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

Optimal. Leaf size=143 \[ -\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}+\frac {\sqrt {c+d x^2} \left (a d (8 b c-a d)+8 b^2 c^2\right )}{8 c^2}-\frac {\left (a d (8 b c-a d)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}}-\frac {a \left (c+d x^2\right )^{3/2} (8 b c-a d)}{8 c^2 x^2} \]

[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

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Rubi [A]  time = 0.16, antiderivative size = 140, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 78, 50, 63, 208} \[ -\frac {a^2 \left (c+d x^2\right )^{3/2}}{4 c x^4}+\frac {1}{8} \sqrt {c+d x^2} \left (\frac {a d (8 b c-a d)}{c^2}+8 b^2\right )-\frac {\left (a d (8 b c-a d)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}}-\frac {a \left (c+d x^2\right )^{3/2} (8 b c-a d)}{8 c^2 x^2} \]

Antiderivative was successfully verified.

[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 50

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 63

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 78

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] || LtQ
[p, n]))))

Rule 89

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 208

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

Rule 446

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*} \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^5} \, dx &=\frac {1}{2} \operatorname {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 {\operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.12, size = 104, normalized size = 0.73 \[ \frac {\sqrt {c+d x^2} \left (-a^2 \left (2 c+d x^2\right )-8 a b c x^2+8 b^2 c x^4\right )}{8 c x^4}-\frac {\left (-a^2 d^2+8 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 225, normalized size = 1.57 \[ \left [-\frac {{\left (8 \, b^{2} c^{2} + 8 \, a b c d - a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, b^{2} c^{2} x^{4} - 2 \, a^{2} c^{2} - {\left (8 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, c^{2} x^{4}}, \frac {{\left (8 \, b^{2} c^{2} + 8 \, a b c d - a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} c^{2} x^{4} - 2 \, a^{2} c^{2} - {\left (8 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, c^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.36, size = 153, normalized size = 1.07 \[ \frac {8 \, \sqrt {d x^{2} + c} b^{2} d + \frac {{\left (8 \, b^{2} c^{2} d + 8 \, a b c d^{2} - a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} + \sqrt {d x^{2} + c} a^{2} c d^{3}}{c d^{2} x^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 207, normalized size = 1.45 \[ \frac {a^{2} d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{8 c^{\frac {3}{2}}}-\frac {a b d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{\sqrt {c}}-b^{2} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )-\frac {\sqrt {d \,x^{2}+c}\, a^{2} d^{2}}{8 c^{2}}+\frac {\sqrt {d \,x^{2}+c}\, a b d}{c}+\sqrt {d \,x^{2}+c}\, b^{2}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d}{8 c^{2} x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b}{c \,x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2}}{4 c \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.07, size = 173, normalized size = 1.21 \[ -b^{2} \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} + \frac {a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {3}{2}}} + \sqrt {d x^{2} + c} b^{2} + \frac {\sqrt {d x^{2} + c} a b d}{c} - \frac {\sqrt {d x^{2} + c} a^{2} d^{2}}{8 \, c^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{c x^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{4 \, c x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.33, size = 137, normalized size = 0.96 \[ b^2\,\sqrt {d\,x^2+c}-\frac {\left (\frac {a^2\,d^2}{8}-a\,b\,c\,d\right )\,\sqrt {d\,x^2+c}+\frac {\left (a^2\,d^2+8\,b\,c\,a\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{8\,c}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (-a^2\,d^2+8\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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