3.2.46 \(\int \frac {c+d x^2+e x^4+f x^6}{x^9 \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=195 \[ \frac {\sqrt {a+b x^2} (7 b c-8 a d)}{48 a^2 x^6}-\frac {\sqrt {a+b x^2} \left (48 a^2 e-40 a b d+35 b^2 c\right )}{192 a^3 x^4}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (-64 a^3 f+48 a^2 b e-40 a b^2 d+35 b^3 c\right )}{128 a^{9/2}}+\frac {\sqrt {a+b x^2} \left (-64 a^3 f+48 a^2 b e-40 a b^2 d+35 b^3 c\right )}{128 a^4 x^2}-\frac {c \sqrt {a+b x^2}}{8 a x^8} \]

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Rubi [A]  time = 0.35, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {1799, 1621, 897, 1157, 385, 199, 208} \begin {gather*} \frac {\sqrt {a+b x^2} \left (48 a^2 b e-64 a^3 f-40 a b^2 d+35 b^3 c\right )}{128 a^4 x^2}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (48 a^2 b e-64 a^3 f-40 a b^2 d+35 b^3 c\right )}{128 a^{9/2}}-\frac {\sqrt {a+b x^2} \left (48 a^2 e-40 a b d+35 b^2 c\right )}{192 a^3 x^4}+\frac {\sqrt {a+b x^2} (7 b c-8 a d)}{48 a^2 x^6}-\frac {c \sqrt {a+b x^2}}{8 a x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^9*Sqrt[a + b*x^2]),x]

[Out]

-(c*Sqrt[a + b*x^2])/(8*a*x^8) + ((7*b*c - 8*a*d)*Sqrt[a + b*x^2])/(48*a^2*x^6) - ((35*b^2*c - 40*a*b*d + 48*a
^2*e)*Sqrt[a + b*x^2])/(192*a^3*x^4) + ((35*b^3*c - 40*a*b^2*d + 48*a^2*b*e - 64*a^3*f)*Sqrt[a + b*x^2])/(128*
a^4*x^2) - (b*(35*b^3*c - 40*a*b^2*d + 48*a^2*b*e - 64*a^3*f)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(9/2))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {c+d x^2+e x^4+f x^6}{x^9 \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (7 b c-8 a d)-4 a e x-4 a f x^2}{x^4 \sqrt {a+b x}} \, dx,x,x^2\right )}{8 a}\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}-\frac {\operatorname {Subst}\left (\int \frac {\frac {\frac {1}{2} b^2 (7 b c-8 a d)+4 a^2 b e-4 a^3 f}{b^2}-\frac {\left (4 a b e-8 a^2 f\right ) x^2}{b^2}-\frac {4 a f x^4}{b^2}}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^4} \, dx,x,\sqrt {a+b x^2}\right )}{4 a b}\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}+\frac {(7 b c-8 a d) \sqrt {a+b x^2}}{48 a^2 x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (-35 b c+40 a d-\frac {48 a^2 e}{b}+\frac {48 a^3 f}{b^2}\right )-\frac {24 a^2 f x^2}{b^2}}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^3} \, dx,x,\sqrt {a+b x^2}\right )}{24 a^2}\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}+\frac {(7 b c-8 a d) \sqrt {a+b x^2}}{48 a^2 x^6}-\frac {\left (35 b^2 c-40 a b d+48 a^2 e\right ) \sqrt {a+b x^2}}{192 a^3 x^4}+\frac {\left (b^2 \left (\frac {24 a^3 f}{b^3}+\frac {3 \left (-35 b c+40 a d-\frac {48 a^2 e}{b}+\frac {48 a^3 f}{b^2}\right )}{2 b}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^2} \, dx,x,\sqrt {a+b x^2}\right )}{96 a^3}\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}+\frac {(7 b c-8 a d) \sqrt {a+b x^2}}{48 a^2 x^6}-\frac {\left (35 b^2 c-40 a b d+48 a^2 e\right ) \sqrt {a+b x^2}}{192 a^3 x^4}+\frac {\left (35 b^3 c-40 a b^2 d+48 a^2 b e-64 a^3 f\right ) \sqrt {a+b x^2}}{128 a^4 x^2}+\frac {\left (35 b^3 c-40 a b^2 d+48 a^2 b e-64 a^3 f\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{128 a^4}\\ &=-\frac {c \sqrt {a+b x^2}}{8 a x^8}+\frac {(7 b c-8 a d) \sqrt {a+b x^2}}{48 a^2 x^6}-\frac {\left (35 b^2 c-40 a b d+48 a^2 e\right ) \sqrt {a+b x^2}}{192 a^3 x^4}+\frac {\left (35 b^3 c-40 a b^2 d+48 a^2 b e-64 a^3 f\right ) \sqrt {a+b x^2}}{128 a^4 x^2}-\frac {b \left (35 b^3 c-40 a b^2 d+48 a^2 b e-64 a^3 f\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.34, size = 140, normalized size = 0.72 \begin {gather*} \frac {b \sqrt {a+b x^2} \left (-\frac {a^4 f}{b x^2}+\frac {a^3 f \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}-2 a^2 b e \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x^2}{a}+1\right )-2 b^3 c \, _2F_1\left (\frac {1}{2},5;\frac {3}{2};\frac {b x^2}{a}+1\right )+2 a b^2 d \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};\frac {b x^2}{a}+1\right )\right )}{2 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^9*Sqrt[a + b*x^2]),x]

[Out]

(b*Sqrt[a + b*x^2]*(-((a^4*f)/(b*x^2)) + (a^3*f*ArcTanh[Sqrt[1 + (b*x^2)/a]])/Sqrt[1 + (b*x^2)/a] - 2*a^2*b*e*
Hypergeometric2F1[1/2, 3, 3/2, 1 + (b*x^2)/a] + 2*a*b^2*d*Hypergeometric2F1[1/2, 4, 3/2, 1 + (b*x^2)/a] - 2*b^
3*c*Hypergeometric2F1[1/2, 5, 3/2, 1 + (b*x^2)/a]))/(2*a^5)

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IntegrateAlgebraic [A]  time = 0.40, size = 172, normalized size = 0.88 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (64 a^3 b f-48 a^2 b^2 e+40 a b^3 d-35 b^4 c\right )}{128 a^{9/2}}+\frac {\sqrt {a+b x^2} \left (-48 a^3 c-64 a^3 d x^2-96 a^3 e x^4-192 a^3 f x^6+56 a^2 b c x^2+80 a^2 b d x^4+144 a^2 b e x^6-70 a b^2 c x^4-120 a b^2 d x^6+105 b^3 c x^6\right )}{384 a^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2 + e*x^4 + f*x^6)/(x^9*Sqrt[a + b*x^2]),x]

[Out]

(Sqrt[a + b*x^2]*(-48*a^3*c + 56*a^2*b*c*x^2 - 64*a^3*d*x^2 - 70*a*b^2*c*x^4 + 80*a^2*b*d*x^4 - 96*a^3*e*x^4 +
105*b^3*c*x^6 - 120*a*b^2*d*x^6 + 144*a^2*b*e*x^6 - 192*a^3*f*x^6))/(384*a^4*x^8) + ((-35*b^4*c + 40*a*b^3*d
- 48*a^2*b^2*e + 64*a^3*b*f)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(9/2))

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fricas [A]  time = 1.40, size = 341, normalized size = 1.75 \begin {gather*} \left [-\frac {3 \, {\left (35 \, b^{4} c - 40 \, a b^{3} d + 48 \, a^{2} b^{2} e - 64 \, a^{3} b f\right )} \sqrt {a} x^{8} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (35 \, a b^{3} c - 40 \, a^{2} b^{2} d + 48 \, a^{3} b e - 64 \, a^{4} f\right )} x^{6} - 48 \, a^{4} c - 2 \, {\left (35 \, a^{2} b^{2} c - 40 \, a^{3} b d + 48 \, a^{4} e\right )} x^{4} + 8 \, {\left (7 \, a^{3} b c - 8 \, a^{4} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{768 \, a^{5} x^{8}}, \frac {3 \, {\left (35 \, b^{4} c - 40 \, a b^{3} d + 48 \, a^{2} b^{2} e - 64 \, a^{3} b f\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (35 \, a b^{3} c - 40 \, a^{2} b^{2} d + 48 \, a^{3} b e - 64 \, a^{4} f\right )} x^{6} - 48 \, a^{4} c - 2 \, {\left (35 \, a^{2} b^{2} c - 40 \, a^{3} b d + 48 \, a^{4} e\right )} x^{4} + 8 \, {\left (7 \, a^{3} b c - 8 \, a^{4} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{384 \, a^{5} x^{8}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^9/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/768*(3*(35*b^4*c - 40*a*b^3*d + 48*a^2*b^2*e - 64*a^3*b*f)*sqrt(a)*x^8*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqr
t(a) + 2*a)/x^2) - 2*(3*(35*a*b^3*c - 40*a^2*b^2*d + 48*a^3*b*e - 64*a^4*f)*x^6 - 48*a^4*c - 2*(35*a^2*b^2*c -
40*a^3*b*d + 48*a^4*e)*x^4 + 8*(7*a^3*b*c - 8*a^4*d)*x^2)*sqrt(b*x^2 + a))/(a^5*x^8), 1/384*(3*(35*b^4*c - 40
*a*b^3*d + 48*a^2*b^2*e - 64*a^3*b*f)*sqrt(-a)*x^8*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(35*a*b^3*c - 40*a^2*
b^2*d + 48*a^3*b*e - 64*a^4*f)*x^6 - 48*a^4*c - 2*(35*a^2*b^2*c - 40*a^3*b*d + 48*a^4*e)*x^4 + 8*(7*a^3*b*c -
8*a^4*d)*x^2)*sqrt(b*x^2 + a))/(a^5*x^8)]

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giac [B]  time = 0.40, size = 361, normalized size = 1.85 \begin {gather*} \frac {\frac {3 \, {\left (35 \, b^{5} c - 40 \, a b^{4} d - 64 \, a^{3} b^{2} f + 48 \, a^{2} b^{3} e\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5} c - 385 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{5} c + 511 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{5} c - 279 \, \sqrt {b x^{2} + a} a^{3} b^{5} c - 120 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a b^{4} d + 440 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} b^{4} d - 584 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} b^{4} d + 264 \, \sqrt {b x^{2} + a} a^{4} b^{4} d - 192 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{3} b^{2} f + 576 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{4} b^{2} f - 576 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{5} b^{2} f + 192 \, \sqrt {b x^{2} + a} a^{6} b^{2} f + 144 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2} b^{3} e - 528 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3} b^{3} e + 624 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4} b^{3} e - 240 \, \sqrt {b x^{2} + a} a^{5} b^{3} e}{a^{4} b^{4} x^{8}}}{384 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^9/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/384*(3*(35*b^5*c - 40*a*b^4*d - 64*a^3*b^2*f + 48*a^2*b^3*e)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4)
+ (105*(b*x^2 + a)^(7/2)*b^5*c - 385*(b*x^2 + a)^(5/2)*a*b^5*c + 511*(b*x^2 + a)^(3/2)*a^2*b^5*c - 279*sqrt(b
*x^2 + a)*a^3*b^5*c - 120*(b*x^2 + a)^(7/2)*a*b^4*d + 440*(b*x^2 + a)^(5/2)*a^2*b^4*d - 584*(b*x^2 + a)^(3/2)*
a^3*b^4*d + 264*sqrt(b*x^2 + a)*a^4*b^4*d - 192*(b*x^2 + a)^(7/2)*a^3*b^2*f + 576*(b*x^2 + a)^(5/2)*a^4*b^2*f
- 576*(b*x^2 + a)^(3/2)*a^5*b^2*f + 192*sqrt(b*x^2 + a)*a^6*b^2*f + 144*(b*x^2 + a)^(7/2)*a^2*b^3*e - 528*(b*x
^2 + a)^(5/2)*a^3*b^3*e + 624*(b*x^2 + a)^(3/2)*a^4*b^3*e - 240*sqrt(b*x^2 + a)*a^5*b^3*e)/(a^4*b^4*x^8))/b

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maple [A]  time = 0.02, size = 320, normalized size = 1.64 \begin {gather*} \frac {b f \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {3 b^{2} e \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}+\frac {5 b^{3} d \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {7}{2}}}-\frac {35 b^{4} c \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {9}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, f}{2 a \,x^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, b e}{8 a^{2} x^{2}}-\frac {5 \sqrt {b \,x^{2}+a}\, b^{2} d}{16 a^{3} x^{2}}+\frac {35 \sqrt {b \,x^{2}+a}\, b^{3} c}{128 a^{4} x^{2}}-\frac {\sqrt {b \,x^{2}+a}\, e}{4 a \,x^{4}}+\frac {5 \sqrt {b \,x^{2}+a}\, b d}{24 a^{2} x^{4}}-\frac {35 \sqrt {b \,x^{2}+a}\, b^{2} c}{192 a^{3} x^{4}}-\frac {\sqrt {b \,x^{2}+a}\, d}{6 a \,x^{6}}+\frac {7 \sqrt {b \,x^{2}+a}\, b c}{48 a^{2} x^{6}}-\frac {\sqrt {b \,x^{2}+a}\, c}{8 a \,x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^6+e*x^4+d*x^2+c)/x^9/(b*x^2+a)^(1/2),x)

[Out]

-1/8*c*(b*x^2+a)^(1/2)/a/x^8+7/48*c/a^2*b/x^6*(b*x^2+a)^(1/2)-35/192*c/a^3*b^2/x^4*(b*x^2+a)^(1/2)+35/128*c/a^
4*b^3/x^2*(b*x^2+a)^(1/2)-35/128*c/a^(9/2)*b^4*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-1/2*f/a/x^2*(b*x^2+a)^(1/
2)+1/2*f*b/a^(3/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-1/6*d/a/x^6*(b*x^2+a)^(1/2)+5/24*d/a^2*b/x^4*(b*x^2+a
)^(1/2)-5/16*d/a^3*b^2/x^2*(b*x^2+a)^(1/2)+5/16*d/a^(7/2)*b^3*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-1/4*e/a/x^
4*(b*x^2+a)^(1/2)+3/8*e/a^2*b/x^2*(b*x^2+a)^(1/2)-3/8*e/a^(5/2)*b^2*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 1.35, size = 275, normalized size = 1.41 \begin {gather*} -\frac {35 \, b^{4} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {9}{2}}} + \frac {5 \, b^{3} d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {7}{2}}} - \frac {3 \, b^{2} e \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {b f \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} + \frac {35 \, \sqrt {b x^{2} + a} b^{3} c}{128 \, a^{4} x^{2}} - \frac {5 \, \sqrt {b x^{2} + a} b^{2} d}{16 \, a^{3} x^{2}} + \frac {3 \, \sqrt {b x^{2} + a} b e}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} f}{2 \, a x^{2}} - \frac {35 \, \sqrt {b x^{2} + a} b^{2} c}{192 \, a^{3} x^{4}} + \frac {5 \, \sqrt {b x^{2} + a} b d}{24 \, a^{2} x^{4}} - \frac {\sqrt {b x^{2} + a} e}{4 \, a x^{4}} + \frac {7 \, \sqrt {b x^{2} + a} b c}{48 \, a^{2} x^{6}} - \frac {\sqrt {b x^{2} + a} d}{6 \, a x^{6}} - \frac {\sqrt {b x^{2} + a} c}{8 \, a x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^9/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-35/128*b^4*c*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(9/2) + 5/16*b^3*d*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 3/8*b
^2*e*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 1/2*b*f*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) + 35/128*sqrt(b*x^2
+ a)*b^3*c/(a^4*x^2) - 5/16*sqrt(b*x^2 + a)*b^2*d/(a^3*x^2) + 3/8*sqrt(b*x^2 + a)*b*e/(a^2*x^2) - 1/2*sqrt(b*
x^2 + a)*f/(a*x^2) - 35/192*sqrt(b*x^2 + a)*b^2*c/(a^3*x^4) + 5/24*sqrt(b*x^2 + a)*b*d/(a^2*x^4) - 1/4*sqrt(b*
x^2 + a)*e/(a*x^4) + 7/48*sqrt(b*x^2 + a)*b*c/(a^2*x^6) - 1/6*sqrt(b*x^2 + a)*d/(a*x^6) - 1/8*sqrt(b*x^2 + a)*
c/(a*x^8)

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mupad [B]  time = 2.91, size = 277, normalized size = 1.42 \begin {gather*} \frac {511\,c\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a^2\,x^8}-\frac {93\,c\,\sqrt {b\,x^2+a}}{128\,a\,x^8}-\frac {385\,c\,{\left (b\,x^2+a\right )}^{5/2}}{384\,a^3\,x^8}+\frac {35\,c\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^4\,x^8}-\frac {11\,d\,\sqrt {b\,x^2+a}}{16\,a\,x^6}+\frac {5\,d\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a^2\,x^6}-\frac {5\,d\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^3\,x^6}-\frac {5\,e\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,e\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {f\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {b\,f\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {3\,b^2\,e\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}+\frac {b^4\,c\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,35{}\mathrm {i}}{128\,a^{9/2}}-\frac {b^3\,d\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x^9*(a + b*x^2)^(1/2)),x)

[Out]

(511*c*(a + b*x^2)^(3/2))/(384*a^2*x^8) - (93*c*(a + b*x^2)^(1/2))/(128*a*x^8) - (385*c*(a + b*x^2)^(5/2))/(38
4*a^3*x^8) + (35*c*(a + b*x^2)^(7/2))/(128*a^4*x^8) - (11*d*(a + b*x^2)^(1/2))/(16*a*x^6) + (5*d*(a + b*x^2)^(
3/2))/(6*a^2*x^6) - (5*d*(a + b*x^2)^(5/2))/(16*a^3*x^6) - (5*e*(a + b*x^2)^(1/2))/(8*a*x^4) + (3*e*(a + b*x^2
)^(3/2))/(8*a^2*x^4) - (f*(a + b*x^2)^(1/2))/(2*a*x^2) + (b*f*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(3/2)) +
(b^4*c*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*35i)/(128*a^(9/2)) - (b^3*d*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5
i)/(16*a^(7/2)) - (3*b^2*e*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(5/2))

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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((f*x**6+e*x**4+d*x**2+c)/x**9/(b*x**2+a)**(1/2),x)

[Out]

Timed out

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