3.1.57 \(\int \frac {A+B x+C x^2}{x^3 (a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=219 \[ \frac {(9 A b-2 a C) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{11/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}} \]

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Rubi [A]  time = 0.48, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1805, 1807, 807, 266, 63, 208} \begin {gather*} -\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}+\frac {(9 A b-2 a C) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{11/2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(x^3*(a + b*x^2)^(9/2)),x]

[Out]

-(a*((A*b)/a - C) + b*B*x)/(7*a^2*(a + b*x^2)^(7/2)) - (7*(2*A*b - a*C) + 13*b*B*x)/(35*a^3*(a + b*x^2)^(5/2))
- (35*(3*A*b - a*C) + 87*b*B*x)/(105*a^4*(a + b*x^2)^(3/2)) - (35*(4*A*b - a*C) + 93*b*B*x)/(35*a^5*Sqrt[a +
b*x^2]) - (A*Sqrt[a + b*x^2])/(2*a^5*x^2) - (B*Sqrt[a + b*x^2])/(a^5*x) + ((9*A*b - 2*a*C)*ArcTanh[Sqrt[a + b*
x^2]/Sqrt[a]])/(2*a^(11/2))

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]

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 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 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 1807

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

Rubi steps

\begin {align*} \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2\right )^{9/2}} \, dx &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {-7 A-7 B x+7 \left (\frac {A b}{a}-C\right ) x^2+\frac {6 b B x^3}{a}}{x^3 \left (a+b x^2\right )^{7/2}} \, dx}{7 a}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}+\frac {\int \frac {35 A+35 B x-35 \left (\frac {2 A b}{a}-C\right ) x^2-\frac {52 b B x^3}{a}}{x^3 \left (a+b x^2\right )^{5/2}} \, dx}{35 a^2}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {-105 A-105 B x+105 \left (\frac {3 A b}{a}-C\right ) x^2+\frac {174 b B x^3}{a}}{x^3 \left (a+b x^2\right )^{3/2}} \, dx}{105 a^3}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}+\frac {\int \frac {105 A+105 B x-105 \left (\frac {4 A b}{a}-C\right ) x^2}{x^3 \sqrt {a+b x^2}} \, dx}{105 a^4}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {\int \frac {-210 a B+105 (9 A b-2 a C) x}{x^2 \sqrt {a+b x^2}} \, dx}{210 a^5}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x}-\frac {(9 A b-2 a C) \int \frac {1}{x \sqrt {a+b x^2}} \, dx}{2 a^5}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x}-\frac {(9 A b-2 a C) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{4 a^5}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x}-\frac {(9 A b-2 a C) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 a^5 b}\\ &=-\frac {a \left (\frac {A b}{a}-C\right )+b B x}{7 a^2 \left (a+b x^2\right )^{7/2}}-\frac {7 (2 A b-a C)+13 b B x}{35 a^3 \left (a+b x^2\right )^{5/2}}-\frac {35 (3 A b-a C)+87 b B x}{105 a^4 \left (a+b x^2\right )^{3/2}}-\frac {35 (4 A b-a C)+93 b B x}{35 a^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^5 x^2}-\frac {B \sqrt {a+b x^2}}{a^5 x}+\frac {(9 A b-2 a C) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 178, normalized size = 0.81 \begin {gather*} \frac {\frac {a^5 \left (-105 A-210 B x+352 C x^2\right )}{x^2}-4 a^4 b (396 A+7 x (60 B-29 C x))+14 a^3 b^2 x^2 (10 x (5 C x-24 B)-261 A)+42 a^2 b^3 x^4 (x (5 C x-64 B)-75 A)-3 a b^4 x^6 (315 A+256 B x)+\frac {105 \left (a+b x^2\right )^4 (9 A b-2 a C) \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}}{210 a^6 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(x^3*(a + b*x^2)^(9/2)),x]

[Out]

(-3*a*b^4*x^6*(315*A + 256*B*x) + (a^5*(-105*A - 210*B*x + 352*C*x^2))/x^2 - 4*a^4*b*(396*A + 7*x*(60*B - 29*C
*x)) + 42*a^2*b^3*x^4*(-75*A + x*(-64*B + 5*C*x)) + 14*a^3*b^2*x^2*(-261*A + 10*x*(-24*B + 5*C*x)) + (105*(9*A
*b - 2*a*C)*(a + b*x^2)^4*ArcTanh[Sqrt[1 + (b*x^2)/a]])/Sqrt[1 + (b*x^2)/a])/(210*a^6*(a + b*x^2)^(7/2))

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IntegrateAlgebraic [A]  time = 1.57, size = 202, normalized size = 0.92 \begin {gather*} \frac {(2 a C-9 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{11/2}}+\frac {-105 a^4 A-210 a^4 B x+352 a^4 C x^2-1584 a^3 A b x^2-1680 a^3 b B x^3+812 a^3 b C x^4-3654 a^2 A b^2 x^4-3360 a^2 b^2 B x^5+700 a^2 b^2 C x^6-3150 a A b^3 x^6-2688 a b^3 B x^7+210 a b^3 C x^8-945 A b^4 x^8-768 b^4 B x^9}{210 a^5 x^2 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x + C*x^2)/(x^3*(a + b*x^2)^(9/2)),x]

[Out]

(-105*a^4*A - 210*a^4*B*x - 1584*a^3*A*b*x^2 + 352*a^4*C*x^2 - 1680*a^3*b*B*x^3 - 3654*a^2*A*b^2*x^4 + 812*a^3
*b*C*x^4 - 3360*a^2*b^2*B*x^5 - 3150*a*A*b^3*x^6 + 700*a^2*b^2*C*x^6 - 2688*a*b^3*B*x^7 - 945*A*b^4*x^8 + 210*
a*b^3*C*x^8 - 768*b^4*B*x^9)/(210*a^5*x^2*(a + b*x^2)^(7/2)) + ((-9*A*b + 2*a*C)*ArcTanh[(Sqrt[b]*x - Sqrt[a +
b*x^2])/Sqrt[a]])/a^(11/2)

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fricas [A]  time = 1.17, size = 688, normalized size = 3.14 \begin {gather*} \left [-\frac {105 \, {\left ({\left (2 \, C a b^{4} - 9 \, A b^{5}\right )} x^{10} + 4 \, {\left (2 \, C a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{8} + 6 \, {\left (2 \, C a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{6} + 4 \, {\left (2 \, C a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{4} + {\left (2 \, C a^{5} - 9 \, A a^{4} b\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (768 \, B a b^{4} x^{9} + 2688 \, B a^{2} b^{3} x^{7} + 3360 \, B a^{3} b^{2} x^{5} + 1680 \, B a^{4} b x^{3} - 105 \, {\left (2 \, C a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{8} + 210 \, B a^{5} x - 350 \, {\left (2 \, C a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{6} + 105 \, A a^{5} - 406 \, {\left (2 \, C a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{4} - 176 \, {\left (2 \, C a^{5} - 9 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{420 \, {\left (a^{6} b^{4} x^{10} + 4 \, a^{7} b^{3} x^{8} + 6 \, a^{8} b^{2} x^{6} + 4 \, a^{9} b x^{4} + a^{10} x^{2}\right )}}, \frac {105 \, {\left ({\left (2 \, C a b^{4} - 9 \, A b^{5}\right )} x^{10} + 4 \, {\left (2 \, C a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{8} + 6 \, {\left (2 \, C a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{6} + 4 \, {\left (2 \, C a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{4} + {\left (2 \, C a^{5} - 9 \, A a^{4} b\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (768 \, B a b^{4} x^{9} + 2688 \, B a^{2} b^{3} x^{7} + 3360 \, B a^{3} b^{2} x^{5} + 1680 \, B a^{4} b x^{3} - 105 \, {\left (2 \, C a^{2} b^{3} - 9 \, A a b^{4}\right )} x^{8} + 210 \, B a^{5} x - 350 \, {\left (2 \, C a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{6} + 105 \, A a^{5} - 406 \, {\left (2 \, C a^{4} b - 9 \, A a^{3} b^{2}\right )} x^{4} - 176 \, {\left (2 \, C a^{5} - 9 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{210 \, {\left (a^{6} b^{4} x^{10} + 4 \, a^{7} b^{3} x^{8} + 6 \, a^{8} b^{2} x^{6} + 4 \, a^{9} b x^{4} + a^{10} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/x^3/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

[-1/420*(105*((2*C*a*b^4 - 9*A*b^5)*x^10 + 4*(2*C*a^2*b^3 - 9*A*a*b^4)*x^8 + 6*(2*C*a^3*b^2 - 9*A*a^2*b^3)*x^6
+ 4*(2*C*a^4*b - 9*A*a^3*b^2)*x^4 + (2*C*a^5 - 9*A*a^4*b)*x^2)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a
) + 2*a)/x^2) + 2*(768*B*a*b^4*x^9 + 2688*B*a^2*b^3*x^7 + 3360*B*a^3*b^2*x^5 + 1680*B*a^4*b*x^3 - 105*(2*C*a^2
*b^3 - 9*A*a*b^4)*x^8 + 210*B*a^5*x - 350*(2*C*a^3*b^2 - 9*A*a^2*b^3)*x^6 + 105*A*a^5 - 406*(2*C*a^4*b - 9*A*a
^3*b^2)*x^4 - 176*(2*C*a^5 - 9*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^6*b^4*x^10 + 4*a^7*b^3*x^8 + 6*a^8*b^2*x^6 +
4*a^9*b*x^4 + a^10*x^2), 1/210*(105*((2*C*a*b^4 - 9*A*b^5)*x^10 + 4*(2*C*a^2*b^3 - 9*A*a*b^4)*x^8 + 6*(2*C*a^3
*b^2 - 9*A*a^2*b^3)*x^6 + 4*(2*C*a^4*b - 9*A*a^3*b^2)*x^4 + (2*C*a^5 - 9*A*a^4*b)*x^2)*sqrt(-a)*arctan(sqrt(-a
)/sqrt(b*x^2 + a)) - (768*B*a*b^4*x^9 + 2688*B*a^2*b^3*x^7 + 3360*B*a^3*b^2*x^5 + 1680*B*a^4*b*x^3 - 105*(2*C*
a^2*b^3 - 9*A*a*b^4)*x^8 + 210*B*a^5*x - 350*(2*C*a^3*b^2 - 9*A*a^2*b^3)*x^6 + 105*A*a^5 - 406*(2*C*a^4*b - 9*
A*a^3*b^2)*x^4 - 176*(2*C*a^5 - 9*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^6*b^4*x^10 + 4*a^7*b^3*x^8 + 6*a^8*b^2*x^6
+ 4*a^9*b*x^4 + a^10*x^2)]

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giac [A]  time = 0.48, size = 325, normalized size = 1.48 \begin {gather*} -\frac {{\left ({\left ({\left ({\left (3 \, {\left ({\left (\frac {93 \, B b^{4} x}{a^{5}} - \frac {35 \, {\left (C a^{24} b^{6} - 4 \, A a^{23} b^{7}\right )}}{a^{28} b^{3}}\right )} x + \frac {308 \, B b^{3}}{a^{4}}\right )} x - \frac {35 \, {\left (10 \, C a^{25} b^{5} - 39 \, A a^{24} b^{6}\right )}}{a^{28} b^{3}}\right )} x + \frac {1050 \, B b^{2}}{a^{3}}\right )} x - \frac {14 \, {\left (29 \, C a^{26} b^{4} - 108 \, A a^{25} b^{5}\right )}}{a^{28} b^{3}}\right )} x + \frac {420 \, B b}{a^{2}}\right )} x - \frac {2 \, {\left (88 \, C a^{27} b^{3} - 291 \, A a^{26} b^{4}\right )}}{a^{28} b^{3}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {{\left (2 \, C a - 9 \, A b\right )} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{5}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/x^3/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

-1/105*(((((3*((93*B*b^4*x/a^5 - 35*(C*a^24*b^6 - 4*A*a^23*b^7)/(a^28*b^3))*x + 308*B*b^3/a^4)*x - 35*(10*C*a^
25*b^5 - 39*A*a^24*b^6)/(a^28*b^3))*x + 1050*B*b^2/a^3)*x - 14*(29*C*a^26*b^4 - 108*A*a^25*b^5)/(a^28*b^3))*x
+ 420*B*b/a^2)*x - 2*(88*C*a^27*b^3 - 291*A*a^26*b^4)/(a^28*b^3))/(b*x^2 + a)^(7/2) + (2*C*a - 9*A*b)*arctan(-
(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^5) + ((sqrt(b)*x - sqrt(b*x^2 + a))^3*A*b + 2*(sqrt(b)*x -
sqrt(b*x^2 + a))^2*B*a*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A*a*b - 2*B*a^2*sqrt(b))/(((sqrt(b)*x - sqrt(b
*x^2 + a))^2 - a)^2*a^5)

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maple [A]  time = 0.02, size = 288, normalized size = 1.32 \begin {gather*} -\frac {8 B b x}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2}}-\frac {9 A b}{14 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2}}-\frac {48 B b x}{35 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}+\frac {C}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a}-\frac {9 A b}{10 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}-\frac {B}{\left (b \,x^{2}+a \right )^{\frac {7}{2}} a x}-\frac {64 B b x}{35 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{4}}+\frac {C}{5 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{2}}-\frac {A}{2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a \,x^{2}}-\frac {3 A b}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{4}}-\frac {128 B b x}{35 \sqrt {b \,x^{2}+a}\, a^{5}}+\frac {C}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}+\frac {9 A b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {11}{2}}}-\frac {C \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {9}{2}}}-\frac {9 A b}{2 \sqrt {b \,x^{2}+a}\, a^{5}}+\frac {C}{\sqrt {b \,x^{2}+a}\, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/x^3/(b*x^2+a)^(9/2),x)

[Out]

-1/2*A/a/x^2/(b*x^2+a)^(7/2)-9/14*A/a^2*b/(b*x^2+a)^(7/2)-9/10*A/a^3*b/(b*x^2+a)^(5/2)-3/2*A/a^4*b/(b*x^2+a)^(
3/2)-9/2*A/a^5*b/(b*x^2+a)^(1/2)+9/2*A/a^(11/2)*b*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-B/a/x/(b*x^2+a)^(7/2)-
8/7*B/a^2*b*x/(b*x^2+a)^(7/2)-48/35*B/a^3*b*x/(b*x^2+a)^(5/2)-64/35*B/a^4*b*x/(b*x^2+a)^(3/2)-128/35*B/a^5*b*x
/(b*x^2+a)^(1/2)+1/7*C/a/(b*x^2+a)^(7/2)+1/5*C/a^2/(b*x^2+a)^(5/2)+1/3*C/a^3/(b*x^2+a)^(3/2)+C/a^4/(b*x^2+a)^(
1/2)-C/a^(9/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 1.49, size = 265, normalized size = 1.21 \begin {gather*} -\frac {128 \, B b x}{35 \, \sqrt {b x^{2} + a} a^{5}} - \frac {64 \, B b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {48 \, B b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3}} - \frac {8 \, B b x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2}} - \frac {C \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {9}{2}}} + \frac {9 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {11}{2}}} + \frac {C}{\sqrt {b x^{2} + a} a^{4}} + \frac {C}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {C}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {C}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {9 \, A b}{2 \, \sqrt {b x^{2} + a} a^{5}} - \frac {3 \, A b}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {9 \, A b}{10 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3}} - \frac {9 \, A b}{14 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2}} - \frac {B}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} a x} - \frac {A}{2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/x^3/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

-128/35*B*b*x/(sqrt(b*x^2 + a)*a^5) - 64/35*B*b*x/((b*x^2 + a)^(3/2)*a^4) - 48/35*B*b*x/((b*x^2 + a)^(5/2)*a^3
) - 8/7*B*b*x/((b*x^2 + a)^(7/2)*a^2) - C*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(9/2) + 9/2*A*b*arcsinh(a/(sqrt(a*b)
*abs(x)))/a^(11/2) + C/(sqrt(b*x^2 + a)*a^4) + 1/3*C/((b*x^2 + a)^(3/2)*a^3) + 1/5*C/((b*x^2 + a)^(5/2)*a^2) +
1/7*C/((b*x^2 + a)^(7/2)*a) - 9/2*A*b/(sqrt(b*x^2 + a)*a^5) - 3/2*A*b/((b*x^2 + a)^(3/2)*a^4) - 9/10*A*b/((b*
x^2 + a)^(5/2)*a^3) - 9/14*A*b/((b*x^2 + a)^(7/2)*a^2) - B/((b*x^2 + a)^(7/2)*a*x) - 1/2*A/((b*x^2 + a)^(7/2)*
a*x^2)

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mupad [B]  time = 2.52, size = 279, normalized size = 1.27 \begin {gather*} \frac {\frac {C}{7\,a}+\frac {C\,{\left (b\,x^2+a\right )}^2}{3\,a^3}+\frac {C\,{\left (b\,x^2+a\right )}^3}{a^4}+\frac {C\,\left (b\,x^2+a\right )}{5\,a^2}}{{\left (b\,x^2+a\right )}^{7/2}}-\frac {\frac {A\,b}{7\,a}+\frac {9\,A\,b\,\left (b\,x^2+a\right )}{35\,a^2}+\frac {3\,A\,b\,{\left (b\,x^2+a\right )}^2}{5\,a^3}+\frac {3\,A\,b\,{\left (b\,x^2+a\right )}^3}{a^4}-\frac {9\,A\,b\,{\left (b\,x^2+a\right )}^4}{2\,a^5}}{a\,{\left (b\,x^2+a\right )}^{7/2}-{\left (b\,x^2+a\right )}^{9/2}}-\frac {\frac {B}{a^4}+\frac {128\,B\,b\,x^2}{35\,a^5}}{x\,\sqrt {b\,x^2+a}}-\frac {C\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{9/2}}+\frac {9\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{11/2}}-\frac {29\,B\,b\,x}{35\,a^4\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {13\,B\,b\,x}{35\,a^3\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {B\,b\,x}{7\,a^2\,{\left (b\,x^2+a\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x + C*x^2)/(x^3*(a + b*x^2)^(9/2)),x)

[Out]

(C/(7*a) + (C*(a + b*x^2)^2)/(3*a^3) + (C*(a + b*x^2)^3)/a^4 + (C*(a + b*x^2))/(5*a^2))/(a + b*x^2)^(7/2) - ((
A*b)/(7*a) + (9*A*b*(a + b*x^2))/(35*a^2) + (3*A*b*(a + b*x^2)^2)/(5*a^3) + (3*A*b*(a + b*x^2)^3)/a^4 - (9*A*b
*(a + b*x^2)^4)/(2*a^5))/(a*(a + b*x^2)^(7/2) - (a + b*x^2)^(9/2)) - (B/a^4 + (128*B*b*x^2)/(35*a^5))/(x*(a +
b*x^2)^(1/2)) - (C*atanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(9/2) + (9*A*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(
11/2)) - (29*B*b*x)/(35*a^4*(a + b*x^2)^(3/2)) - (13*B*b*x)/(35*a^3*(a + b*x^2)^(5/2)) - (B*b*x)/(7*a^2*(a + b
*x^2)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/x**3/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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