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

Optimal. Leaf size=90 \[ -\frac {d (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{5/2}}+\frac {x^2 \sqrt {c+\frac {d}{x^2}} (4 b c-3 a d)}{8 c^2}+\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{4 c} \]

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Rubi [A]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {x^2 \sqrt {c+\frac {d}{x^2}} (4 b c-3 a d)}{8 c^2}-\frac {d (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{5/2}}+\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 95, normalized size = 1.06 \begin {gather*} \frac {\sqrt {c} x \left (c x^2+d\right ) \left (2 a c x^2-3 a d+4 b c\right )+d \sqrt {c x^2+d} (3 a d-4 b c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2+d}}\right )}{8 c^{5/2} x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 88, normalized size = 0.98 \begin {gather*} \frac {\left (3 a d^2-4 b c d\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{8 c^{5/2}}+\frac {\sqrt {\frac {c x^2+d}{x^2}} \left (2 a c x^4-3 a d x^2+4 b c x^2\right )}{8 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 192, normalized size = 2.13 \begin {gather*} \left [-\frac {{\left (4 \, b c d - 3 \, a d^{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^{2} x^{4} + {\left (4 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, c^{3}}, \frac {{\left (4 \, b c d - 3 \, a d^{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^{2} x^{4} + {\left (4 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, c^{3}}\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)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 113, normalized size = 1.26 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + d x^{2}} {\left (\frac {2 \, a x^{2}}{c} + \frac {4 \, b c - 3 \, a d}{c^{2}}\right )} + \frac {{\left (4 \, b c d - 3 \, a d^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + d x^{2}}\right )} \sqrt {c} - d \right |}\right )}{16 \, c^{\frac {5}{2}}} - \frac {4 \, b c d \log \left ({\left | d \right |}\right ) - 3 \, a d^{2} \log \left ({\left | d \right |}\right )}{16 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 129, normalized size = 1.43 \begin {gather*} \frac {\sqrt {c \,x^{2}+d}\, \left (2 \sqrt {c \,x^{2}+d}\, a \,c^{\frac {5}{2}} x^{3}+3 a c \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-4 b \,c^{2} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-3 \sqrt {c \,x^{2}+d}\, a \,c^{\frac {3}{2}} d x +4 \sqrt {c \,x^{2}+d}\, b \,c^{\frac {5}{2}} x \right )}{8 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, c^{\frac {7}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x)

[Out]

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

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maxima [B]  time = 1.17, size = 178, normalized size = 1.98 \begin {gather*} \frac {1}{4} \, b {\left (\frac {2 \, \sqrt {c + \frac {d}{x^{2}}} d}{{\left (c + \frac {d}{x^{2}}\right )} c - c^{2}} + \frac {d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} - \frac {1}{16} \, a {\left (\frac {3 \, d^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} - 5 \, \sqrt {c + \frac {d}{x^{2}}} c d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} c^{3} + c^{4}}\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)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.35, size = 99, normalized size = 1.10 \begin {gather*} \frac {5\,a\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8\,c}-\frac {3\,a\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8\,c^2}+\frac {b\,x^2\,\sqrt {c+\frac {d}{x^2}}}{2\,c}-\frac {b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,c^{3/2}}+\frac {3\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b/x^2))/(c + d/x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 71.06, size = 150, normalized size = 1.67 \begin {gather*} \frac {a x^{5}}{4 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a \sqrt {d} x^{3}}{8 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {3 a d^{\frac {3}{2}} x}{8 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 a d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 c^{\frac {5}{2}}} + \frac {b \sqrt {d} x \sqrt {\frac {c x^{2}}{d} + 1}}{2 c} - \frac {b d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*x**3/(c+d/x**2)**(1/2),x)

[Out]

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

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