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

Optimal. Leaf size=134 \[ \frac {a}{3 b \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}+\frac {a d+2 b c}{3 b \sqrt {c+d x^3} (b c-a d)^2}-\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 \sqrt {b} (b c-a d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 134, 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, 51, 63, 208} \[ \frac {a}{3 b \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}+\frac {a d+2 b c}{3 b \sqrt {c+d x^3} (b c-a d)^2}-\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 \sqrt {b} (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {a}{3 b (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}+\frac {(2 b c+a d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{6 b (b c-a d)}\\ &=\frac {2 b c+a d}{3 b (b c-a d)^2 \sqrt {c+d x^3}}+\frac {a}{3 b (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}+\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 c-a d)^2}\\ &=\frac {2 b c+a d}{3 b (b c-a d)^2 \sqrt {c+d x^3}}+\frac {a}{3 b (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}+\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 d (b c-a d)^2}\\ &=\frac {2 b c+a d}{3 b (b c-a d)^2 \sqrt {c+d x^3}}+\frac {a}{3 b (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {(2 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 \sqrt {b} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 91, normalized size = 0.68 \[ \frac {\left (a+b x^3\right ) (a d+2 b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^3+c\right )}{b c-a d}\right )+a (b c-a d)}{3 b \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(b*c - a*d) + (2*b*c + a*d)*(a + b*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3))/(b*c - a*d)])/(3*b*
(b*c - a*d)^2*(a + b*x^3)*Sqrt[c + d*x^3])

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fricas [B]  time = 1.39, size = 630, normalized size = 4.70 \[ \left [\frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{6} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\right )} x^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + 2 \, {\left (3 \, a b^{2} c^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{6} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{3}\right )}}, \frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{6} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\right )} x^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) + {\left (3 \, a b^{2} c^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{6} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 181, normalized size = 1.35 \[ \frac {\frac {{\left (2 \, b c d + a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (d x^{3} + c\right )} b c d - 2 \, b c^{2} d + {\left (d x^{3} + c\right )} a d^{2} + 2 \, a c d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{3} + c} b c + \sqrt {d x^{3} + c} a d\right )}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 958, normalized size = 7.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-2/3/(a*d-b*c)/((x^3+c/d)*d)^(1/2)-1/3*I*b/d^2*2^(1/2)*sum(1/(-a*d+b*c)/(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)))-a/b*(-1/3/(a*d-b*c)^2*(d*x^3+c)^(1/2)/(b*x
^3+a)*b-2/3/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)*d+1/2*I*b/d*2^(1/2)*sum(1/(a*d-b*c)^3*(-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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^2/(d*x^3+c)^(3/2),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 = 7.70, size = 247, normalized size = 1.84 \[ -\frac {\sqrt {d\,x^3+c}\,\left (x^3\,\left (\frac {3\,b\,d\,\left (a\,d+b\,c\right )-b\,d\,\left (a\,d+2\,b\,c\right )}{3\,\left (a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d\right )}-\frac {b\,d\,\left (a\,d+b\,c\right )}{a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d}\right )-\frac {a\,b\,c\,d}{a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d}\right )}{b\,d\,x^6+\left (a\,d+b\,c\right )\,x^3+a\,c}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (a\,d+2\,b\,c\right )\,1{}\mathrm {i}}{6\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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