\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^5} \, dx\) [603]

Optimal result

Integrand size = 24, antiderivative size = 143 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=\frac {\left (8 b^2 c^2+a d (8 b c-a d)\right ) \sqrt {c+d x^2}}{8 c^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {a (8 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2+a d (8 b c-a d)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 79, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {\left (a d (8 b c-a d)+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}}+\frac {1}{8} \sqrt {c+d x^2} \left (\frac {a d (8 b c-a d)}{c^2}+8 b^2\right )-\frac {a \left (c+d x^2\right )^{3/2} (8 b c-a d)}{8 c^2 x^2} \]

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \sqrt {c+d x}}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}+\frac {\text {Subst}\left (\int \frac {\left (\frac {1}{2} a (8 b c-a d)+2 b^2 c x\right ) \sqrt {c+d x}}{x^2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {a (8 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{8} \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {a (8 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^2}+\frac {1}{16} \left (c \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {1}{8} \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {a (8 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^2}+\frac {\left (c \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right )\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 {1}{8} \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}-\frac {a (8 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^2}-\frac {1}{8} \sqrt {c} \left (8 b^2+\frac {a d (8 b c-a d)}{c^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=\frac {\sqrt {c+d x^2} \left (-2 a^2 c-8 a b c x^2-a^2 d x^2+8 b^2 c x^4\right )}{8 c x^4}+\frac {\left (-8 b^2 c^2-8 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {-x^{4} \left (a^{2} d^{2}-8 a b c d -8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\sqrt {d \,x^{2}+c}\, \left (\left (-8 b^{2} x^{4}+8 a b \,x^{2}+2 a^{2}\right ) c^{\frac {3}{2}}+\sqrt {c}\, a^{2} d \,x^{2}\right )}{8 c^{\frac {3}{2}} x^{4}}\)	\(98\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (a d \,x^{2}+8 c b \,x^{2}+2 a c \right )}{8 x^{4} c}-\frac {-8 b^{2} c \sqrt {d \,x^{2}+c}+\frac {\left (-a^{2} d^{2}+8 a b c d +8 b^{2} c^{2}\right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}}{8 c}\)	\(109\)
default	\(b^{2} \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 c \,x^{4}}-\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )}{2 c}\right )}{4 c}\right )+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )}{2 c}\right )\)	\(200\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=\left [-\frac {{\left (8 \, b^{2} c^{2} + 8 \, a b c d - a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, b^{2} c^{2} x^{4} - 2 \, a^{2} c^{2} - {\left (8 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, c^{2} x^{4}}, \frac {{\left (8 \, b^{2} c^{2} + 8 \, a b c d - a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} c^{2} x^{4} - 2 \, a^{2} c^{2} - {\left (8 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, c^{2} x^{4}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 53.20 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=- \frac {a^{2} c}{4 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} \sqrt {d}}{8 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{\frac {3}{2}}}{8 c x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{8 c^{\frac {3}{2}}} - \frac {a b \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{x} - \frac {a b d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{\sqrt {c}} - b^{2} \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {b^{2} c}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {b^{2} \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=-b^{2} \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} + \frac {a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {3}{2}}} + \sqrt {d x^{2} + c} b^{2} + \frac {\sqrt {d x^{2} + c} a b d}{c} - \frac {\sqrt {d x^{2} + c} a^{2} d^{2}}{8 \, c^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{c x^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{4 \, c x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=\frac {8 \, \sqrt {d x^{2} + c} b^{2} d + \frac {{\left (8 \, b^{2} c^{2} d + 8 \, a b c d^{2} - a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} + \sqrt {d x^{2} + c} a^{2} c d^{3}}{c d^{2} x^{4}}}{8 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx=b^2\,\sqrt {d\,x^2+c}-\frac {\left (\frac {a^2\,d^2}{8}-a\,b\,c\,d\right )\,\sqrt {d\,x^2+c}+\frac {\left (a^2\,d^2+8\,b\,c\,a\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{8\,c}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (-a^2\,d^2+8\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{3/2}} \]

[In]

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

[Out]

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