3.464 \(\int \frac{\sqrt{c+d x^3}}{x^4 (a+b x^3)^2} \, dx\)

Optimal. Leaf size=161 \[ -\frac{2 b \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{\sqrt{b} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^3 \sqrt{b c-a d}}+\frac{(4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3 \sqrt{c}}-\frac{\sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.217879, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 99, 151, 156, 63, 208} \[ -\frac{2 b \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{\sqrt{b} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^3 \sqrt{b c-a d}}+\frac{(4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3 \sqrt{c}}-\frac{\sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.228915, size = 190, normalized size = 1.18 \[ \frac{\sqrt{c} \left (a \left (a+2 b x^3\right ) \sqrt{c+d x^3} (b c-a d)+\sqrt{b} x^3 \left (a+b x^3\right ) (4 b c-3 a d) \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )\right )-x^3 \left (a+b x^3\right ) \left (a^2 d^2-5 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3 \sqrt{c} x^3 \left (a+b x^3\right ) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*x^3*(a + b*x^3)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]) + Sqrt[c]*(a*(b*c - a*d
)*(a + 2*b*x^3)*Sqrt[c + d*x^3] + Sqrt[b]*(4*b*c - 3*a*d)*Sqrt[b*c - a*d]*x^3*(a + b*x^3)*ArcTanh[(Sqrt[b]*Sqr
t[c + d*x^3])/Sqrt[b*c - a*d]]))/(3*a^3*Sqrt[c]*(-(b*c) + a*d)*x^3*(a + b*x^3))

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Maple [C]  time = 0.013, size = 978, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^3*b^2*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)
^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^
(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1
/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c
)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^
(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-
d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*
c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a^2*(-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))+b^2/a^2*(-1/3*(d*x^3+c)^(1/2)/b/(b*x^3+a)-1/6*I/d/b*2^(1/2)*sum(1/(a*d-b*c)*(-d^2*c)^(1/3)*(1/2*
I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(
-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(
-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^
2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-
d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*
3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^
2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))-2/a^3*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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{3} + c}}{{\left (b x^{3} + a\right )}^{2} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.80802, size = 1858, normalized size = 11.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1641, size = 258, normalized size = 1.6 \begin{align*} -\frac{1}{3} \, d^{3}{\left (\frac{2 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} b - 2 \, \sqrt{d x^{3} + c} b c + \sqrt{d x^{3} + c} a d}{{\left ({\left (d x^{3} + c\right )}^{2} b - 2 \,{\left (d x^{3} + c\right )} b c + b c^{2} +{\left (d x^{3} + c\right )} a d - a c d\right )} a^{2} d^{2}} - \frac{{\left (4 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{3} d^{3}} + \frac{{\left (4 \, b c - a d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c} d^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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