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

Optimal. Leaf size=88 \[ \frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 b d}-\frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^3}}{\sqrt {b} \sqrt {c+d x^3}}\right )}{3 b^{3/2} d^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {446, 80, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x^3} \sqrt {c+d x^3}}{3 b d}-\frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^3}}{\sqrt {b} \sqrt {c+d x^3}}\right )}{3 b^{3/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*Sqrt[c + d*x^3])/(3*b*d) - ((b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^3])/(Sqrt[b]*Sqrt[c + d
*x^3])])/(3*b^(3/2)*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 80

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

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Mathematica [A]  time = 0.27, size = 123, normalized size = 1.40 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x^3} \left (c+d x^3\right )-\sqrt {b c-a d} (a d+b c) \sqrt {\frac {b \left (c+d x^3\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^3}}{\sqrt {b c-a d}}\right )}{3 b^2 d^{3/2} \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.30, size = 122, normalized size = 1.39 \begin {gather*} \frac {(-a d-b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {d} \sqrt {a+b x^3}}\right )}{3 b^{3/2} d^{3/2}}-\frac {\sqrt {c+d x^3} (a d-b c)}{3 b d \sqrt {a+b x^3} \left (\frac {b \left (c+d x^3\right )}{a+b x^3}-d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 256, normalized size = 2.91 \begin {gather*} \left [\frac {4 \, \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} b d + {\left (b c + a d\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{6} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{3} - 4 \, {\left (2 \, b d x^{3} + b c + a d\right )} \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} \sqrt {b d}\right )}{12 \, b^{2} d^{2}}, \frac {2 \, \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} b d + {\left (b c + a d\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{3} + b c + a d\right )} \sqrt {b x^{3} + a} \sqrt {d x^{3} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{6} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{3}\right )}}\right )}{6 \, b^{2} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*sqrt(b*x^3 + a)*sqrt(d*x^3 + c)*b*d + (b*c + a*d)*sqrt(b*d)*log(8*b^2*d^2*x^6 + b^2*c^2 + 6*a*b*c*d +
a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^3 - 4*(2*b*d*x^3 + b*c + a*d)*sqrt(b*x^3 + a)*sqrt(d*x^3 + c)*sqrt(b*d)))/(
b^2*d^2), 1/6*(2*sqrt(b*x^3 + a)*sqrt(d*x^3 + c)*b*d + (b*c + a*d)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^3 + b*c + a*
d)*sqrt(b*x^3 + a)*sqrt(d*x^3 + c)*sqrt(-b*d)/(b^2*d^2*x^6 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^3)))/(b^2*d^2)]

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giac [A]  time = 0.21, size = 104, normalized size = 1.18 \begin {gather*} \frac {\frac {{\left (b c + a d\right )} \log \left ({\left | -\sqrt {b x^{3} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d} + \frac {\sqrt {b x^{3} + a} \sqrt {b^{2} c + {\left (b x^{3} + a\right )} b d - a b d}}{b d}}{3 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\sqrt {b \,x^{3}+a}\, \sqrt {d \,x^{3}+c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(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 zero or nonzero?

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mupad [B]  time = 9.25, size = 283, normalized size = 3.22 \begin {gather*} \frac {\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,\left (\frac {2\,a\,d}{3}+\frac {2\,b\,c}{3}\right )}{d^3\,\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}+\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\frac {2\,a\,d}{3}+\frac {2\,b\,c}{3}\right )}{b\,d^2\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3}-\frac {8\,\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^2}{3\,d^2\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^4}{{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^4}+\frac {b^2}{d^2}-\frac {2\,b\,{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^2}{d\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^2}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}\right )\,\left (a\,d+b\,c\right )}{3\,b^{3/2}\,d^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x^3)^(1/2) - a^(1/2))*((2*a*d)/3 + (2*b*c)/3))/(d^3*((c + d*x^3)^(1/2) - c^(1/2))) + (((a + b*x^3)^(
1/2) - a^(1/2))^3*((2*a*d)/3 + (2*b*c)/3))/(b*d^2*((c + d*x^3)^(1/2) - c^(1/2))^3) - (8*a^(1/2)*c^(1/2)*((a +
b*x^3)^(1/2) - a^(1/2))^2)/(3*d^2*((c + d*x^3)^(1/2) - c^(1/2))^2))/(((a + b*x^3)^(1/2) - a^(1/2))^4/((c + d*x
^3)^(1/2) - c^(1/2))^4 + b^2/d^2 - (2*b*((a + b*x^3)^(1/2) - a^(1/2))^2)/(d*((c + d*x^3)^(1/2) - c^(1/2))^2))
- (2*atanh((d^(1/2)*((a + b*x^3)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x^3)^(1/2) - c^(1/2))))*(a*d + b*c))/(3*b^
(3/2)*d^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\sqrt {a + b x^{3}} \sqrt {c + d x^{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/(sqrt(a + b*x**3)*sqrt(c + d*x**3)), x)

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