3.2.44 \(\int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=114 \[ \frac {\sqrt {a+b x^2} (3 b c-4 a d)}{8 a^2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (8 a^2 e-4 a b d+3 b^2 c\right )}{8 a^{5/2}}-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {f \sqrt {a+b x^2}}{b} \]

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Rubi [A]  time = 0.23, antiderivative size = 114, 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, 388, 208} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (8 a^2 e-4 a b d+3 b^2 c\right )}{8 a^{5/2}}+\frac {\sqrt {a+b x^2} (3 b c-4 a d)}{8 a^2 x^2}-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {f \sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^5*Sqrt[a + b*x^2]),x]

[Out]

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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^5 \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {a+b x^2}}{4 a x^4}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (3 b c-4 a d)-2 a e x-2 a f x^2}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {c \sqrt {a+b x^2}}{4 a x^4}-\frac {\operatorname {Subst}\left (\int \frac {\frac {\frac {1}{2} b^2 (3 b c-4 a d)+2 a^2 b e-2 a^3 f}{b^2}-\frac {\left (2 a b e-4 a^2 f\right ) x^2}{b^2}-\frac {2 a f x^4}{b^2}}{\left (-\frac {a}{b}+\frac {x^2}{b}\right )^2} \, dx,x,\sqrt {a+b x^2}\right )}{2 a b}\\ &=-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 b c-4 a d) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (-3 b c+4 a d-\frac {8 a^2 e}{b}+\frac {8 a^3 f}{b^2}\right )-\frac {4 a^2 f x^2}{b^2}}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{4 a^2}\\ &=\frac {f \sqrt {a+b x^2}}{b}-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 b c-4 a d) \sqrt {a+b x^2}}{8 a^2 x^2}+\frac {\left (3 b c-4 a d+\frac {8 a^2 e}{b}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{8 a^2}\\ &=\frac {f \sqrt {a+b x^2}}{b}-\frac {c \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 b c-4 a d) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {\left (3 b^2 c-4 a b d+8 a^2 e\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 141, normalized size = 1.24 \begin {gather*} -\frac {b^2 c \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 d \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 {e \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {f \sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^5*Sqrt[a + b*x^2]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 102, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x^2} \left (8 a^2 f x^4-2 a b c-4 a b d x^2+3 b^2 c x^2\right )}{8 a^2 b x^4}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \left (-8 a^2 e+4 a b d-3 b^2 c\right )}{8 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2 + e*x^4 + f*x^6)/(x^5*Sqrt[a + b*x^2]),x]

[Out]

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

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fricas [A]  time = 1.08, size = 221, normalized size = 1.94 \begin {gather*} \left [\frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c + {\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{3} b x^{4}}, \frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c + {\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{3} b x^{4}}\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^5/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.40, size = 141, normalized size = 1.24 \begin {gather*} \frac {8 \, \sqrt {b x^{2} + a} f + \frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} c - 5 \, \sqrt {b x^{2} + a} a b^{3} c - 4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2} d + 4 \, \sqrt {b x^{2} + a} a^{2} b^{2} d}{a^{2} b^{2} x^{4}}}{8 \, 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^5/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/8*(8*sqrt(b*x^2 + a)*f + (3*b^3*c - 4*a*b^2*d + 8*a^2*b*e)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) +
(3*(b*x^2 + a)^(3/2)*b^3*c - 5*sqrt(b*x^2 + a)*a*b^3*c - 4*(b*x^2 + a)^(3/2)*a*b^2*d + 4*sqrt(b*x^2 + a)*a^2*
b^2*d)/(a^2*b^2*x^4))/b

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maple [A]  time = 0.01, size = 162, normalized size = 1.42 \begin {gather*} -\frac {e \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}}+\frac {b d \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {3 b^{2} c \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, f}{b}-\frac {\sqrt {b \,x^{2}+a}\, d}{2 a \,x^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, b c}{8 a^{2} x^{2}}-\frac {\sqrt {b \,x^{2}+a}\, c}{4 a \,x^{4}} \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^5/(b*x^2+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.38, size = 128, normalized size = 1.12 \begin {gather*} -\frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {b d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {e \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} f}{b} + \frac {3 \, \sqrt {b x^{2} + a} b c}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} d}{2 \, a x^{2}} - \frac {\sqrt {b x^{2} + a} c}{4 \, a x^{4}} \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^5/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.19, size = 133, normalized size = 1.17 \begin {gather*} \frac {f\,\sqrt {b\,x^2+a}}{b}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {5\,c\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,c\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {d\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {b\,d\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {3\,b^2\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}} \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^5*(a + b*x^2)^(1/2)),x)

[Out]

(f*(a + b*x^2)^(1/2))/b - (e*atanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(1/2) - (5*c*(a + b*x^2)^(1/2))/(8*a*x^4) + (
3*c*(a + b*x^2)^(3/2))/(8*a^2*x^4) - (d*(a + b*x^2)^(1/2))/(2*a*x^2) + (b*d*atanh((a + b*x^2)^(1/2)/a^(1/2)))/
(2*a^(3/2)) - (3*b^2*c*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**5/(b*x**2+a)**(1/2),x)

[Out]

Timed out

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