\(\int \frac {(a+b x^2)^2}{x^5 \sqrt {c+d x^2}} \, dx\) [645]

Optimal result

Integrand size = 24, antiderivative size = 106 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}-\frac {a (8 b c-3 a d) \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{5/2}} \]

[Out]

-1/8*(3*a^2*d^2-8*a*b*c*d+8*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(5/2)-1/4*a^2*(d*x^2+c)^(1/2)/c/x^4-1/
8*a*(-3*a*d+8*b*c)*(d*x^2+c)^(1/2)/c^2/x^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 106, 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 = {457, 91, 79, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {\left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{5/2}}-\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}-\frac {a \sqrt {c+d x^2} (8 b c-3 a d)}{8 c^2 x^2} \]

[In]

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

[Out]

-1/4*(a^2*Sqrt[c + d*x^2])/(c*x^4) - (a*(8*b*c - 3*a*d)*Sqrt[c + d*x^2])/(8*c^2*x^2) - ((8*b^2*c^2 - 8*a*b*c*d
+ 3*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^(5/2))

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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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 {(a+b x)^2}{x^3 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (8 b c-3 a d)+2 b^2 c x}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}-\frac {a (8 b c-3 a d) \sqrt {c+d x^2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2-\frac {a d (8 b c-3 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}-\frac {a (8 b c-3 a d) \sqrt {c+d x^2}}{8 c^2 x^2}+\frac {\left (8 b^2-\frac {a d (8 b c-3 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 d} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{4 c x^4}-\frac {a (8 b c-3 a d) \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {a \sqrt {c+d x^2} \left (2 a c+8 b c x^2-3 a d x^2\right )}{8 c^2 x^4}+\frac {\left (-8 b^2 c^2+8 a b c d-3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{5/2}} \]

[In]

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

[Out]

-1/8*(a*Sqrt[c + d*x^2]*(2*a*c + 8*b*c*x^2 - 3*a*d*x^2))/(c^2*x^4) + ((-8*b^2*c^2 + 8*a*b*c*d - 3*a^2*d^2)*Arc
Tanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^(5/2))

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {-\frac {3 x^{4} \left (a^{2} d^{2}-\frac {8}{3} a b c d +\frac {8}{3} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{8}+\frac {3 \left (\frac {2 \left (-4 b \,x^{2}-a \right ) c^{\frac {3}{2}}}{3}+a \sqrt {c}\, d \,x^{2}\right ) \sqrt {d \,x^{2}+c}\, a}{8}}{c^{\frac {5}{2}} x^{4}}\)	\(87\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (-3 a d \,x^{2}+8 c b \,x^{2}+2 a c \right )}{8 c^{2} x^{4}}-\frac {\left (3 a^{2} d^{2}-8 a b c d +8 b^{2} c^{2}\right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{8 c^{\frac {5}{2}}}\)	\(90\)
default	\(-\frac {b^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}+a^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{4 c \,x^{4}}-\frac {3 d \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+2 a b \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(159\)

[In]

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

[Out]

3/8*(-x^4*(a^2*d^2-8/3*a*b*c*d+8/3*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))+(2/3*(-4*b*x^2-a)*c^(3/2)+a*c^(1/
2)*d*x^2)*(d*x^2+c)^(1/2)*a)/c^(5/2)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=\left [\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} + {\left (8 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, c^{3} x^{4}}, \frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, a^{2} c^{2} + {\left (8 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, c^{3} x^{4}}\right ] \]

[In]

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

[Out]

[1/16*((8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2)*sqrt(c)*x^4*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2
*(2*a^2*c^2 + (8*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c^3*x^4), 1/8*((8*b^2*c^2 - 8*a*b*c*d + 3*a^2*d^2
)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (2*a^2*c^2 + (8*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(
c^3*x^4)]

Sympy [A] (verification not implemented)

Time = 42.37 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=- \frac {a^{2}}{4 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} \sqrt {d}}{8 c x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {3 a^{2} d^{\frac {3}{2}}}{8 c^{2} x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{8 c^{\frac {5}{2}}} - \frac {a b \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{c x} + \frac {a b d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{c^{\frac {3}{2}}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{\sqrt {c}} \]

[In]

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

[Out]

-a**2/(4*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) + a**2*sqrt(d)/(8*c*x**3*sqrt(c/(d*x**2) + 1)) + 3*a**2*d**(3/2)/(
8*c**2*x*sqrt(c/(d*x**2) + 1)) - 3*a**2*d**2*asinh(sqrt(c)/(sqrt(d)*x))/(8*c**(5/2)) - a*b*sqrt(d)*sqrt(c/(d*x
**2) + 1)/(c*x) + a*b*d*asinh(sqrt(c)/(sqrt(d)*x))/c**(3/2) - b**2*asinh(sqrt(c)/(sqrt(d)*x))/sqrt(c)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {b^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} + \frac {a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {5}{2}}} - \frac {\sqrt {d x^{2} + c} a b}{c x^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a^{2} d}{8 \, c^{2} x^{2}} - \frac {\sqrt {d x^{2} + c} a^{2}}{4 \, c x^{4}} \]

[In]

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

[Out]

-b^2*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) + a*b*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) - 3/8*a^2*d^2*arcsinh
(c/(sqrt(c*d)*abs(x)))/c^(5/2) - sqrt(d*x^2 + c)*a*b/(c*x^2) + 3/8*sqrt(d*x^2 + c)*a^2*d/(c^2*x^2) - 1/4*sqrt(
d*x^2 + c)*a^2/(c*x^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=\frac {\frac {{\left (8 \, b^{2} c^{2} d - 8 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} - 3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} + 5 \, \sqrt {d x^{2} + c} a^{2} c d^{3}}{c^{2} d^{2} x^{4}}}{8 \, d} \]

[In]

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

[Out]

1/8*((8*b^2*c^2*d - 8*a*b*c*d^2 + 3*a^2*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^2) - (8*(d*x^2 + c)^
(3/2)*a*b*c*d^2 - 8*sqrt(d*x^2 + c)*a*b*c^2*d^2 - 3*(d*x^2 + c)^(3/2)*a^2*d^3 + 5*sqrt(d*x^2 + c)*a^2*c*d^3)/(
c^2*d^2*x^4))/d

Mupad [B] (verification not implemented)

Time = 5.57 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {\frac {\left (5\,a^2\,d^2-8\,a\,b\,c\,d\right )\,\sqrt {d\,x^2+c}}{8\,c}-\frac {\left (3\,a^2\,d^2-8\,a\,b\,c\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{8\,c^2}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (3\,a^2\,d^2-8\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{5/2}} \]

[In]

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

[Out]

- (((5*a^2*d^2 - 8*a*b*c*d)*(c + d*x^2)^(1/2))/(8*c) - ((3*a^2*d^2 - 8*a*b*c*d)*(c + d*x^2)^(3/2))/(8*c^2))/((
c + d*x^2)^2 - 2*c*(c + d*x^2) + c^2) - (atanh((c + d*x^2)^(1/2)/c^(1/2))*(3*a^2*d^2 + 8*b^2*c^2 - 8*a*b*c*d))
/(8*c^(5/2))