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

Optimal. Leaf size=66 \[ \frac {a x^3 \left (c+\frac {d}{x^2}\right )^{3/2}}{3 c}+b x \sqrt {c+\frac {d}{x^2}}-b \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right ) \]

[Out]

1/3*a*(c+d/x^2)^(3/2)*x^3/c-b*arctanh(d^(1/2)/x/(c+d/x^2)^(1/2))*d^(1/2)+b*x*(c+d/x^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 66, 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 = {451, 242, 277, 217, 206} \[ \frac {a x^3 \left (c+\frac {d}{x^2}\right )^{3/2}}{3 c}+b x \sqrt {c+\frac {d}{x^2}}-b \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 451

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 84, normalized size = 1.27 \[ \frac {x \sqrt {c+\frac {d}{x^2}} \left (\sqrt {c x^2+d} \left (a \left (c x^2+d\right )+3 b c\right )-3 b c \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )\right )}{3 c \sqrt {c x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 156, normalized size = 2.36 \[ \left [\frac {3 \, b c \sqrt {d} \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (a c x^{3} + {\left (3 \, b c + a d\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{6 \, c}, \frac {3 \, b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (a c x^{3} + {\left (3 \, b c + a d\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3 \, c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 116, normalized size = 1.76 \[ \frac {b d \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-d}} - \frac {{\left (3 \, b c d \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + 3 \, b c \sqrt {-d} \sqrt {d} + a \sqrt {-d} d^{\frac {3}{2}}\right )} \mathrm {sgn}\relax (x)}{3 \, c \sqrt {-d}} + \frac {{\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{2} \mathrm {sgn}\relax (x) + 3 \, \sqrt {c x^{2} + d} b c^{3} \mathrm {sgn}\relax (x)}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d*arctan(sqrt(c*x^2 + d)/sqrt(-d))*sgn(x)/sqrt(-d) - 1/3*(3*b*c*d*arctan(sqrt(d)/sqrt(-d)) + 3*b*c*sqrt(-d)*
sqrt(d) + a*sqrt(-d)*d^(3/2))*sgn(x)/(c*sqrt(-d)) + 1/3*((c*x^2 + d)^(3/2)*a*c^2*sgn(x) + 3*sqrt(c*x^2 + d)*b*
c^3*sgn(x))/c^3

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maple [A]  time = 0.06, size = 83, normalized size = 1.26 \[ -\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (3 b c \sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-3 \sqrt {c \,x^{2}+d}\, b c -\left (c \,x^{2}+d \right )^{\frac {3}{2}} a \right ) x}{3 \sqrt {c \,x^{2}+d}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 75, normalized size = 1.14 \[ \frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3}}{3 \, c} + \frac {1}{2} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.72, size = 80, normalized size = 1.21 \[ b\,x\,\sqrt {c+\frac {d}{x^2}}+\frac {a\,x\,\sqrt {c+\frac {d}{x^2}}\,\left (c\,x^2+d\right )}{3\,c}+\frac {b\,\sqrt {d}\,\mathrm {asin}\left (\frac {\sqrt {d}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}\,\sqrt {\frac {d}{c\,x^2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x*(c + d/x^2)^(1/2) + (a*x*(c + d/x^2)^(1/2)*(d + c*x^2))/(3*c) + (b*d^(1/2)*asin((d^(1/2)*1i)/(c^(1/2)*x))*
(c + d/x^2)^(1/2)*1i)/(c^(1/2)*(d/(c*x^2) + 1)^(1/2))

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sympy [A]  time = 3.27, size = 107, normalized size = 1.62 \[ \frac {a \sqrt {d} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3} + \frac {a d^{\frac {3}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c} + \frac {b \sqrt {c} x}{\sqrt {1 + \frac {d}{c x^{2}}}} - b \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {b d}{\sqrt {c} x \sqrt {1 + \frac {d}{c x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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