3.5.63 \(\int \frac {\sqrt {c+d x^3}}{x (a+b x^3)^2} \, dx\) [463]

Optimal. Leaf size=121 \[ \frac {\sqrt {c+d x^3}}{3 a \left (a+b x^3\right )}-\frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2}+\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b} \sqrt {b c-a d}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 121, 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 = {457, 101, 162, 65, 214} \begin {gather*} \frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b} \sqrt {b c-a d}}-\frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2}+\frac {\sqrt {c+d x^3}}{3 a \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 101

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 162

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 214

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

Rule 457

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

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Mathematica [A]
time = 0.46, size = 111, normalized size = 0.92 \begin {gather*} \frac {\frac {a \sqrt {c+d x^3}}{a+b x^3}+\frac {(-2 b c+a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} \sqrt {-b c+a d}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.43, size = 934, normalized size = 7.72

method	result	size
default	\(\text {Expression too large to display}\)	\(934\)
elliptic	\(\text {Expression too large to display}\)	\(1656\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/a^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*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1
/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/
3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)
*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(
2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/
3))^(1/2),1/2*b/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d
^2)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^
(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))-b/a*(-1/3*(d*x^3+c)^(1/2)/b/(b*x^3+a)-1/6*I/b/d*2^(1/2)*sum(1/(a*d-b*
c)*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/
d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(
1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d
^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(
1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/2*b/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alph
a^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-c*
d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a^2*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^3 + c)/((b*x^3 + a)^2*x), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (97) = 194\).
time = 1.94, size = 856, normalized size = 7.07 \begin {gather*} \left [-\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {b^{2} c - a b d} \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 ) - 2 \, {\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b^{2} c - a^{4} b d + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a^{3} b^{2} c - a^{4} b d + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )}}, \frac {4 \, {\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - {\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {b^{2} c - a b d} \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 ) + 2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b^{2} c - a^{4} b d + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - 2 \, {\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a^{3} b^{2} c - a^{4} b d + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d x^{3}}}{x \left (a + b x^{3}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c + d*x**3)/(x*(a + b*x**3)**2), x)

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Giac [A]
time = 1.32, size = 114, normalized size = 0.94 \begin {gather*} \frac {\sqrt {d x^{3} + c} d}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} a} - \frac {{\left (2 \, b c - a 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 {2 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 8.28, size = 182, normalized size = 1.50 \begin {gather*} \frac {\sqrt {c}\,\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 )}{3\,a^2}-\frac {\left (\frac {b\,d}{3\,\left (b^2\,c-a\,b\,d\right )}-\frac {b^2\,c}{3\,a\,\left (b^2\,c-a\,b\,d\right )}\right )\,\sqrt {d\,x^3+c}}{b\,x^3+a}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {d\,x^3+c}\,\sqrt {a\,b\,d-b^2\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,a^2\,\sqrt {a\,b\,d-b^2\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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