3.2.49 \(\int \frac {c+d x^2+e x^4+f x^6}{x^7 \sqrt {a+b x^2}} \, dx\) [149]

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

[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 = 0.20, 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 = {1813, 1635, 911, 1171, 393, 214} \begin {gather*} \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} \end {gather*}

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]

-1/6*(c*Sqrt[a + b*x^2])/(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]/Sqr
t[a]])/(16*a^(7/2))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 393

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 911

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1635

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 1813

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} \text {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 {\text {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 {\text {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 {\text {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 ) \text {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 [A]
time = 0.24, size = 126, normalized size = 0.86 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-8 a^2 c+10 a b c x^2-12 a^2 d x^2-15 b^2 c x^4+18 a b d x^4-24 a^2 e x^4\right )}{48 a^3 x^6}+\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 {gather*}

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]

(Sqrt[a + b*x^2]*(-8*a^2*c + 10*a*b*c*x^2 - 12*a^2*d*x^2 - 15*b^2*c*x^4 + 18*a*b*d*x^4 - 24*a^2*e*x^4))/(48*a^
3*x^6) + ((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))

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Maple [A]
time = 0.13, size = 250, normalized size = 1.71

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (24 a^{2} e \,x^{4}-18 a b d \,x^{4}+15 b^{2} c \,x^{4}+12 a^{2} d \,x^{2}-10 a b c \,x^{2}+8 a^{2} c \right )}{48 a^{3} x^{6}}-\frac {f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b e}{2 a^{\frac {3}{2}}}-\frac {3 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b^{2} d}{8 a^{\frac {5}{2}}}+\frac {5 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b^{3} c}{16 a^{\frac {7}{2}}}\)	\(192\)
default	\(d \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+c \left (-\frac {\sqrt {b \,x^{2}+a}}{6 a \,x^{6}}-\frac {5 b \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )}{6 a}\right )+e \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )-\frac {f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}\)	\(250\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 195, normalized size = 1.34 \begin {gather*} \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 {f \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) e}{2 \, a^{\frac {3}{2}}} - \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}} \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^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) - f*arcsinh
(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/2*b*arcsinh(a/(sqrt(a*b)*abs(x)))*e/a^(3/2) - 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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Fricas [A]
time = 1.15, size = 281, normalized size = 1.92 \begin {gather*} \left [\frac {3 \, {\left (8 \, a^{2} b x^{6} e + {\left (5 \, b^{3} c - 6 \, a b^{2} d - 16 \, a^{3} f\right )} x^{6}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (24 \, a^{3} x^{4} e + 3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d\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 (8 \, a^{2} b x^{6} e + {\left (5 \, b^{3} c - 6 \, a b^{2} d - 16 \, a^{3} f\right )} x^{6}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, a^{3} x^{4} e + 3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d\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 ] \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^7/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(8*a^2*b*x^6*e + (5*b^3*c - 6*a*b^2*d - 16*a^3*f)*x^6)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a
) + 2*a)/x^2) - 2*(24*a^3*x^4*e + 3*(5*a*b^2*c - 6*a^2*b*d)*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*(8*a^2*b*x^6*e + (5*b^3*c - 6*a*b^2*d - 16*a^3*f)*x^6)*sqrt(-a)*arctan(sqrt(-a
)/sqrt(b*x^2 + a)) + (24*a^3*x^4*e + 3*(5*a*b^2*c - 6*a^2*b*d)*x^4 + 8*a^3*c - 2*(5*a^2*b*c - 6*a^3*d)*x^2)*sq
rt(b*x^2 + a))/(a^4*x^6)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (141) = 282\).
time = 85.98, size = 303, normalized size = 2.08 \begin {gather*} - \frac {c}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {d}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {\sqrt {b} c}{24 a x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {\sqrt {b} d}{8 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} - \frac {5 b^{\frac {3}{2}} c}{48 a^{2} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b^{\frac {3}{2}} d}{8 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 b^{\frac {5}{2}} c}{16 a^{3} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {f \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{\sqrt {a}} + \frac {b e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} - \frac {3 b^{2} d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {5}{2}}} + \frac {5 b^{3} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {7}{2}}} \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**7/(b*x**2+a)**(1/2),x)

[Out]

-c/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - d/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + sqrt(b)*c/(24*a*x**5*sqrt
(a/(b*x**2) + 1)) + sqrt(b)*d/(8*a*x**3*sqrt(a/(b*x**2) + 1)) - sqrt(b)*e*sqrt(a/(b*x**2) + 1)/(2*a*x) - 5*b**
(3/2)*c/(48*a**2*x**3*sqrt(a/(b*x**2) + 1)) + 3*b**(3/2)*d/(8*a**2*x*sqrt(a/(b*x**2) + 1)) - 5*b**(5/2)*c/(16*
a**3*x*sqrt(a/(b*x**2) + 1)) - f*asinh(sqrt(a)/(sqrt(b)*x))/sqrt(a) + b*e*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3/
2)) - 3*b**2*d*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2)) + 5*b**3*c*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(7/2))

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Giac [A]
time = 1.08, size = 232, normalized size = 1.59 \begin {gather*} -\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} \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^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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Mupad [B]
time = 2.54, size = 199, normalized size = 1.36 \begin {gather*} \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}} \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^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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