}-\frac {\sqrt {c+d x^3}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 99

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

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

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Mathematica [A]  time = 0.18, size = 107, normalized size = 0.93 \[ \frac {\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}-2 \sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )-\frac {a \sqrt {c+d x^3}}{x^3}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 513, normalized size = 4.46 \[ \left [\frac {2 \, \sqrt {b^{2} c - a b d} c x^{3} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - {\left (2 \, b c - a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c x^{3}}, \frac {4 \, \sqrt {-b^{2} c + a b d} c x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, b c - a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c x^{3}}, -\frac {{\left (2 \, b c - a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - \sqrt {b^{2} c - a b d} c x^{3} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + \sqrt {d x^{3} + c} a c}{3 \, a^{2} c x^{3}}, \frac {2 \, \sqrt {-b^{2} c + a b d} c x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, b c - a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - \sqrt {d x^{3} + c} a c}{3 \, a^{2} c x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 107, normalized size = 0.93 \[ \frac {2 \, {\left (b^{2} c - a b d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{3} + c}}{3 \, a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.25, size = 518, normalized size = 4.50 \[ \frac {\left (\frac {2 \sqrt {d \,x^{3}+c}}{3 b}+\frac {i \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 b \,d^{2} \sqrt {d \,x^{3}+c}}\right ) b^{2}}{a^{2}}+\frac {-\frac {d \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}}{a}-\frac {\left (-\frac {2 \sqrt {c}\, \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}+\frac {2 \sqrt {d \,x^{3}+c}}{3}\right ) b}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*(d*x^3+c)^(1/2)/x^3-1/3*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2))+1/a^2*b^2*(2/3*(d*x^3+c)^(1/2)/b
+1/3*I/b/d^2*2^(1/2)*sum((-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)))-1/a^2*b*(2/3*(d*x^3+c)^(1/2)-2/3*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 \[ \int \frac {\sqrt {d x^{3} + c}}{{\left (b x^{3} + a\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.13, size = 137, normalized size = 1.19 \[ \frac {\ln \left (\frac {a\,d-2\,b\,c+2\,\sqrt {d\,x^3+c}\,\sqrt {b^2\,c-a\,b\,d}-b\,d\,x^3}{b\,x^3+a}\right )\,\sqrt {b^2\,c-a\,b\,d}}{3\,a^2}-\frac {\sqrt {d\,x^3+c}}{3\,a\,x^3}+\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )\,\left (a\,d-2\,b\,c\right )}{6\,a^2\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d x^{3}}}{x^{4} \left (a + b x^{3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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