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

Optimal. Leaf size=88 \[ \frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}}+\frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 87, 63, 208} \begin {gather*} \frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{\sqrt {c+d x^2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}}+\frac {(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

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Mathematica [C]  time = 0.04, size = 67, normalized size = 0.76 \begin {gather*} \frac {a^2 d^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {d x^2}{c}+1\right )-b c \left (2 a d+2 b c+3 b d x^2\right )}{3 c d^2 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*c*(2*b*c + 2*a*d + 3*b*d*x^2)) + a^2*d^2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (d*x^2)/c])/(3*c*d^2*(c + d
*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 0.11, size = 99, normalized size = 1.12 \begin {gather*} \frac {4 a^2 c d^2+3 a^2 d^3 x^2-2 a b c^2 d-2 b^2 c^3-3 b^2 c^2 d x^2}{3 c^2 d^2 \left (c+d x^2\right )^{3/2}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.17, size = 316, normalized size = 3.59 \begin {gather*} \left [\frac {3 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{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 (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}, \frac {3 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.44, size = 102, normalized size = 1.16 \begin {gather*} \frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {3 \, {\left (d x^{2} + c\right )} b^{2} c^{2} - b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, {\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^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 120, normalized size = 1.36 \begin {gather*} -\frac {b^{2} x^{2}}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {a^{2}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c}-\frac {2 a b}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}-\frac {2 b^{2} c}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{2}}-\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {5}{2}}}+\frac {a^{2}}{\sqrt {d \,x^{2}+c}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.93, size = 108, normalized size = 1.23 \begin {gather*} -\frac {b^{2} x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} + \frac {a^{2}}{\sqrt {d x^{2} + c} c^{2}} + \frac {a^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {2 \, b^{2} c}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 90, normalized size = 1.02 \begin {gather*} \frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{3\,c}+\frac {\left (a^2\,d^2-b^2\,c^2\right )\,\left (d\,x^2+c\right )}{c^2}}{d^2\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 44.36, size = 87, normalized size = 0.99 \begin {gather*} \frac {a^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{c^{2} \sqrt {- c}} + \frac {\left (a d - b c\right )^{2}}{3 c d^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}} + \frac {\left (a d - b c\right ) \left (a d + b c\right )}{c^{2} d^{2} \sqrt {c + d x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*atan(sqrt(c + d*x**2)/sqrt(-c))/(c**2*sqrt(-c)) + (a*d - b*c)**2/(3*c*d**2*(c + d*x**2)**(3/2)) + (a*d -
b*c)*(a*d + b*c)/(c**2*d**2*sqrt(c + d*x**2))

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