3.6.83 \(\int (a+\frac {b}{x^2}) \sqrt {c+\frac {d}{x^2}} x^5 \, dx\)

Optimal. Leaf size=123 \[ -\frac {d^2 (2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{5/2}}+\frac {d x^2 \sqrt {c+\frac {d}{x^2}} (2 b c-a d)}{16 c^2}+\frac {x^4 \sqrt {c+\frac {d}{x^2}} (2 b c-a d)}{8 c}+\frac {a x^6 \left (c+\frac {d}{x^2}\right )^{3/2}}{6 c} \]

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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 = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 78, 47, 51, 63, 208} \begin {gather*} -\frac {d^2 (2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{5/2}}+\frac {d x^2 \sqrt {c+\frac {d}{x^2}} (2 b c-a d)}{16 c^2}+\frac {x^4 \sqrt {c+\frac {d}{x^2}} (2 b c-a d)}{8 c}+\frac {a x^6 \left (c+\frac {d}{x^2}\right )^{3/2}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c - a*d)*Sqrt[c + d/x^2]*x^2)/(16*c^2) + ((2*b*c - a*d)*Sqrt[c + d/x^2]*x^4)/(8*c) + (a*(c + d/x^2)^(3
/2)*x^6)/(6*c) - (d^2*(2*b*c - a*d)*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/(16*c^(5/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 ) \sqrt {c+\frac {d}{x^2}} x^5 \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x) \sqrt {c+d x}}{x^4} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^6}{6 c}-\frac {\left (3 b c-\frac {3 a d}{2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^3} \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=\frac {(2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^4}{8 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^6}{6 c}-\frac {(d (2 b c-a d)) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {c+d x}} \, dx,x,\frac {1}{x^2}\right )}{16 c}\\ &=\frac {d (2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c^2}+\frac {(2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^4}{8 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^6}{6 c}+\frac {\left (d^2 (2 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^2}\\ &=\frac {d (2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c^2}+\frac {(2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^4}{8 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^6}{6 c}+\frac {(d (2 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^2}\\ &=\frac {d (2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^2}{16 c^2}+\frac {(2 b c-a d) \sqrt {c+\frac {d}{x^2}} x^4}{8 c}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^6}{6 c}-\frac {d^2 (2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 121, normalized size = 0.98 \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+2 c d x^2-3 d^2\right )+6 b c \left (2 c x^2+d\right )\right )+3 d^{3/2} (a d-2 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )\right )}{48 c^{5/2} \sqrt {\frac {c x^2}{d}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 111, normalized size = 0.90 \begin {gather*} \frac {\left (a d^3-2 b c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{16 c^{5/2}}+\frac {\sqrt {\frac {c x^2+d}{x^2}} \left (8 a c^2 x^6+2 a c d x^4-3 a d^2 x^2+12 b c^2 x^4+6 b c d x^2\right )}{48 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(d + c*x^2)/x^2]*(6*b*c*d*x^2 - 3*a*d^2*x^2 + 12*b*c^2*x^4 + 2*a*c*d*x^4 + 8*a*c^2*x^6))/(48*c^2) + ((-2
*b*c*d^2 + a*d^3)*ArcTanh[Sqrt[(d + c*x^2)/x^2]/Sqrt[c]])/(16*c^(5/2))

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fricas [A]  time = 0.44, size = 242, normalized size = 1.97 \begin {gather*} \left [-\frac {3 \, {\left (2 \, 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} + a c^{2} d\right )} x^{4} + 3 \, {\left (2 \, b c^{2} d - a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{96 \, c^{3}}, \frac {3 \, {\left (2 \, 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} + a c^{2} d\right )} x^{4} + 3 \, {\left (2 \, b c^{2} d - a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{48 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^5*(c+d/x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(2*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 + a*c^2*d)*x^4 + 3*(2*b*c^2*d - a*c*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/c^3, 1/48*(3*(2*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 + a*c^2*d)*x
^4 + 3*(2*b*c^2*d - a*c*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/c^3]

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giac [A]  time = 0.19, size = 143, normalized size = 1.16 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, a x^{2} \mathrm {sgn}\relax (x) + \frac {6 \, b c^{4} \mathrm {sgn}\relax (x) + a c^{3} d \mathrm {sgn}\relax (x)}{c^{4}}\right )} x^{2} + \frac {3 \, {\left (2 \, b c^{3} d \mathrm {sgn}\relax (x) - a c^{2} d^{2} \mathrm {sgn}\relax (x)\right )}}{c^{4}}\right )} \sqrt {c x^{2} + d} x + \frac {{\left (2 \, 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 {5}{2}}} - \frac {{\left (2 \, 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 {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^5*(c+d/x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 162, normalized size = 1.32 \begin {gather*} \frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{\frac {3}{2}} x^{3}+3 a \,d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-6 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 -6 \sqrt {c \,x^{2}+d}\, b \,c^{\frac {3}{2}} d x -6 \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 \right ) x}{48 \sqrt {c \,x^{2}+d}\, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((c*x^2+d)/x^2)^(1/2)*x*(8*c^(3/2)*(c*x^2+d)^(3/2)*x^3*a-6*c^(1/2)*(c*x^2+d)^(3/2)*x*a*d+12*c^(3/2)*(c*x^
2+d)^(3/2)*x*b+3*c^(1/2)*(c*x^2+d)^(1/2)*x*a*d^2-6*c^(3/2)*(c*x^2+d)^(1/2)*x*b*d+3*ln(c^(1/2)*x+(c*x^2+d)^(1/2
))*a*d^3-6*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*b*c*d^2)/(c*x^2+d)^(1/2)/c^(5/2)

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maxima [B]  time = 1.36, size = 243, normalized size = 1.98 \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 {5}{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^{2} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{3} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} c^{4} - c^{5}}\right )} a + \frac {1}{16} \, {\left (\frac {d^{2} \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 ({\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} + \sqrt {c + \frac {d}{x^{2}}} c d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} c - 2 \, {\left (c + \frac {d}{x^{2}}\right )} c^{2} + c^{3}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 5.71, size = 134, normalized size = 1.09 \begin {gather*} \frac {a\,x^6\,\sqrt {c+\frac {d}{x^2}}}{16}+\frac {b\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8}+\frac {a\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{6\,c}-\frac {a\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{5/2}}{16\,c^2}+\frac {b\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8\,c}-\frac {b\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,c^{3/2}}-\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^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 72.29, size = 226, normalized size = 1.84 \begin {gather*} \frac {a c x^{7}}{6 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {5 a \sqrt {d} x^{5}}{24 \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a d^{\frac {3}{2}} x^{3}}{48 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a d^{\frac {5}{2}} x}{16 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {a d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{16 c^{\frac {5}{2}}} + \frac {b c x^{5}}{4 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 b \sqrt {d} x^{3}}{8 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {b d^{\frac {3}{2}} x}{8 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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