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

Optimal. Leaf size=138 \[ \frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}+\frac{-a C+A b+b B x}{7 a b \left (a+b x^2\right )^{7/2}} \]

[Out]

(A*b - a*C + b*B*x)/(7*a*b*(a + b*x^2)^(7/2)) + (7*A + 6*B*x)/(35*a^2*(a + b*x^2)^(5/2)) + (35*A + 24*B*x)/(10
5*a^3*(a + b*x^2)^(3/2)) + (35*A + 16*B*x)/(35*a^4*Sqrt[a + b*x^2]) - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(
9/2)

________________________________________________________________________________________

Rubi [A]  time = 0.162066, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1805, 823, 12, 266, 63, 208} \[ \frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}+\frac{-a C+A b+b B x}{7 a b \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b - a*C + b*B*x)/(7*a*b*(a + b*x^2)^(7/2)) + (7*A + 6*B*x)/(35*a^2*(a + b*x^2)^(5/2)) + (35*A + 24*B*x)/(10
5*a^3*(a + b*x^2)^(3/2)) + (35*A + 16*B*x)/(35*a^4*Sqrt[a + b*x^2]) - (A*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \left (a+b x^2\right )^{9/2}} \, dx &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{-7 A-6 B x}{x \left (a+b x^2\right )^{7/2}} \, dx}{7 a}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{35 a A b+24 a b B x}{x \left (a+b x^2\right )^{5/2}} \, dx}{35 a^3 b}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{-105 a^2 A b^2-48 a^2 b^2 B x}{x \left (a+b x^2\right )^{3/2}} \, dx}{105 a^5 b^2}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{\int \frac{105 a^3 A b^3}{x \sqrt{a+b x^2}} \, dx}{105 a^7 b^3}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{A \int \frac{1}{x \sqrt{a+b x^2}} \, dx}{a^4}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{A \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{2 a^4}\\ &=\frac{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}+\frac{A \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{A b-a C+b B x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac{7 A+6 B x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac{35 A+24 B x}{105 a^3 \left (a+b x^2\right )^{3/2}}+\frac{35 A+16 B x}{35 a^4 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.156086, size = 120, normalized size = 0.87 \[ \frac{14 a^2 b^2 x^2 (29 A+15 B x)+a^3 b (176 A+105 B x)-15 a^4 C+14 a b^3 x^4 (25 A+12 B x)+3 b^4 x^6 (35 A+16 B x)}{105 a^4 b \left (a+b x^2\right )^{7/2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^4*C + 14*a*b^3*x^4*(25*A + 12*B*x) + 14*a^2*b^2*x^2*(29*A + 15*B*x) + 3*b^4*x^6*(35*A + 16*B*x) + a^3*b
*(176*A + 105*B*x))/(105*a^4*b*(a + b*x^2)^(7/2)) - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(9/2)

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 169, normalized size = 1.2 \begin{align*} -{\frac{C}{7\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{Bx}{7\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{6\,Bx}{35\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,Bx}{35\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,Bx}{35\,{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{A}{7\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{A}{5\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{A}{3\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{A}{{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*C/b/(b*x^2+a)^(7/2)+1/7*B*x/a/(b*x^2+a)^(7/2)+6/35*B/a^2*x/(b*x^2+a)^(5/2)+8/35*B/a^3*x/(b*x^2+a)^(3/2)+1
6/35*B/a^4*x/(b*x^2+a)^(1/2)+1/7*A/a/(b*x^2+a)^(7/2)+1/5*A/a^2/(b*x^2+a)^(5/2)+1/3*A/a^3/(b*x^2+a)^(3/2)+A/a^4
/(b*x^2+a)^(1/2)-A/a^(9/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)

________________________________________________________________________________________

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/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.85621, size = 1040, normalized size = 7.54 \begin{align*} \left [\frac{105 \,{\left (A b^{5} x^{8} + 4 \, A a b^{4} x^{6} + 6 \, A a^{2} b^{3} x^{4} + 4 \, A a^{3} b^{2} x^{2} + A a^{4} b\right )} \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (48 \, B a b^{4} x^{7} + 105 \, A a b^{4} x^{6} + 168 \, B a^{2} b^{3} x^{5} + 350 \, A a^{2} b^{3} x^{4} + 210 \, B a^{3} b^{2} x^{3} + 406 \, A a^{3} b^{2} x^{2} + 105 \, B a^{4} b x - 15 \, C a^{5} + 176 \, A a^{4} b\right )} \sqrt{b x^{2} + a}}{210 \,{\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}, \frac{105 \,{\left (A b^{5} x^{8} + 4 \, A a b^{4} x^{6} + 6 \, A a^{2} b^{3} x^{4} + 4 \, A a^{3} b^{2} x^{2} + A a^{4} b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (48 \, B a b^{4} x^{7} + 105 \, A a b^{4} x^{6} + 168 \, B a^{2} b^{3} x^{5} + 350 \, A a^{2} b^{3} x^{4} + 210 \, B a^{3} b^{2} x^{3} + 406 \, A a^{3} b^{2} x^{2} + 105 \, B a^{4} b x - 15 \, C a^{5} + 176 \, A a^{4} b\right )} \sqrt{b x^{2} + a}}{105 \,{\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/210*(105*(A*b^5*x^8 + 4*A*a*b^4*x^6 + 6*A*a^2*b^3*x^4 + 4*A*a^3*b^2*x^2 + A*a^4*b)*sqrt(a)*log(-(b*x^2 - 2*
sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(48*B*a*b^4*x^7 + 105*A*a*b^4*x^6 + 168*B*a^2*b^3*x^5 + 350*A*a^2*b^3*
x^4 + 210*B*a^3*b^2*x^3 + 406*A*a^3*b^2*x^2 + 105*B*a^4*b*x - 15*C*a^5 + 176*A*a^4*b)*sqrt(b*x^2 + a))/(a^5*b^
5*x^8 + 4*a^6*b^4*x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b), 1/105*(105*(A*b^5*x^8 + 4*A*a*b^4*x^6 + 6*A*a^
2*b^3*x^4 + 4*A*a^3*b^2*x^2 + A*a^4*b)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (48*B*a*b^4*x^7 + 105*A*a*b
^4*x^6 + 168*B*a^2*b^3*x^5 + 350*A*a^2*b^3*x^4 + 210*B*a^3*b^2*x^3 + 406*A*a^3*b^2*x^2 + 105*B*a^4*b*x - 15*C*
a^5 + 176*A*a^4*b)*sqrt(b*x^2 + a))/(a^5*b^5*x^8 + 4*a^6*b^4*x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b)]

________________________________________________________________________________________

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/(b*x**2+a)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.2207, size = 205, normalized size = 1.49 \begin{align*} \frac{{\left ({\left ({\left ({\left (3 \,{\left ({\left (\frac{16 \, B b^{3} x}{a^{4}} + \frac{35 \, A b^{3}}{a^{4}}\right )} x + \frac{56 \, B b^{2}}{a^{3}}\right )} x + \frac{350 \, A b^{2}}{a^{3}}\right )} x + \frac{210 \, B b}{a^{2}}\right )} x + \frac{406 \, A b}{a^{2}}\right )} x + \frac{105 \, B}{a}\right )} x - \frac{15 \, C a^{14} b^{2} - 176 \, A a^{13} b^{3}}{a^{14} b^{3}}}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} + \frac{2 \, A \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(((((3*((16*B*b^3*x/a^4 + 35*A*b^3/a^4)*x + 56*B*b^2/a^3)*x + 350*A*b^2/a^3)*x + 210*B*b/a^2)*x + 406*A*
b/a^2)*x + 105*B/a)*x - (15*C*a^14*b^2 - 176*A*a^13*b^3)/(a^14*b^3))/(b*x^2 + a)^(7/2) + 2*A*arctan(-(sqrt(b)*
x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^4)