\(\int \frac {1}{x^4 (a+b x^2) \sqrt {c+d x^2}} \, dx\) [712]

Optimal result

Integrand size = 24, antiderivative size = 110 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {(3 b c+2 a d) \sqrt {c+d x^2}}{3 a^2 c^2 x}+\frac {b^2 \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {491, 597, 12, 385, 211} \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {b^2 \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} \sqrt {b c-a d}}+\frac {\sqrt {c+d x^2} (2 a d+3 b c)}{3 a^2 c^2 x}-\frac {\sqrt {c+d x^2}}{3 a c x^3} \]

[In]

Int[1/(x^4*(a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

-1/3*Sqrt[c + d*x^2]/(a*c*x^3) + ((3*b*c + 2*a*d)*Sqrt[c + d*x^2])/(3*a^2*c^2*x) + (b^2*ArcTan[(Sqrt[b*c - a*d
]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(5/2)*Sqrt[b*c - a*d])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 491

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {\int \frac {-3 b c-2 a d-2 b d x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{3 a c} \\ & = -\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {(3 b c+2 a d) \sqrt {c+d x^2}}{3 a^2 c^2 x}-\frac {\int -\frac {3 b^2 c^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{3 a^2 c^2} \\ & = -\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {(3 b c+2 a d) \sqrt {c+d x^2}}{3 a^2 c^2 x}+\frac {b^2 \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{a^2} \\ & = -\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {(3 b c+2 a d) \sqrt {c+d x^2}}{3 a^2 c^2 x}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{a^2} \\ & = -\frac {\sqrt {c+d x^2}}{3 a c x^3}+\frac {(3 b c+2 a d) \sqrt {c+d x^2}}{3 a^2 c^2 x}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} \sqrt {b c-a d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\sqrt {c+d x^2} \left (-a c+3 b c x^2+2 a d x^2\right )}{3 a^2 c^2 x^3}-\frac {b^2 \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{5/2} \sqrt {b c-a d}} \]

[In]

Integrate[1/(x^4*(a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

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

Maple [A] (verified)

Time = 3.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{2}+c}\, \left (-2 a d \,x^{2}-3 c b \,x^{2}+a c \right )}{3 x^{3}}+\frac {b^{2} c^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}}{a^{2} c^{2}}\)	\(87\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-2 a d \,x^{2}-3 c b \,x^{2}+a c \right )}{3 c^{2} a^{2} x^{3}}+\frac {b^{2} \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\right )}{a^{2}}\)	\(352\)
default	\(\frac {-\frac {\sqrt {d \,x^{2}+c}}{3 c \,x^{3}}+\frac {2 d \sqrt {d \,x^{2}+c}}{3 c^{2} x}}{a}+\frac {b \sqrt {d \,x^{2}+c}}{a^{2} c x}-\frac {b^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 a^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {b^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 a^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\)	\(378\)

[In]

int(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (92) = 184\).

Time = 0.31 (sec) , antiderivative size = 414, normalized size of antiderivative = 3.76 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\left [-\frac {3 \, \sqrt {-a b c + a^{2} d} b^{2} c^{2} x^{3} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a^{2} b c^{2} - a^{3} c d - {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{3}}, \frac {3 \, \sqrt {a b c - a^{2} d} b^{2} c^{2} x^{3} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (a^{2} b c^{2} - a^{3} c d - {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{3}}\right ] \]

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}\, dx \]

[In]

integrate(1/x**4/(b*x**2+a)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/(x**4*(a + b*x**2)*sqrt(c + d*x**2)), x)

Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*sqrt(d*x^2 + c)*x^4), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (92) = 184\).

Time = 1.02 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.77 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {1}{3} \, d^{\frac {5}{2}} {\left (\frac {3 \, b^{2} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a^{2} d^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + 3 \, b c^{2} + 2 \, a c d\right )}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2} d^{2}}\right )} \]

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int \frac {1}{x^4\,\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}} \,d x \]

[In]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(1/2)),x)

[Out]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(1/2)), x)