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

Optimal. Leaf size=108 \[ -\frac{d}{\sqrt{c+d x^3} (b c-a d)^2}-\frac{1}{3 \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)}+\frac{\sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

[Out]

-(d/((b*c - a*d)^2*Sqrt[c + d*x^3])) - 1/(3*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) + (Sqrt[b]*d*ArcTanh[(Sqr
t[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

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Rubi [A]  time = 0.0913992, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {444, 51, 63, 208} \[ -\frac{d}{\sqrt{c+d x^3} (b c-a d)^2}-\frac{1}{3 \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)}+\frac{\sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d/((b*c - a*d)^2*Sqrt[c + d*x^3])) - 1/(3*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) + (Sqrt[b]*d*ArcTanh[(Sqr
t[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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{x^2}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac{1}{3 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{d \operatorname{Subst}\left (\int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{2 (b c-a d)}\\ &=-\frac{d}{(b c-a d)^2 \sqrt{c+d x^3}}-\frac{1}{3 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{(b d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{2 (b c-a d)^2}\\ &=-\frac{d}{(b c-a d)^2 \sqrt{c+d x^3}}-\frac{1}{3 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{(b c-a d)^2}\\ &=-\frac{d}{(b c-a d)^2 \sqrt{c+d x^3}}-\frac{1}{3 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\frac{\sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0170034, size = 54, normalized size = 0.5 \[ -\frac{2 d \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{b \left (d x^3+c\right )}{a d-b c}\right )}{3 \sqrt{c+d x^3} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.007, size = 485, normalized size = 4.5 \begin{align*} -{\frac{2\,d}{3\, \left ( ad-bc \right ) ^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}-{\frac{b}{3\, \left ( ad-bc \right ) ^{2} \left ( b{x}^{3}+a \right ) }\sqrt{d{x}^{3}+c}}+{\frac{{\frac{i}{2}}b\sqrt{2}}{d}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{1}{ \left ( ad-bc \right ) ^{3}}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*d/(a*d-b*c)^2/((x^3+1/d*c)*d)^(1/2)-1/3*b/(a*d-b*c)^2*(d*x^3+c)^(1/2)/(b*x^3+a)+1/2*I/d*b*2^(1/2)*sum(1/(
a*d-b*c)^3*(-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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.84924, size = 923, normalized size = 8.55 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{6} +{\left (b c d + a d^{2}\right )} x^{3} + a c d\right )} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{3} + a}\right ) - 2 \,{\left (3 \, b d x^{3} + b c + 2 \, a d\right )} \sqrt{d x^{3} + c}}{6 \,{\left ({\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{6} + a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{3}\right )}}, \frac{3 \,{\left (b d^{2} x^{6} +{\left (b c d + a d^{2}\right )} x^{3} + a c d\right )} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a d}}}{b d x^{3} + b c}\right ) -{\left (3 \, b d x^{3} + b c + 2 \, a d\right )} \sqrt{d x^{3} + c}}{3 \,{\left ({\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{6} + a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.10527, size = 203, normalized size = 1.88 \begin{align*} -\frac{1}{3} \, d{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{3 \,{\left (d x^{3} + c\right )} b - 2 \, b c + 2 \, a d}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} b - \sqrt{d x^{3} + c} b c + \sqrt{d x^{3} + c} a d\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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