3.56 \(\int \frac{A+B x+C x^2}{x^2 (a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=188 \[ \frac{35 B-x \left (\frac{93 A b}{a}-16 C\right )}{35 a^4 \sqrt{a+b x^2}}+\frac{35 B-3 x \left (\frac{29 A b}{a}-8 C\right )}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{7 B-x \left (\frac{13 A b}{a}-6 C\right )}{35 a^2 \left (a+b x^2\right )^{5/2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}+\frac{B-x \left (\frac{A b}{a}-C\right )}{7 a \left (a+b x^2\right )^{7/2}} \]

[Out]

(B - ((A*b)/a - C)*x)/(7*a*(a + b*x^2)^(7/2)) + (7*B - ((13*A*b)/a - 6*C)*x)/(35*a^2*(a + b*x^2)^(5/2)) + (35*
B - 3*((29*A*b)/a - 8*C)*x)/(105*a^3*(a + b*x^2)^(3/2)) + (35*B - ((93*A*b)/a - 16*C)*x)/(35*a^4*Sqrt[a + b*x^
2]) - (A*Sqrt[a + b*x^2])/(a^5*x) - (B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(9/2)

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Rubi [A]  time = 0.381314, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1805, 807, 266, 63, 208} \[ \frac{35 B-x \left (\frac{93 A b}{a}-16 C\right )}{35 a^4 \sqrt{a+b x^2}}+\frac{35 B-3 x \left (\frac{29 A b}{a}-8 C\right )}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{7 B-x \left (\frac{13 A b}{a}-6 C\right )}{35 a^2 \left (a+b x^2\right )^{5/2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}+\frac{B-x \left (\frac{A b}{a}-C\right )}{7 a \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(x^2*(a + b*x^2)^(9/2)),x]

[Out]

(B - ((A*b)/a - C)*x)/(7*a*(a + b*x^2)^(7/2)) + (7*B - ((13*A*b)/a - 6*C)*x)/(35*a^2*(a + b*x^2)^(5/2)) + (35*
B - 3*((29*A*b)/a - 8*C)*x)/(105*a^3*(a + b*x^2)^(3/2)) + (35*B - ((93*A*b)/a - 16*C)*x)/(35*a^4*Sqrt[a + b*x^
2]) - (A*Sqrt[a + b*x^2])/(a^5*x) - (B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(9/2)

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{x^2 \left (a+b x^2\right )^{9/2}} \, dx &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{-7 A-7 B x+6 \left (\frac{A b}{a}-C\right ) x^2}{x^2 \left (a+b x^2\right )^{7/2}} \, dx}{7 a}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{35 A+35 B x-4 \left (\frac{13 A b}{a}-6 C\right ) x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{-105 A-105 B x+6 \left (\frac{29 A b}{a}-8 C\right ) x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx}{105 a^3}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 B-\left (\frac{93 A b}{a}-16 C\right ) x}{35 a^4 \sqrt{a+b x^2}}+\frac{\int \frac{105 A+105 B x}{x^2 \sqrt{a+b x^2}} \, dx}{105 a^4}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 B-\left (\frac{93 A b}{a}-16 C\right ) x}{35 a^4 \sqrt{a+b x^2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}+\frac{B \int \frac{1}{x \sqrt{a+b x^2}} \, dx}{a^4}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 B-\left (\frac{93 A b}{a}-16 C\right ) x}{35 a^4 \sqrt{a+b x^2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}+\frac{B \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{2 a^4}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 B-\left (\frac{93 A b}{a}-16 C\right ) x}{35 a^4 \sqrt{a+b x^2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}+\frac{B \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{a^4 b}\\ &=\frac{B-\left (\frac{A b}{a}-C\right ) x}{7 a \left (a+b x^2\right )^{7/2}}+\frac{7 B-\left (\frac{13 A b}{a}-6 C\right ) x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 B-3 \left (\frac{29 A b}{a}-8 C\right ) x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 B-\left (\frac{93 A b}{a}-16 C\right ) x}{35 a^4 \sqrt{a+b x^2}}-\frac{A \sqrt{a+b x^2}}{a^5 x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.15615, size = 158, normalized size = 0.84 \[ \frac{14 a^2 b^2 x^4 (x (25 B+12 C x)-120 A)+14 a^3 b x^2 (x (29 B+15 C x)-60 A)+a^4 (x (176 B+105 C x)-105 A)+3 a b^3 x^6 (x (35 B+16 C x)-448 A)-105 \sqrt{a} B x \left (a+b x^2\right )^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-384 A b^4 x^8}{105 a^5 x \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(x^2*(a + b*x^2)^(9/2)),x]

[Out]

(-384*A*b^4*x^8 + 14*a^2*b^2*x^4*(-120*A + x*(25*B + 12*C*x)) + 14*a^3*b*x^2*(-60*A + x*(29*B + 15*C*x)) + 3*a
*b^3*x^6*(-448*A + x*(35*B + 16*C*x)) + a^4*(-105*A + x*(176*B + 105*C*x)) - 105*Sqrt[a]*B*x*(a + b*x^2)^(7/2)
*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(105*a^5*x*(a + b*x^2)^(7/2))

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Maple [A]  time = 0.009, size = 240, normalized size = 1.3 \begin{align*}{\frac{Cx}{7\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{6\,Cx}{35\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,Cx}{35\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,Cx}{35\,{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{B}{7\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{B}{5\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{B}{3\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{B}{{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{B\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{9}{2}}}}-{\frac{A}{ax} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{8\,Abx}{7\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{48\,Abx}{35\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}-{\frac{64\,Abx}{35\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{128\,Abx}{35\,{a}^{5}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/x^2/(b*x^2+a)^(9/2),x)

[Out]

1/7*C*x/a/(b*x^2+a)^(7/2)+6/35*C/a^2*x/(b*x^2+a)^(5/2)+8/35*C/a^3*x/(b*x^2+a)^(3/2)+16/35*C/a^4*x/(b*x^2+a)^(1
/2)+1/7*B/a/(b*x^2+a)^(7/2)+1/5*B/a^2/(b*x^2+a)^(5/2)+1/3*B/a^3/(b*x^2+a)^(3/2)+B/a^4/(b*x^2+a)^(1/2)-B/a^(9/2
)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-A/a/x/(b*x^2+a)^(7/2)-8/7*A*b/a^2*x/(b*x^2+a)^(7/2)-48/35*A*b/a^3*x/(b
*x^2+a)^(5/2)-64/35*A*b/a^4*x/(b*x^2+a)^(3/2)-128/35*A*b/a^5*x/(b*x^2+a)^(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((C*x^2+B*x+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88288, size = 1172, normalized size = 6.23 \begin{align*} \left [\frac{105 \,{\left (B b^{4} x^{9} + 4 \, B a b^{3} x^{7} + 6 \, B a^{2} b^{2} x^{5} + 4 \, B a^{3} b x^{3} + B a^{4} x\right )} \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (105 \, B a b^{3} x^{7} + 350 \, B a^{2} b^{2} x^{5} + 48 \,{\left (C a b^{3} - 8 \, A b^{4}\right )} x^{8} + 406 \, B a^{3} b x^{3} + 168 \,{\left (C a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{6} + 176 \, B a^{4} x - 105 \, A a^{4} + 210 \,{\left (C a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{4} + 105 \,{\left (C a^{4} - 8 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{210 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}, \frac{105 \,{\left (B b^{4} x^{9} + 4 \, B a b^{3} x^{7} + 6 \, B a^{2} b^{2} x^{5} + 4 \, B a^{3} b x^{3} + B a^{4} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (105 \, B a b^{3} x^{7} + 350 \, B a^{2} b^{2} x^{5} + 48 \,{\left (C a b^{3} - 8 \, A b^{4}\right )} x^{8} + 406 \, B a^{3} b x^{3} + 168 \,{\left (C a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{6} + 176 \, B a^{4} x - 105 \, A a^{4} + 210 \,{\left (C a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{4} + 105 \,{\left (C a^{4} - 8 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{105 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

[1/210*(105*(B*b^4*x^9 + 4*B*a*b^3*x^7 + 6*B*a^2*b^2*x^5 + 4*B*a^3*b*x^3 + B*a^4*x)*sqrt(a)*log(-(b*x^2 - 2*sq
rt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(105*B*a*b^3*x^7 + 350*B*a^2*b^2*x^5 + 48*(C*a*b^3 - 8*A*b^4)*x^8 + 406*
B*a^3*b*x^3 + 168*(C*a^2*b^2 - 8*A*a*b^3)*x^6 + 176*B*a^4*x - 105*A*a^4 + 210*(C*a^3*b - 8*A*a^2*b^2)*x^4 + 10
5*(C*a^4 - 8*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x
), 1/105*(105*(B*b^4*x^9 + 4*B*a*b^3*x^7 + 6*B*a^2*b^2*x^5 + 4*B*a^3*b*x^3 + B*a^4*x)*sqrt(-a)*arctan(sqrt(-a)
/sqrt(b*x^2 + a)) + (105*B*a*b^3*x^7 + 350*B*a^2*b^2*x^5 + 48*(C*a*b^3 - 8*A*b^4)*x^8 + 406*B*a^3*b*x^3 + 168*
(C*a^2*b^2 - 8*A*a*b^3)*x^6 + 176*B*a^4*x - 105*A*a^4 + 210*(C*a^3*b - 8*A*a^2*b^2)*x^4 + 105*(C*a^4 - 8*A*a^3
*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/x**2/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17356, size = 323, normalized size = 1.72 \begin{align*} \frac{{\left ({\left ({\left ({\left (3 \,{\left (x{\left (\frac{35 \, B b^{3}}{a^{4}} + \frac{{\left (16 \, C a^{20} b^{6} - 93 \, A a^{19} b^{7}\right )} x}{a^{24} b^{3}}\right )} + \frac{28 \,{\left (2 \, C a^{21} b^{5} - 11 \, A a^{20} b^{6}\right )}}{a^{24} b^{3}}\right )} x + \frac{350 \, B b^{2}}{a^{3}}\right )} x + \frac{210 \,{\left (C a^{22} b^{4} - 5 \, A a^{21} b^{5}\right )}}{a^{24} b^{3}}\right )} x + \frac{406 \, B b}{a^{2}}\right )} x + \frac{105 \,{\left (C a^{23} b^{3} - 4 \, A a^{22} b^{4}\right )}}{a^{24} b^{3}}\right )} x + \frac{176 \, B}{a}}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} + \frac{2 \, B \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{4}} + \frac{2 \, A \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

1/105*(((((3*(x*(35*B*b^3/a^4 + (16*C*a^20*b^6 - 93*A*a^19*b^7)*x/(a^24*b^3)) + 28*(2*C*a^21*b^5 - 11*A*a^20*b
^6)/(a^24*b^3))*x + 350*B*b^2/a^3)*x + 210*(C*a^22*b^4 - 5*A*a^21*b^5)/(a^24*b^3))*x + 406*B*b/a^2)*x + 105*(C
*a^23*b^3 - 4*A*a^22*b^4)/(a^24*b^3))*x + 176*B/a)/(b*x^2 + a)^(7/2) + 2*B*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a
))/sqrt(-a))/(sqrt(-a)*a^4) + 2*A*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a^4)