3.4.59 \(\int \frac {x^5 \sqrt {c+d x^3}}{a+b x^3} \, dx\) [359]

Optimal. Leaf size=93 \[ -\frac {2 a \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d}+\frac {2 a \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 81, 52, 65, 214} \begin {gather*} \frac {2 a \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}}-\frac {2 a \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*Sqrt[c + d*x^3])/(a + b*x^3),x]

[Out]

(-2*a*Sqrt[c + d*x^3])/(3*b^2) + (2*(c + d*x^3)^(3/2))/(9*b*d) + (2*a*Sqrt[b*c - a*d]*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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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^5 \sqrt {c+d x^3}}{a+b x^3} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x \sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )\\ &=\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d}-\frac {a \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b}\\ &=-\frac {2 a \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d}-\frac {(a (b c-a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 b^2}\\ &=-\frac {2 a \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d}-\frac {(2 a (b c-a d)) \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 a \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d}+\frac {2 a \sqrt {b c-a d} \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 = 0.14, size = 88, normalized size = 0.95 \begin {gather*} \frac {2 \sqrt {c+d x^3} \left (-3 a d+b \left (c+d x^3\right )\right )}{9 b^2 d}+\frac {2 a \sqrt {-b c+a d} \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^5*Sqrt[c + d*x^3])/(a + b*x^3),x]

[Out]

(2*Sqrt[c + d*x^3]*(-3*a*d + b*(c + d*x^3)))/(9*b^2*d) + (2*a*Sqrt[-(b*c) + a*d]*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 = 0.39, size = 458, normalized size = 4.92 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^3+c)^(1/2)/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

2/9*(d*x^3+c)^(3/2)/b/d-a/b*(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^5*(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 = 4.38, size = 195, normalized size = 2.10 \begin {gather*} \left [\frac {3 \, a d \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} + b c - 3 \, a d\right )} \sqrt {d x^{3} + c}}{9 \, b^{2} d}, \frac {2 \, {\left (3 \, a d \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} + b c - 3 \, a d\right )} \sqrt {d x^{3} + c}\right )}}{9 \, b^{2} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 7.33, size = 95, normalized size = 1.02 \begin {gather*} \frac {2 \left (- \frac {a d^{2} \sqrt {c + d x^{3}}}{3 b^{2}} + \frac {a d^{2} \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^{3} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b}\right )}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(d*x**3+c)**(1/2)/(b*x**3+a),x)

[Out]

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

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Giac [A]
time = 0.59, size = 96, normalized size = 1.03 \begin {gather*} -\frac {2 \, {\left (a b c - a^{2} 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^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 3 \, \sqrt {d x^{3} + c} a b d^{3}\right )}}{9 \, b^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.06, size = 136, normalized size = 1.46 \begin {gather*} \frac {2\,{\left (d\,x^3+c\right )}^{3/2}}{9\,b\,d}-\frac {2\,a\,\sqrt {d\,x^3+c}}{3\,b^2}+\frac {a\,\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^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(c + d*x^3)^(1/2))/(a + b*x^3),x)

[Out]

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

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