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

Optimal. Leaf size=99 \[ \frac {a \sqrt {c+d x^3}}{3 b \left (a+b x^3\right ) (b c-a d)}-\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 78, 63, 208} \begin {gather*} \frac {a \sqrt {c+d x^3}}{3 b \left (a+b x^3\right ) (b c-a d)}-\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 98, normalized size = 0.99 \begin {gather*} \frac {\frac {a \sqrt {b} \sqrt {c+d x^3}}{\left (a+b x^3\right ) (b c-a d)}+\frac {(a d-2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}}{3 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 109, normalized size = 1.10 \begin {gather*} \frac {(2 b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 b^{3/2} (a d-b c)^{3/2}}+\frac {a \sqrt {c+d x^3}}{3 b \left (a+b x^3\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.11, size = 348, normalized size = 3.52 \begin {gather*} \left [\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - 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 (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3}\right )}}, \frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - 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 (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 116, normalized size = 1.17 \begin {gather*} \frac {\frac {\sqrt {d x^{3} + c} a d^{2}}{{\left (b^{2} c - a b d\right )} {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 892, normalized size = 9.01

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I/b/d^2*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=R
ootOf(_Z^3*b+a))-a/b*(1/3/(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)-1/6*I/d*2^(1/2)*sum(1/(a*d-b*c)^2*(-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/(b*x^3+a)^2/(d*x^3+c)^(1/2),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 = 6.85, size = 111, normalized size = 1.12 \begin {gather*} \frac {2\,a\,\sqrt {d\,x^3+c}}{3\,\left (b\,x^3+a\right )\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {\ln \left (\frac {2\,b\,c-a\,d+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 (a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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