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

Optimal. Leaf size=112 \[ -\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (4 a d+b c)}{4 c x}-\frac {3 \sqrt {c+\frac {d}{x^2}} (4 a d+b c)}{8 x}-\frac {3 c (4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 \sqrt {d}}+\frac {a x \left (c+\frac {d}{x^2}\right )^{5/2}}{c} \]

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Rubi [A]  time = 0.06, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {375, 453, 195, 217, 206} \begin {gather*} -\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (4 a d+b c)}{4 c x}-\frac {3 \sqrt {c+\frac {d}{x^2}} (4 a d+b c)}{8 x}-\frac {3 c (4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 \sqrt {d}}+\frac {a x \left (c+\frac {d}{x^2}\right )^{5/2}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*(c + d/x^2)^(3/2),x]

[Out]

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

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 375

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

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

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Mathematica [C]  time = 0.03, size = 68, normalized size = 0.61 \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} \left (c x^2+d\right )^2 \left (c x^4 (4 a d+b c) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x^2}{d}+1\right )-5 b d^2\right )}{20 d^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*(d + c*x^2)^2*(-5*b*d^2 + c*(b*c + 4*a*d)*x^4*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/d]))
/(20*d^3*x^3)

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IntegrateAlgebraic [A]  time = 0.22, size = 107, normalized size = 0.96 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\sqrt {c x^2+d} \left (8 a c x^4-4 a d x^2-5 b c x^2-2 b d\right )}{8 x^4}-\frac {3 \left (4 a c d+b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{8 \sqrt {d}}\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]

[Out]

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

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fricas [A]  time = 0.44, size = 216, normalized size = 1.93 \begin {gather*} \left [\frac {3 \, {\left (b c^{2} + 4 \, a c d\right )} \sqrt {d} x^{3} \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 (8 \, a c d x^{4} - 2 \, b d^{2} - {\left (5 \, b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, d x^{3}}, \frac {3 \, {\left (b c^{2} + 4 \, a c d\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (8 \, a c d x^{4} - 2 \, b d^{2} - {\left (5 \, b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, d x^{3}}\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, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 145, normalized size = 1.29 \begin {gather*} \frac {8 \, \sqrt {c x^{2} + d} a c^{2} \mathrm {sgn}\relax (x) + \frac {3 \, {\left (b c^{3} \mathrm {sgn}\relax (x) + 4 \, a c^{2} d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {5 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{3} \mathrm {sgn}\relax (x) + 4 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{2} d \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + d} b c^{3} d \mathrm {sgn}\relax (x) - 4 \, \sqrt {c x^{2} + d} a c^{2} d^{2} \mathrm {sgn}\relax (x)}{c^{2} x^{4}}}{8 \, 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, algorithm="giac")

[Out]

1/8*(8*sqrt(c*x^2 + d)*a*c^2*sgn(x) + 3*(b*c^3*sgn(x) + 4*a*c^2*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/sqr
t(-d) - (5*(c*x^2 + d)^(3/2)*b*c^3*sgn(x) + 4*(c*x^2 + d)^(3/2)*a*c^2*d*sgn(x) - 3*sqrt(c*x^2 + d)*b*c^3*d*sgn
(x) - 4*sqrt(c*x^2 + d)*a*c^2*d^2*sgn(x))/(c^2*x^4))/c

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maple [B]  time = 0.06, size = 213, normalized size = 1.90 \begin {gather*} -\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (12 a c \,d^{\frac {5}{2}} x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+3 b \,c^{2} d^{\frac {3}{2}} x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-12 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{4}-3 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{4}-4 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a c d \,x^{4}-\left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{2} x^{4}+4 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a d \,x^{2}+\left (c \,x^{2}+d \right )^{\frac {5}{2}} b c \,x^{2}+2 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b d \right )}{8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{2} x} \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)

[Out]

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

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maxima [B]  time = 1.42, size = 207, normalized size = 1.85 \begin {gather*} \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 )} a + \frac {1}{16} \, {\left (\frac {3 \, c^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{\sqrt {d}} - \frac {2 \, {\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} d x^{2} + d^{2}}\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, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.86, size = 78, normalized size = 0.70 \begin {gather*} \frac {a\,x\,{\left (c\,x^2+d\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {d}{c\,x^2}\right )}{{\left (\frac {d}{c}+x^2\right )}^{3/2}}-\frac {b\,{\left (c\,x^2+d\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {d}{c\,x^2}\right )}{x\,{\left (\frac {d}{c}+x^2\right )}^{3/2}} \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)

[Out]

(a*x*(d + c*x^2)^(3/2)*hypergeom([-3/2, -1/2], 1/2, -d/(c*x^2)))/(d/c + x^2)^(3/2) - (b*(d + c*x^2)^(3/2)*hype
rgeom([-3/2, 1/2], 3/2, -d/(c*x^2)))/(x*(d/c + x^2)^(3/2))

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sympy [B]  time = 11.78, size = 216, normalized size = 1.93 \begin {gather*} \frac {a c^{\frac {3}{2}} x}{\sqrt {1 + \frac {d}{c x^{2}}}} - \frac {a \sqrt {c} d \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} + \frac {a \sqrt {c} d}{x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 a c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2} - \frac {b c^{\frac {3}{2}} \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} - \frac {b c^{\frac {3}{2}}}{8 x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b \sqrt {c} d}{8 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 \sqrt {d}} - \frac {b d^{2}}{4 \sqrt {c} x^{5} \sqrt {1 + \frac {d}{c x^{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)

[Out]

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

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