3.7.11 \(\int (a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2} x^4 \, dx\)

Optimal. Leaf size=86 \[ \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )+b d x \sqrt {c+\frac {d}{x^2}}+\frac {1}{3} b x^3 \left (c+\frac {d}{x^2}\right )^{3/2} \]

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Rubi [A]  time = 0.06, antiderivative size = 86, 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 = {451, 335, 277, 217, 206} \begin {gather*} \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )+\frac {1}{3} b x^3 \left (c+\frac {d}{x^2}\right )^{3/2}+b d x \sqrt {c+\frac {d}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*(c + d/x^2)^(3/2)*x^4,x]

[Out]

b*d*Sqrt[c + d/x^2]*x + (b*(c + d/x^2)^(3/2)*x^3)/3 + (a*(c + d/x^2)^(5/2)*x^5)/(5*c) - b*d^(3/2)*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 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 335

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

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

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Mathematica [A]  time = 0.15, size = 81, normalized size = 0.94 \begin {gather*} \frac {1}{15} x \sqrt {c+\frac {d}{x^2}} \left (\frac {3 a \left (c x^2+d\right )^2}{c}-\frac {15 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{\sqrt {c x^2+d}}+5 b \left (c x^2+4 d\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)*(c + d/x^2)^(3/2)*x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 107, normalized size = 1.24 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\sqrt {c x^2+d} \left (3 a c^2 x^4+6 a c d x^2+3 a d^2+5 b c^2 x^2+20 b c d\right )}{15 c}-b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )\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^4,x]

[Out]

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

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fricas [A]  time = 0.45, size = 203, normalized size = 2.36 \begin {gather*} \left [\frac {15 \, b c d^{\frac {3}{2}} \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 (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} + 6 \, a c d\right )} x^{3} + {\left (20 \, b c d + 3 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{30 \, c}, \frac {15 \, b c \sqrt {-d} d \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} + 6 \, a c d\right )} x^{3} + {\left (20 \, b c d + 3 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, 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^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 140, normalized size = 1.63 \begin {gather*} \frac {b d^{2} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-d}} - \frac {{\left (15 \, b c d^{2} \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + 20 \, b c \sqrt {-d} d^{\frac {3}{2}} + 3 \, a \sqrt {-d} d^{\frac {5}{2}}\right )} \mathrm {sgn}\relax (x)}{15 \, c \sqrt {-d}} + \frac {3 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{4} \mathrm {sgn}\relax (x) + 5 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{5} \mathrm {sgn}\relax (x) + 15 \, \sqrt {c x^{2} + d} b c^{5} d \mathrm {sgn}\relax (x)}{15 \, c^{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^4,x, algorithm="giac")

[Out]

b*d^2*arctan(sqrt(c*x^2 + d)/sqrt(-d))*sgn(x)/sqrt(-d) - 1/15*(15*b*c*d^2*arctan(sqrt(d)/sqrt(-d)) + 20*b*c*sq
rt(-d)*d^(3/2) + 3*a*sqrt(-d)*d^(5/2))*sgn(x)/(c*sqrt(-d)) + 1/15*(3*(c*x^2 + d)^(5/2)*a*c^4*sgn(x) + 5*(c*x^2
+ d)^(3/2)*b*c^5*sgn(x) + 15*sqrt(c*x^2 + d)*b*c^5*d*sgn(x))/c^5

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maple [A]  time = 0.06, size = 99, normalized size = 1.15 \begin {gather*} \frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-15 b c \,d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+15 \sqrt {c \,x^{2}+d}\, b c d +5 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b c +3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \right ) x^{3}}{15 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c} \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^4,x)

[Out]

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

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maxima [A]  time = 1.35, size = 91, normalized size = 1.06 \begin {gather*} \frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5}}{5 \, c} + \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 )} 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^4,x, algorithm="maxima")

[Out]

1/5*a*(c + d/x^2)^(5/2)*x^5/c + 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))))*b

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

[Out]

int(x^4*(a + b/x^2)*(c + d/x^2)^(3/2), x)

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sympy [B]  time = 5.25, size = 184, normalized size = 2.14 \begin {gather*} \frac {a c \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {2 a d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {a d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{5 c} + \frac {b \sqrt {c} d x}{\sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c \sqrt {d} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3} + \frac {b d^{\frac {3}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3} - b d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {b d^{2}}{\sqrt {c} x \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**4,x)

[Out]

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

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