3.7.17 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=262 \[ \frac {A b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {5 a A b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {10 a^2 A b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b}+\frac {a^5 A \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^4 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

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Rubi [A]  time = 0.08, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {770, 80, 43} \begin {gather*} \frac {5 a^4 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 A b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {5 a A b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {A b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^5 A \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5 (A+B x)}{x} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b}+\frac {\left (A \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {\left (a b+b^2 x\right )^5}{x} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b}+\frac {\left (A \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (5 a^4 b^6+\frac {a^5 b^5}{x}+10 a^3 b^7 x+10 a^2 b^8 x^2+5 a b^9 x^3+b^{10} x^4\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {5 a^4 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 A b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {5 a A b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {A b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b}+\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 122, normalized size = 0.47 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (60 a^5 A \log (x)+x \left (60 a^5 B+150 a^4 b (2 A+B x)+100 a^3 b^2 x (3 A+2 B x)+50 a^2 b^3 x^2 (4 A+3 B x)+15 a b^4 x^3 (5 A+4 B x)+2 b^5 x^4 (6 A+5 B x)\right )\right )}{60 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(x*(60*a^5*B + 150*a^4*b*(2*A + B*x) + 100*a^3*b^2*x*(3*A + 2*B*x) + 50*a^2*b^3*x^2*(4*A +
3*B*x) + 15*a*b^4*x^3*(5*A + 4*B*x) + 2*b^5*x^4*(6*A + 5*B*x)) + 60*a^5*A*Log[x]))/(60*(a + b*x))

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IntegrateAlgebraic [A]  time = 0.81, size = 485, normalized size = 1.85 \begin {gather*} \frac {1}{2} a^5 A \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {a^5 A \sqrt {b^2} \log \left (b \sqrt {a^2+2 a b x+b^2 x^2}-a b-\sqrt {b^2} b x\right )}{2 b}+\frac {\left (a^5 (-A) b-a^5 A \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}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (10 a^5 B+137 a^4 A b+50 a^4 b B x+163 a^3 A b^2 x+100 a^3 b^2 B x^2+137 a^2 A b^3 x^2+100 a^2 b^3 B x^3+63 a A b^4 x^3+50 a b^4 B x^4+12 A b^5 x^4+10 b^5 B x^5\right )}{120 b}+\frac {-60 a^5 \sqrt {b^2} B x-300 a^4 A b \sqrt {b^2} x-150 a^4 b \sqrt {b^2} B x^2-300 a^3 A \left (b^2\right )^{3/2} x^2-200 a^3 \left (b^2\right )^{3/2} B x^3-200 a^2 A b^3 \sqrt {b^2} x^3-150 a^2 b^3 \sqrt {b^2} B x^4-75 a A b^4 \sqrt {b^2} x^4-60 a b^4 \sqrt {b^2} B x^5-12 A b^5 \sqrt {b^2} x^5-10 b^5 \sqrt {b^2} B x^6}{120 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x,x]

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(137*a^4*A*b + 10*a^5*B + 163*a^3*A*b^2*x + 50*a^4*b*B*x + 137*a^2*A*b^3*x^2 +
100*a^3*b^2*B*x^2 + 63*a*A*b^4*x^3 + 100*a^2*b^3*B*x^3 + 12*A*b^5*x^4 + 50*a*b^4*B*x^4 + 10*b^5*B*x^5))/(120*b
) + (-300*a^4*A*b*Sqrt[b^2]*x - 60*a^5*Sqrt[b^2]*B*x - 300*a^3*A*(b^2)^(3/2)*x^2 - 150*a^4*b*Sqrt[b^2]*B*x^2 -
200*a^2*A*b^3*Sqrt[b^2]*x^3 - 200*a^3*(b^2)^(3/2)*B*x^3 - 75*a*A*b^4*Sqrt[b^2]*x^4 - 150*a^2*b^3*Sqrt[b^2]*B*
x^4 - 12*A*b^5*Sqrt[b^2]*x^5 - 60*a*b^4*Sqrt[b^2]*B*x^5 - 10*b^5*Sqrt[b^2]*B*x^6)/(120*b) + (a^5*A*Log[-a - Sq
rt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 + ((-(a^5*A*b) - a^5*A*Sqrt[b^2])*Log[a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(2*b) - (a^5*A*Sqrt[b^2]*Log[-(a*b) - b*Sqrt[b^2]*x + b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]]
)/(2*b)

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fricas [A]  time = 0.43, size = 114, normalized size = 0.44 \begin {gather*} \frac {1}{6} \, B b^{5} x^{6} + A a^{5} \log \relax (x) + \frac {1}{5} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + \frac {5}{4} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + \frac {10}{3} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + \frac {5}{2} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} + 5 \, A a^{4} b\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*b^5*x^6 + A*a^5*log(x) + 1/5*(5*B*a*b^4 + A*b^5)*x^5 + 5/4*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 10/3*(B*a^3*b^2
+ A*a^2*b^3)*x^3 + 5/2*(B*a^4*b + 2*A*a^3*b^2)*x^2 + (B*a^5 + 5*A*a^4*b)*x

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giac [A]  time = 0.21, size = 190, normalized size = 0.73 \begin {gather*} \frac {1}{6} \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + A a^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*b^5*x^6*sgn(b*x + a) + B*a*b^4*x^5*sgn(b*x + a) + 1/5*A*b^5*x^5*sgn(b*x + a) + 5/2*B*a^2*b^3*x^4*sgn(b*x
+ a) + 5/4*A*a*b^4*x^4*sgn(b*x + a) + 10/3*B*a^3*b^2*x^3*sgn(b*x + a) + 10/3*A*a^2*b^3*x^3*sgn(b*x + a) + 5/2
*B*a^4*b*x^2*sgn(b*x + a) + 5*A*a^3*b^2*x^2*sgn(b*x + a) + B*a^5*x*sgn(b*x + a) + 5*A*a^4*b*x*sgn(b*x + a) + A
*a^5*log(abs(x))*sgn(b*x + a)

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maple [A]  time = 0.06, size = 139, normalized size = 0.53 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (10 B \,b^{5} x^{6}+12 A \,b^{5} x^{5}+60 B a \,b^{4} x^{5}+75 A a \,b^{4} x^{4}+150 B \,a^{2} b^{3} x^{4}+200 A \,a^{2} b^{3} x^{3}+200 B \,a^{3} b^{2} x^{3}+300 A \,a^{3} b^{2} x^{2}+150 B \,a^{4} b \,x^{2}+60 A \,a^{5} \ln \relax (x )+300 A \,a^{4} b x +60 B \,a^{5} x \right )}{60 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(10*B*b^5*x^6+12*x^5*A*b^5+60*x^5*B*a*b^4+75*x^4*A*a*b^4+150*x^4*B*a^2*b^3+200*A*a^2*b^
3*x^3+200*B*a^3*b^2*x^3+300*x^2*A*a^3*b^2+150*x^2*B*a^4*b+60*A*a^5*ln(x)+300*x*A*a^4*b+60*x*B*a^5)/(b*x+a)^5

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maxima [A]  time = 0.53, size = 236, normalized size = 0.90 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{4} + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a b x + \frac {7}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2} + \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x + \frac {1}{5} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*A*a^5*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*A*a^5*log(2*a*b*x/abs(x) + 2*a^2/ab
s(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^3*b*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^4 + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*A*a*b*x + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/
2)*B*x + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x, x)

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