\(\int \frac {x^3}{(a+b x^2)^2 \sqrt {c+d x^2}} \, dx\) [759]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 99 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {a \sqrt {c+d x^2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} (b c-a d)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 79, 65, 214} \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {a \sqrt {c+d x^2}}{2 b \left (a+b x^2\right ) (b c-a d)}-\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} (b c-a d)^{3/2}} \]

[In]

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

[Out]

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

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 79

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] || L
tQ[p, n]))))

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}{(a+b x)^2 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {a \sqrt {c+d x^2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-a d) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b (b c-a d)} \\ & = \frac {a \sqrt {c+d x^2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-a d) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 b d (b c-a d)} \\ & = \frac {a \sqrt {c+d x^2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} (b c-a d)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {\frac {a \sqrt {b} \sqrt {c+d x^2}}{(b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}}{2 b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84

method result size
pseudoelliptic \(\frac {-\frac {a \sqrt {d \,x^{2}+c}}{b \,x^{2}+a}+\frac {\left (a d -2 b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{2 \left (a d -b c \right ) b}\) \(83\)
default \(-\frac {\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^{2} \sqrt {-\frac {a d -b c}{b}}}-\frac {\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^{2} \sqrt {-\frac {a d -b c}{b}}}+\frac {\sqrt {-a b}\, \left (\frac {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}}}{\left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {d \sqrt {-a b}\, \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 )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 b^{3}}-\frac {\sqrt {-a b}\, \left (\frac {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}}}{\left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {d \sqrt {-a b}\, \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 )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 b^{3}}\) \(816\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (83) = 166\).

Time = 0.28 (sec) , antiderivative size = 450, normalized size of antiderivative = 4.55 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\left [\frac {{\left (2 \, a b c - a^{2} d + {\left (2 \, b^{2} c - a b d\right )} x^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2}\right )}}, -\frac {{\left (2 \, a b c - a^{2} d + {\left (2 \, b^{2} c - a b d\right )} x^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\int \frac {x^{3}}{\left (a + b x^{2}\right )^{2} \sqrt {c + d x^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^3/(b*x^2+a)^2/(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 = 116, normalized size of antiderivative = 1.17 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {\frac {\sqrt {d x^{2} + c} a d^{2}}{{\left (b^{2} c - a b d\right )} {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )\,\left (a\,d-2\,b\,c\right )}{2\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {a\,d\,\sqrt {d\,x^2+c}}{2\,b\,\left (a\,d-b\,c\right )\,\left (b\,\left (d\,x^2+c\right )+a\,d-b\,c\right )} \]

[In]

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

[Out]

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