3.1.96 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^7} \, dx\)

Optimal. Leaf size=61 \[ -\frac {\left (b x^2+c x^4\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^6}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \]

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Rubi [A]  time = 0.16, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2034, 792, 650} \begin {gather*} -\frac {\left (b x^2+c x^4\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^6}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(5*b*x^8) - ((5*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(15*b^2*x^6)

Rule 650

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))/((p + 1)*(2*c*d - b*e)), 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] && EqQ[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 {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}+\frac {\left (-4 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^3} \, dx,x,x^2\right )}{5 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}-\frac {(5 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.72 \begin {gather*} -\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (3 A b-2 A c x^2+5 b B x^2\right )}{15 b^2 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]

[Out]

-1/15*((x^2*(b + c*x^2))^(3/2)*(3*A*b + 5*b*B*x^2 - 2*A*c*x^2))/(b^2*x^8)

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IntegrateAlgebraic [A]  time = 0.34, size = 66, normalized size = 1.08 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 A b^2-A b c x^2+2 A c^2 x^4-5 b^2 B x^2-5 b B c x^4\right )}{15 b^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*A*b^2 - 5*b^2*B*x^2 - A*b*c*x^2 - 5*b*B*c*x^4 + 2*A*c^2*x^4))/(15*b^2*x^6)

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fricas [A]  time = 0.41, size = 59, normalized size = 0.97 \begin {gather*} -\frac {{\left ({\left (5 \, B b c - 2 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="fricas")

[Out]

-1/15*((5*B*b*c - 2*A*c^2)*x^4 + 3*A*b^2 + (5*B*b^2 + A*b*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^2*x^6)

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giac [B]  time = 0.92, size = 250, normalized size = 4.10 \begin {gather*} \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B c^{\frac {3}{2}} \mathrm {sgn}\relax (x) - 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{2} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{3} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{2} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 5 \, B b^{4} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) - 2 \, A b^{3} c^{\frac {5}{2}} \mathrm {sgn}\relax (x)\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*c^(3/2)*sgn(x) - 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b*c^(3/2)*sgn
(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*c^(5/2)*sgn(x) + 20*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^2*c^(3/2)*s
gn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b*c^(5/2)*sgn(x) - 10*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^3*c^(3/
2)*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^2*c^(5/2)*sgn(x) + 5*B*b^4*c^(3/2)*sgn(x) - 2*A*b^3*c^(5/2)
*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5

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maple [A]  time = 0.05, size = 48, normalized size = 0.79 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+5 B b \,x^{2}+3 A b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x)

[Out]

-1/15*(c*x^2+b)*(-2*A*c*x^2+5*B*b*x^2+3*A*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^6

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maxima [B]  time = 1.50, size = 111, normalized size = 1.82 \begin {gather*} -\frac {1}{3} \, B {\left (\frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{2}} + \frac {\sqrt {c x^{4} + b x^{2}}}{x^{4}}\right )} + \frac {1}{15} \, A {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{x^{6}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/3*B*(sqrt(c*x^4 + b*x^2)*c/(b*x^2) + sqrt(c*x^4 + b*x^2)/x^4) + 1/15*A*(2*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^2)
- sqrt(c*x^4 + b*x^2)*c/(b*x^4) - 3*sqrt(c*x^4 + b*x^2)/x^6)

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mupad [B]  time = 0.47, size = 113, normalized size = 1.85 \begin {gather*} \frac {\left (A\,c^2+B\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{5\,b^2\,x^2}-\frac {\left (5\,B\,b^2+A\,c\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^4}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {c\,\left (A\,c+8\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^7,x)

[Out]

((A*c^2 + B*b*c)*(b*x^2 + c*x^4)^(1/2))/(5*b^2*x^2) - ((5*B*b^2 + A*b*c)*(b*x^2 + c*x^4)^(1/2))/(15*b^2*x^4) -
(A*(b*x^2 + c*x^4)^(1/2))/(5*x^6) - (c*(A*c + 8*B*b)*(b*x^2 + c*x^4)^(1/2))/(15*b^2*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**7,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**7, x)

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