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

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 = 0.346043, 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 \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}}+\frac{2 c^2}{3 d^2 \sqrt{c+d x^3} (b c-a d)}+\frac{2 \sqrt{c+d x^3}}{3 b d^2} \]

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))

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Rubi in Sympy [A]  time = 50.2725, size = 138, normalized size = 1.29 \[ \frac{2 a^{2} \sqrt{c + d x^{3}}}{3 b \left (a d - b c\right )^{2}} - \frac{2 a^{2} \operatorname{atan}{\left (\frac{\sqrt{b} \sqrt{c + d x^{3}}}{\sqrt{a d - b c}} \right )}}{3 b^{\frac{3}{2}} \left (a d - b c\right )^{\frac{3}{2}}} - \frac{2 c^{2}}{3 d^{2} \sqrt{c + d x^{3}} \left (a d - b c\right )} - \frac{2 c \sqrt{c + d x^{3}} \left (2 a d - b c\right )}{3 d^{2} \left (a d - b c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.492404, size = 99, normalized size = 0.93 \[ \frac{1}{3} \left (\frac{2 \left (\frac{c^2}{b c-a d}+\frac{c+d x^3}{b}\right )}{d^2 \sqrt{c+d x^3}}-\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{b^{3/2} (b c-a d)^{3/2}}\right ) \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [C]  time = 0.055, size = 527, normalized size = 4.9 \[{\frac{1}{{b}^{2}} \left ( b \left ({\frac{2\,c}{3\,{d}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+{\frac{2}{3\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) +{\frac{2\,a}{3\,d}{\frac{1}{\sqrt{d{x}^{3}+c}}}} \right ) }+{\frac{{a}^{2}}{{b}^{2}} \left ( -{\frac{2}{3\,ad-3\,bc}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}-{\frac{{\frac{i}{3}}b\sqrt{2}}{{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{1}{ \left ( -ad+bc \right ) \left ( ad-bc \right ) }\sqrt [3]{-c{d}^{2}}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-c{d}^{2}}+\sqrt [3]{-c{d}^{2}} \right ) } \right ){\frac{1}{\sqrt [3]{-c{d}^{2}}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-c{d}^{2}}} \right ) \left ( -3\,\sqrt [3]{-c{d}^{2}}+i\sqrt{3}\sqrt [3]{-c{d}^{2}} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-c{d}^{2}}+\sqrt [3]{-c{d}^{2}} \right ) } \right ){\frac{1}{\sqrt [3]{-c{d}^{2}}}}}} \left ( i\sqrt [3]{-c{d}^{2}}{\it \_alpha}\,\sqrt{3}d+2\,{{\it \_alpha}}^{2}{d}^{2}-i \left ( -c{d}^{2} \right ) ^{{\frac{2}{3}}}\sqrt{3}-\sqrt [3]{-c{d}^{2}}{\it \_alpha}\,d- \left ( -c{d}^{2} \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{id\sqrt{3} \left ( x+{\frac{1}{2\,d}\sqrt [3]{-c{d}^{2}}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-c{d}^{2}}} \right ){\frac{1}{\sqrt [3]{-c{d}^{2}}}}}}},{\frac{b}{2\, \left ( ad-bc \right ) d} \left ( 2\,i{{\it \_alpha}}^{2}\sqrt [3]{-c{d}^{2}}\sqrt{3}d-i{\it \_alpha}\, \left ( -c{d}^{2} \right ) ^{{\frac{2}{3}}}\sqrt{3}+i\sqrt{3}cd-3\,{\it \_alpha}\, \left ( -c{d}^{2} \right ) ^{2/3}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-c{d}^{2}} \left ( -{\frac{3}{2\,d}\sqrt [3]{-c{d}^{2}}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-c{d}^{2}}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \right ) } \]

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+c/d)*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+c/d)*d)^(1/2)-1/3*I/d^2*b*2^(1/2)*sum(1/
(-a*d+b*c)/(a*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)
+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(
1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/
3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alp
ha*3^(1/2)*d+2*_alpha^2*d^2-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(1/3)*_alpha*d-(-c
*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(
-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/2*b/d*(2*I*_alpha^2*(-c*d^2)^(1
/3)*3^(1/2)*d-I*_alpha*(-c*d^2)^(2/3)*3^(1/2)+I*3^(1/2)*c*d-3*_alpha*(-c*d^2)^(2
/3)-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^
(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

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 [A]  time = 0.229162, size = 1, normalized size = 0.01 \[ \left [-\frac{\sqrt{d x^{3} + c} a^{2} d^{2} \log \left (\frac{{\left (b d x^{3} + 2 \, b c - a d\right )} \sqrt{b^{2} c - a b d} + 2 \, \sqrt{d x^{3} + c}{\left (b^{2} c - a b d\right )}}{b x^{3} + a}\right ) - 2 \,{\left ({\left (b c d - a d^{2}\right )} x^{3} + 2 \, b c^{2} - a c d\right )} \sqrt{b^{2} c - a b d}}{3 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} \sqrt{d x^{3} + c} \sqrt{b^{2} c - a b d}}, -\frac{2 \,{\left (\sqrt{d x^{3} + c} a^{2} d^{2} \arctan \left (-\frac{b c - a d}{\sqrt{d x^{3} + c} \sqrt{-b^{2} c + a b d}}\right ) -{\left ({\left (b c d - a d^{2}\right )} x^{3} + 2 \, b c^{2} - a c d\right )} \sqrt{-b^{2} c + a b d}\right )}}{3 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} \sqrt{d x^{3} + c} \sqrt{-b^{2} c + a b d}}\right ] \]

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

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

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/XCAS [A]  time = 0.217434, size = 139, normalized size = 1.3 \[ \frac{2 \, a^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \,{\left (b^{2} c - a b d\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \, c^{2}}{3 \,{\left (b c d^{2} - a d^{3}\right )} \sqrt{d x^{3} + c}} + \frac{2 \, \sqrt{d x^{3} + c}}{3 \, b d^{2}} \]

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)