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

Optimal. Leaf size=121 \[ \frac {x \left (c+\frac {d}{x^2}\right )^{3/2} (2 a d+3 b c)}{3 c}-\frac {d \sqrt {c+\frac {d}{x^2}} (2 a d+3 b c)}{2 c x}-\frac {1}{2} \sqrt {d} (2 a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )+\frac {a x^3 \left (c+\frac {d}{x^2}\right )^{5/2}}{3 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {453, 242, 277, 195, 217, 206} \begin {gather*} \frac {x \left (c+\frac {d}{x^2}\right )^{3/2} (2 a d+3 b c)}{3 c}-\frac {d \sqrt {c+\frac {d}{x^2}} (2 a d+3 b c)}{2 c x}-\frac {1}{2} \sqrt {d} (2 a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )+\frac {a x^3 \left (c+\frac {d}{x^2}\right )^{5/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 105, normalized size = 0.87 \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} \left (\sqrt {c x^2+d} \left (2 a c x^4+8 a d x^2+6 b c x^2-3 b d\right )-3 \sqrt {d} x^2 (2 a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )\right )}{6 x \sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 109, normalized size = 0.90 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {1}{2} \left (-2 a d^{3/2}-3 b c \sqrt {d}\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )+\frac {\sqrt {c x^2+d} \left (2 a c x^4+8 a d x^2+6 b c x^2-3 b d\right )}{6 x^2}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 190, normalized size = 1.57 \begin {gather*} \left [\frac {3 \, {\left (3 \, b c + 2 \, a d\right )} \sqrt {d} x \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 (2 \, a c x^{4} + 2 \, {\left (3 \, b c + 4 \, a d\right )} x^{2} - 3 \, b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{12 \, x}, \frac {3 \, {\left (3 \, b c + 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a c x^{4} + 2 \, {\left (3 \, b c + 4 \, a d\right )} x^{2} - 3 \, b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{6 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 115, normalized size = 0.95 \begin {gather*} \frac {2 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c \mathrm {sgn}\relax (x) + 6 \, \sqrt {c x^{2} + d} b c^{2} \mathrm {sgn}\relax (x) + 6 \, \sqrt {c x^{2} + d} a c d \mathrm {sgn}\relax (x) - \frac {3 \, \sqrt {c x^{2} + d} b c d \mathrm {sgn}\relax (x)}{x^{2}} + \frac {3 \, {\left (3 \, b c^{2} d \mathrm {sgn}\relax (x) + 2 \, a c d^{2} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}}}{6 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(c*x^2 + d)^(3/2)*a*c*sgn(x) + 6*sqrt(c*x^2 + d)*b*c^2*sgn(x) + 6*sqrt(c*x^2 + d)*a*c*d*sgn(x) - 3*sqrt
(c*x^2 + d)*b*c*d*sgn(x)/x^2 + 3*(3*b*c^2*d*sgn(x) + 2*a*c*d^2*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/sqrt(-
d))/c

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maple [A]  time = 0.06, size = 170, normalized size = 1.40 \begin {gather*} -\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (6 a \,d^{\frac {5}{2}} x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+9 b c \,d^{\frac {3}{2}} x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-6 \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{2}-9 \sqrt {c \,x^{2}+d}\, b c d \,x^{2}-2 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a d \,x^{2}-3 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b c \,x^{2}+3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \right ) x}{6 \left (c \,x^{2}+d \right )^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((c*x^2+d)/x^2)^(3/2)*x*(-2*(c*x^2+d)^(3/2)*a*d*x^2-3*(c*x^2+d)^(3/2)*b*c*x^2+6*d^(5/2)*ln(2*(d+(c*x^2+d)
^(1/2)*d^(1/2))/x)*x^2*a+9*d^(3/2)*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^2*b*c+3*(c*x^2+d)^(5/2)*b-6*(c*x^2+d)
^(1/2)*x^2*a*d^2-9*(c*x^2+d)^(1/2)*x^2*b*c*d)/(c*x^2+d)^(3/2)/d

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maxima [A]  time = 1.36, size = 163, normalized size = 1.35 \begin {gather*} \frac {1}{6} \, {\left (2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3} + 6 \, \sqrt {c + \frac {d}{x^{2}}} d x + 3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )\right )} a + \frac {1}{4} \, {\left (4 \, \sqrt {c + \frac {d}{x^{2}}} c x - \frac {2 \, \sqrt {c + \frac {d}{x^{2}}} c d x}{{\left (c + \frac {d}{x^{2}}\right )} x^{2} - d} + 3 \, c \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(c + d/x^2)^(3/2)*x^3 + 6*sqrt(c + d/x^2)*d*x + 3*d^(3/2)*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d
/x^2)*x + sqrt(d))))*a + 1/4*(4*sqrt(c + d/x^2)*c*x - 2*sqrt(c + d/x^2)*c*d*x/((c + d/x^2)*x^2 - d) + 3*c*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 [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\left (a+\frac {b}{x^2}\right )\,{\left (c+\frac {d}{x^2}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.65, size = 202, normalized size = 1.67 \begin {gather*} \frac {a \sqrt {c} d x}{\sqrt {1 + \frac {d}{c x^{2}}}} + \frac {a c \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} - a d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {a d^{2}}{\sqrt {c} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{\frac {3}{2}} x}{\sqrt {1 + \frac {d}{c x^{2}}}} - \frac {b \sqrt {c} d \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} + \frac {b \sqrt {c} d}{x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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