3.7.27 \(\int \frac {(a+\frac {b}{x^2}) x^3}{(c+\frac {d}{x^2})^{3/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac {3 d (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{7/2}}+\frac {3 d (4 b c-5 a d)}{8 c^3 \sqrt {c+\frac {d}{x^2}}}+\frac {x^2 (4 b c-5 a d)}{8 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {a x^4}{4 c \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.08, antiderivative size = 120, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 78, 51, 63, 208} \begin {gather*} \frac {3 x^2 \sqrt {c+\frac {d}{x^2}} (4 b c-5 a d)}{8 c^3}-\frac {x^2 (4 b c-5 a d)}{4 c^2 \sqrt {c+\frac {d}{x^2}}}-\frac {3 d (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{7/2}}+\frac {a x^4}{4 c \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*x^3)/(c + d/x^2)^(3/2),x]

[Out]

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

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Mathematica [A]  time = 0.19, size = 111, normalized size = 0.94 \begin {gather*} \frac {\sqrt {c} x \left (a \left (2 c^2 x^4-5 c d x^2-15 d^2\right )+4 b c \left (c x^2+3 d\right )\right )+3 d^{3/2} \sqrt {\frac {c x^2}{d}+1} (5 a d-4 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{8 c^{7/2} x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x^2)*x^3)/(c + d/x^2)^(3/2),x]

[Out]

(Sqrt[c]*x*(4*b*c*(3*d + c*x^2) + a*(-15*d^2 - 5*c*d*x^2 + 2*c^2*x^4)) + 3*d^(3/2)*(-4*b*c + 5*a*d)*Sqrt[1 + (
c*x^2)/d]*ArcSinh[(Sqrt[c]*x)/Sqrt[d]])/(8*c^(7/2)*Sqrt[c + d/x^2]*x)

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IntegrateAlgebraic [A]  time = 0.18, size = 119, normalized size = 1.01 \begin {gather*} \frac {\sqrt {\frac {c x^2+d}{x^2}} \left (2 a c^2 x^6-5 a c d x^4-15 a d^2 x^2+4 b c^2 x^4+12 b c d x^2\right )}{8 c^3 \left (c x^2+d\right )}-\frac {3 \left (4 b c d-5 a d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{8 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b/x^2)*x^3)/(c + d/x^2)^(3/2),x]

[Out]

(Sqrt[(d + c*x^2)/x^2]*(12*b*c*d*x^2 - 15*a*d^2*x^2 + 4*b*c^2*x^4 - 5*a*c*d*x^4 + 2*a*c^2*x^6))/(8*c^3*(d + c*
x^2)) - (3*(4*b*c*d - 5*a*d^2)*ArcTanh[Sqrt[(d + c*x^2)/x^2]/Sqrt[c]])/(8*c^(7/2))

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fricas [A]  time = 0.46, size = 304, normalized size = 2.58 \begin {gather*} \left [-\frac {3 \, {\left (4 \, b c d^{2} - 5 \, a d^{3} + {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left (2 \, a c^{3} x^{6} + {\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \, {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, {\left (c^{5} x^{2} + c^{4} d\right )}}, \frac {3 \, {\left (4 \, b c d^{2} - 5 \, a d^{3} + {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a c^{3} x^{6} + {\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \, {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, {\left (c^{5} x^{2} + c^{4} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^3/(c+d/x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^3/(c+d/x^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Unable to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,1,2]%%%}+%%%{%%{[4,0]:[1,0,%%
%{-1,[1]%%%}]%%},[1,1,3]%%%}+%%%{-2,[0,1,4]%%%} / %%%{%%%{1,[2]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0
,%%%{-1,[1]%%%}]%%},[1,0,1]%%%}+%%%{%%%{1,[1]%%%},[0,0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.06, size = 140, normalized size = 1.19 \begin {gather*} \frac {\left (c \,x^{2}+d \right ) \left (2 a \,c^{\frac {7}{2}} x^{5}-5 a \,c^{\frac {5}{2}} d \,x^{3}+4 b \,c^{\frac {7}{2}} x^{3}-15 a \,c^{\frac {3}{2}} d^{2} x +12 b \,c^{\frac {5}{2}} d x +15 \sqrt {c \,x^{2}+d}\, a c \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-12 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )\right )}{8 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} c^{\frac {9}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*x^3/(c+d/x^2)^(3/2),x)

[Out]

1/8*(c*x^2+d)*(2*c^(7/2)*x^5*a-5*c^(5/2)*x^3*a*d+4*c^(7/2)*x^3*b-15*c^(3/2)*x*a*d^2+12*c^(5/2)*x*b*d+15*(c*x^2
+d)^(1/2)*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*a*c*d^2-12*(c*x^2+d)^(1/2)*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*b*c^2*d)/((c*
x^2+d)/x^2)^(3/2)/x^3/c^(9/2)

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maxima [B]  time = 1.38, size = 215, normalized size = 1.82 \begin {gather*} -\frac {1}{16} \, a {\left (\frac {2 \, {\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} - 25 \, {\left (c + \frac {d}{x^{2}}\right )} c d^{2} + 8 \, c^{2} d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} - 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{4} + \sqrt {c + \frac {d}{x^{2}}} c^{5}} + \frac {15 \, d^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )} + \frac {1}{4} \, b {\left (\frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )} d - 2 \, c d\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} - \sqrt {c + \frac {d}{x^{2}}} c^{3}} + \frac {3 \, d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^3/(c+d/x^2)^(3/2),x, algorithm="maxima")

[Out]

-1/16*a*(2*(15*(c + d/x^2)^2*d^2 - 25*(c + d/x^2)*c*d^2 + 8*c^2*d^2)/((c + d/x^2)^(5/2)*c^3 - 2*(c + d/x^2)^(3
/2)*c^4 + sqrt(c + d/x^2)*c^5) + 15*d^2*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(7/2))
+ 1/4*b*(2*(3*(c + d/x^2)*d - 2*c*d)/((c + d/x^2)^(3/2)*c^2 - sqrt(c + d/x^2)*c^3) + 3*d*log((sqrt(c + d/x^2)
- sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(5/2))

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mupad [B]  time = 6.34, size = 134, normalized size = 1.14 \begin {gather*} \frac {a\,x^4}{4\,c\,\sqrt {c+\frac {d}{x^2}}}-\frac {15\,a\,d^2}{8\,c^3\,\sqrt {c+\frac {d}{x^2}}}+\frac {b\,x^2}{2\,c\,\sqrt {c+\frac {d}{x^2}}}-\frac {3\,b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,c^{5/2}}+\frac {15\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,c^{7/2}}+\frac {3\,b\,d}{2\,c^2\,\sqrt {c+\frac {d}{x^2}}}-\frac {5\,a\,d\,x^2}{8\,c^2\,\sqrt {c+\frac {d}{x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b/x^2))/(c + d/x^2)^(3/2),x)

[Out]

(a*x^4)/(4*c*(c + d/x^2)^(1/2)) - (15*a*d^2)/(8*c^3*(c + d/x^2)^(1/2)) + (b*x^2)/(2*c*(c + d/x^2)^(1/2)) - (3*
b*d*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(2*c^(5/2)) + (15*a*d^2*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(8*c^(7/2)) +
(3*b*d)/(2*c^2*(c + d/x^2)^(1/2)) - (5*a*d*x^2)/(8*c^2*(c + d/x^2)^(1/2))

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sympy [A]  time = 103.90, size = 177, normalized size = 1.50 \begin {gather*} a \left (\frac {x^{5}}{4 c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {5 \sqrt {d} x^{3}}{8 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {15 d^{\frac {3}{2}} x}{8 c^{3} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {15 d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 c^{\frac {7}{2}}}\right ) + b \left (\frac {x^{3}}{2 c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 \sqrt {d} x}{2 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 c^{\frac {5}{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*x**3/(c+d/x**2)**(3/2),x)

[Out]

a*(x**5/(4*c*sqrt(d)*sqrt(c*x**2/d + 1)) - 5*sqrt(d)*x**3/(8*c**2*sqrt(c*x**2/d + 1)) - 15*d**(3/2)*x/(8*c**3*
sqrt(c*x**2/d + 1)) + 15*d**2*asinh(sqrt(c)*x/sqrt(d))/(8*c**(7/2))) + b*(x**3/(2*c*sqrt(d)*sqrt(c*x**2/d + 1)
) + 3*sqrt(d)*x/(2*c**2*sqrt(c*x**2/d + 1)) - 3*d*asinh(sqrt(c)*x/sqrt(d))/(2*c**(5/2)))

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