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

Optimal. Leaf size=105 \[ -\frac{x (b c-a d) (2 a d+3 b c)}{3 c^2 d^2 \sqrt{c+d x^2}}-\frac{x \left (a+b x^2\right ) (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}}+\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{d^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0497085, antiderivative size = 105, 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 = {413, 385, 217, 206} \[ -\frac{x (b c-a d) (2 a d+3 b c)}{3 c^2 d^2 \sqrt{c+d x^2}}-\frac{x \left (a+b x^2\right ) (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}}+\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 413

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

Rule 385

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

Mathematica [A]  time = 0.122668, size = 101, normalized size = 0.96 \[ \frac{x \left (a^2 d^2 \left (3 c+2 d x^2\right )+2 a b c d^2 x^2-b^2 c^2 \left (3 c+4 d x^2\right )\right )}{3 c^2 d^2 \left (c+d x^2\right )^{3/2}}+\frac{b^2 \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 136, normalized size = 1.3 \begin{align*} -{\frac{{b}^{2}{x}^{3}}{3\,d} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}-{\frac{{b}^{2}x}{{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{{b}^{2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}}-{\frac{2\,abx}{3\,d} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,abx}{3\,cd}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{{a}^{2}x}{3\,c} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{a}^{2}x}{3\,{c}^{2}}{\frac{1}{\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)^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37971, size = 655, normalized size = 6.24 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) - 2 \,{\left (2 \,{\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{3} + 3 \,{\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{6 \,{\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}}, -\frac{3 \,{\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (2 \,{\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{3} + 3 \,{\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{3 \,{\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.15513, size = 142, normalized size = 1.35 \begin{align*} -\frac{x{\left (\frac{2 \,{\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{2}}{c^{2} d^{3}} + \frac{3 \,{\left (b^{2} c^{3} d - a^{2} c d^{3}\right )}}{c^{2} d^{3}}\right )}}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} - \frac{b^{2} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{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)^(5/2),x, algorithm="giac")

[Out]

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