3.6.99 \(\int (a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2} x^5 \, dx\)

Optimal. Leaf size=123 \[ \frac {d^2 (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{3/2}}+\frac {d x^2 \sqrt {c+\frac {d}{x^2}} (6 b c-a d)}{16 c}+\frac {x^4 \left (c+\frac {d}{x^2}\right )^{3/2} (6 b c-a d)}{24 c}+\frac {a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{6 c} \]

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Rubi [A]  time = 0.09, antiderivative size = 123, 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, 78, 47, 63, 208} \begin {gather*} \frac {d^2 (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{3/2}}+\frac {x^4 \left (c+\frac {d}{x^2}\right )^{3/2} (6 b c-a d)}{24 c}+\frac {d x^2 \sqrt {c+\frac {d}{x^2}} (6 b c-a d)}{16 c}+\frac {a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*(c + d/x^2)^(3/2)*x^5,x]

[Out]

(d*(6*b*c - a*d)*Sqrt[c + d/x^2]*x^2)/(16*c) + ((6*b*c - a*d)*(c + d/x^2)^(3/2)*x^4)/(24*c) + (a*(c + d/x^2)^(
5/2)*x^6)/(6*c) + (d^2*(6*b*c - a*d)*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/(16*c^(3/2))

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 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^5 \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x) (c+d x)^{3/2}}{x^4} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^6}{6 c}-\frac {\left (3 b c-\frac {a d}{2}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^3} \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=\frac {(6 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^4}{24 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^6}{6 c}-\frac {(d (6 b c-a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^2} \, dx,x,\frac {1}{x^2}\right )}{16 c}\\ &=\frac {d (6 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c}+\frac {(6 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^4}{24 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^6}{6 c}-\frac {\left (d^2 (6 b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,\frac {1}{x^2}\right )}{32 c}\\ &=\frac {d (6 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c}+\frac {(6 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^4}{24 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^6}{6 c}-\frac {(d (6 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 )}{16 c}\\ &=\frac {d (6 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c}+\frac {(6 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^4}{24 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^6}{6 c}+\frac {d^2 (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 123, normalized size = 1.00 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\sqrt {c} x \sqrt {\frac {c x^2}{d}+1} \left (a \left (8 c^2 x^4+14 c d x^2+3 d^2\right )+6 b c \left (2 c x^2+5 d\right )\right )-3 d^{3/2} (a d-6 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )\right )}{48 c^{3/2} \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^5,x]

[Out]

(Sqrt[c + d/x^2]*x*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/d]*(6*b*c*(5*d + 2*c*x^2) + a*(3*d^2 + 14*c*d*x^2 + 8*c^2*x^4))
- 3*d^(3/2)*(-6*b*c + a*d)*ArcSinh[(Sqrt[c]*x)/Sqrt[d]]))/(48*c^(3/2)*Sqrt[1 + (c*x^2)/d])

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IntegrateAlgebraic [A]  time = 0.19, size = 112, normalized size = 0.91 \begin {gather*} \frac {\left (6 b c d^2-a d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{16 c^{3/2}}+\frac {\sqrt {\frac {c x^2+d}{x^2}} \left (8 a c^2 x^6+14 a c d x^4+3 a d^2 x^2+12 b c^2 x^4+30 b c d x^2\right )}{48 c} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b/x^2)*(c + d/x^2)^(3/2)*x^5,x]

[Out]

(Sqrt[(d + c*x^2)/x^2]*(30*b*c*d*x^2 + 3*a*d^2*x^2 + 12*b*c^2*x^4 + 14*a*c*d*x^4 + 8*a*c^2*x^6))/(48*c) + ((6*
b*c*d^2 - a*d^3)*ArcTanh[Sqrt[(d + c*x^2)/x^2]/Sqrt[c]])/(16*c^(3/2))

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fricas [A]  time = 0.45, size = 243, normalized size = 1.98 \begin {gather*} \left [-\frac {3 \, {\left (6 \, b c d^{2} - a d^{3}\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 (8 \, a c^{3} x^{6} + 2 \, {\left (6 \, b c^{3} + 7 \, a c^{2} d\right )} x^{4} + 3 \, {\left (10 \, b c^{2} d + a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{96 \, c^{2}}, -\frac {3 \, {\left (6 \, b c d^{2} - a d^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (8 \, a c^{3} x^{6} + 2 \, {\left (6 \, b c^{3} + 7 \, a c^{2} d\right )} x^{4} + 3 \, {\left (10 \, b c^{2} d + a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{48 \, c^{2}}\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^5,x, algorithm="fricas")

[Out]

[-1/96*(3*(6*b*c*d^2 - a*d^3)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*(8*a*c^3*x^6
+ 2*(6*b*c^3 + 7*a*c^2*d)*x^4 + 3*(10*b*c^2*d + a*c*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/c^2, -1/48*(3*(6*b*c*d^2
- a*d^3)*sqrt(-c)*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (8*a*c^3*x^6 + 2*(6*b*c^3 + 7*a*c^
2*d)*x^4 + 3*(10*b*c^2*d + a*c*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/c^2]

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giac [A]  time = 0.23, size = 144, normalized size = 1.17 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, a c x^{2} \mathrm {sgn}\relax (x) + \frac {6 \, b c^{5} \mathrm {sgn}\relax (x) + 7 \, a c^{4} d \mathrm {sgn}\relax (x)}{c^{4}}\right )} x^{2} + \frac {3 \, {\left (10 \, b c^{4} d \mathrm {sgn}\relax (x) + a c^{3} d^{2} \mathrm {sgn}\relax (x)\right )}}{c^{4}}\right )} \sqrt {c x^{2} + d} x - \frac {{\left (6 \, b c d^{2} \mathrm {sgn}\relax (x) - a d^{3} \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right )}{16 \, c^{\frac {3}{2}}} + \frac {{\left (6 \, b c d^{2} \log \left ({\left | d \right |}\right ) - a d^{3} \log \left ({\left | d \right |}\right )\right )} \mathrm {sgn}\relax (x)}{32 \, c^{\frac {3}{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^5,x, algorithm="giac")

[Out]

1/48*(2*(4*a*c*x^2*sgn(x) + (6*b*c^5*sgn(x) + 7*a*c^4*d*sgn(x))/c^4)*x^2 + 3*(10*b*c^4*d*sgn(x) + a*c^3*d^2*sg
n(x))/c^4)*sqrt(c*x^2 + d)*x - 1/16*(6*b*c*d^2*sgn(x) - a*d^3*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))/c
^(3/2) + 1/32*(6*b*c*d^2*log(abs(d)) - a*d^3*log(abs(d)))*sgn(x)/c^(3/2)

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maple [A]  time = 0.06, size = 162, normalized size = 1.32 \begin {gather*} \frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-3 a \,d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )+18 b c \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-3 \sqrt {c \,x^{2}+d}\, a \sqrt {c}\, d^{2} x +18 \sqrt {c \,x^{2}+d}\, b \,c^{\frac {3}{2}} d x -2 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \sqrt {c}\, d x +12 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{\frac {3}{2}} x +8 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \sqrt {c}\, x \right ) x^{3}}{48 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {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^5,x)

[Out]

1/48*((c*x^2+d)/x^2)^(3/2)*x^3*(8*c^(1/2)*(c*x^2+d)^(5/2)*x*a-2*(c*x^2+d)^(3/2)*a*c^(1/2)*d*x+12*(c*x^2+d)^(3/
2)*b*c^(3/2)*x-3*(c*x^2+d)^(1/2)*a*c^(1/2)*d^2*x+18*(c*x^2+d)^(1/2)*b*c^(3/2)*d*x-3*a*d^3*ln(c^(1/2)*x+(c*x^2+
d)^(1/2))+18*b*c*d^2*ln(c^(1/2)*x+(c*x^2+d)^(1/2)))/(c*x^2+d)^(3/2)/c^(3/2)

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maxima [B]  time = 1.30, size = 240, normalized size = 1.95 \begin {gather*} \frac {1}{96} \, {\left (\frac {3 \, d^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{3} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c d^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d^{3}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} c - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} c^{3} - c^{4}}\right )} a - \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 )} 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^5,x, algorithm="maxima")

[Out]

1/96*(3*d^3*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(3/2) + 2*(3*(c + d/x^2)^(5/2)*d^3
+ 8*(c + d/x^2)^(3/2)*c*d^3 - 3*sqrt(c + d/x^2)*c^2*d^3)/((c + d/x^2)^3*c - 3*(c + d/x^2)^2*c^2 + 3*(c + d/x^2
)*c^3 - c^4))*a - 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))*b

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mupad [B]  time = 5.78, size = 130, normalized size = 1.06 \begin {gather*} \frac {a\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{6}+\frac {5\,b\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8}+\frac {a\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{5/2}}{16\,c}+\frac {3\,b\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,\sqrt {c}}-\frac {a\,c\,x^6\,\sqrt {c+\frac {d}{x^2}}}{16}-\frac {3\,b\,c\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8}+\frac {a\,d^3\,\mathrm {atan}\left (\frac {\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{16\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b/x^2)*(c + d/x^2)^(3/2),x)

[Out]

(a*x^6*(c + d/x^2)^(3/2))/6 + (5*b*x^4*(c + d/x^2)^(3/2))/8 + (a*x^6*(c + d/x^2)^(5/2))/(16*c) + (a*d^3*atan((
(c + d/x^2)^(1/2)*1i)/c^(1/2))*1i)/(16*c^(3/2)) + (3*b*d^2*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(8*c^(1/2)) - (a*
c*x^6*(c + d/x^2)^(1/2))/16 - (3*b*c*x^4*(c + d/x^2)^(1/2))/8

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sympy [B]  time = 100.22, size = 253, normalized size = 2.06 \begin {gather*} \frac {a c^{2} x^{7}}{6 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {11 a c \sqrt {d} x^{5}}{24 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {17 a d^{\frac {3}{2}} x^{3}}{48 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {a d^{\frac {5}{2}} x}{16 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{16 c^{\frac {3}{2}}} + \frac {b c^{2} x^{5}}{4 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 b c \sqrt {d} x^{3}}{8 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {b d^{\frac {3}{2}} x \sqrt {\frac {c x^{2}}{d} + 1}}{2} + \frac {b d^{\frac {3}{2}} x}{8 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 \sqrt {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**5,x)

[Out]

a*c**2*x**7/(6*sqrt(d)*sqrt(c*x**2/d + 1)) + 11*a*c*sqrt(d)*x**5/(24*sqrt(c*x**2/d + 1)) + 17*a*d**(3/2)*x**3/
(48*sqrt(c*x**2/d + 1)) + a*d**(5/2)*x/(16*c*sqrt(c*x**2/d + 1)) - a*d**3*asinh(sqrt(c)*x/sqrt(d))/(16*c**(3/2
)) + b*c**2*x**5/(4*sqrt(d)*sqrt(c*x**2/d + 1)) + 3*b*c*sqrt(d)*x**3/(8*sqrt(c*x**2/d + 1)) + b*d**(3/2)*x*sqr
t(c*x**2/d + 1)/2 + b*d**(3/2)*x/(8*sqrt(c*x**2/d + 1)) + 3*b*d**2*asinh(sqrt(c)*x/sqrt(d))/(8*sqrt(c))

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