3.6.98 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x} \, dx\)

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

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Rubi [A]  time = 0.06, antiderivative size = 182, 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 {3 a^2 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {A b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^3 A \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*A*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (3*a*A*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a +
b*x)) + (A*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (B*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(4*b) + (a^3*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 )^{3/2}}{x} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3 (A+B x)}{x} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 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 )^3}{x} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {B (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b}+\frac {\left (A \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (3 a^2 b^4+\frac {a^3 b^3}{x}+3 a b^5 x+b^6 x^2\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {3 a^2 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {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)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b}+\frac {a^3 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 = 0.04, size = 83, normalized size = 0.46 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (12 a^3 A \log (x)+x \left (12 a^3 B+18 a^2 b (2 A+B x)+6 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )\right )}{12 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(x*(12*a^3*B + 18*a^2*b*(2*A + B*x) + 6*a*b^2*x*(3*A + 2*B*x) + b^3*x^2*(4*A + 3*B*x)) + 12
*a^3*A*Log[x]))/(12*(a + b*x))

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IntegrateAlgebraic [A]  time = 0.72, size = 361, normalized size = 1.98 \begin {gather*} \frac {1}{2} a^3 A \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {a^3 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^3 (-A) b-a^3 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 (3 a^3 B+22 a^2 A b+9 a^2 b B x+14 a A b^2 x+9 a b^2 B x^2+4 A b^3 x^2+3 b^3 B x^3\right )}{24 b}+\frac {-12 a^3 \sqrt {b^2} B x-36 a^2 A b \sqrt {b^2} x-18 a^2 b \sqrt {b^2} B x^2-18 a A \left (b^2\right )^{3/2} x^2-12 a \left (b^2\right )^{3/2} B x^3-4 A b^3 \sqrt {b^2} x^3-3 b^3 \sqrt {b^2} B x^4}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x,x]

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(22*a^2*A*b + 3*a^3*B + 14*a*A*b^2*x + 9*a^2*b*B*x + 4*A*b^3*x^2 + 9*a*b^2*B*x^
2 + 3*b^3*B*x^3))/(24*b) + (-36*a^2*A*b*Sqrt[b^2]*x - 12*a^3*Sqrt[b^2]*B*x - 18*a*A*(b^2)^(3/2)*x^2 - 18*a^2*b
*Sqrt[b^2]*B*x^2 - 4*A*b^3*Sqrt[b^2]*x^3 - 12*a*(b^2)^(3/2)*B*x^3 - 3*b^3*Sqrt[b^2]*B*x^4)/(24*b) + (a^3*A*Log
[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 + ((-(a^3*A*b) - a^3*A*Sqrt[b^2])*Log[a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b) - (a^3*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 = 0.42, size = 68, normalized size = 0.37 \begin {gather*} \frac {1}{4} \, B b^{3} x^{4} + A a^{3} \log \relax (x) + \frac {1}{3} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + \frac {3}{2} \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + {\left (B a^{3} + 3 \, A a^{2} 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)^(3/2)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 118, normalized size = 0.65 \begin {gather*} \frac {1}{4} \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + A a^{3} \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)^(3/2)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 91, normalized size = 0.50 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (3 B \,b^{3} x^{4}+4 A \,b^{3} x^{3}+12 B a \,b^{2} x^{3}+18 A a \,b^{2} x^{2}+18 B \,a^{2} b \,x^{2}+12 A \,a^{3} \ln \relax (x )+36 A \,a^{2} b x +12 B \,a^{3} x \right )}{12 \left (b x +a \right )^{3}} \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)^(3/2)/x,x)

[Out]

1/12*((b*x+a)^2)^(3/2)*(3*b^3*B*x^4+4*A*b^3*x^3+12*x^3*B*a*b^2+18*x^2*A*a*b^2+18*B*a^2*b*x^2+12*A*a^3*ln(x)+36
*x*A*a^2*b+12*B*a^3*x)/(b*x+a)^3

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maxima [A]  time = 0.48, size = 186, normalized size = 1.02 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{3} \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 b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2} + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x + \frac {1}{3} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a}{4 \, 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)^(3/2)/x,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*A*a^3*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*A*a^3*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*b*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2 + 1/4*(b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*B*x + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{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)^(3/2))/x,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/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 {3}{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)**(3/2)/x,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x, x)

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