3.358 \(\int \frac{x^8 \sqrt{c+d x^3}}{a+b x^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 a^2 \sqrt{c+d x^3}}{3 b^3}-\frac{2 a^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}-\frac{2 \left (c+d x^3\right )^{3/2} (a d+b c)}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2} \]

[Out]

(2*a^2*Sqrt[c + d*x^3])/(3*b^3) - (2*(b*c + a*d)*(c + d*x^3)^(3/2))/(9*b^2*d^2) + (2*(c + d*x^3)^(5/2))/(15*b*
d^2) - (2*a^2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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Rubi [A]  time = 0.129485, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 88, 50, 63, 208} \[ \frac{2 a^2 \sqrt{c+d x^3}}{3 b^3}-\frac{2 a^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}-\frac{2 \left (c+d x^3\right )^{3/2} (a d+b c)}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[c + d*x^3])/(3*b^3) - (2*(b*c + a*d)*(c + d*x^3)^(3/2))/(9*b^2*d^2) + (2*(c + d*x^3)^(5/2))/(15*b*
d^2) - (2*a^2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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^8 \sqrt{c+d x^3}}{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{(-b c-a d) \sqrt{c+d x}}{b^2 d}+\frac{a^2 \sqrt{c+d x}}{b^2 (a+b x)}+\frac{(c+d x)^{3/2}}{b d}\right ) \, dx,x,x^3\right )\\ &=-\frac{2 (b c+a d) \left (c+d x^3\right )^{3/2}}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2}+\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b^2}\\ &=\frac{2 a^2 \sqrt{c+d x^3}}{3 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{3/2}}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2}+\frac{\left (a^2 (b c-a d)\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 b^3}\\ &=\frac{2 a^2 \sqrt{c+d x^3}}{3 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{3/2}}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2}+\frac{\left (2 a^2 (b c-a d)\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^3 d}\\ &=\frac{2 a^2 \sqrt{c+d x^3}}{3 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{3/2}}{9 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d^2}-\frac{2 a^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.265231, size = 121, normalized size = 0.97 \[ \frac{2 \sqrt{c+d x^3} \left (15 a^2 d^2-5 a b d \left (c+d x^3\right )+b^2 \left (-2 c^2+c d x^3+3 d^2 x^6\right )\right )}{45 b^3 d^2}-\frac{2 a^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x^3]*(15*a^2*d^2 - 5*a*b*d*(c + d*x^3) + b^2*(-2*c^2 + c*d*x^3 + 3*d^2*x^6)))/(45*b^3*d^2) - (2*
a^2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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Maple [C]  time = 0.054, size = 514, normalized size = 4.1 \begin{align*}{\frac{1}{{b}^{2}} \left ( b \left ({\frac{2\,{x}^{6}}{15}\sqrt{d{x}^{3}+c}}+{\frac{2\,c{x}^{3}}{45\,d}\sqrt{d{x}^{3}+c}}-{\frac{4\,{c}^{2}}{45\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) -{\frac{2\,a}{9\,d} \left ( d{x}^{3}+c \right ) ^{{\frac{3}{2}}}} \right ) }+{\frac{{a}^{2}}{{b}^{2}} \left ({\frac{2}{3\,b}\sqrt{d{x}^{3}+c}}+{\frac{{\frac{i}{3}}\sqrt{2}}{b{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \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{{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*(d*x^3+c)^(1/2)/(b*x^3+a),x)

[Out]

1/b^2*(b*(2/15*x^6*(d*x^3+c)^(1/2)+2/45/d*c*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)-2/9*a/d*(d*x^3+c
)^(3/2))+a^2/b^2*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-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*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36972, size = 603, normalized size = 4.82 \begin{align*} \left [\frac{15 \, a^{2} d^{2} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right ) + 2 \,{\left (3 \, b^{2} d^{2} x^{6} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} +{\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{45 \, b^{3} d^{2}}, -\frac{2 \,{\left (15 \, a^{2} d^{2} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) -{\left (3 \, b^{2} d^{2} x^{6} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} +{\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, b^{3} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="fricas")

[Out]

[1/45*(15*a^2*d^2*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b
*x^3 + a)) + 2*(3*b^2*d^2*x^6 - 2*b^2*c^2 - 5*a*b*c*d + 15*a^2*d^2 + (b^2*c*d - 5*a*b*d^2)*x^3)*sqrt(d*x^3 + c
))/(b^3*d^2), -2/45*(15*a^2*d^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d
)) - (3*b^2*d^2*x^6 - 2*b^2*c^2 - 5*a*b*c*d + 15*a^2*d^2 + (b^2*c*d - 5*a*b*d^2)*x^3)*sqrt(d*x^3 + c))/(b^3*d^
2)]

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Sympy [A]  time = 50.9361, size = 128, normalized size = 1.02 \begin{align*} \frac{2 \left (\frac{a^{2} d^{3} \sqrt{c + d x^{3}}}{3 b^{3}} - \frac{a^{2} d^{3} \left (a d - b c\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{3 b^{4} \sqrt{\frac{a d - b c}{b}}} + \frac{d \left (c + d x^{3}\right )^{\frac{5}{2}}}{15 b} + \frac{\left (c + d x^{3}\right )^{\frac{3}{2}} \left (- a d^{2} - b c d\right )}{9 b^{2}}\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(d*x**3+c)**(1/2)/(b*x**3+a),x)

[Out]

2*(a**2*d**3*sqrt(c + d*x**3)/(3*b**3) - a**2*d**3*(a*d - b*c)*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b
**4*sqrt((a*d - b*c)/b)) + d*(c + d*x**3)**(5/2)/(15*b) + (c + d*x**3)**(3/2)*(-a*d**2 - b*c*d)/(9*b**2))/d**3

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Giac [A]  time = 1.13744, size = 188, normalized size = 1.5 \begin{align*} \frac{2 \,{\left (a^{2} b c - a^{3} d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \, \sqrt{-b^{2} c + a b d} b^{3}} + \frac{2 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} b^{4} d^{8} - 5 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} b^{4} c d^{8} - 5 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} a b^{3} d^{9} + 15 \, \sqrt{d x^{3} + c} a^{2} b^{2} d^{10}\right )}}{45 \, b^{5} d^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="giac")

[Out]

2/3*(a^2*b*c - a^3*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 2/45*(3*(d*x
^3 + c)^(5/2)*b^4*d^8 - 5*(d*x^3 + c)^(3/2)*b^4*c*d^8 - 5*(d*x^3 + c)^(3/2)*a*b^3*d^9 + 15*sqrt(d*x^3 + c)*a^2
*b^2*d^10)/(b^5*d^10)