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

Optimal. Leaf size=131 \[ \frac {-\frac {5 a^2 d}{c}+4 a b-\frac {2 b^2 c}{d}}{6 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}-\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 131, 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 {-\frac {5 a^2 d}{c}+4 a b-\frac {2 b^2 c}{d}}{6 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}-\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*b - (2*b^2*c)/d - (5*a^2*d)/c)/(6*c*(c + d*x^2)^(3/2)) - a^2/(2*c*x^2*(c + d*x^2)^(3/2)) + (a*(4*b*c - 5*
a*d))/(2*c^3*Sqrt[c + d*x^2]) - (a*(4*b*c - 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*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^3 \left (c+d x^2\right )^{5/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^2 (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (4 b c-5 a d)+b^2 c x}{x (c+d x)^{5/2}} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {(a (4 b c-5 a d)) \operatorname {Subst}\left (\int \frac {1}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 c^2}\\ &=-\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}+\frac {(a (4 b c-5 a d)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c^3}\\ &=-\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}+\frac {(a (4 b c-5 a d)) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 c^3 d}\\ &=-\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}-\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 105, normalized size = 0.80 \begin {gather*} \frac {-c \left (a^2 d \left (3 c+5 d x^2\right )-4 a b c d x^2+2 b^2 c^2 x^2\right )-3 a d x^2 \left (c+d x^2\right ) (5 a d-4 b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^2}{c}+1\right )}{6 c^3 d x^2 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(c*(2*b^2*c^2*x^2 - 4*a*b*c*d*x^2 + a^2*d*(3*c + 5*d*x^2))) - 3*a*d*(-4*b*c + 5*a*d)*x^2*(c + d*x^2)*Hyperge
ometric2F1[-1/2, 1, 1/2, 1 + (d*x^2)/c])/(6*c^3*d*x^2*(c + d*x^2)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 130, normalized size = 0.99 \begin {gather*} \frac {-3 a^2 c^2 d-20 a^2 c d^2 x^2-15 a^2 d^3 x^4+16 a b c^2 d x^2+12 a b c d^2 x^4-2 b^2 c^3 x^2}{6 c^3 d x^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (5 a^2 d-4 a b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*c^2*d - 2*b^2*c^3*x^2 + 16*a*b*c^2*d*x^2 - 20*a^2*c*d^2*x^2 + 12*a*b*c*d^2*x^4 - 15*a^2*d^3*x^4)/(6*c^
3*d*x^2*(c + d*x^2)^(3/2)) + ((-4*a*b*c + 5*a^2*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 1.10, size = 426, normalized size = 3.25 \begin {gather*} \left [-\frac {3 \, {\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + {\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, a^{2} c^{3} d - 3 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}, \frac {3 \, {\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + {\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (3 \, a^{2} c^{3} d - 3 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*((4*a*b*c*d^3 - 5*a^2*d^4)*x^6 + 2*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 + (4*a*b*c^3*d - 5*a^2*c^2*d^2)
*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3*a^2*c^3*d - 3*(4*a*b*c^2*d^2 - 5*a^2*
c*d^3)*x^4 + 2*(b^2*c^4 - 8*a*b*c^3*d + 10*a^2*c^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c^4*d^3*x^6 + 2*c^5*d^2*x^4 + c
^6*d*x^2), 1/6*(3*((4*a*b*c*d^3 - 5*a^2*d^4)*x^6 + 2*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 + (4*a*b*c^3*d - 5*a^2*
c^2*d^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (3*a^2*c^3*d - 3*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 +
2*(b^2*c^4 - 8*a*b*c^3*d + 10*a^2*c^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c^4*d^3*x^6 + 2*c^5*d^2*x^4 + c^6*d*x^2)]

________________________________________________________________________________________

giac [A]  time = 0.37, size = 128, normalized size = 0.98 \begin {gather*} \frac {{\left (4 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, \sqrt {-c} c^{3}} - \frac {\sqrt {d x^{2} + c} a^{2}}{2 \, c^{3} x^{2}} - \frac {b^{2} c^{3} - 6 \, {\left (d x^{2} + c\right )} a b c d - 2 \, a b c^{2} d + 6 \, {\left (d x^{2} + c\right )} a^{2} d^{2} + a^{2} c d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*b*c - 5*a^2*d)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/2*sqrt(d*x^2 + c)*a^2/(c^3*x^2) -
1/3*(b^2*c^3 - 6*(d*x^2 + c)*a*b*c*d - 2*a*b*c^2*d + 6*(d*x^2 + c)*a^2*d^2 + a^2*c*d^2)/((d*x^2 + c)^(3/2)*c^3
*d)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 169, normalized size = 1.29 \begin {gather*} -\frac {5 a^{2} d}{6 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}}+\frac {2 a b}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c}-\frac {b^{2}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {5 a^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 c^{\frac {7}{2}}}-\frac {2 a b \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {5}{2}}}-\frac {5 a^{2} d}{2 \sqrt {d \,x^{2}+c}\, c^{3}}+\frac {2 a b}{\sqrt {d \,x^{2}+c}\, c^{2}}-\frac {a^{2}}{2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2/d/(d*x^2+c)^(3/2)-1/2*a^2/c/x^2/(d*x^2+c)^(3/2)-5/6*a^2*d/c^2/(d*x^2+c)^(3/2)-5/2*a^2*d/c^3/(d*x^2+c)
^(1/2)+5/2*a^2*d/c^(7/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+2/3*a*b/c/(d*x^2+c)^(3/2)+2*a*b/c^2/(d*x^2+c)^(
1/2)-2*a*b/c^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 146, normalized size = 1.11 \begin {gather*} -\frac {2 \, a b \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} + \frac {5 \, a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {7}{2}}} + \frac {2 \, a b}{\sqrt {d x^{2} + c} c^{2}} + \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {b^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {5 \, a^{2} d}{2 \, \sqrt {d x^{2} + c} c^{3}} - \frac {5 \, a^{2} d}{6 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {a^{2}}{2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) + 5/2*a^2*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(7/2) + 2*a*b/(sqrt(d
*x^2 + c)*c^2) + 2/3*a*b/((d*x^2 + c)^(3/2)*c) - 1/3*b^2/((d*x^2 + c)^(3/2)*d) - 5/2*a^2*d/(sqrt(d*x^2 + c)*c^
3) - 5/6*a^2*d/((d*x^2 + c)^(3/2)*c^2) - 1/2*a^2/((d*x^2 + c)^(3/2)*c*x^2)

________________________________________________________________________________________

mupad [B]  time = 0.90, size = 147, normalized size = 1.12 \begin {gather*} \frac {a\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (5\,a\,d-4\,b\,c\right )}{2\,c^{7/2}}-\frac {\frac {\left (d\,x^2+c\right )\,\left (-5\,a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{3\,c^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{3\,c}+\frac {d\,{\left (d\,x^2+c\right )}^2\,\left (5\,a^2\,d-4\,a\,b\,c\right )}{2\,c^3}}{d\,{\left (d\,x^2+c\right )}^{5/2}-c\,d\,{\left (d\,x^2+c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________