3.658 \(\int \frac{1}{x^4 (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=291 \[ \frac{1155 b \left (a+b x^2\right )}{128 a^6 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{385 \left (a+b x^2\right )}{128 a^5 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1155 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

231/(128*a^4*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*x^3*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
+ 11/(48*a^2*x^3*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 33/(64*a^3*x^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4]) - (385*(a + b*x^2))/(128*a^5*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (1155*b*(a + b*x^2))/(128*
a^6*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (1155*b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(13/2)*
Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.124814, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1112, 290, 325, 205} \[ \frac{1155 b \left (a+b x^2\right )}{128 a^6 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{385 \left (a+b x^2\right )}{128 a^5 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1155 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

231/(128*a^4*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*x^3*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
+ 11/(48*a^2*x^3*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 33/(64*a^3*x^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4]) - (385*(a + b*x^2))/(128*a^5*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (1155*b*(a + b*x^2))/(128*
a^6*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (1155*b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(13/2)*
Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^5} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (11 b^3 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^4} \, dx}{8 a \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (33 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^3} \, dx}{16 a^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (231 b \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^2} \, dx}{64 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (1155 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )} \, dx}{128 a^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{385 \left (a+b x^2\right )}{128 a^5 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (1155 b \left (a b+b^2 x^2\right )\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{128 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{385 \left (a+b x^2\right )}{128 a^5 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1155 b \left (a+b x^2\right )}{128 a^6 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (1155 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{128 a^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{231}{128 a^4 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a x^3 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{11}{48 a^2 x^3 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{33}{64 a^3 x^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{385 \left (a+b x^2\right )}{128 a^5 x^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1155 b \left (a+b x^2\right )}{128 a^6 x \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1155 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0479967, size = 127, normalized size = 0.44 \[ \frac{\sqrt{a} \left (16863 a^2 b^3 x^6+9207 a^3 b^2 x^4+1408 a^4 b x^2-128 a^5+12705 a b^4 x^8+3465 b^5 x^{10}\right )+3465 b^{3/2} x^3 \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{384 a^{13/2} x^3 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

(Sqrt[a]*(-128*a^5 + 1408*a^4*b*x^2 + 9207*a^3*b^2*x^4 + 16863*a^2*b^3*x^6 + 12705*a*b^4*x^8 + 3465*b^5*x^10)
+ 3465*b^(3/2)*x^3*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(384*a^(13/2)*x^3*(a + b*x^2)^3*Sqrt[(a + b*x^2)
^2])

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Maple [A]  time = 0.224, size = 211, normalized size = 0.7 \begin{align*}{\frac{b{x}^{2}+a}{384\,{a}^{6}{x}^{3}} \left ( 3465\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{11}{b}^{6}+3465\,\sqrt{ab}{x}^{10}{b}^{5}+13860\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{9}a{b}^{5}+12705\,\sqrt{ab}{x}^{8}a{b}^{4}+20790\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{7}{a}^{2}{b}^{4}+16863\,\sqrt{ab}{x}^{6}{a}^{2}{b}^{3}+13860\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}{a}^{3}{b}^{3}+9207\,\sqrt{ab}{x}^{4}{a}^{3}{b}^{2}+3465\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}{a}^{4}{b}^{2}+1408\,\sqrt{ab}{x}^{2}{a}^{4}b-128\,\sqrt{ab}{a}^{5} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/384*(3465*arctan(b*x/(a*b)^(1/2))*x^11*b^6+3465*(a*b)^(1/2)*x^10*b^5+13860*arctan(b*x/(a*b)^(1/2))*x^9*a*b^5
+12705*(a*b)^(1/2)*x^8*a*b^4+20790*arctan(b*x/(a*b)^(1/2))*x^7*a^2*b^4+16863*(a*b)^(1/2)*x^6*a^2*b^3+13860*arc
tan(b*x/(a*b)^(1/2))*x^5*a^3*b^3+9207*(a*b)^(1/2)*x^4*a^3*b^2+3465*arctan(b*x/(a*b)^(1/2))*x^3*a^4*b^2+1408*(a
*b)^(1/2)*x^2*a^4*b-128*(a*b)^(1/2)*a^5)*(b*x^2+a)/(a*b)^(1/2)/x^3/a^6/((b*x^2+a)^2)^(5/2)

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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(1/x^4/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38203, size = 815, normalized size = 2.8 \begin{align*} \left [\frac{6930 \, b^{5} x^{10} + 25410 \, a b^{4} x^{8} + 33726 \, a^{2} b^{3} x^{6} + 18414 \, a^{3} b^{2} x^{4} + 2816 \, a^{4} b x^{2} - 256 \, a^{5} + 3465 \,{\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right )}{768 \,{\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}, \frac{3465 \, b^{5} x^{10} + 12705 \, a b^{4} x^{8} + 16863 \, a^{2} b^{3} x^{6} + 9207 \, a^{3} b^{2} x^{4} + 1408 \, a^{4} b x^{2} - 128 \, a^{5} + 3465 \,{\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{384 \,{\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

[1/768*(6930*b^5*x^10 + 25410*a*b^4*x^8 + 33726*a^2*b^3*x^6 + 18414*a^3*b^2*x^4 + 2816*a^4*b*x^2 - 256*a^5 + 3
465*(b^5*x^11 + 4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b
/a) - a)/(b*x^2 + a)))/(a^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x^3), 1/384*(3465*b^
5*x^10 + 12705*a*b^4*x^8 + 16863*a^2*b^3*x^6 + 9207*a^3*b^2*x^4 + 1408*a^4*b*x^2 - 128*a^5 + 3465*(b^5*x^11 +
4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^6*b^4*x^11 + 4*a^7*
b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(1/(x**4*((a + b*x**2)**2)**(5/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x