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

Optimal. Leaf size=116 \[ -\frac{2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 \sqrt{b}}+\frac{\sqrt{c} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2}-\frac{c \sqrt{c+d x^3}}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.145402, antiderivative size = 116, normalized size of antiderivative = 1., 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, 98, 156, 63, 208} \[ -\frac{2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 \sqrt{b}}+\frac{\sqrt{c} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2}-\frac{c \sqrt{c+d x^3}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.106637, size = 108, normalized size = 0.93 \[ \frac{-\frac{2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}+\sqrt{c} (2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )-\frac{a c \sqrt{c+d x^3}}{x^3}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.01, size = 620, normalized size = 5.3 \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),x)

[Out]

b^2/a^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))*Ell
ipticPi(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*(-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)))-1/a^2*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 )} 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.14663, size = 1220, normalized size = 10.52 \begin{align*} \left [-\frac{2 \,{\left (b c - a d\right )} x^{3} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right ) +{\left (2 \, b c - 3 \, 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} x^{3}}, -\frac{4 \,{\left (b c - a d\right )} x^{3} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (2 \, b c - 3 \, 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} x^{3}}, -\frac{{\left (2 \, b c - 3 \, a d\right )} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) +{\left (b c - a d\right )} x^{3} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right ) + \sqrt{d x^{3} + c} a c}{3 \, a^{2} x^{3}}, -\frac{2 \,{\left (b c - a d\right )} x^{3} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (2 \, b c - 3 \, 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} x^{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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.12009, size = 182, normalized size = 1.57 \begin{align*} \frac{1}{3} \, d^{2}{\left (\frac{2 \,{\left (b^{2} c^{2} - 2 \, 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^{2} d^{2}} - \frac{{\left (2 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} d^{2}} - \frac{\sqrt{d x^{3} + c} c}{a d^{2} x^{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),x, algorithm="giac")

[Out]

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