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

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

[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))

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Rubi [A]  time = 0.0942834, antiderivative size = 123, normalized size of antiderivative = 1., 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} \[ -\frac{d^2 (2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{16 c^{5/2}}+\frac{d x^2 \sqrt{c+\frac{d}{x^2}} (2 b c-a d)}{16 c^2}+\frac{x^4 \sqrt{c+\frac{d}{x^2}} (2 b c-a d)}{8 c}+\frac{a x^6 \left (c+\frac{d}{x^2}\right )^{3/2}}{6 c} \]

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 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]]

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 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 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]

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*}

Mathematica [A]  time = 0.198508, size = 121, normalized size = 0.98 \[ \frac{x \sqrt{c+\frac{d}{x^2}} \left (\sqrt{c} x \sqrt{\frac{c x^2}{d}+1} \left (a \left (8 c^2 x^4+2 c d x^2-3 d^2\right )+6 b c \left (2 c x^2+d\right )\right )+3 d^{3/2} (a d-2 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )\right )}{48 c^{5/2} \sqrt{\frac{c x^2}{d}+1}} \]

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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Maple [A]  time = 0.017, size = 162, normalized size = 1.3 \begin{align*}{\frac{x}{48}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 8\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{3}a-6\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{3/2}xad+12\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{3/2}xb+3\,\sqrt{c}\sqrt{c{x}^{2}+d}xa{d}^{2}-6\,{c}^{3/2}\sqrt{c{x}^{2}+d}xbd+3\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) a{d}^{3}-6\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) bc{d}^{2} \right ){\frac{1}{\sqrt{c{x}^{2}+d}}}{c}^{-{\frac{5}{2}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.66413, size = 533, normalized size = 4.33 \begin{align*} \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{align*}

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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Sympy [B]  time = 37.8391, size = 226, normalized size = 1.84 \begin{align*} \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{align*}

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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Giac [A]  time = 1.14942, size = 193, normalized size = 1.57 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, a x^{2} \mathrm{sgn}\left (x\right ) + \frac{6 \, b c^{4} \mathrm{sgn}\left (x\right ) + a c^{3} d \mathrm{sgn}\left (x\right )}{c^{4}}\right )} x^{2} + \frac{3 \,{\left (2 \, b c^{3} d \mathrm{sgn}\left (x\right ) - a c^{2} d^{2} \mathrm{sgn}\left (x\right )\right )}}{c^{4}}\right )} \sqrt{c x^{2} + d} x + \frac{{\left (2 \, b c d^{2} \mathrm{sgn}\left (x\right ) - a d^{3} \mathrm{sgn}\left (x\right )\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}\left (x\right )}{32 \, c^{\frac{5}{2}}} \end{align*}

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)