\(\int \frac {x^5}{(a+b x^2) \sqrt {c+d x^2}} \, dx\) [703]

Optimal result

Integrand size = 24, antiderivative size = 100 \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {(b c+a d) \sqrt {c+d x^2}}{b^2 d^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2} \sqrt {b c-a d}} \]

[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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 90, 65, 214} \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2} (a d+b c)}{b^2 d^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d^2} \]

[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 65

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 90

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 214

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

Rule 457

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {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 \text {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 \text {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*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\sqrt {c+d x^2} \left (-2 b c-3 a d+b d x^2\right )}{3 b^2 d^2}+\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{5/2} \sqrt {-b c+a d}} \]

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

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {\arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) a^{2} d^{2}-\sqrt {d \,x^{2}+c}\, \left (\left (-\frac {b \,x^{2}}{3}+a \right ) d +\frac {2 b c}{3}\right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{2} d^{2}}\)	\(90\)
risch	\(-\frac {\left (-b d \,x^{2}+3 a d +2 b c \right ) \sqrt {d \,x^{2}+c}}{3 d^{2} b^{2}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b^{3} \sqrt {-\frac {a d -b c}{b}}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b^{3} \sqrt {-\frac {a d -b c}{b}}}\)	\(339\)
default	\(\frac {\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}}{b}-\frac {a \sqrt {d \,x^{2}+c}}{b^{2} d}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b^{3} \sqrt {-\frac {a d -b c}{b}}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b^{3} \sqrt {-\frac {a d -b c}{b}}}\)	\(361\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (84) = 168\).

Time = 0.29 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.90 \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\left [\frac {3 \, \sqrt {b^{2} c - a b d} a^{2} d^{2} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}, -\frac {3 \, \sqrt {-b^{2} c + a b d} a^{2} d^{2} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int \frac {x^{5}}{\left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[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 detail

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05 \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{4} - 3 \, \sqrt {d x^{2} + c} b^{2} c d^{4} - 3 \, \sqrt {d x^{2} + c} a b d^{5}}{3 \, b^{3} d^{6}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b\,d^2}-\left (\frac {2\,c}{b\,d^2}+\frac {a\,d^3-b\,c\,d^2}{b^2\,d^4}\right )\,\sqrt {d\,x^2+c}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{b^{5/2}\,\sqrt {a\,d-b\,c}} \]

[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))