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

Optimal. Leaf size=88 \[ \frac{\frac{a^2}{c^2}-\frac{b^2}{d^2}}{\sqrt{c+d x^2}}-\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{c^{5/2}}+\frac{(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0920913, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {446, 87, 63, 208} \[ \frac{\frac{a^2}{c^2}-\frac{b^2}{d^2}}{\sqrt{c+d x^2}}-\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{c^{5/2}}+\frac{(b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c - a*d)^2/(3*c*d^2*(c + d*x^2)^(3/2)) + (a^2/c^2 - b^2/d^2)/Sqrt[c + d*x^2] - (a^2*ArcTanh[Sqrt[c + d*x^2]
/Sqrt[c]])/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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

Mathematica [C]  time = 0.0446822, size = 67, normalized size = 0.76 \[ \frac{a^2 d^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d x^2}{c}+1\right )-b c \left (2 a d+2 b c+3 b d x^2\right )}{3 c d^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*c*(2*b*c + 2*a*d + 3*b*d*x^2)) + a^2*d^2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (d*x^2)/c])/(3*c*d^2*(c + d
*x^2)^(3/2))

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Maple [A]  time = 0.011, size = 120, normalized size = 1.4 \begin{align*} -{\frac{{b}^{2}{x}^{2}}{d} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{b}^{2}c}{3\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,ab}{3\,d} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{2}}{3\,c} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{2}}{{c}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{{a}^{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*x^2/d/(d*x^2+c)^(3/2)-2/3*b^2*c/d^2/(d*x^2+c)^(3/2)-2/3*a*b/d/(d*x^2+c)^(3/2)+1/3*a^2/c/(d*x^2+c)^(3/2)+a
^2/c^2/(d*x^2+c)^(1/2)-a^2/c^(5/2)*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/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46352, size = 651, normalized size = 7.4 \begin{align*} \left [\frac{3 \,{\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{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 (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \,{\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{6 \,{\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}, \frac{3 \,{\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) -{\left (2 \, b^{2} c^{4} + 2 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} + 3 \,{\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{3 \,{\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 23.7789, size = 87, normalized size = 0.99 \begin{align*} \frac{a^{2} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{- c}} \right )}}{c^{2} \sqrt{- c}} + \frac{\left (a d - b c\right )^{2}}{3 c d^{2} \left (c + d x^{2}\right )^{\frac{3}{2}}} + \frac{\left (a d - b c\right ) \left (a d + b c\right )}{c^{2} d^{2} \sqrt{c + d x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*atan(sqrt(c + d*x**2)/sqrt(-c))/(c**2*sqrt(-c)) + (a*d - b*c)**2/(3*c*d**2*(c + d*x**2)**(3/2)) + (a*d -
b*c)*(a*d + b*c)/(c**2*d**2*sqrt(c + d*x**2))

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Giac [A]  time = 1.1615, size = 138, normalized size = 1.57 \begin{align*} \frac{a^{2} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{2}} - \frac{3 \,{\left (d x^{2} + c\right )} b^{2} c^{2} - b^{2} c^{3} + 2 \, a b c^{2} d - 3 \,{\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^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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