### 3.200 $$\int \frac{x^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=171 $-\frac{a^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

(4*a)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^4/(4*b^5*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*a^3)/(3
*b^5*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a^2)/(b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^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 = 0.0781741, antiderivative size = 171, 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} $-\frac{a^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*a)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^4/(4*b^5*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*a^3)/(3
*b^5*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a^2)/(b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{x^4}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac{a^4}{b^9 (a+b x)^5}-\frac{4 a^3}{b^9 (a+b x)^4}+\frac{6 a^2}{b^9 (a+b x)^3}-\frac{4 a}{b^9 (a+b x)^2}+\frac{1}{b^9 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{4 a}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 a^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(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 = 0.0256027, size = 73, normalized size = 0.43 $\frac{a \left (88 a^2 b x+25 a^3+108 a b^2 x^2+48 b^3 x^3\right )+12 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a*(25*a^3 + 88*a^2*b*x + 108*a*b^2*x^2 + 48*b^3*x^3) + 12*(a + b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[
(a + b*x)^2])

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Maple [A]  time = 0.223, size = 123, normalized size = 0.7 \begin{align*}{\frac{ \left ( 12\,\ln \left ( bx+a \right ){x}^{4}{b}^{4}+48\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}+72\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}+48\,a{b}^{3}{x}^{3}+48\,\ln \left ( bx+a \right ) x{a}^{3}b+108\,{x}^{2}{a}^{2}{b}^{2}+12\,{a}^{4}\ln \left ( bx+a \right ) +88\,x{a}^{3}b+25\,{a}^{4} \right ) \left ( bx+a \right ) }{12\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

1/12*(12*ln(b*x+a)*x^4*b^4+48*ln(b*x+a)*x^3*a*b^3+72*ln(b*x+a)*x^2*a^2*b^2+48*a*b^3*x^3+48*ln(b*x+a)*x*a^3*b+1
08*x^2*a^2*b^2+12*a^4*ln(b*x+a)+88*x*a^3*b+25*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(5/2)

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Maxima [A]  time = 1.18393, size = 124, normalized size = 0.73 \begin{align*} \frac{48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{12 \,{\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} + \frac{\log \left (b x + a\right )}{b^{5}} \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)^(5/2),x, algorithm="maxima")

[Out]

1/12*(48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6
*x + a^4*b^5) + log(b*x + a)/b^5

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Fricas [A]  time = 1.61907, size = 271, normalized size = 1.58 \begin{align*} \frac{48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4} + 12 \,{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{12 \,{\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

1/12*(48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4 + 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3
*b*x + a^4)*log(b*x + a))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*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{5}{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)**(5/2),x)

[Out]

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

[Out]

sage0*x