3.4.58 \(\int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx\) [358]

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

[Out]

-2/9*(a*d+b*c)*(d*x^3+c)^(3/2)/b^2/d^2+2/15*(d*x^3+c)^(5/2)/b/d^2-2/3*a^2*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*
d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(7/2)+2/3*a^2*(d*x^3+c)^(1/2)/b^3

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Rubi [A]
time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 = {457, 90, 52, 65, 214} \begin {gather*} -\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 a^2 \sqrt {c+d x^3}}{3 b^3}-\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} \end {gather*}

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 52

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 65

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 90

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

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]]

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 121, normalized size = 0.97 \begin {gather*} \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} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2}} \end {gather*}

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]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(3*b^(7/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.38, size = 514, normalized size = 4.11 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/b*(2/15*x^6*(d*x^3+c)^(1/2)+2/45*c/d*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)-2/9*a/b^2*(d*x^3+c)^(
3/2)/d+a^2/b^2*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-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)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-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*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d
-3*(-c*d^2)^(2/3)*_alpha-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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [A]
time = 3.23, size = 280, normalized size = 2.24 \begin {gather*} \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 {gather*}

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 = 12.26, size = 128, normalized size = 1.02 \begin {gather*} \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 {gather*}

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 = 0.52, size = 139, normalized size = 1.11 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 6.17, size = 176, normalized size = 1.41 \begin {gather*} \frac {2\,a^2\,\sqrt {d\,x^3+c}}{3\,b^3}+\frac {2\,{\left (d\,x^3+c\right )}^{5/2}}{15\,b\,d^2}-\frac {2\,a\,{\left (d\,x^3+c\right )}^{3/2}}{9\,b^2\,d}-\frac {2\,c\,{\left (d\,x^3+c\right )}^{3/2}}{9\,b\,d^2}+\frac {a^2\,\ln \left (\frac {a^2\,d^2\,1{}\mathrm {i}+b^2\,c^2\,2{}\mathrm {i}-2\,\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}-a\,b\,d^2\,x^3\,1{}\mathrm {i}+b^2\,c\,d\,x^3\,1{}\mathrm {i}-a\,b\,c\,d\,3{}\mathrm {i}}{2\,b\,x^3+2\,a}\right )\,\sqrt {a\,d-b\,c}\,1{}\mathrm {i}}{3\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*(c + d*x^3)^(1/2))/(3*b^3) + (2*(c + d*x^3)^(5/2))/(15*b*d^2) - (2*a*(c + d*x^3)^(3/2))/(9*b^2*d) - (2*
c*(c + d*x^3)^(3/2))/(9*b*d^2) + (a^2*log((a^2*d^2*1i + b^2*c^2*2i - 2*b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(
3/2) - a*b*d^2*x^3*1i + b^2*c*d*x^3*1i - a*b*c*d*3i)/(2*a + 2*b*x^3))*(a*d - b*c)^(1/2)*1i)/(3*b^(7/2))

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