3.149 \(\int \frac {c+d x^2+e x^4+f x^6}{x^7 \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {\sqrt {a+b x^2} (5 b c-6 a d)}{24 a^2 x^4}-\frac {\sqrt {a+b x^2} \left (8 a^2 e-6 a b d+5 b^2 c\right )}{16 a^3 x^2}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (-16 a^3 f+8 a^2 b e-6 a b^2 d+5 b^3 c\right )}{16 a^{7/2}}-\frac {c \sqrt {a+b x^2}}{6 a x^6} \]

[Out]

1/16*(-16*a^3*f+8*a^2*b*e-6*a*b^2*d+5*b^3*c)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(7/2)-1/6*c*(b*x^2+a)^(1/2)/a/
x^6+1/24*(-6*a*d+5*b*c)*(b*x^2+a)^(1/2)/a^2/x^4-1/16*(8*a^2*e-6*a*b*d+5*b^2*c)*(b*x^2+a)^(1/2)/a^3/x^2

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Rubi [A]  time = 0.28, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1799, 1621, 897, 1157, 385, 208} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (8 a^2 b e-16 a^3 f-6 a b^2 d+5 b^3 c\right )}{16 a^{7/2}}-\frac {\sqrt {a+b x^2} \left (8 a^2 e-6 a b d+5 b^2 c\right )}{16 a^3 x^2}+\frac {\sqrt {a+b x^2} (5 b c-6 a d)}{24 a^2 x^4}-\frac {c \sqrt {a+b x^2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^7*Sqrt[a + b*x^2]),x]

[Out]

-(c*Sqrt[a + b*x^2])/(6*a*x^6) + ((5*b*c - 6*a*d)*Sqrt[a + b*x^2])/(24*a^2*x^4) - ((5*b^2*c - 6*a*b*d + 8*a^2*
e)*Sqrt[a + b*x^2])/(16*a^3*x^2) + ((5*b^3*c - 6*a*b^2*d + 8*a^2*b*e - 16*a^3*f)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[
a]])/(16*a^(7/2))

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^7 \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x^4 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {a+b x^2}}{6 a x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (5 b c-6 a d)-3 a e x-3 a f x^2}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac {c \sqrt {a+b x^2}}{6 a x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {\frac {1}{2} b^2 (5 b c-6 a d)+3 a^2 b e-3 a^3 f}{b^2}-\frac {\left (3 a b e-6 a^2 f\right ) x^2}{b^2}-\frac {3 a f x^4}{b^2}}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^3} \, dx,x,\sqrt {a+b x^2}\right )}{3 a b}\\ &=-\frac {c \sqrt {a+b x^2}}{6 a x^6}+\frac {(5 b c-6 a d) \sqrt {a+b x^2}}{24 a^2 x^4}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{2} \left (5 b c-6 a d+\frac {8 a^2 e}{b}-\frac {8 a^3 f}{b^2}\right )-\frac {12 a^2 f x^2}{b^2}}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^2} \, dx,x,\sqrt {a+b x^2}\right )}{12 a^2}\\ &=-\frac {c \sqrt {a+b x^2}}{6 a x^6}+\frac {(5 b c-6 a d) \sqrt {a+b x^2}}{24 a^2 x^4}-\frac {\left (5 b^2 c-6 a b d+8 a^2 e\right ) \sqrt {a+b x^2}}{16 a^3 x^2}+\frac {\left (b^2 \left (\frac {12 a^3 f}{b^3}-\frac {3 \left (5 b c-6 a d+\frac {8 a^2 e}{b}-\frac {8 a^3 f}{b^2}\right )}{2 b}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{24 a^3}\\ &=-\frac {c \sqrt {a+b x^2}}{6 a x^6}+\frac {(5 b c-6 a d) \sqrt {a+b x^2}}{24 a^2 x^4}-\frac {\left (5 b^2 c-6 a b d+8 a^2 e\right ) \sqrt {a+b x^2}}{16 a^3 x^2}+\frac {\left (5 b^3 c-6 a b^2 d+8 a^2 b e-16 a^3 f\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 1.02, size = 162, normalized size = 1.11 \[ \frac {b^3 c \sqrt {a+b x^2} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};\frac {b x^2}{a}+1\right )}{a^4}-\frac {b^2 d \sqrt {a+b x^2} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x^2}{a}+1\right )}{a^3}-\frac {b e \sqrt {a+b x^2} \left (\frac {a}{b x^2}-\frac {\tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{2 a^2}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^7*Sqrt[a + b*x^2]),x]

[Out]

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

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fricas [A]  time = 0.58, size = 261, normalized size = 1.79 \[ \left [-\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} d + 8 \, a^{2} b e - 16 \, a^{3} f\right )} \sqrt {a} x^{6} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d + 8 \, a^{3} e\right )} x^{4} + 8 \, a^{3} c - 2 \, {\left (5 \, a^{2} b c - 6 \, a^{3} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, a^{4} x^{6}}, -\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} d + 8 \, a^{2} b e - 16 \, a^{3} f\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d + 8 \, a^{3} e\right )} x^{4} + 8 \, a^{3} c - 2 \, {\left (5 \, a^{2} b c - 6 \, a^{3} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a^{4} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^7/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*b^3*c - 6*a*b^2*d + 8*a^2*b*e - 16*a^3*f)*sqrt(a)*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2
*a)/x^2) + 2*(3*(5*a*b^2*c - 6*a^2*b*d + 8*a^3*e)*x^4 + 8*a^3*c - 2*(5*a^2*b*c - 6*a^3*d)*x^2)*sqrt(b*x^2 + a)
)/(a^4*x^6), -1/48*(3*(5*b^3*c - 6*a*b^2*d + 8*a^2*b*e - 16*a^3*f)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a
)) + (3*(5*a*b^2*c - 6*a^2*b*d + 8*a^3*e)*x^4 + 8*a^3*c - 2*(5*a^2*b*c - 6*a^3*d)*x^2)*sqrt(b*x^2 + a))/(a^4*x
^6)]

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giac [A]  time = 0.39, size = 232, normalized size = 1.59 \[ -\frac {\frac {3 \, {\left (5 \, b^{4} c - 6 \, a b^{3} d - 16 \, a^{3} b f + 8 \, a^{2} b^{2} e\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4} c - 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{4} c + 33 \, \sqrt {b x^{2} + a} a^{2} b^{4} c - 18 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{3} d + 48 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{3} d - 30 \, \sqrt {b x^{2} + a} a^{3} b^{3} d + 24 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} b^{2} e - 48 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} b^{2} e + 24 \, \sqrt {b x^{2} + a} a^{4} b^{2} e}{a^{3} b^{3} x^{6}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^7/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/48*(3*(5*b^4*c - 6*a*b^3*d - 16*a^3*b*f + 8*a^2*b^2*e)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + (1
5*(b*x^2 + a)^(5/2)*b^4*c - 40*(b*x^2 + a)^(3/2)*a*b^4*c + 33*sqrt(b*x^2 + a)*a^2*b^4*c - 18*(b*x^2 + a)^(5/2)
*a*b^3*d + 48*(b*x^2 + a)^(3/2)*a^2*b^3*d - 30*sqrt(b*x^2 + a)*a^3*b^3*d + 24*(b*x^2 + a)^(5/2)*a^2*b^2*e - 48
*(b*x^2 + a)^(3/2)*a^3*b^2*e + 24*sqrt(b*x^2 + a)*a^4*b^2*e)/(a^3*b^3*x^6))/b

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maple [A]  time = 0.01, size = 238, normalized size = 1.63 \[ -\frac {f \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}}+\frac {b e \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {3 b^{2} d \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}+\frac {5 b^{3} c \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {7}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, e}{2 a \,x^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, b d}{8 a^{2} x^{2}}-\frac {5 \sqrt {b \,x^{2}+a}\, b^{2} c}{16 a^{3} x^{2}}-\frac {\sqrt {b \,x^{2}+a}\, d}{4 a \,x^{4}}+\frac {5 \sqrt {b \,x^{2}+a}\, b c}{24 a^{2} x^{4}}-\frac {\sqrt {b \,x^{2}+a}\, c}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^6+e*x^4+d*x^2+c)/x^7/(b*x^2+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.39, size = 193, normalized size = 1.32 \[ \frac {5 \, b^{3} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {7}{2}}} - \frac {3 \, b^{2} d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {b e \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {f \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} - \frac {5 \, \sqrt {b x^{2} + a} b^{2} c}{16 \, a^{3} x^{2}} + \frac {3 \, \sqrt {b x^{2} + a} b d}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} e}{2 \, a x^{2}} + \frac {5 \, \sqrt {b x^{2} + a} b c}{24 \, a^{2} x^{4}} - \frac {\sqrt {b x^{2} + a} d}{4 \, a x^{4}} - \frac {\sqrt {b x^{2} + a} c}{6 \, a x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^7/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.54, size = 199, normalized size = 1.36 \[ \frac {5\,c\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a^2\,x^6}-\frac {11\,c\,\sqrt {b\,x^2+a}}{16\,a\,x^6}-\frac {f\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {5\,c\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^3\,x^6}-\frac {5\,d\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,d\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {e\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {b\,e\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {3\,b^2\,d\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {b^3\,c\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x^7*(a + b*x^2)^(1/2)),x)

[Out]

(5*c*(a + b*x^2)^(3/2))/(6*a^2*x^6) - (11*c*(a + b*x^2)^(1/2))/(16*a*x^6) - (f*atanh((a + b*x^2)^(1/2)/a^(1/2)
))/a^(1/2) - (5*c*(a + b*x^2)^(5/2))/(16*a^3*x^6) - (5*d*(a + b*x^2)^(1/2))/(8*a*x^4) + (3*d*(a + b*x^2)^(3/2)
)/(8*a^2*x^4) - (e*(a + b*x^2)^(1/2))/(2*a*x^2) + (b*e*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(3/2)) - (b^3*c*
atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(16*a^(7/2)) - (3*b^2*d*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 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**6+e*x**4+d*x**2+c)/x**7/(b*x**2+a)**(1/2),x)

[Out]

Timed out

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