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

Optimal. Leaf size=181 \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac{\left (c+d x^2\right )^{3/2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{24 c^2}+\frac{\sqrt{c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac{\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 \sqrt{c}}-\frac{a \left (c+d x^2\right )^{5/2} (a d+8 b c)}{8 c^2 x^2} \]

[Out]

((8*b^2*c^2 + 3*a*d*(8*b*c + a*d))*Sqrt[c + d*x^2])/(8*c) + ((8*b^2*c^2 + 3*a*d*(8*b*c + a*d))*(c + d*x^2)^(3/
2))/(24*c^2) - (a^2*(c + d*x^2)^(5/2))/(4*c*x^4) - (a*(8*b*c + a*d)*(c + d*x^2)^(5/2))/(8*c^2*x^2) - ((8*b^2*c
^2 + 3*a*d*(8*b*c + a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*Sqrt[c])

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Rubi [A]  time = 0.207048, antiderivative size = 178, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 89, 78, 50, 63, 208} \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac{1}{24} \left (c+d x^2\right )^{3/2} \left (\frac{3 a d (a d+8 b c)}{c^2}+8 b^2\right )+\frac{\sqrt{c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac{\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 \sqrt{c}}-\frac{a \left (c+d x^2\right )^{5/2} (a d+8 b c)}{8 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^2*c^2 + 3*a*d*(8*b*c + a*d))*Sqrt[c + d*x^2])/(8*c) + ((8*b^2 + (3*a*d*(8*b*c + a*d))/c^2)*(c + d*x^2)^(
3/2))/24 - (a^2*(c + d*x^2)^(5/2))/(4*c*x^4) - (a*(8*b*c + a*d)*(c + d*x^2)^(5/2))/(8*c^2*x^2) - ((8*b^2*c^2 +
3*a*d*(8*b*c + a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*Sqrt[c])

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (c+d x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{1}{2} a (8 b c+a d)+2 b^2 c x\right ) (c+d x)^{3/2}}{x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac{a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac{1}{16} \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{24} \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac{a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac{1}{16} \left (c \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{8} c \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{1}{24} \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac{a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac{1}{16} \left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{1}{8} c \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{1}{24} \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac{a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac{\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{8 d}\\ &=\frac{1}{8} c \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{1}{24} \left (8 b^2+\frac{3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac{a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}-\frac{\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0969294, size = 116, normalized size = 0.64 \[ \frac{1}{24} \left (\frac{\sqrt{c+d x^2} \left (-3 a^2 \left (2 c+5 d x^2\right )-24 a b x^2 \left (c-2 d x^2\right )+8 b^2 x^4 \left (4 c+d x^2\right )\right )}{x^4}-\frac{3 \left (3 a^2 d^2+24 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c + d*x^2]*(-24*a*b*x^2*(c - 2*d*x^2) + 8*b^2*x^4*(4*c + d*x^2) - 3*a^2*(2*c + 5*d*x^2)))/x^4 - (3*(8*b
^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/Sqrt[c])/24

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Maple [A]  time = 0.011, size = 256, normalized size = 1.4 \begin{align*}{\frac{{b}^{2}}{3} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{b}^{2}{c}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) +{b}^{2}\sqrt{d{x}^{2}+c}c-{\frac{{a}^{2}}{4\,c{x}^{4}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{a}^{2}d}{8\,{c}^{2}{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}{d}^{2}}{8\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{a}^{2}{d}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){\frac{1}{\sqrt{c}}}}+{\frac{3\,{a}^{2}{d}^{2}}{8\,c}\sqrt{d{x}^{2}+c}}-{\frac{ab}{c{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{abd}{c} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-3\,abd\sqrt{c}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) +3\,abd\sqrt{d{x}^{2}+c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51112, size = 605, normalized size = 3.34 \begin{align*} \left [\frac{3 \,{\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt{c} x^{4} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (8 \, b^{2} c d x^{6} + 16 \,{\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \,{\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{48 \, c x^{4}}, \frac{3 \,{\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (8 \, b^{2} c d x^{6} + 16 \,{\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \,{\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{24 \, c x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/48*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*sqrt(c)*x^4*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2)
+ 2*(8*b^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 +
c))/(c*x^4), 1/24*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (8*b
^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c*x
^4)]

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Sympy [A]  time = 89.4264, size = 332, normalized size = 1.83 \begin{align*} - \frac{a^{2} c^{2}}{4 \sqrt{d} x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{3 a^{2} c \sqrt{d}}{8 x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{a^{2} d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{2 x} - \frac{a^{2} d^{\frac{3}{2}}}{8 x \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{3 a^{2} d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{8 \sqrt{c}} - 3 a b \sqrt{c} d \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )} - \frac{a b c \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{x} + \frac{2 a b c \sqrt{d}}{x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{2 a b d^{\frac{3}{2}} x}{\sqrt{\frac{c}{d x^{2}} + 1}} - b^{2} c^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )} + \frac{b^{2} c^{2}}{\sqrt{d} x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{b^{2} c \sqrt{d} x}{\sqrt{\frac{c}{d x^{2}} + 1}} + b^{2} d \left (\begin{cases} \frac{\sqrt{c} x^{2}}{2} & \text{for}\: d = 0 \\\frac{\left (c + d x^{2}\right )^{\frac{3}{2}}}{3 d} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c**2/(4*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a**2*c*sqrt(d)/(8*x**3*sqrt(c/(d*x**2) + 1)) - a**2*d**(3
/2)*sqrt(c/(d*x**2) + 1)/(2*x) - a**2*d**(3/2)/(8*x*sqrt(c/(d*x**2) + 1)) - 3*a**2*d**2*asinh(sqrt(c)/(sqrt(d)
*x))/(8*sqrt(c)) - 3*a*b*sqrt(c)*d*asinh(sqrt(c)/(sqrt(d)*x)) - a*b*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/x + 2*a*b*c
*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*a*b*d**(3/2)*x/sqrt(c/(d*x**2) + 1) - b**2*c**(3/2)*asinh(sqrt(c)/(sqrt(
d)*x)) + b**2*c**2/(sqrt(d)*x*sqrt(c/(d*x**2) + 1)) + b**2*c*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + b**2*d*Piecewise
((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True))

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Giac [A]  time = 1.16235, size = 246, normalized size = 1.36 \begin{align*} \frac{8 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} d + 24 \, \sqrt{d x^{2} + c} b^{2} c d + 48 \, \sqrt{d x^{2} + c} a b d^{2} + \frac{3 \,{\left (8 \, b^{2} c^{2} d + 24 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{3 \,{\left (8 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c d^{2} - 8 \, \sqrt{d x^{2} + c} a b c^{2} d^{2} + 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} d^{3} - 3 \, \sqrt{d x^{2} + c} a^{2} c d^{3}\right )}}{d^{2} x^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^5,x, algorithm="giac")

[Out]

1/24*(8*(d*x^2 + c)^(3/2)*b^2*d + 24*sqrt(d*x^2 + c)*b^2*c*d + 48*sqrt(d*x^2 + c)*a*b*d^2 + 3*(8*b^2*c^2*d + 2
4*a*b*c*d^2 + 3*a^2*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - 3*(8*(d*x^2 + c)^(3/2)*a*b*c*d^2 - 8*sqrt
(d*x^2 + c)*a*b*c^2*d^2 + 5*(d*x^2 + c)^(3/2)*a^2*d^3 - 3*sqrt(d*x^2 + c)*a^2*c*d^3)/(d^2*x^4))/d