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

Optimal. Leaf size=75 \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{\sqrt{c}}-\frac{b \sqrt{c+d x^2} (b c-2 a d)}{d^2}+\frac{b^2 \left (c+d x^2\right )^{3/2}}{3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0706001, antiderivative size = 75, 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, 88, 63, 208} \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{\sqrt{c}}-\frac{b \sqrt{c+d x^2} (b c-2 a d)}{d^2}+\frac{b^2 \left (c+d x^2\right )^{3/2}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b*c - 2*a*d)*Sqrt[c + d*x^2])/d^2) + (b^2*(c + d*x^2)^(3/2))/(3*d^2) - (a^2*ArcTanh[Sqrt[c + d*x^2]/Sqrt
[c]])/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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A]  time = 0.0624067, size = 63, normalized size = 0.84 \[ \frac{b \sqrt{c+d x^2} \left (6 a d-2 b c+b d x^2\right )}{3 d^2}-\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 87, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}{x}^{2}}{3\,d}\sqrt{d{x}^{2}+c}}-{\frac{2\,{b}^{2}c}{3\,{d}^{2}}\sqrt{d{x}^{2}+c}}+2\,{\frac{\sqrt{d{x}^{2}+c}ab}{d}}-{{a}^{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33734, size = 365, normalized size = 4.87 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{c} d^{2} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt{d x^{2} + c}}{6 \, c d^{2}}, \frac{3 \, a^{2} \sqrt{-c} d^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt{d x^{2} + c}}{3 \, c d^{2}}\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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 25.2199, size = 76, normalized size = 1.01 \begin{align*} \frac{a^{2} \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{c}} \sqrt{c + d x^{2}}} \right )}}{c \sqrt{- \frac{1}{c}}} + \frac{b^{2} \left (c + d x^{2}\right )^{\frac{3}{2}}}{3 d^{2}} + \frac{b \sqrt{c + d x^{2}} \left (2 a d - b c\right )}{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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.13875, size = 111, normalized size = 1.48 \begin{align*} \frac{a^{2} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} d^{4} - 3 \, \sqrt{d x^{2} + c} b^{2} c d^{4} + 6 \, \sqrt{d x^{2} + c} a b d^{5}}{3 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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