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

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

[Out]

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

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Rubi [A]  time = 0.0503894, 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} \[ \frac{x (b c-a d) (3 a d+2 b c)}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}}+\frac{x \left (c+d x^2\right ) (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 4.13494, size = 214, normalized size = 2.04 \[ \frac{\sqrt{\frac{b x^2}{a}+1} \left (-16 b^3 x^6 \left (c+d x^2\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},-\frac{b x^2}{a}\right )+\frac{7 a^2 \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \left (\sqrt{-\frac{b x^2 \left (a+b x^2\right )}{a^2}} \left (2 b x^2-3 a\right )+3 a \sin ^{-1}\left (\sqrt{-\frac{b x^2}{a}}\right )\right )}{\sqrt{-\frac{b x^2}{a}}}-32 b^3 x^6 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{b x^2}{a}\right )\right )}{168 a^3 b^2 x^3 \sqrt{a+b x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + (b*x^2)/a]*((7*a^2*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*(Sqrt[-((b*x^2*(a + b*x^2))/a^2)]*(-3*a + 2*b*x
^2) + 3*a*ArcSin[Sqrt[-((b*x^2)/a)]]))/Sqrt[-((b*x^2)/a)] - 32*b^3*x^6*(2*c^2 + 3*c*d*x^2 + d^2*x^4)*Hypergeom
etric2F1[3/2, 7/2, 9/2, -((b*x^2)/a)] - 16*b^3*x^6*(c + d*x^2)^2*HypergeometricPFQ[{3/2, 2, 7/2}, {1, 9/2}, -(
(b*x^2)/a)]))/(168*a^3*b^2*x^3*Sqrt[a + b*x^2])

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Maple [A]  time = 0.006, size = 136, normalized size = 1.3 \begin{align*} -{\frac{{d}^{2}{x}^{3}}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{{d}^{2}x}{{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{{d}^{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}-{\frac{2\,cdx}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,cdx}{3\,ab}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{{c}^{2}x}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{c}^{2}x}{3\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65094, size = 655, normalized size = 6.24 \begin{align*} \left [\frac{3 \,{\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \,{\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \,{\left (a b^{3} c^{2} - a^{3} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac{3 \,{\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \,{\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \,{\left (a b^{3} c^{2} - a^{3} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16602, size = 139, normalized size = 1.32 \begin{align*} \frac{x{\left (\frac{2 \,{\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{2}}{a^{2} b^{3}} + \frac{3 \,{\left (a b^{3} c^{2} - a^{3} b d^{2}\right )}}{a^{2} b^{3}}\right )}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} - \frac{d^{2} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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