3.4.70 \(\int \frac {x^2 (c+d x^3)^{3/2}}{a+b x^3} \, dx\) [370]

Optimal. Leaf size=96 \[ \frac {2 (b c-a d) \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}-\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 = {455, 52, 65, 214} \begin {gather*} -\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}}+\frac {2 \sqrt {c+d x^3} (b c-a d)}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 455

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]

Rubi steps

\begin {align*} \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )\\ &=\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}+\frac {(b c-a d) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b}\\ &=\frac {2 (b c-a d) \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}+\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 b^2}\\ &=\frac {2 (b c-a d) \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}+\frac {\left (2 (b c-a d)^2\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^2 d}\\ &=\frac {2 (b c-a d) \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}-\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 85, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {c+d x^3} \left (4 b c-3 a d+b d x^3\right )}{9 b^2}+\frac {2 (-b c+a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x^3]*(4*b*c - 3*a*d + b*d*x^3))/(9*b^2) + (2*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3
])/Sqrt[-(b*c) + a*d]])/(3*b^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.33, size = 507, normalized size = 5.28 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*c*d/b)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1/2)*sum((-a
^2*d^2+2*a*b*c*d-b^2*c^2)/(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)
*_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)),_al
pha=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^2*(d*x^3+c)^(3/2)/(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 = 3.03, size = 204, normalized size = 2.12 \begin {gather*} \left [-\frac {3 \, {\left (b c - a d\right )} \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 (b d x^{3} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{3} + c}}{9 \, b^{2}}, -\frac {2 \, {\left (3 \, {\left (b c - a d\right )} \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 (b d x^{3} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{3} + c}\right )}}{9 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/9*(3*(b*c - a*d)*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*(b*d*x^3 + 4*b*c - 3*a*d)*sqrt(d*x^3 + c))/b^2, -2/9*(3*(b*c - a*d)*sqrt(-(b*c - a*d)/b)*arc
tan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (b*d*x^3 + 4*b*c - 3*a*d)*sqrt(d*x^3 + c))/b^2]

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Sympy [A]
time = 14.37, size = 90, normalized size = 0.94 \begin {gather*} \frac {2 \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b} + \frac {\sqrt {c + d x^{3}} \left (- 2 a d + 2 b c\right )}{3 b^{2}} + \frac {2 \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{3} \sqrt {\frac {a d - b c}{b}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.33, size = 113, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x^{3} + c} b^{2} c - 3 \, \sqrt {d x^{3} + c} a b d\right )}}{9 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2)
+ 2/9*((d*x^3 + c)^(3/2)*b^2 + 3*sqrt(d*x^3 + c)*b^2*c - 3*sqrt(d*x^3 + c)*a*b*d)/b^3

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Mupad [B]
time = 5.91, size = 143, normalized size = 1.49 \begin {gather*} \frac {2\,d\,x^3\,\sqrt {d\,x^3+c}}{9\,b}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {8\,c\,d}{3\,b}\right )}{3\,d}+\frac {\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d-\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((a^2*d^2 + 2*b^2*c^2 - b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/2)*2i - a*b*d^2*x^3 + b^2*c*d*x^3 - 3*a*b
*c*d)/(a + b*x^3))*(a*d - b*c)^(3/2)*1i)/(3*b^(5/2)) - ((c + d*x^3)^(1/2)*((2*a*d^2)/b^2 - (8*c*d)/(3*b)))/(3*
d) + (2*d*x^3*(c + d*x^3)^(1/2))/(9*b)

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