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

Optimal. Leaf size=115 \[ \frac {x^2 \left (c+\frac {d}{x^2}\right )^{3/2} (a d+4 b c)}{8 c}-\frac {3 d \sqrt {c+\frac {d}{x^2}} (a d+4 b c)}{8 c}+\frac {3 d (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {a x^4 \left (c+\frac {d}{x^2}\right )^{5/2}}{4 c} \]

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Rubi [A]  time = 0.08, antiderivative size = 115, 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 = {446, 78, 47, 50, 63, 208} \begin {gather*} \frac {x^2 \left (c+\frac {d}{x^2}\right )^{3/2} (a d+4 b c)}{8 c}-\frac {3 d \sqrt {c+\frac {d}{x^2}} (a d+4 b c)}{8 c}+\frac {3 d (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {a x^4 \left (c+\frac {d}{x^2}\right )^{5/2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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

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

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Mathematica [A]  time = 0.25, size = 89, normalized size = 0.77 \begin {gather*} \frac {1}{8} \sqrt {c+\frac {d}{x^2}} \left (\frac {3 \sqrt {d} x (a d+4 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {\frac {c x^2}{d}+1}}+2 a c x^4+5 a d x^2+4 b c x^2-8 b d\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*(-8*b*d + 4*b*c*x^2 + 5*a*d*x^2 + 2*a*c*x^4 + (3*Sqrt[d]*(4*b*c + a*d)*x*ArcSinh[(Sqrt[c]*x)/
Sqrt[d]])/(Sqrt[c]*Sqrt[1 + (c*x^2)/d])))/8

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IntegrateAlgebraic [A]  time = 0.14, size = 88, normalized size = 0.77 \begin {gather*} \frac {3 \left (a d^2+4 b c d\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {1}{8} \sqrt {\frac {c x^2+d}{x^2}} \left (2 a c x^4+5 a d x^2+4 b c x^2-8 b d\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 203, normalized size = 1.77 \begin {gather*} \left [\frac {3 \, {\left (4 \, b c d + 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} - 8 \, b c d + {\left (4 \, b c^{2} + 5 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, c}, -\frac {3 \, {\left (4 \, b c d + 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} - 8 \, b c d + {\left (4 \, b c^{2} + 5 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, c}\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^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 126, normalized size = 1.10 \begin {gather*} \frac {2 \, b \sqrt {c} d^{2} \mathrm {sgn}\relax (x)}{{\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d} + \frac {1}{8} \, {\left (2 \, a c x^{2} \mathrm {sgn}\relax (x) + \frac {4 \, b c^{3} \mathrm {sgn}\relax (x) + 5 \, a c^{2} d \mathrm {sgn}\relax (x)}{c^{2}}\right )} \sqrt {c x^{2} + d} x - \frac {3 \, {\left (4 \, b c^{\frac {3}{2}} d \mathrm {sgn}\relax (x) + a \sqrt {c} d^{2} \mathrm {sgn}\relax (x)\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right )}{16 \, 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^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 174, normalized size = 1.51 \begin {gather*} \frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (3 a \,d^{3} x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )+12 b c \,d^{2} x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )+3 \sqrt {c \,x^{2}+d}\, a \sqrt {c}\, d^{2} x^{2}+12 \sqrt {c \,x^{2}+d}\, b \,c^{\frac {3}{2}} d \,x^{2}+2 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \sqrt {c}\, d \,x^{2}+8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{\frac {3}{2}} x^{2}-8 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \sqrt {c}\right ) x^{2}}{8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {c}\, 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^3,x)

[Out]

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

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maxima [A]  time = 1.33, size = 171, normalized size = 1.49 \begin {gather*} -\frac {1}{16} \, {\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 )}{\sqrt {c}} - \frac {2 \, {\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} c + c^{2}}\right )} a + \frac {1}{4} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} c x^{2} - 3 \, \sqrt {c} d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) - 4 \, \sqrt {c + \frac {d}{x^{2}}} d\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^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.70, size = 105, normalized size = 0.91 \begin {gather*} \frac {5\,a\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8}-b\,d\,\sqrt {c+\frac {d}{x^2}}+\frac {3\,b\,\sqrt {c}\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2}+\frac {3\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,\sqrt {c}}-\frac {3\,a\,c\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8}+\frac {b\,c\,x^2\,\sqrt {c+\frac {d}{x^2}}}{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)^(3/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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**3,x)

[Out]

Timed out

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