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

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

[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))

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Rubi [A]  time = 0.0901921, antiderivative size = 123, normalized size of antiderivative = 1., 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} \[ \frac{d^2 (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{16 c^{3/2}}+\frac{x^4 \left (c+\frac{d}{x^2}\right )^{3/2} (6 b c-a d)}{24 c}+\frac{d x^2 \sqrt{c+\frac{d}{x^2}} (6 b c-a d)}{16 c}+\frac{a x^6 \left (c+\frac{d}{x^2}\right )^{5/2}}{6 c} \]

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 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 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 ) \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*}

Mathematica [A]  time = 0.118012, size = 123, normalized size = 1. \[ \frac{x \sqrt{c+\frac{d}{x^2}} \left (\sqrt{c} x \sqrt{\frac{c x^2}{d}+1} \left (a \left (8 c^2 x^4+14 c d x^2+3 d^2\right )+6 b c \left (2 c x^2+5 d\right )\right )-3 d^{3/2} (a d-6 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )\right )}{48 c^{3/2} \sqrt{\frac{c x^2}{d}+1}} \]

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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Maple [A]  time = 0.007, size = 162, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}}{48} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 8\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{5/2}xa-2\,\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}+18\,{c}^{3/2}\sqrt{c{x}^{2}+d}xbd-3\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) a{d}^{3}+18\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) bc{d}^{2} \right ) \left ( c{x}^{2}+d \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{3}{2}}}} \end{align*}

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^(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+18*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+18*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*b*c*d^2)/(c*x^2+d)^(3/2)/c^(3/2)

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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)*(c+d/x^2)^(3/2)*x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47824, size = 543, normalized size = 4.41 \begin{align*} \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{align*}

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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Sympy [B]  time = 52.0316, size = 253, normalized size = 2.06 \begin{align*} \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{align*}

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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Giac [A]  time = 1.16028, size = 194, normalized size = 1.58 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, a c x^{2} \mathrm{sgn}\left (x\right ) + \frac{6 \, b c^{5} \mathrm{sgn}\left (x\right ) + 7 \, a c^{4} d \mathrm{sgn}\left (x\right )}{c^{4}}\right )} x^{2} + \frac{3 \,{\left (10 \, b c^{4} d \mathrm{sgn}\left (x\right ) + a c^{3} d^{2} \mathrm{sgn}\left (x\right )\right )}}{c^{4}}\right )} \sqrt{c x^{2} + d} x - \frac{{\left (6 \, 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{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}\left (x\right )}{32 \, c^{\frac{3}{2}}} \end{align*}

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)