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

Optimal. Leaf size=258 \[ \frac {x^4 (a+b x) (A b-a B)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x^2 (a+b x) (A b-a B)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^3 (a+b x) (A b-a B)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^5 (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^4 (a+b x) (A b-a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 x (a+b x) (A b-a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 77} \begin {gather*} \frac {x^4 (a+b x) (A b-a B)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^3 (a+b x) (A b-a B)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x^2 (a+b x) (A b-a B)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 x (a+b x) (A b-a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^4 (a+b x) (A b-a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^5 (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 116, normalized size = 0.45 \begin {gather*} \frac {(a+b x) \left (b x \left (60 a^4 B-30 a^3 b (2 A+B x)+10 a^2 b^2 x (3 A+2 B x)-5 a b^3 x^2 (4 A+3 B x)+3 b^4 x^3 (5 A+4 B x)\right )-60 a^4 (a B-A b) \log (a+b x)\right )}{60 b^6 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*(b*x*(60*a^4*B - 30*a^3*b*(2*A + B*x) + 10*a^2*b^2*x*(3*A + 2*B*x) - 5*a*b^3*x^2*(4*A + 3*B*x) + 3*
b^4*x^3*(5*A + 4*B*x)) - 60*a^4*(-(A*b) + a*B)*Log[a + b*x]))/(60*b^6*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 0.94, size = 378, normalized size = 1.47 \begin {gather*} \frac {\left (a^5 \sqrt {b^2} B+a^5 b B-a^4 A b^2-a^4 A b \sqrt {b^2}\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b^7}+\frac {\left (a^5 \sqrt {b^2} B+a^5 (-b) B+a^4 A b^2-a^4 A b \sqrt {b^2}\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b^7}+\frac {-60 a^4 B x+60 a^3 A b x+30 a^3 b B x^2-30 a^2 A b^2 x^2-20 a^2 b^2 B x^3+20 a A b^3 x^3+15 a b^3 B x^4-15 A b^4 x^4-12 b^4 B x^5}{120 b^4 \sqrt {b^2}}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (137 a^4 B-125 a^3 A b-77 a^3 b B x+65 a^2 A b^2 x+47 a^2 b^2 B x^2-35 a A b^3 x^2-27 a b^3 B x^3+15 A b^4 x^3+12 b^4 B x^4\right )}{120 b^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-125*a^3*A*b + 137*a^4*B + 65*a^2*A*b^2*x - 77*a^3*b*B*x - 35*a*A*b^3*x^2 + 47
*a^2*b^2*B*x^2 + 15*A*b^4*x^3 - 27*a*b^3*B*x^3 + 12*b^4*B*x^4))/(120*b^6) + (60*a^3*A*b*x - 60*a^4*B*x - 30*a^
2*A*b^2*x^2 + 30*a^3*b*B*x^2 + 20*a*A*b^3*x^3 - 20*a^2*b^2*B*x^3 - 15*A*b^4*x^4 + 15*a*b^3*B*x^4 - 12*b^4*B*x^
5)/(120*b^4*Sqrt[b^2]) + ((-(a^4*A*b^2) - a^4*A*b*Sqrt[b^2] + a^5*b*B + a^5*Sqrt[b^2]*B)*Log[-a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^7) + ((a^4*A*b^2 - a^4*A*b*Sqrt[b^2] - a^5*b*B + a^5*Sqrt[b^2]*B)*Log[a
- Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^7)

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fricas [A]  time = 0.41, size = 117, normalized size = 0.45 \begin {gather*} \frac {12 \, B b^{5} x^{5} - 15 \, {\left (B a b^{4} - A b^{5}\right )} x^{4} + 20 \, {\left (B a^{2} b^{3} - A a b^{4}\right )} x^{3} - 30 \, {\left (B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{2} + 60 \, {\left (B a^{4} b - A a^{3} b^{2}\right )} x - 60 \, {\left (B a^{5} - A a^{4} b\right )} \log \left (b x + a\right )}{60 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*b^5*x^5 - 15*(B*a*b^4 - A*b^5)*x^4 + 20*(B*a^2*b^3 - A*a*b^4)*x^3 - 30*(B*a^3*b^2 - A*a^2*b^3)*x^2
+ 60*(B*a^4*b - A*a^3*b^2)*x - 60*(B*a^5 - A*a^4*b)*log(b*x + a))/b^6

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giac [A]  time = 0.16, size = 185, normalized size = 0.72 \begin {gather*} \frac {12 \, B b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) - 15 \, B a b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, A b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 20 \, B a^{2} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 20 \, A a b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) - 30 \, B a^{3} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 30 \, A a^{2} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 60 \, B a^{4} x \mathrm {sgn}\left (b x + a\right ) - 60 \, A a^{3} b x \mathrm {sgn}\left (b x + a\right )}{60 \, b^{5}} - \frac {{\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) - A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*b^4*x^5*sgn(b*x + a) - 15*B*a*b^3*x^4*sgn(b*x + a) + 15*A*b^4*x^4*sgn(b*x + a) + 20*B*a^2*b^2*x^3*s
gn(b*x + a) - 20*A*a*b^3*x^3*sgn(b*x + a) - 30*B*a^3*b*x^2*sgn(b*x + a) + 30*A*a^2*b^2*x^2*sgn(b*x + a) + 60*B
*a^4*x*sgn(b*x + a) - 60*A*a^3*b*x*sgn(b*x + a))/b^5 - (B*a^5*sgn(b*x + a) - A*a^4*b*sgn(b*x + a))*log(abs(b*x
+ a))/b^6

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maple [A]  time = 0.06, size = 138, normalized size = 0.53 \begin {gather*} \frac {\left (b x +a \right ) \left (12 B \,b^{5} x^{5}+15 A \,b^{5} x^{4}-15 B a \,b^{4} x^{4}-20 A a \,b^{4} x^{3}+20 B \,a^{2} b^{3} x^{3}+30 A \,a^{2} b^{3} x^{2}-30 B \,a^{3} b^{2} x^{2}+60 A \,a^{4} b \ln \left (b x +a \right )-60 A \,a^{3} b^{2} x -60 B \,a^{5} \ln \left (b x +a \right )+60 B \,a^{4} b x \right )}{60 \sqrt {\left (b x +a \right )^{2}}\, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(b*x+a)*(12*B*b^5*x^5+15*A*x^4*b^5-15*B*x^4*a*b^4-20*A*x^3*a*b^4+20*B*x^3*a^2*b^3+30*A*x^2*a^2*b^3-30*B*x
^2*a^3*b^2+60*A*ln(b*x+a)*a^4*b-60*A*a^3*b^2*x-60*B*ln(b*x+a)*a^5+60*B*a^4*b*x)/((b*x+a)^2)^(1/2)/b^6

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maxima [A]  time = 0.57, size = 272, normalized size = 1.05 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B x^{4}}{5 \, b^{2}} - \frac {9 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a x^{3}}{20 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A x^{3}}{4 \, b^{2}} - \frac {77 \, B a^{3} x^{2}}{60 \, b^{4}} + \frac {13 \, A a^{2} x^{2}}{12 \, b^{3}} + \frac {47 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} x^{2}}{60 \, b^{4}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a x^{2}}{12 \, b^{3}} + \frac {77 \, B a^{4} x}{30 \, b^{5}} - \frac {13 \, A a^{3} x}{6 \, b^{4}} - \frac {B a^{5} \log \left (x + \frac {a}{b}\right )}{b^{6}} + \frac {A a^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {47 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4}}{30 \, b^{6}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3}}{6 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*x^4/b^2 - 9/20*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*x^3/b^3 + 1/4*sqrt(b^2*x^
2 + 2*a*b*x + a^2)*A*x^3/b^2 - 77/60*B*a^3*x^2/b^4 + 13/12*A*a^2*x^2/b^3 + 47/60*sqrt(b^2*x^2 + 2*a*b*x + a^2)
*B*a^2*x^2/b^4 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*x^2/b^3 + 77/30*B*a^4*x/b^5 - 13/6*A*a^3*x/b^4 - B*a^5
*log(x + a/b)/b^6 + A*a^4*log(x + a/b)/b^5 - 47/30*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^4/b^6 + 7/6*sqrt(b^2*x^2
+ 2*a*b*x + a^2)*A*a^3/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 109, normalized size = 0.42 \begin {gather*} \frac {B x^{5}}{5 b} - \frac {a^{4} \left (- A b + B a\right ) \log {\left (a + b x \right )}}{b^{6}} + x^{4} \left (\frac {A}{4 b} - \frac {B a}{4 b^{2}}\right ) + x^{3} \left (- \frac {A a}{3 b^{2}} + \frac {B a^{2}}{3 b^{3}}\right ) + x^{2} \left (\frac {A a^{2}}{2 b^{3}} - \frac {B a^{3}}{2 b^{4}}\right ) + x \left (- \frac {A a^{3}}{b^{4}} + \frac {B a^{4}}{b^{5}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x**5/(5*b) - a**4*(-A*b + B*a)*log(a + b*x)/b**6 + x**4*(A/(4*b) - B*a/(4*b**2)) + x**3*(-A*a/(3*b**2) + B*a
**2/(3*b**3)) + x**2*(A*a**2/(2*b**3) - B*a**3/(2*b**4)) + x*(-A*a**3/b**4 + B*a**4/b**5)

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