3.392 \(\int \frac{1}{x^4 (a+b x^3) (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=158 \[ -\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}+\frac{(3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{5/2}}-\frac{d (b c-3 a d)}{3 a c^2 \sqrt{c+d x^3} (b c-a d)}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}} \]
[Out]
-(d*(b*c - 3*a*d))/(3*a*c^2*(b*c - a*d)*Sqrt[c + d*x^3]) - 1/(3*a*c*x^3*Sqrt[c + d*x^3]) + ((2*b*c + 3*a*d)*Ar
cTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(5/2)) - (2*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]
])/(3*a^2*(b*c - a*d)^(3/2))
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Rubi [A] time = 0.222325, antiderivative size = 158, 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, 103, 152, 156, 63, 208} \[ -\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}+\frac{(3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{5/2}}-\frac{d (b c-3 a d)}{3 a c^2 \sqrt{c+d x^3} (b c-a d)}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
[Out]
-(d*(b*c - 3*a*d))/(3*a*c^2*(b*c - a*d)*Sqrt[c + d*x^3]) - 1/(3*a*c*x^3*Sqrt[c + d*x^3]) + ((2*b*c + 3*a*d)*Ar
cTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(5/2)) - (2*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]
])/(3*a^2*(b*c - a*d)^(3/2))
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 103
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])
Rule 152
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] && IntegersQ[2*m, 2*n, 2*p]
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{1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac{1}{3 a c x^3 \sqrt{c+d x^3}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (2 b c+3 a d)+\frac{3 b d x}{2}}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{3 a c}\\ &=-\frac{d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt{c+d x^3}}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}}+\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{4} (b c-a d) (2 b c+3 a d)-\frac{1}{4} b d (b c-3 a d) x}{x (a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 a c^2 (b c-a d)}\\ &=-\frac{d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt{c+d x^3}}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 a^2 (b c-a d)}-\frac{(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 c^2}\\ &=-\frac{d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt{c+d x^3}}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}}+\frac{\left (2 b^3\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 (b c-a d)}-\frac{(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 c^2 d}\\ &=-\frac{d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt{c+d x^3}}-\frac{1}{3 a c x^3 \sqrt{c+d x^3}}+\frac{(2 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{5/2}}-\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.0453529, size = 117, normalized size = 0.74 \[ \frac{2 b^2 c^2 x^3 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^3+c\right )}{b c-a d}\right )+(a d-b c) \left (x^3 (3 a d+2 b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3}{c}+1\right )+a c\right )}{3 a^2 c^2 x^3 \sqrt{c+d x^3} (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
[Out]
(2*b^2*c^2*x^3*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3))/(b*c - a*d)] + (-(b*c) + a*d)*(a*c + (2*b*c + 3
*a*d)*x^3*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^3)/c]))/(3*a^2*c^2*(b*c - a*d)*x^3*Sqrt[c + d*x^3])
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Maple [C] time = 0.01, size = 575, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^4/(b*x^3+a)/(d*x^3+c)^(3/2),x)
[Out]
b^2/a^2*(-2/3/(a*d-b*c)/((x^3+1/d*c)*d)^(1/2)-1/3*I/d^2*b*2^(1/2)*sum(1/(-a*d+b*c)/(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)))+1/a*(-2/3*d/c^2/((x^3+1/d*c)*d
)^(1/2)-1/3*(d*x^3+c)^(1/2)/c^2/x^3+d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2))-1/a^2*b*(2/3/c/((x^3+1/d*c)*d)
^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}{\left (d x^{3} + c\right )}^{\frac{3}{2}} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^4/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^4), x)
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Fricas [B] time = 6.0326, size = 2290, normalized size = 14.49 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^4/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
[-1/6*(2*(b^2*c^3*d*x^6 + b^2*c^4*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*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*
d - 3*a^2*c*d^2)*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(a*b*c^3 - a^2*c^2*d + (a
*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3),
-1/6*(4*(b^2*c^3*d*x^6 + b^2*c^4*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)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x
^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c
*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/3*(((2*b^2*c^2
*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*s
qrt(-c)/c) + (b^2*c^3*d*x^6 + b^2*c^4*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)) + (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x
^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/3*(2*(b^2*c^3*d*x^6 + b^2*c^4*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)) + ((2*b^2*c^2
*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*s
qrt(-c)/c) + (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^
2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x**4/(b*x**3+a)/(d*x**3+c)**(3/2),x)
[Out]
Integral(1/(x**4*(a + b*x**3)*(c + d*x**3)**(3/2)), x)
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Giac [A] time = 1.16323, size = 248, normalized size = 1.57 \begin{align*} \frac{1}{3} \,{\left (\frac{2 \, b^{3} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{{\left (d x^{3} + c\right )} b c - 3 \,{\left (d x^{3} + c\right )} a d + 2 \, a c d}{{\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )}{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} - \sqrt{d x^{3} + c} c\right )}} - \frac{{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{2} d^{2}}\right )} d^{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^4/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
1/3*(2*b^3*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b*c*d^2 - a^3*d^3)*sqrt(-b^2*c + a*b*d)) - ((d
*x^3 + c)*b*c - 3*(d*x^3 + c)*a*d + 2*a*c*d)/((a*b*c^3*d - a^2*c^2*d^2)*((d*x^3 + c)^(3/2) - sqrt(d*x^3 + c)*c
)) - (2*b*c + 3*a*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^2*sqrt(-c)*c^2*d^2))*d^2