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

Optimal. Leaf size=133 \[ -\frac{8 c^2 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{105 b^4 x^2}+\frac{4 c \sqrt{b x^2+c x^4} (7 b B-6 A c)}{105 b^3 x^4}-\frac{\sqrt{b x^2+c x^4} (7 b B-6 A c)}{35 b^2 x^6}-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8} \]

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(7*b*x^8) - ((7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(35*b^2*x^6) + (4*c*(7*b*B - 6*A*c)
*Sqrt[b*x^2 + c*x^4])/(105*b^3*x^4) - (8*c^2*(7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(105*b^4*x^2)

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Rubi [A]  time = 0.253481, antiderivative size = 133, normalized size of antiderivative = 1., 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, 650} \[ -\frac{8 c^2 \sqrt{b x^2+c x^4} (7 b B-6 A c)}{105 b^4 x^2}+\frac{4 c \sqrt{b x^2+c x^4} (7 b B-6 A c)}{105 b^3 x^4}-\frac{\sqrt{b x^2+c x^4} (7 b B-6 A c)}{35 b^2 x^6}-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(7*b*x^8) - ((7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(35*b^2*x^6) + (4*c*(7*b*B - 6*A*c)
*Sqrt[b*x^2 + c*x^4])/(105*b^3*x^4) - (8*c^2*(7*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(105*b^4*x^2)

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^7 \sqrt{b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^4 \sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8}+\frac{\left (-4 (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{b x+c x^2}} \, dx,x,x^2\right )}{7 b}\\ &=-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8}-\frac{(7 b B-6 A c) \sqrt{b x^2+c x^4}}{35 b^2 x^6}-\frac{(2 c (7 b B-6 A c)) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{b x+c x^2}} \, dx,x,x^2\right )}{35 b^2}\\ &=-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8}-\frac{(7 b B-6 A c) \sqrt{b x^2+c x^4}}{35 b^2 x^6}+\frac{4 c (7 b B-6 A c) \sqrt{b x^2+c x^4}}{105 b^3 x^4}+\frac{\left (4 c^2 (7 b B-6 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{b x+c x^2}} \, dx,x,x^2\right )}{105 b^3}\\ &=-\frac{A \sqrt{b x^2+c x^4}}{7 b x^8}-\frac{(7 b B-6 A c) \sqrt{b x^2+c x^4}}{35 b^2 x^6}+\frac{4 c (7 b B-6 A c) \sqrt{b x^2+c x^4}}{105 b^3 x^4}-\frac{8 c^2 (7 b B-6 A c) \sqrt{b x^2+c x^4}}{105 b^4 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0307912, size = 89, normalized size = 0.67 \[ -\frac{\sqrt{x^2 \left (b+c x^2\right )} \left (3 A \left (-6 b^2 c x^2+5 b^3+8 b c^2 x^4-16 c^3 x^6\right )+7 b B x^2 \left (3 b^2-4 b c x^2+8 c^2 x^4\right )\right )}{105 b^4 x^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x^2*(b + c*x^2)]*(7*b*B*x^2*(3*b^2 - 4*b*c*x^2 + 8*c^2*x^4) + 3*A*(5*b^3 - 6*b^2*c*x^2 + 8*b*c^2*x^4 -
16*c^3*x^6)))/(105*b^4*x^8)

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Maple [A]  time = 0.006, size = 94, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -48\,A{c}^{3}{x}^{6}+56\,B{x}^{6}b{c}^{2}+24\,Ab{c}^{2}{x}^{4}-28\,B{x}^{4}{b}^{2}c-18\,A{b}^{2}c{x}^{2}+21\,B{x}^{2}{b}^{3}+15\,A{b}^{3} \right ) }{105\,{x}^{6}{b}^{4}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(c*x^2+b)*(-48*A*c^3*x^6+56*B*b*c^2*x^6+24*A*b*c^2*x^4-28*B*b^2*c*x^4-18*A*b^2*c*x^2+21*B*b^3*x^2+15*A*
b^3)/x^6/b^4/(c*x^4+b*x^2)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.15395, size = 190, normalized size = 1.43 \begin{align*} -\frac{{\left (8 \,{\left (7 \, B b c^{2} - 6 \, A c^{3}\right )} x^{6} - 4 \,{\left (7 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{4} + 15 \, A b^{3} + 3 \,{\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{105 \, b^{4} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(8*(7*B*b*c^2 - 6*A*c^3)*x^6 - 4*(7*B*b^2*c - 6*A*b*c^2)*x^4 + 15*A*b^3 + 3*(7*B*b^3 - 6*A*b^2*c)*x^2)*
sqrt(c*x^4 + b*x^2)/(b^4*x^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16977, size = 140, normalized size = 1.05 \begin{align*} -\frac{21 \, B b{\left (c + \frac{b}{x^{2}}\right )}^{\frac{5}{2}} + 15 \, A{\left (c + \frac{b}{x^{2}}\right )}^{\frac{7}{2}} - 70 \, B b{\left (c + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} c - 63 \, A{\left (c + \frac{b}{x^{2}}\right )}^{\frac{5}{2}} c + 105 \, B b \sqrt{c + \frac{b}{x^{2}}} c^{2} + 105 \, A{\left (c + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} c^{2} - 105 \, A \sqrt{c + \frac{b}{x^{2}}} c^{3}}{105 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(21*B*b*(c + b/x^2)^(5/2) + 15*A*(c + b/x^2)^(7/2) - 70*B*b*(c + b/x^2)^(3/2)*c - 63*A*(c + b/x^2)^(5/2
)*c + 105*B*b*sqrt(c + b/x^2)*c^2 + 105*A*(c + b/x^2)^(3/2)*c^2 - 105*A*sqrt(c + b/x^2)*c^3)/b^4