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

Optimal. Leaf size=131 \[ \frac{-\frac{5 a^2 d}{c}+4 a b-\frac{2 b^2 c}{d}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{a (4 b c-5 a d)}{2 c^3 \sqrt{c+d x^2}}-\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{7/2}} \]

[Out]

(4*a*b - (2*b^2*c)/d - (5*a^2*d)/c)/(6*c*(c + d*x^2)^(3/2)) - a^2/(2*c*x^2*(c + d*x^2)^(3/2)) + (a*(4*b*c - 5*
a*d))/(2*c^3*Sqrt[c + d*x^2]) - (a*(4*b*c - 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(7/2))

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Rubi [A]  time = 0.117171, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 89, 78, 51, 63, 208} \[ \frac{-\frac{5 a^2 d}{c}+4 a b-\frac{2 b^2 c}{d}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{a (4 b c-5 a d)}{2 c^3 \sqrt{c+d x^2}}-\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*b - (2*b^2*c)/d - (5*a^2*d)/c)/(6*c*(c + d*x^2)^(3/2)) - a^2/(2*c*x^2*(c + d*x^2)^(3/2)) + (a*(4*b*c - 5*
a*d))/(2*c^3*Sqrt[c + d*x^2]) - (a*(4*b*c - 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(7/2))

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 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 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 \frac{\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^2 (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a (4 b c-5 a d)+b^2 c x}{x (c+d x)^{5/2}} \, dx,x,x^2\right )}{2 c}\\ &=\frac{4 a b-\frac{2 b^2 c}{d}-\frac{5 a^2 d}{c}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{(a (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{4 a b-\frac{2 b^2 c}{d}-\frac{5 a^2 d}{c}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{a (4 b c-5 a d)}{2 c^3 \sqrt{c+d x^2}}+\frac{(a (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{4 c^3}\\ &=\frac{4 a b-\frac{2 b^2 c}{d}-\frac{5 a^2 d}{c}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{a (4 b c-5 a d)}{2 c^3 \sqrt{c+d x^2}}+\frac{(a (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{2 c^3 d}\\ &=\frac{4 a b-\frac{2 b^2 c}{d}-\frac{5 a^2 d}{c}}{6 c \left (c+d x^2\right )^{3/2}}-\frac{a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac{a (4 b c-5 a d)}{2 c^3 \sqrt{c+d x^2}}-\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0452251, size = 105, normalized size = 0.8 \[ \frac{-c \left (a^2 d \left (3 c+5 d x^2\right )-4 a b c d x^2+2 b^2 c^2 x^2\right )-3 a d x^2 \left (c+d x^2\right ) (5 a d-4 b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )}{6 c^3 d x^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(c*(2*b^2*c^2*x^2 - 4*a*b*c*d*x^2 + a^2*d*(3*c + 5*d*x^2))) - 3*a*d*(-4*b*c + 5*a*d)*x^2*(c + d*x^2)*Hyperge
ometric2F1[-1/2, 1, 1/2, 1 + (d*x^2)/c])/(6*c^3*d*x^2*(c + d*x^2)^(3/2))

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Maple [A]  time = 0.012, size = 169, normalized size = 1.3 \begin{align*} -{\frac{{b}^{2}}{3\,d} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,ab}{3\,c} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{ab}{{c}^{2}\sqrt{d{x}^{2}+c}}}-2\,{\frac{ab}{{c}^{5/2}}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) }-{\frac{{a}^{2}}{2\,c{x}^{2}} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,{a}^{2}d}{6\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,{a}^{2}d}{2\,{c}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{5\,{a}^{2}d}{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2/d/(d*x^2+c)^(3/2)+2/3*a*b/c/(d*x^2+c)^(3/2)+2*a*b/c^2/(d*x^2+c)^(1/2)-2*a*b/c^(5/2)*ln((2*c+2*c^(1/2)
*(d*x^2+c)^(1/2))/x)-1/2*a^2/c/x^2/(d*x^2+c)^(3/2)-5/6*a^2*d/c^2/(d*x^2+c)^(3/2)-5/2*a^2*d/c^3/(d*x^2+c)^(1/2)
+5/2*a^2*d/c^(7/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.44415, size = 894, normalized size = 6.82 \begin{align*} \left [-\frac{3 \,{\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \,{\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} +{\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{c} \log \left (-\frac{d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (3 \, a^{2} c^{3} d - 3 \,{\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \,{\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{12 \,{\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}, \frac{3 \,{\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \,{\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} +{\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) -{\left (3 \, a^{2} c^{3} d - 3 \,{\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \,{\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{6 \,{\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*((4*a*b*c*d^3 - 5*a^2*d^4)*x^6 + 2*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 + (4*a*b*c^3*d - 5*a^2*c^2*d^2)
*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3*a^2*c^3*d - 3*(4*a*b*c^2*d^2 - 5*a^2*
c*d^3)*x^4 + 2*(b^2*c^4 - 8*a*b*c^3*d + 10*a^2*c^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c^4*d^3*x^6 + 2*c^5*d^2*x^4 + c
^6*d*x^2), 1/6*(3*((4*a*b*c*d^3 - 5*a^2*d^4)*x^6 + 2*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 + (4*a*b*c^3*d - 5*a^2*
c^2*d^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (3*a^2*c^3*d - 3*(4*a*b*c^2*d^2 - 5*a^2*c*d^3)*x^4 +
2*(b^2*c^4 - 8*a*b*c^3*d + 10*a^2*c^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c^4*d^3*x^6 + 2*c^5*d^2*x^4 + c^6*d*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{x^{3} \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1442, size = 173, normalized size = 1.32 \begin{align*} \frac{{\left (4 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{2 \, \sqrt{-c} c^{3}} - \frac{\sqrt{d x^{2} + c} a^{2}}{2 \, c^{3} x^{2}} - \frac{b^{2} c^{3} - 6 \,{\left (d x^{2} + c\right )} a b c d - 2 \, a b c^{2} d + 6 \,{\left (d x^{2} + c\right )} a^{2} d^{2} + a^{2} c d^{2}}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*b*c - 5*a^2*d)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/2*sqrt(d*x^2 + c)*a^2/(c^3*x^2) -
1/3*(b^2*c^3 - 6*(d*x^2 + c)*a*b*c*d - 2*a*b*c^2*d + 6*(d*x^2 + c)*a^2*d^2 + a^2*c*d^2)/((d*x^2 + c)^(3/2)*c^3
*d)