3.647 \(\int \frac {(a+b x^2)^2}{x^7 \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=151 \[ \frac {d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{7/2}}-\frac {\sqrt {c+d x^2} \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{16 c^3 x^2}-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a \sqrt {c+d x^2} (12 b c-5 a d)}{24 c^2 x^4} \]

[Out]

1/16*d*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(7/2)-1/6*a^2*(d*x^2+c)^(1/2)/c/x^6
-1/24*a*(-5*a*d+12*b*c)*(d*x^2+c)^(1/2)/c^2/x^4-1/16*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*(d*x^2+c)^(1/2)/c^3/x^2

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Rubi [A]  time = 0.16, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 78, 51, 63, 208} \[ -\frac {\sqrt {c+d x^2} \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{16 c^3 x^2}+\frac {d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{7/2}}-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a \sqrt {c+d x^2} (12 b c-5 a d)}{24 c^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sqrt[c + d*x^2])/(6*c*x^6) - (a*(12*b*c - 5*a*d)*Sqrt[c + d*x^2])/(24*c^2*x^4) - ((8*b^2*c^2 - 12*a*b*c*
d + 5*a^2*d^2)*Sqrt[c + d*x^2])/(16*c^3*x^2) + (d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]
/Sqrt[c]])/(16*c^(7/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^4 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (12 b c-5 a d)+3 b^2 c x}{x^3 \sqrt {c+d x}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}+\frac {1}{16} \left (8 b^2-\frac {a d (12 b c-5 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {\left (d \left (-8 b^2+\frac {a d (12 b c-5 a d)}{c^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{32 c}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {\left (-8 b^2+\frac {a d (12 b c-5 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{16 c}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {d \left (8 b^2-\frac {a d (12 b c-5 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 135, normalized size = 0.89 \[ \frac {\sqrt {c+d x^2} \left (\frac {3 d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\sqrt {\frac {d x^2}{c}+1}\right )}{\sqrt {\frac {d x^2}{c}+1}}-\frac {c \left (a^2 \left (8 c^2-10 c d x^2+15 d^2 x^4\right )+12 a b c x^2 \left (2 c-3 d x^2\right )+24 b^2 c^2 x^4\right )}{x^6}\right )}{48 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-((c*(24*b^2*c^2*x^4 + 12*a*b*c*x^2*(2*c - 3*d*x^2) + a^2*(8*c^2 - 10*c*d*x^2 + 15*d^2*x^4))
)/x^6) + (3*d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[1 + (d*x^2)/c]])/Sqrt[1 + (d*x^2)/c]))/(48*c^4
)

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fricas [A]  time = 0.61, size = 279, normalized size = 1.85 \[ \left [\frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt {c} x^{6} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c^{4} x^{6}}, -\frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c^{4} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x
^2) - 2*(8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d
*x^2 + c))/(c^4*x^6), -1/48*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^
2 + c)) + (8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt
(d*x^2 + c))/(c^4*x^6)]

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giac [A]  time = 0.34, size = 241, normalized size = 1.60 \[ -\frac {\frac {3 \, {\left (8 \, b^{2} c^{2} d^{2} - 12 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{3}} + \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d^{2} - 36 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} + 96 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{3} - 60 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} + 15 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} - 40 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 33 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{c^{3} d^{3} x^{6}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(8*b^2*c^2*d^2 - 12*a*b*c*d^3 + 5*a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) + (24*(d*x
^2 + c)^(5/2)*b^2*c^2*d^2 - 48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 - 36*(d*x^2 + c)
^(5/2)*a*b*c*d^3 + 96*(d*x^2 + c)^(3/2)*a*b*c^2*d^3 - 60*sqrt(d*x^2 + c)*a*b*c^3*d^3 + 15*(d*x^2 + c)^(5/2)*a^
2*d^4 - 40*(d*x^2 + c)^(3/2)*a^2*c*d^4 + 33*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(c^3*d^3*x^6))/d

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maple [A]  time = 0.01, size = 224, normalized size = 1.48 \[ \frac {5 a^{2} d^{3} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{16 c^{\frac {7}{2}}}-\frac {3 a b \,d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{4 c^{\frac {5}{2}}}+\frac {b^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 c^{\frac {3}{2}}}-\frac {5 \sqrt {d \,x^{2}+c}\, a^{2} d^{2}}{16 c^{3} x^{2}}+\frac {3 \sqrt {d \,x^{2}+c}\, a b d}{4 c^{2} x^{2}}-\frac {\sqrt {d \,x^{2}+c}\, b^{2}}{2 c \,x^{2}}+\frac {5 \sqrt {d \,x^{2}+c}\, a^{2} d}{24 c^{2} x^{4}}-\frac {\sqrt {d \,x^{2}+c}\, a b}{2 c \,x^{4}}-\frac {\sqrt {d \,x^{2}+c}\, a^{2}}{6 c \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 190, normalized size = 1.26 \[ \frac {b^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {3}{2}}} - \frac {3 \, a b d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{4 \, c^{\frac {5}{2}}} + \frac {5 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, c^{\frac {7}{2}}} - \frac {\sqrt {d x^{2} + c} b^{2}}{2 \, c x^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a b d}{4 \, c^{2} x^{2}} - \frac {5 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {\sqrt {d x^{2} + c} a b}{2 \, c x^{4}} + \frac {5 \, \sqrt {d x^{2} + c} a^{2} d}{24 \, c^{2} x^{4}} - \frac {\sqrt {d x^{2} + c} a^{2}}{6 \, c x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 207, normalized size = 1.37 \[ \frac {\frac {{\left (d\,x^2+c\right )}^{5/2}\,\left (5\,a^2\,d^3-12\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c^3}-\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (5\,a^2\,d^3-12\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{6\,c^2}+\frac {\sqrt {d\,x^2+c}\,\left (11\,a^2\,d^3-20\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c}}{3\,c\,{\left (d\,x^2+c\right )}^2-3\,c^2\,\left (d\,x^2+c\right )-{\left (d\,x^2+c\right )}^3+c^3}+\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (5\,a^2\,d^2-12\,a\,b\,c\,d+8\,b^2\,c^2\right )}{16\,c^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c + d*x^2)^(5/2)*(5*a^2*d^3 + 8*b^2*c^2*d - 12*a*b*c*d^2))/(16*c^3) - ((c + d*x^2)^(3/2)*(5*a^2*d^3 + 6*b^2
*c^2*d - 12*a*b*c*d^2))/(6*c^2) + ((c + d*x^2)^(1/2)*(11*a^2*d^3 + 8*b^2*c^2*d - 20*a*b*c*d^2))/(16*c))/(3*c*(
c + d*x^2)^2 - 3*c^2*(c + d*x^2) - (c + d*x^2)^3 + c^3) + (d*atanh((c + d*x^2)^(1/2)/c^(1/2))*(5*a^2*d^2 + 8*b
^2*c^2 - 12*a*b*c*d))/(16*c^(7/2))

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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((b*x**2+a)**2/x**7/(d*x**2+c)**(1/2),x)

[Out]

Timed out

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