### 3.190 $$\int \frac{x^4}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=172 $-\frac{a^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 a^2 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

(4*a^3)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^4/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*x*(a
+ b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^2*(a + b*x))/(2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*a^2*(
a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0727534, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 a^2 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(4*a^3)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^4/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*x*(a
+ b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^2*(a + b*x))/(2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*a^2*(
a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 646

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{x^4}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (-\frac{3 a}{b^7}+\frac{x}{b^6}+\frac{a^4}{b^7 (a+b x)^3}-\frac{4 a^3}{b^7 (a+b x)^2}+\frac{6 a^2}{b^7 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{4 a^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 a^2 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0266041, size = 83, normalized size = 0.48 $\frac{-11 a^2 b^2 x^2+2 a^3 b x+12 a^2 (a+b x)^2 \log (a+b x)+7 a^4-4 a b^3 x^3+b^4 x^4}{2 b^5 (a+b x) \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(7*a^4 + 2*a^3*b*x - 11*a^2*b^2*x^2 - 4*a*b^3*x^3 + b^4*x^4 + 12*a^2*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x
)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.229, size = 101, normalized size = 0.6 \begin{align*}{\frac{ \left ({b}^{4}{x}^{4}+12\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}-4\,a{b}^{3}{x}^{3}+24\,\ln \left ( bx+a \right ) x{a}^{3}b-11\,{x}^{2}{a}^{2}{b}^{2}+12\,{a}^{4}\ln \left ( bx+a \right ) +2\,x{a}^{3}b+7\,{a}^{4} \right ) \left ( bx+a \right ) }{2\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(b^4*x^4+12*ln(b*x+a)*x^2*a^2*b^2-4*a*b^3*x^3+24*ln(b*x+a)*x*a^3*b-11*x^2*a^2*b^2+12*a^4*ln(b*x+a)+2*x*a^3
*b+7*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [A]  time = 1.62069, size = 223, normalized size = 1.3 \begin{align*} \frac{x^{3}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{5 \, a x^{2}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac{6 \, a^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{2}} + \frac{9 \, a^{4}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{12 \, a^{3} x}{{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{5 \, a^{3}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac{5 \, a^{4}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b^{4}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*a^2*log(x + a/
b)/((b^2)^(3/2)*b^2) + 9*a^4/((b^2)^(7/2)*(x + a/b)^2) + 12*a^3*x/((b^2)^(5/2)*b*(x + a/b)^2) - 5*a^3/(sqrt(b^
2*x^2 + 2*a*b*x + a^2)*b^5) + 5/2*a^4/((b^2)^(3/2)*b^4*(x + a/b)^2)

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Fricas [A]  time = 1.58875, size = 200, normalized size = 1.16 \begin{align*} \frac{b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4} + 12 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4 + 12*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*log(b*x +
a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**4/((a + b*x)**2)**(3/2), 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(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x