3.181 \(\int \frac{x^4}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=182 \[ -\frac{a^3 x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x^3 (a+b x)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^4 (a+b x)}{4 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^4 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-((a^3*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^2*x^2*(a + b*x))/(2*b^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (a*x^3*(a + b*x))/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^4*(a + b*x))/(4*b*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) + (a^4*(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 = 0.0616439, antiderivative size = 182, 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} \[ -\frac{a^3 x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x^3 (a+b x)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^4 (a+b x)}{4 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^4 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

-((a^3*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^2*x^2*(a + b*x))/(2*b^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (a*x^3*(a + b*x))/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^4*(a + b*x))/(4*b*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) + (a^4*(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}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^4}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a^3}{b^5}+\frac{a^2 x}{b^4}-\frac{a x^2}{b^3}+\frac{x^3}{b^2}+\frac{a^4}{b^5 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a^3 x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x^3 (a+b x)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^4 (a+b x)}{4 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^4 (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 = 0.0252535, size = 68, normalized size = 0.37 \[ \frac{(a+b x) \left (b x \left (6 a^2 b x-12 a^3-4 a b^2 x^2+3 b^3 x^3\right )+12 a^4 \log (a+b x)\right )}{12 b^5 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*(b*x*(-12*a^3 + 6*a^2*b*x - 4*a*b^2*x^2 + 3*b^3*x^3) + 12*a^4*Log[a + b*x]))/(12*b^5*Sqrt[(a + b*x)
^2])

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Maple [A]  time = 0.246, size = 67, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 3\,{b}^{4}{x}^{4}-4\,a{b}^{3}{x}^{3}+6\,{x}^{2}{a}^{2}{b}^{2}+12\,{a}^{4}\ln \left ( bx+a \right ) -12\,x{a}^{3}b \right ) }{12\,{b}^{5}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.25644, size = 201, normalized size = 1.1 \begin{align*} \frac{13 \, a^{4} \log \left (x + \frac{a}{b}\right )}{6 \,{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{13 \, a^{3} x}{6 \,{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{13 \, a^{2} x^{2}}{12 \, \sqrt{b^{2}} b^{2}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} x^{3}}{4 \, b^{2}} - \frac{7 \, a^{4} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{6 \, b^{4}} - \frac{7 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} a x^{2}}{12 \, b^{3}} + \frac{7 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}}{6 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/6*a^4*log(x + a/b)/(b^2)^(5/2) - 13/6*a^3*x/((b^2)^(3/2)*b) + 13/12*a^2*x^2/(sqrt(b^2)*b^2) + 1/4*sqrt(b^2*
x^2 + 2*a*b*x + a^2)*x^3/b^2 - 7/6*a^4*sqrt(b^(-2))*log(x + a/b)/b^4 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*x^
2/b^3 + 7/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3/b^5

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Fricas [A]  time = 1.72371, size = 117, normalized size = 0.64 \begin{align*} \frac{3 \, b^{4} x^{4} - 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 12 \, a^{3} b x + 12 \, a^{4} \log \left (b x + a\right )}{12 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.06767, size = 49, normalized size = 0.27 \begin{align*} \frac{a^{4} \log{\left (a + b x \right )}}{b^{5}} - \frac{a^{3} x}{b^{4}} + \frac{a^{2} x^{2}}{2 b^{3}} - \frac{a x^{3}}{3 b^{2}} + \frac{x^{4}}{4 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*log(a + b*x)/b**5 - a**3*x/b**4 + a**2*x**2/(2*b**3) - a*x**3/(3*b**2) + x**4/(4*b)

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Giac [A]  time = 1.35437, size = 112, normalized size = 0.62 \begin{align*} \frac{a^{4} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{5}} + \frac{3 \, b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) - 4 \, a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) - 12 \, a^{3} x \mathrm{sgn}\left (b x + a\right )}{12 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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