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

Optimal. Leaf size=76 \[ a c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {1}{3} a \left (c+\frac {d}{x^2}\right )^{3/2}-a c \sqrt {c+\frac {d}{x^2}}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d} \]

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Rubi [A]  time = 0.05, antiderivative size = 76, normalized size of antiderivative = 1.00, 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, 80, 50, 63, 208} \begin {gather*} a c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {1}{3} a \left (c+\frac {d}{x^2}\right )^{3/2}-a c \sqrt {c+\frac {d}{x^2}}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

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Mathematica [C]  time = 0.07, size = 90, normalized size = 1.18 \begin {gather*} -\frac {\sqrt {c+\frac {d}{x^2}} \left (5 a d^2 x^2 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{d}\right )+3 b \left (c x^2+d\right )^2 \sqrt {\frac {c x^2}{d}+1}\right )}{15 d x^4 \sqrt {\frac {c x^2}{d}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[c + d/x^2]*(3*b*(d + c*x^2)^2*Sqrt[1 + (c*x^2)/d] + 5*a*d^2*x^2*Hypergeometric2F1[-3/2, -3/2, -1/2
, -((c*x^2)/d)]))/(d*x^4*Sqrt[1 + (c*x^2)/d])

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IntegrateAlgebraic [A]  time = 0.11, size = 96, normalized size = 1.26 \begin {gather*} \frac {\sqrt {\frac {c x^2+d}{x^2}} \left (-20 a c d x^4-5 a d^2 x^2-3 b c^2 x^4-6 b c d x^2-3 b d^2\right )}{15 d x^4}+a c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(d + c*x^2)/x^2]*(-3*b*d^2 - 6*b*c*d*x^2 - 5*a*d^2*x^2 - 3*b*c^2*x^4 - 20*a*c*d*x^4))/(15*d*x^4) + a*c^(
3/2)*ArcTanh[Sqrt[(d + c*x^2)/x^2]/Sqrt[c]]

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fricas [A]  time = 0.43, size = 213, normalized size = 2.80 \begin {gather*} \left [\frac {15 \, a c^{\frac {3}{2}} d x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} + {\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{30 \, d x^{4}}, -\frac {15 \, a \sqrt {-c} c d x^{4} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} + {\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, d x^{4}}\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,x, algorithm="fricas")

[Out]

[1/30*(15*a*c^(3/2)*d*x^4*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*((3*b*c^2 + 20*a*c*d)*x^
4 + 3*b*d^2 + (6*b*c*d + 5*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d*x^4), -1/15*(15*a*sqrt(-c)*c*d*x^4*arctan(sqr
t(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + ((3*b*c^2 + 20*a*c*d)*x^4 + 3*b*d^2 + (6*b*c*d + 5*a*d^2)*x^2)*
sqrt((c*x^2 + d)/x^2))/(d*x^4)]

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giac [B]  time = 1.07, size = 254, normalized size = 3.34 \begin {gather*} -\frac {1}{2} \, a c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {3}{2}} d \mathrm {sgn}\relax (x) - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {3}{2}} d^{2} \mathrm {sgn}\relax (x) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {5}{2}} d^{2} \mathrm {sgn}\relax (x) + 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {3}{2}} d^{3} \mathrm {sgn}\relax (x) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {3}{2}} d^{4} \mathrm {sgn}\relax (x) + 3 \, b c^{\frac {5}{2}} d^{4} \mathrm {sgn}\relax (x) + 20 \, a c^{\frac {3}{2}} d^{5} \mathrm {sgn}\relax (x)\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{5}} \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,x, algorithm="giac")

[Out]

-1/2*a*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + d))^2)*sgn(x) + 2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(5/2
)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(3/2)*d*sgn(x) - 90*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(3/2
)*d^2*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(5/2)*d^2*sgn(x) + 110*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a
*c^(3/2)*d^3*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(3/2)*d^4*sgn(x) + 3*b*c^(5/2)*d^4*sgn(x) + 20*a*
c^(3/2)*d^5*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5

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maple [B]  time = 0.07, size = 153, normalized size = 2.01 \begin {gather*} \frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (15 a \,c^{2} d^{2} x^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )+15 \sqrt {c \,x^{2}+d}\, a \,c^{\frac {5}{2}} d \,x^{6}+10 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{\frac {5}{2}} x^{6}-10 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{\frac {3}{2}} x^{4}-5 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \sqrt {c}\, d \,x^{2}-3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \sqrt {c}\, d \right )}{15 \left (c \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {c}\, d^{2} x^{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,x)

[Out]

1/15*((c*x^2+d)/x^2)^(3/2)*(10*c^(5/2)*(c*x^2+d)^(3/2)*x^6*a+15*c^(5/2)*(c*x^2+d)^(1/2)*x^6*a*d-10*c^(3/2)*(c*
x^2+d)^(5/2)*x^4*a+15*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x^5*a*c^2*d^2-5*(c*x^2+d)^(5/2)*a*c^(1/2)*d*x^2-3*(c*x^2+d
)^(5/2)*b*c^(1/2)*d)/x^2/(c*x^2+d)^(3/2)/d^2/c^(1/2)

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maxima [A]  time = 1.31, size = 80, normalized size = 1.05 \begin {gather*} -\frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) + 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} + 6 \, \sqrt {c + \frac {d}{x^{2}}} c\right )} a \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,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.83, size = 72, normalized size = 0.95 \begin {gather*} a\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {a\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{3}-a\,c\,\sqrt {c+\frac {d}{x^2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}\,{\left (c\,x^2+d\right )}^2}{5\,d\,x^4} \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,x)

[Out]

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

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sympy [A]  time = 54.14, size = 73, normalized size = 0.96 \begin {gather*} - \frac {a c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x^{2}}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - a c \sqrt {c + \frac {d}{x^{2}}} - \frac {a \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {b \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5 d} \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,x)

[Out]

-a*c**2*atan(sqrt(c + d/x**2)/sqrt(-c))/sqrt(-c) - a*c*sqrt(c + d/x**2) - a*(c + d/x**2)**(3/2)/3 - b*(c + d/x
**2)**(5/2)/(5*d)

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