3.2.52 \(\int \frac {A+B x^2}{x^5 (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {8 c^2 \left (b+2 c x^2\right ) (7 b B-8 A c)}{35 b^5 \sqrt {b x^2+c x^4}}+\frac {2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.27, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 613} \begin {gather*} -\frac {8 c^2 \left (b+2 c x^2\right ) (7 b B-8 A c)}{35 b^5 \sqrt {b x^2+c x^4}}+\frac {2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

-A/(7*b*x^6*Sqrt[b*x^2 + c*x^4]) - (7*b*B - 8*A*c)/(35*b^2*x^4*Sqrt[b*x^2 + c*x^4]) + (2*c*(7*b*B - 8*A*c))/(3
5*b^3*x^2*Sqrt[b*x^2 + c*x^4]) - (8*c^2*(7*b*B - 8*A*c)*(b + 2*c*x^2))/(35*b^5*Sqrt[b*x^2 + c*x^4])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
 + c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
 b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
 /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}}+\frac {\left (\frac {1}{2} (b B-2 A c)-3 (-b B+A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{7 b}\\ &=-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}-\frac {(3 c (7 b B-8 A c)) \operatorname {Subst}\left (\int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{35 b^2}\\ &=-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}+\frac {2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt {b x^2+c x^4}}+\frac {\left (4 c^2 (7 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{35 b^3}\\ &=-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}+\frac {2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt {b x^2+c x^4}}-\frac {8 c^2 (7 b B-8 A c) \left (b+2 c x^2\right )}{35 b^5 \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.54 \begin {gather*} \frac {x^2 \left (b^3-2 b^2 c x^2+8 b c^2 x^4+16 c^3 x^6\right ) (8 A c-7 b B)-5 A b^4}{35 b^5 x^6 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-5*A*b^4 + (-7*b*B + 8*A*c)*x^2*(b^3 - 2*b^2*c*x^2 + 8*b*c^2*x^4 + 16*c^3*x^6))/(35*b^5*x^6*Sqrt[x^2*(b + c*x
^2)])

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IntegrateAlgebraic [A]  time = 0.51, size = 123, normalized size = 0.89 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-5 A b^4+8 A b^3 c x^2-16 A b^2 c^2 x^4+64 A b c^3 x^6+128 A c^4 x^8-7 b^4 B x^2+14 b^3 B c x^4-56 b^2 B c^2 x^6-112 b B c^3 x^8\right )}{35 b^5 x^8 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-5*A*b^4 - 7*b^4*B*x^2 + 8*A*b^3*c*x^2 + 14*b^3*B*c*x^4 - 16*A*b^2*c^2*x^4 - 56*b^2*B*c^
2*x^6 + 64*A*b*c^3*x^6 - 112*b*B*c^3*x^8 + 128*A*c^4*x^8))/(35*b^5*x^8*(b + c*x^2))

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fricas [A]  time = 0.43, size = 121, normalized size = 0.88 \begin {gather*} -\frac {{\left (16 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{8} + 8 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{6} + 5 \, A b^{4} - 2 \, {\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{4} + {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, {\left (b^{5} c x^{10} + b^{6} x^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/35*(16*(7*B*b*c^3 - 8*A*c^4)*x^8 + 8*(7*B*b^2*c^2 - 8*A*b*c^3)*x^6 + 5*A*b^4 - 2*(7*B*b^3*c - 8*A*b^2*c^2)*
x^4 + (7*B*b^4 - 8*A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*c*x^10 + b^6*x^8)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/((c*x^4 + b*x^2)^(3/2)*x^5), x)

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maple [A]  time = 0.05, size = 118, normalized size = 0.86 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-128 A \,c^{4} x^{8}+112 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+56 B \,b^{2} c^{2} x^{6}+16 A \,b^{2} c^{2} x^{4}-14 B \,b^{3} c \,x^{4}-8 A \,b^{3} c \,x^{2}+7 B \,b^{4} x^{2}+5 A \,b^{4}\right )}{35 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/35*(c*x^2+b)*(-128*A*c^4*x^8+112*B*b*c^3*x^8-64*A*b*c^3*x^6+56*B*b^2*c^2*x^6+16*A*b^2*c^2*x^4-14*B*b^3*c*x^
4-8*A*b^3*c*x^2+7*B*b^4*x^2+5*A*b^4)/x^4/b^5/(c*x^4+b*x^2)^(3/2)

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maxima [A]  time = 1.48, size = 208, normalized size = 1.51 \begin {gather*} -\frac {1}{5} \, B {\left (\frac {16 \, c^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{4}} + \frac {8 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} - \frac {2 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{4}}\right )} + \frac {1}{35} \, A {\left (\frac {128 \, c^{4} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{5}} + \frac {64 \, c^{3}}{\sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {16 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3} x^{2}} + \frac {8 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{4}} - \frac {5}{\sqrt {c x^{4} + b x^{2}} b x^{6}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

-1/5*B*(16*c^3*x^2/(sqrt(c*x^4 + b*x^2)*b^4) + 8*c^2/(sqrt(c*x^4 + b*x^2)*b^3) - 2*c/(sqrt(c*x^4 + b*x^2)*b^2*
x^2) + 1/(sqrt(c*x^4 + b*x^2)*b*x^4)) + 1/35*A*(128*c^4*x^2/(sqrt(c*x^4 + b*x^2)*b^5) + 64*c^3/(sqrt(c*x^4 + b
*x^2)*b^4) - 16*c^2/(sqrt(c*x^4 + b*x^2)*b^3*x^2) + 8*c/(sqrt(c*x^4 + b*x^2)*b^2*x^4) - 5/(sqrt(c*x^4 + b*x^2)
*b*x^6))

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mupad [B]  time = 0.56, size = 173, normalized size = 1.25 \begin {gather*} -\frac {\left (7\,B\,b^2-13\,A\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^6}-\frac {\left (x^2\,\left (\frac {58\,A\,c^4-42\,B\,b\,c^3}{35\,b^5}-\frac {2\,c^3\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^5}\right )-\frac {c^2\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^4}\right )\,\sqrt {c\,x^4+b\,x^2}}{x^2\,\left (c\,x^2+b\right )}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{7\,b^2\,x^8}-\frac {c\,\left (29\,A\,c-21\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

- ((7*B*b^2 - 13*A*b*c)*(b*x^2 + c*x^4)^(1/2))/(35*b^4*x^6) - ((x^2*((58*A*c^4 - 42*B*b*c^3)/(35*b^5) - (2*c^3
*(93*A*c - 77*B*b))/(35*b^5)) - (c^2*(93*A*c - 77*B*b))/(35*b^4))*(b*x^2 + c*x^4)^(1/2))/(x^2*(b + c*x^2)) - (
A*(b*x^2 + c*x^4)^(1/2))/(7*b^2*x^8) - (c*(29*A*c - 21*B*b)*(b*x^2 + c*x^4)^(1/2))/(35*b^4*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral((A + B*x**2)/(x**5*(x**2*(b + c*x**2))**(3/2)), x)

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