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

Optimal. Leaf size=107 \[ \frac{\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{5/2}}-\frac{3 b x \sqrt{c+d x^2} (b c-2 a d)}{8 d^2}+\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0594074, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {416, 388, 217, 206} \[ \frac{\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{5/2}}-\frac{3 b x \sqrt{c+d x^2} (b c-2 a d)}{8 d^2}+\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/Sqrt[c + d*x^2],x]

[Out]

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

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{\sqrt{c+d x^2}} \, dx &=\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d}+\frac{\int \frac{-a (b c-4 a d)-3 b (b c-2 a d) x^2}{\sqrt{c+d x^2}} \, dx}{4 d}\\ &=-\frac{3 b (b c-2 a d) x \sqrt{c+d x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d}-\frac{(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{8 d^2}\\ &=-\frac{3 b (b c-2 a d) x \sqrt{c+d x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d}-\frac{(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{8 d^2}\\ &=-\frac{3 b (b c-2 a d) x \sqrt{c+d x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right ) \sqrt{c+d x^2}}{4 d}+\frac{\left (3 b^2 c^2-8 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0515335, size = 91, normalized size = 0.85 \[ \frac{\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )+b \sqrt{d} x \sqrt{c+d x^2} \left (8 a d-3 b c+2 b d x^2\right )}{8 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 131, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}{x}^{3}}{4\,d}\sqrt{d{x}^{2}+c}}-{\frac{3\,{b}^{2}cx}{8\,{d}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{3\,{b}^{2}{c}^{2}}{8}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}}+{\frac{abx}{d}\sqrt{d{x}^{2}+c}}-{abc\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}}+{{a}^{2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){\frac{1}{\sqrt{d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^3/d*(d*x^2+c)^(1/2)-3/8*b^2*c/d^2*x*(d*x^2+c)^(1/2)+3/8*b^2*c^2/d^(5/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2)
)+a*b*x/d*(d*x^2+c)^(1/2)-a*b*c/d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+a^2*ln(x*d^(1/2)+(d*x^2+c)^(1/2))/d^(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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38382, size = 440, normalized size = 4.11 \begin{align*} \left [\frac{{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (2 \, b^{2} d^{2} x^{3} -{\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{16 \, d^{3}}, -\frac{{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (2 \, b^{2} d^{2} x^{3} -{\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{8 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 6.20184, size = 238, normalized size = 2.22 \begin{align*} a^{2} \left (\begin{cases} \frac{\sqrt{- \frac{c}{d}} \operatorname{asin}{\left (x \sqrt{- \frac{d}{c}} \right )}}{\sqrt{c}} & \text{for}\: c > 0 \wedge d < 0 \\\frac{\sqrt{\frac{c}{d}} \operatorname{asinh}{\left (x \sqrt{\frac{d}{c}} \right )}}{\sqrt{c}} & \text{for}\: c > 0 \wedge d > 0 \\\frac{\sqrt{- \frac{c}{d}} \operatorname{acosh}{\left (x \sqrt{- \frac{d}{c}} \right )}}{\sqrt{- c}} & \text{for}\: d > 0 \wedge c < 0 \end{cases}\right ) + \frac{a b \sqrt{c} x \sqrt{1 + \frac{d x^{2}}{c}}}{d} - \frac{a b c \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{d^{\frac{3}{2}}} - \frac{3 b^{2} c^{\frac{3}{2}} x}{8 d^{2} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{b^{2} \sqrt{c} x^{3}}{8 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{8 d^{\frac{5}{2}}} + \frac{b^{2} x^{5}}{4 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*Piecewise((sqrt(-c/d)*asin(x*sqrt(-d/c))/sqrt(c), (c > 0) & (d < 0)), (sqrt(c/d)*asinh(x*sqrt(d/c))/sqrt(
c), (c > 0) & (d > 0)), (sqrt(-c/d)*acosh(x*sqrt(-d/c))/sqrt(-c), (d > 0) & (c < 0))) + a*b*sqrt(c)*x*sqrt(1 +
d*x**2/c)/d - a*b*c*asinh(sqrt(d)*x/sqrt(c))/d**(3/2) - 3*b**2*c**(3/2)*x/(8*d**2*sqrt(1 + d*x**2/c)) - b**2*
sqrt(c)*x**3/(8*d*sqrt(1 + d*x**2/c)) + 3*b**2*c**2*asinh(sqrt(d)*x/sqrt(c))/(8*d**(5/2)) + b**2*x**5/(4*sqrt(
c)*sqrt(1 + d*x**2/c))

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Giac [A]  time = 1.12756, size = 123, normalized size = 1.15 \begin{align*} \frac{1}{8} \,{\left (\frac{2 \, b^{2} x^{2}}{d} - \frac{3 \, b^{2} c d - 8 \, a b d^{2}}{d^{3}}\right )} \sqrt{d x^{2} + c} x - \frac{{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{8 \, d^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*b^2*x^2/d - (3*b^2*c*d - 8*a*b*d^2)/d^3)*sqrt(d*x^2 + c)*x - 1/8*(3*b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*lo
g(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)