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

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

[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))

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Rubi [A]  time = 0.350011, antiderivative size = 195, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{a+b x^2} \left (48 a^2 b e-64 a^3 f-40 a b^2 d+35 b^3 c\right )}{128 a^4 x^2}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \left (48 a^2 b e-64 a^3 f-40 a b^2 d+35 b^3 c\right )}{128 a^{9/2}}-\frac{\sqrt{a+b x^2} \left (48 a^2 e-40 a b d+35 b^2 c\right )}{192 a^3 x^4}+\frac{\sqrt{a+b x^2} (7 b c-8 a d)}{48 a^2 x^6}-\frac{c \sqrt{a+b x^2}}{8 a x^8} \]

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 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]

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 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 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 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]

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*}

Mathematica [C]  time = 0.328437, size = 140, normalized size = 0.72 \[ \frac{b \sqrt{a+b x^2} \left (-2 a^2 b e \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{b x^2}{a}+1\right )-\frac{a^4 f}{b x^2}+\frac{a^3 f \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{\sqrt{\frac{b x^2}{a}+1}}-2 b^3 c \, _2F_1\left (\frac{1}{2},5;\frac{3}{2};\frac{b x^2}{a}+1\right )+2 a b^2 d \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};\frac{b x^2}{a}+1\right )\right )}{2 a^5} \]

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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Maple [A]  time = 0.014, size = 320, normalized size = 1.6 \begin{align*} -{\frac{c}{8\,a{x}^{8}}\sqrt{b{x}^{2}+a}}+{\frac{7\,bc}{48\,{a}^{2}{x}^{6}}\sqrt{b{x}^{2}+a}}-{\frac{35\,{b}^{2}c}{192\,{a}^{3}{x}^{4}}\sqrt{b{x}^{2}+a}}+{\frac{35\,{b}^{3}c}{128\,{a}^{4}{x}^{2}}\sqrt{b{x}^{2}+a}}-{\frac{35\,c{b}^{4}}{128}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{9}{2}}}}-{\frac{e}{4\,a{x}^{4}}\sqrt{b{x}^{2}+a}}+{\frac{3\,be}{8\,{a}^{2}{x}^{2}}\sqrt{b{x}^{2}+a}}-{\frac{3\,e{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{f}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{bf}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{d}{6\,a{x}^{6}}\sqrt{b{x}^{2}+a}}+{\frac{5\,bd}{24\,{a}^{2}{x}^{4}}\sqrt{b{x}^{2}+a}}-{\frac{5\,{b}^{2}d}{16\,{a}^{3}{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{5\,d{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}} \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05362, size = 803, normalized size = 4.12 \begin{align*} \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{align*}

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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Sympy [B]  time = 176.308, size = 444, normalized size = 2.28 \begin{align*} - \frac{c}{8 \sqrt{b} x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{d}{6 \sqrt{b} x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{e}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{\sqrt{b} c}{48 a x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{\sqrt{b} d}{24 a x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{\sqrt{b} e}{8 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{\sqrt{b} f \sqrt{\frac{a}{b x^{2}} + 1}}{2 a x} - \frac{7 b^{\frac{3}{2}} c}{192 a^{2} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 b^{\frac{3}{2}} d}{48 a^{2} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 b^{\frac{3}{2}} e}{8 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{35 b^{\frac{5}{2}} c}{384 a^{3} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 b^{\frac{5}{2}} d}{16 a^{3} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{35 b^{\frac{7}{2}} c}{128 a^{4} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{3}{2}}} - \frac{3 b^{2} e \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 a^{\frac{5}{2}}} + \frac{5 b^{3} d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{16 a^{\frac{7}{2}}} - \frac{35 b^{4} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{128 a^{\frac{9}{2}}} \end{align*}

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]

-c/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - d/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - e/(4*sqrt(b)*x**5*sqrt(a/
(b*x**2) + 1)) + sqrt(b)*c/(48*a*x**7*sqrt(a/(b*x**2) + 1)) + sqrt(b)*d/(24*a*x**5*sqrt(a/(b*x**2) + 1)) + sqr
t(b)*e/(8*a*x**3*sqrt(a/(b*x**2) + 1)) - sqrt(b)*f*sqrt(a/(b*x**2) + 1)/(2*a*x) - 7*b**(3/2)*c/(192*a**2*x**5*
sqrt(a/(b*x**2) + 1)) - 5*b**(3/2)*d/(48*a**2*x**3*sqrt(a/(b*x**2) + 1)) + 3*b**(3/2)*e/(8*a**2*x*sqrt(a/(b*x*
*2) + 1)) + 35*b**(5/2)*c/(384*a**3*x**3*sqrt(a/(b*x**2) + 1)) - 5*b**(5/2)*d/(16*a**3*x*sqrt(a/(b*x**2) + 1))
+ 35*b**(7/2)*c/(128*a**4*x*sqrt(a/(b*x**2) + 1)) + b*f*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3/2)) - 3*b**2*e*as
inh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2)) + 5*b**3*d*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(7/2)) - 35*b**4*c*asinh(sq
rt(a)/(sqrt(b)*x))/(128*a**(9/2))

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Giac [B]  time = 1.22429, size = 487, normalized size = 2.5 \begin{align*} \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{align*}

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