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

Optimal. Leaf size=100 \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2} (a d+b c)}{b^2 d^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 88, 63, 208} \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2} (a d+b c)}{b^2 d^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

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Mathematica [A]  time = 0.13, size = 89, normalized size = 0.89 \[ \frac {\sqrt {c+d x^2} \left (-3 a d-2 b c+b d x^2\right )}{3 b^2 d^2}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2} \sqrt {b c-a d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.33, size = 390, normalized size = 3.90 \[ \left [\frac {3 \, \sqrt {b^{2} c - a b d} a^{2} d^{2} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}, -\frac {3 \, \sqrt {-b^{2} c + a b d} a^{2} d^{2} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 105, normalized size = 1.05 \[ \frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{4} - 3 \, \sqrt {d x^{2} + c} b^{2} c d^{4} - 3 \, \sqrt {d x^{2} + c} a b d^{5}}{3 \, b^{3} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 362, normalized size = 3.62 \[ -\frac {a^{2} \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{3}}+\frac {\sqrt {d \,x^{2}+c}\, x^{2}}{3 b d}-\frac {\sqrt {d \,x^{2}+c}\, a}{b^{2} d}-\frac {2 \sqrt {d \,x^{2}+c}\, c}{3 b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2+a)/(d*x^2+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 = 0.69, size = 100, normalized size = 1.00 \[ \frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b\,d^2}-\left (\frac {2\,c}{b\,d^2}+\frac {a\,d^3-b\,c\,d^2}{b^2\,d^4}\right )\,\sqrt {d\,x^2+c}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{b^{5/2}\,\sqrt {a\,d-b\,c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(3/2)/(3*b*d^2) - ((2*c)/(b*d^2) + (a*d^3 - b*c*d^2)/(b^2*d^4))*(c + d*x^2)^(1/2) + (a^2*atan((b^(
1/2)*(c + d*x^2)^(1/2))/(a*d - b*c)^(1/2)))/(b^(5/2)*(a*d - b*c)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/((a + b*x**2)*sqrt(c + d*x**2)), x)

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