3.4.2 \(\int \frac {x^5 \sqrt {c+d x^3}}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=136 \[ -\frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}+\frac {\sqrt {c+d x^3} (2 b c-3 a d)}{3 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \]

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Rubi [A]  time = 0.11, antiderivative size = 136, 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 = {446, 78, 50, 63, 208} \begin {gather*} \frac {\sqrt {c+d x^3} (2 b c-3 a d)}{3 b^2 (b c-a d)}-\frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}+\frac {a \left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]

[Out]

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

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 78

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

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]

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]]

Rubi steps

\begin {align*} \int \frac {x^5 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x \sqrt {c+d x}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=\frac {a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {(2 b c-3 a d) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b (b c-a d)}\\ &=\frac {(2 b c-3 a d) \sqrt {c+d x^3}}{3 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {(2 b c-3 a d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{6 b^2}\\ &=\frac {(2 b c-3 a d) \sqrt {c+d x^3}}{3 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {(2 b c-3 a d) \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^2 d}\\ &=\frac {(2 b c-3 a d) \sqrt {c+d x^3}}{3 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 117, normalized size = 0.86 \begin {gather*} \frac {\frac {(2 b c-3 a d) \left (\sqrt {b} \sqrt {c+d x^3}-\sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )\right )}{b^{3/2}}+\frac {a \left (c+d x^3\right )^{3/2}}{a+b x^3}}{3 b (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 108, normalized size = 0.79 \begin {gather*} \frac {(3 a d-2 b c) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 b^{5/2} \sqrt {a d-b c}}+\frac {\left (3 a+2 b x^3\right ) \sqrt {c+d x^3}}{3 b^2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^5*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]

[Out]

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

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fricas [A]  time = 0.63, size = 334, normalized size = 2.46 \begin {gather*} \left [-\frac {{\left ({\left (2 \, b^{2} c - 3 \, a b d\right )} x^{3} + 2 \, a b c - 3 \, a^{2} d\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \, {\left (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a b^{4} c - a^{2} b^{3} d + {\left (b^{5} c - a b^{4} d\right )} x^{3}\right )}}, \frac {{\left ({\left (2 \, b^{2} c - 3 \, a b d\right )} x^{3} + 2 \, a b c - 3 \, a^{2} d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) + {\left (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a b^{4} c - a^{2} b^{3} d + {\left (b^{5} c - a b^{4} d\right )} x^{3}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[-1/6*(((2*b^2*c - 3*a*b*d)*x^3 + 2*a*b*c - 3*a^2*d)*sqrt(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d
*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3 + a)) - 2*(3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c - a*b^2*d)*x^3)*sqrt(d*x^3 +
c))/(a*b^4*c - a^2*b^3*d + (b^5*c - a*b^4*d)*x^3), 1/3*(((2*b^2*c - 3*a*b*d)*x^3 + 2*a*b*c - 3*a^2*d)*sqrt(-b
^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) + (3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c
- a*b^2*d)*x^3)*sqrt(d*x^3 + c))/(a*b^4*c - a^2*b^3*d + (b^5*c - a*b^4*d)*x^3)]

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giac [A]  time = 0.17, size = 102, normalized size = 0.75 \begin {gather*} \frac {\sqrt {d x^{3} + c} a d}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{2}} + \frac {{\left (2 \, b c - 3 \, a 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 \, \sqrt {d x^{3} + c}}{3 \, b^{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)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.26, size = 897, normalized size = 6.60

result 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)^2,x)

[Out]

1/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*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d
^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1
/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alph
a^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*Ell
ipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)
,1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/
3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(
1/2)),_alpha=RootOf(_Z^3*b+a)))-a/b*(-1/3*(d*x^3+c)^(1/2)/(b*x^3+a)/b-1/6*I/d/b*2^(1/2)*sum(1/(a*d-b*c)*(-c*d^
2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/
d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)
)/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*
_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1
/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d
-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*
(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(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)^2,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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 6.09, size = 152, normalized size = 1.12 \begin {gather*} \frac {2\,\sqrt {d\,x^3+c}}{3\,b^2}-\frac {a\,\left (\frac {2\,a\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {2\,b\,c}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )\,\sqrt {d\,x^3+c}}{b\,\left (b\,x^3+a\right )}+\frac {\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (3\,a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{5/2}\,\sqrt {a\,d-b\,c}} \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)^2,x)

[Out]

(2*(c + d*x^3)^(1/2))/(3*b^2) + (log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/
(a + b*x^3))*(3*a*d - 2*b*c)*1i)/(6*b^(5/2)*(a*d - b*c)^(1/2)) - (a*((2*a*d)/(3*(2*b^2*c - 2*a*b*d)) - (2*b*c)
/(3*(2*b^2*c - 2*a*b*d)))*(c + d*x^3)^(1/2))/(b*(a + b*x^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**2,x)

[Out]

Timed out

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