3.7.39 \(\int \frac {(a+b x^2)^2}{x^5 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}}+\frac {8 b^2 c^2-3 a d (8 b c-5 a d)}{8 c^3 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 145, normalized size of antiderivative = 1.00, 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, 51, 63, 208} \begin {gather*} -\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}+\frac {8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt {c+d x^2}}-\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]

[Out]

(8*b^2 - (3*a*d*(8*b*c - 5*a*d))/c^2)/(8*c*Sqrt[c + d*x^2]) - a^2/(4*c*x^4*Sqrt[c + d*x^2]) - (a*(8*b*c - 5*a*
d))/(8*c^2*x^2*Sqrt[c + d*x^2]) - ((8*b^2*c^2 - 3*a*d*(8*b*c - 5*a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^
(7/2))

Rule 51

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

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Mathematica [C]  time = 0.04, size = 89, normalized size = 0.61 \begin {gather*} \frac {x^4 \left (15 a^2 d^2-24 a b c d+8 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^2}{c}+1\right )+a c \left (-2 a c+5 a d x^2-8 b c x^2\right )}{8 c^3 x^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]

[Out]

(a*c*(-2*a*c - 8*b*c*x^2 + 5*a*d*x^2) + (8*b^2*c^2 - 24*a*b*c*d + 15*a^2*d^2)*x^4*Hypergeometric2F1[-1/2, 1, 1
/2, 1 + (d*x^2)/c])/(8*c^3*x^4*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.27, size = 132, normalized size = 0.91 \begin {gather*} \frac {\left (-15 a^2 d^2+24 a b c d-8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}}+\frac {-2 a^2 c^2+5 a^2 c d x^2+15 a^2 d^2 x^4-8 a b c^2 x^2-24 a b c d x^4+8 b^2 c^2 x^4}{8 c^3 x^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]

[Out]

(-2*a^2*c^2 - 8*a*b*c^2*x^2 + 5*a^2*c*d*x^2 + 8*b^2*c^2*x^4 - 24*a*b*c*d*x^4 + 15*a^2*d^2*x^4)/(8*c^3*x^4*Sqrt
[c + d*x^2]) + ((-8*b^2*c^2 + 24*a*b*c*d - 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^(7/2))

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fricas [A]  time = 1.00, size = 364, normalized size = 2.51 \begin {gather*} \left [\frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {c} \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^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}, \frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, a^{2} c^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(((8*b^2*c^2*d - 24*a*b*c*d^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(c)
*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2
)*x^4 + (8*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^4*d*x^6 + c^5*x^4), 1/8*(((8*b^2*c^2*d - 24*a*b*c*d
^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)
) - (2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4 + (8*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c
))/(c^4*d*x^6 + c^5*x^4)]

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giac [A]  time = 0.35, size = 163, normalized size = 1.12 \begin {gather*} \frac {{\left (8 \, b^{2} c^{2} - 24 \, a b c d + 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{3}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt {d x^{2} + c} c^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d - 8 \, \sqrt {d x^{2} + c} a b c^{2} d - 7 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 9 \, \sqrt {d x^{2} + c} a^{2} c d^{2}}{8 \, c^{3} d^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/8*(8*b^2*c^2 - 24*a*b*c*d + 15*a^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) + (b^2*c^2 - 2*a*b*c
*d + a^2*d^2)/(sqrt(d*x^2 + c)*c^3) - 1/8*(8*(d*x^2 + c)^(3/2)*a*b*c*d - 8*sqrt(d*x^2 + c)*a*b*c^2*d - 7*(d*x^
2 + c)^(3/2)*a^2*d^2 + 9*sqrt(d*x^2 + c)*a^2*c*d^2)/(c^3*d^2*x^4)

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maple [A]  time = 0.01, size = 211, normalized size = 1.46 \begin {gather*} -\frac {15 a^{2} d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{8 c^{\frac {7}{2}}}+\frac {3 a b d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {5}{2}}}-\frac {b^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {3}{2}}}+\frac {15 a^{2} d^{2}}{8 \sqrt {d \,x^{2}+c}\, c^{3}}-\frac {3 a b d}{\sqrt {d \,x^{2}+c}\, c^{2}}+\frac {b^{2}}{\sqrt {d \,x^{2}+c}\, c}+\frac {5 a^{2} d}{8 \sqrt {d \,x^{2}+c}\, c^{2} x^{2}}-\frac {a b}{\sqrt {d \,x^{2}+c}\, c \,x^{2}}-\frac {a^{2}}{4 \sqrt {d \,x^{2}+c}\, c \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x)

[Out]

-1/4*a^2/c/x^4/(d*x^2+c)^(1/2)+5/8*a^2*d/c^2/x^2/(d*x^2+c)^(1/2)+15/8*a^2*d^2/c^3/(d*x^2+c)^(1/2)-15/8*a^2*d^2
/c^(7/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-a*b/c/x^2/(d*x^2+c)^(1/2)-3*a*b*d/c^2/(d*x^2+c)^(1/2)+3*a*b*d/c
^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+b^2/c/(d*x^2+c)^(1/2)-b^2/c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/
2))/x)

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maxima [A]  time = 0.85, size = 177, normalized size = 1.22 \begin {gather*} -\frac {b^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {3 \, a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} - \frac {15 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {7}{2}}} + \frac {b^{2}}{\sqrt {d x^{2} + c} c} - \frac {3 \, a b d}{\sqrt {d x^{2} + c} c^{2}} + \frac {15 \, a^{2} d^{2}}{8 \, \sqrt {d x^{2} + c} c^{3}} - \frac {a b}{\sqrt {d x^{2} + c} c x^{2}} + \frac {5 \, a^{2} d}{8 \, \sqrt {d x^{2} + c} c^{2} x^{2}} - \frac {a^{2}}{4 \, \sqrt {d x^{2} + c} c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

-b^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) + 3*a*b*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) - 15/8*a^2*d^2*arcs
inh(c/(sqrt(c*d)*abs(x)))/c^(7/2) + b^2/(sqrt(d*x^2 + c)*c) - 3*a*b*d/(sqrt(d*x^2 + c)*c^2) + 15/8*a^2*d^2/(sq
rt(d*x^2 + c)*c^3) - a*b/(sqrt(d*x^2 + c)*c*x^2) + 5/8*a^2*d/(sqrt(d*x^2 + c)*c^2*x^2) - 1/4*a^2/(sqrt(d*x^2 +
c)*c*x^4)

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mupad [B]  time = 1.06, size = 179, normalized size = 1.23 \begin {gather*} \frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c}-\frac {\left (d\,x^2+c\right )\,\left (25\,a^2\,d^2-40\,a\,b\,c\,d+16\,b^2\,c^2\right )}{8\,c^2}+\frac {{\left (d\,x^2+c\right )}^2\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^3}}{{\left (d\,x^2+c\right )}^{5/2}-2\,c\,{\left (d\,x^2+c\right )}^{3/2}+c^2\,\sqrt {d\,x^2+c}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x)

[Out]

((a^2*d^2 + b^2*c^2 - 2*a*b*c*d)/c - ((c + d*x^2)*(25*a^2*d^2 + 16*b^2*c^2 - 40*a*b*c*d))/(8*c^2) + ((c + d*x^
2)^2*(15*a^2*d^2 + 8*b^2*c^2 - 24*a*b*c*d))/(8*c^3))/((c + d*x^2)^(5/2) - 2*c*(c + d*x^2)^(3/2) + c^2*(c + d*x
^2)^(1/2)) - (atanh((c + d*x^2)^(1/2)/c^(1/2))*(15*a^2*d^2 + 8*b^2*c^2 - 24*a*b*c*d))/(8*c^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{2}}{x^{5} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**5/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/(x**5*(c + d*x**2)**(3/2)), x)

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