\(\int \frac {\sqrt {c+d x^2}}{x^4 (a+b x^2)} \, dx\) [684]

Optimal result

Integrand size = 24, antiderivative size = 105 \[ \int \frac {\sqrt {c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx=-\frac {\sqrt {c+d x^2}}{3 a x^3}+\frac {(3 b c-a d) \sqrt {c+d x^2}}{3 a^2 c x}+\frac {b \sqrt {b c-a d} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 105, 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 = {486, 597, 12, 385, 211} \[ \int \frac {\sqrt {c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx=\frac {b \sqrt {b c-a d} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}}+\frac {\sqrt {c+d x^2} (3 b c-a d)}{3 a^2 c x}-\frac {\sqrt {c+d x^2}}{3 a x^3} \]

[In]

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

[Out]

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

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 486

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/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[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 x^3}+\frac {\int \frac {-3 b c+a d-2 b d x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{3 a} \\ & = -\frac {\sqrt {c+d x^2}}{3 a x^3}+\frac {(3 b c-a d) \sqrt {c+d x^2}}{3 a^2 c x}-\frac {\int -\frac {3 b c (b c-a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{3 a^2 c} \\ & = -\frac {\sqrt {c+d x^2}}{3 a x^3}+\frac {(3 b c-a d) \sqrt {c+d x^2}}{3 a^2 c x}+\frac {(b (b c-a d)) \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 x^3}+\frac {(3 b c-a d) \sqrt {c+d x^2}}{3 a^2 c x}+\frac {(b (b c-a d)) \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 x^3}+\frac {(3 b c-a d) \sqrt {c+d x^2}}{3 a^2 c x}+\frac {b \sqrt {b c-a d} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx=\frac {\sqrt {c+d x^2} \left (3 b c x^2-a \left (c+d x^2\right )\right )}{3 a^2 c x^3}-\frac {b \sqrt {b c-a d} \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{2}+c}\, \left (a d \,x^{2}-3 c b \,x^{2}+a c \right )}{3 x^{3}}-\frac {b c \left (a d -b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}}{a^{2} c}\)	\(91\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (a d \,x^{2}-3 c b \,x^{2}+a c \right )}{3 c \,a^{2} x^{3}}-\frac {\left (a d -b c \right ) b \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\right )}{a^{2}}\)	\(358\)
default	\(-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 a c \,x^{3}}-\frac {b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{c}\right )}{a^{2}}+\frac {b^{2} \left (\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}}+\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {\frac {d \sqrt {-a b}}{b}+d \left (x -\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\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}}\right )}{b}+\frac {\left (a d -b c \right ) \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 )}{b \sqrt {-\frac {a d -b c}{b}}}\right )}{2 a^{2} \sqrt {-a b}}-\frac {b^{2} \left (\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}}-\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {-\frac {d \sqrt {-a b}}{b}+d \left (x +\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\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}}\right )}{b}+\frac {\left (a d -b c \right ) \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 )}{b \sqrt {-\frac {a d -b c}{b}}}\right )}{2 a^{2} \sqrt {-a b}}\)	\(750\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 325, normalized size of antiderivative = 3.10 \[ \int \frac {\sqrt {c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx=\left [\frac {3 \, b c x^{3} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (3 \, b c - a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{12 \, a^{2} c x^{3}}, \frac {3 \, b c x^{3} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (3 \, b c - a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{6 \, a^{2} c x^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (87) = 174\).

Time = 1.06 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.05 \[ \int \frac {\sqrt {c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx=-\frac {{\left (b^{2} c \sqrt {d} - a b d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a^{2}} - \frac {2 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} + 3 \, b c^{3} \sqrt {d} - a c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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