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

Optimal. Leaf size=170 \[ -\frac{\sqrt{c+d x^3} (2 b c-a d)}{3 a^2 \left (a+b x^3\right )}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^3 \sqrt{b}}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.257946, antiderivative size = 170, 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, 98, 151, 156, 63, 208} \[ -\frac{\sqrt{c+d x^3} (2 b c-a d)}{3 a^2 \left (a+b x^3\right )}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^3 \sqrt{b}}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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{\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x^2 (a+b x)^2} \, dx,x,x^3\right )\\ &=-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} c (4 b c-3 a d)+\frac{1}{2} d (3 b c-2 a d) x}{x (a+b x)^2 \sqrt{c+d x}} \, dx,x,x^3\right )}{3 a}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} c (4 b c-3 a d) (b c-a d)+\frac{1}{2} d (b c-a d) (2 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 c-a d) \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )}-\frac{(c (4 b c-3 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{6 a^3}+\frac{((b c-a d) (4 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{6 a^3}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )}-\frac{(c (4 b c-3 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{((b c-a d) (4 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^3 d}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac{c \sqrt{c+d x^3}}{3 a x^3 \left (a+b x^3\right )}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^3 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.219075, size = 142, normalized size = 0.84 \[ \frac{\frac{a \sqrt{c+d x^3} \left (-a c+a d x^3-2 b c x^3\right )}{x^3 \left (a+b x^3\right )}+\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}}{3 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.011, size = 1093, normalized size = 6.4 \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)^(3/2)/x^4/(b*x^3+a)^2,x)

[Out]

2/a^3*b^2*(2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*d/b*c)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1
/2)*sum((-a^2*d^2+2*a*b*c*d-b^2*c^2)/(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))*E
llipticPi(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*c*(d*x^3+c)^(1/2)/x^3+2/3*d*(d*x^3+c)^(1/2)-c^(1/2)*d*arctanh((d
*x^3+c)^(1/2)/c^(1/2)))+b^2/a^2*(1/3*(a*d-b*c)/b^2*(d*x^3+c)^(1/2)/(b*x^3+a)+2/3*d*(d*x^3+c)^(1/2)/b^2+1/2*I/d
/b^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
)))-2/a^3*b*(2/9*d*x^3*(d*x^3+c)^(1/2)+8/9*c*(d*x^3+c)^(1/2)-2/3*c^(3/2)*arctanh((d*x^3+c)^(1/2)/c^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{{\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)^(3/2)/x^4/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84515, size = 1782, normalized size = 10.48 \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)^(3/2)/x^4/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.15051, size = 300, normalized size = 1.76 \begin{align*} -\frac{1}{3} \, d^{3}{\left (\frac{2 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} b c - 2 \, \sqrt{d x^{3} + c} b c^{2} -{\left (d x^{3} + c\right )}^{\frac{3}{2}} a d + 2 \, \sqrt{d x^{3} + c} a c 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^{2} - 5 \, a b c d + a^{2} d^{2}\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^{2} - 3 \, a c 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)^(3/2)/x^4/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/3*d^3*((2*(d*x^3 + c)^(3/2)*b*c - 2*sqrt(d*x^3 + c)*b*c^2 - (d*x^3 + c)^(3/2)*a*d + 2*sqrt(d*x^3 + c)*a*c*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^2 - 5*a*b*c*d +
a^2*d^2)*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^2 - 3*a*c*d)*
arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^3*sqrt(-c)*d^3))