3.388 \(\int \frac{x^8}{(a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}}+\frac{2 c^2}{3 d^2 \sqrt{c+d x^3} (b c-a d)}+\frac{2 \sqrt{c+d x^3}}{3 b d^2} \]

[Out]

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

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Rubi [A]  time = 0.120525, antiderivative size = 107, 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 = {446, 87, 63, 208} \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}}+\frac{2 c^2}{3 d^2 \sqrt{c+d x^3} (b c-a d)}+\frac{2 \sqrt{c+d x^3}}{3 b d^2} \]

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*c^2)/(3*d^2*(b*c - a*d)*Sqrt[c + d*x^3]) + (2*Sqrt[c + d*x^3])/(3*b*d^2) - (2*a^2*ArcTanh[(Sqrt[b]*Sqrt[c +
d*x^3])/Sqrt[b*c - a*d]])/(3*b^(3/2)*(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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{c^2}{d (-b c+a d) (c+d x)^{3/2}}+\frac{1}{b d \sqrt{c+d x}}+\frac{a^2}{b (b c-a d) (a+b x) \sqrt{c+d x}}\right ) \, dx,x,x^3\right )\\ &=\frac{2 c^2}{3 d^2 (b c-a d) \sqrt{c+d x^3}}+\frac{2 \sqrt{c+d x^3}}{3 b d^2}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 b (b c-a d)}\\ &=\frac{2 c^2}{3 d^2 (b c-a d) \sqrt{c+d x^3}}+\frac{2 \sqrt{c+d x^3}}{3 b d^2}+\frac{\left (2 a^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 b d (b c-a d)}\\ &=\frac{2 c^2}{3 d^2 (b c-a d) \sqrt{c+d x^3}}+\frac{2 \sqrt{c+d x^3}}{3 b d^2}-\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0687899, size = 100, normalized size = 0.93 \[ \frac{2 \left (-a^2 d^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^3+c\right )}{b c-a d}\right )+a^2 d^2+a b d \left (c+d x^3\right )+b^2 (-c) \left (2 c+d x^3\right )\right )}{3 b^2 d^2 \sqrt{c+d x^3} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

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

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Maple [C]  time = 0.037, size = 527, normalized size = 4.9 \begin{align*}{\frac{1}{{b}^{2}} \left ( b \left ({\frac{2\,c}{3\,{d}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+{\frac{2}{3\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) +{\frac{2\,a}{3\,d}{\frac{1}{\sqrt{d{x}^{3}+c}}}} \right ) }+{\frac{{a}^{2}}{{b}^{2}} \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 \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{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{{id\sqrt{3} \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\, \left ( ad-bc \right ) d} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x)

[Out]

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

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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^8/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.46543, size = 896, normalized size = 8.37 \begin{align*} \left [-\frac{{\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt{b^{2} c - a b d} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c} \sqrt{b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \,{\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{3 \,{\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} +{\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{3}\right )}}, \frac{2 \,{\left ({\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt{-b^{2} c + a b d} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-b^{2} c + a b d}}{b d x^{3} + b c}\right ) +{\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}\right )}}{3 \,{\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} +{\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/3*((a^2*d^3*x^3 + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*sqrt(b^2*c
- a*b*d))/(b*x^3 + a)) - 2*(2*b^3*c^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)
*x^3)*sqrt(d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2
*d^5)*x^3), 2/3*((a^2*d^3*x^3 + a^2*c*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b
*d*x^3 + b*c)) + (2*b^3*c^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^3)*sqrt(
d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{\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(x**8/(b*x**3+a)/(d*x**3+c)**(3/2),x)

[Out]

Integral(x**8/((a + b*x**3)*(c + d*x**3)**(3/2)), x)

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Giac [A]  time = 1.13845, size = 139, normalized size = 1.3 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \,{\left (b^{2} c - a b d\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \, c^{2}}{3 \,{\left (b c d^{2} - a d^{3}\right )} \sqrt{d x^{3} + c}} + \frac{2 \, \sqrt{d x^{3} + c}}{3 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

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