3.391 \(\int \frac{1}{x (a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a (b c-a d)^{3/2}}-\frac{2 d}{3 c \sqrt{c+d x^3} (b c-a d)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.112687, antiderivative size = 114, 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, 85, 156, 63, 208} \[ \frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a (b c-a d)^{3/2}}-\frac{2 d}{3 c \sqrt{c+d x^3} (b c-a d)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(-2*d)/(3*c*(b*c - a*d)*Sqrt[c + d*x^3]) - (2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a*c^(3/2)) + (2*b^(3/2)*Arc
Tanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*a*(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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

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

Mathematica [C]  time = 0.0291704, size = 89, normalized size = 0.78 \[ -\frac{2 \left (b c \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^3+c\right )}{b c-a d}\right )+(a d-b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3}{c}+1\right )\right )}{3 a c \sqrt{c+d x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(-2*(b*c*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3))/(b*c - a*d)] + (-(b*c) + a*d)*Hypergeometric2F1[-1/2,
1, 1/2, 1 + (d*x^3)/c]))/(3*a*c*(b*c - a*d)*Sqrt[c + d*x^3])

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Maple [C]  time = 0.022, size = 512, normalized size = 4.5 \begin{align*} -{\frac{b}{a} \left ( -{\frac{2}{3\,ad-3\,bc}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}-{\frac{{\frac{i}{3}}b\sqrt{2}}{{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{1}{ \left ( -ad+bc \right ) \left ( ad-bc \right ) }\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{b}{2\,d \left ( ad-bc \right ) } \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \right ) }+{\frac{1}{a} \left ({\frac{2}{3\,c}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}-{\frac{2}{3}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/a*(-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/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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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Fricas [B]  time = 2.10288, size = 1663, normalized size = 14.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 23.871, size = 104, normalized size = 0.91 \begin{align*} \frac{2 d}{3 c \sqrt{c + d x^{3}} \left (a d - b c\right )} + \frac{2 b \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{3 a \sqrt{\frac{a d - b c}{b}} \left (a d - b c\right )} + \frac{2 \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{- c}} \right )}}{3 a c \sqrt{- c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d/(3*c*sqrt(c + d*x**3)*(a*d - b*c)) + 2*b*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*a*sqrt((a*d - b*c)/
b)*(a*d - b*c)) + 2*atan(sqrt(c + d*x**3)/sqrt(-c))/(3*a*c*sqrt(-c))

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Giac [A]  time = 1.12172, size = 158, normalized size = 1.39 \begin{align*} -\frac{2}{3} \,{\left (\frac{b^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a b c d - a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{1}{\sqrt{d x^{3} + c}{\left (b c^{2} - a c d\right )}} - \frac{\arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a \sqrt{-c} c d}\right )} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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