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

Optimal. Leaf size=200 \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{x (a+b x)}+\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}+\frac {3 a b \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 200, 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, 76} \begin {gather*} -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{x (a+b x)}+\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}+\frac {3 a b \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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^3} \, 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^3} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 (A b+3 a B)+\frac {a^3 A b^3}{x^3}+\frac {a^2 b^3 (3 A b+a B)}{x^2}+\frac {3 a b^4 (A b+a B)}{x}+b^6 B x\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^2 (A b+3 a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {3 a b (A b+a B) \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 = 85, normalized size = 0.42 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-\left (a^3 (A+2 B x)\right )-6 a^2 A b x+6 a b x^2 \log (x) (a B+A b)+6 a b^2 B x^3+b^3 x^3 (2 A+B x)\right )}{2 x^2 (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^3,x]

[Out]

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

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IntegrateAlgebraic [B]  time = 1.58, size = 1061, normalized size = 5.30 \begin {gather*} 3 a A \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 b x a+b^2 x^2}}{a}\right ) b^2-\frac {3}{2} a A \sqrt {b^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b-\frac {3}{2} a A \sqrt {b^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b+\frac {-A \sqrt {a^2+2 b x a+b^2 x^2} \left (-2 x^3 b^4-a x^2 b^3+6 a^2 x b^2+a^3 b\right )-A \sqrt {b^2} \left (-a^4-7 b x a^3-5 b^2 x^2 a^2+3 b^3 x^3 a+2 b^4 x^4\right )}{2 x^2 \left (x b^2+a b\right )-2 \sqrt {b^2} x^2 \sqrt {a^2+2 b x a+b^2 x^2}}+\frac {2 \sqrt {b^2} B a^4-\frac {3}{4} b \sqrt {b^2} B x a^3-6 b^2 B x \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) a^3-3 b \sqrt {b^2} B x \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^3-3 b \sqrt {b^2} B x \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^3-2 b B \sqrt {a^2+2 b x a+b^2 x^2} a^3-\frac {35}{4} \left (b^2\right )^{3/2} B x^2 a^2-6 b^3 B x^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) a^2+6 b \sqrt {b^2} B x \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) a^2-3 \left (b^2\right )^{3/2} B x^2 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2+3 b^2 B x \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2-3 \left (b^2\right )^{3/2} B x^2 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2+3 b^2 B x \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2+\frac {11}{4} b^2 B x \sqrt {a^2+2 b x a+b^2 x^2} a^2-7 b^3 \sqrt {b^2} B x^3 a+6 b^3 B x^2 \sqrt {a^2+2 b x a+b^2 x^2} a-b^4 \sqrt {b^2} B x^4+b^4 B x^3 \sqrt {a^2+2 b x a+b^2 x^2}}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a^3*b + 6*a^2*b^2*x - a*b^3*x^2 - 2*b^4*x^3)) - A*Sqrt[b^2]*(-a^4 - 7*a^3*
b*x - 5*a^2*b^2*x^2 + 3*a*b^3*x^3 + 2*b^4*x^4))/(2*x^2*(a*b + b^2*x) - 2*Sqrt[b^2]*x^2*Sqrt[a^2 + 2*a*b*x + b^
2*x^2]) + 3*a*A*b^2*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - (3*a*A*b*Sqrt[b^2]*Log[-a - S
qrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (3*a*A*b*Sqrt[b^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x +
b^2*x^2]])/2 + (2*a^4*Sqrt[b^2]*B - (3*a^3*b*Sqrt[b^2]*B*x)/4 - (35*a^2*(b^2)^(3/2)*B*x^2)/4 - 7*a*b^3*Sqrt[b
^2]*B*x^3 - b^4*Sqrt[b^2]*B*x^4 - 2*a^3*b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + (11*a^2*b^2*B*x*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/4 + 6*a*b^3*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + b^4*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 6*a^3
*b^2*B*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - 6*a^2*b^3*B*x^2*ArcTanh[(-(Sqrt[b^2]*x)
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 6*a^2*b*Sqrt[b^2]*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2
]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - 3*a^3*b*Sqrt[b^2]*B*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b
^2*x^2]] - 3*a^2*(b^2)^(3/2)*B*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + 3*a^2*b^2*B*x*Sqrt[
a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - 3*a^3*b*Sqrt[b^2]*B*x*Log[a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - 3*a^2*(b^2)^(3/2)*B*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2]] + 3*a^2*b^2*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^
2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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fricas [A]  time = 0.41, size = 74, normalized size = 0.37 \begin {gather*} \frac {B b^{3} x^{4} - A a^{3} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} \log \relax (x) - 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 117, normalized size = 0.58 \begin {gather*} \frac {1}{2} \, B b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a b^{2} x \mathrm {sgn}\left (b x + a\right ) + A b^{3} x \mathrm {sgn}\left (b x + a\right ) + 3 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) + A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 95, normalized size = 0.48 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (B \,b^{3} x^{4}+6 A a \,b^{2} x^{2} \ln \relax (x )+2 A \,b^{3} x^{3}+6 B \,a^{2} b \,x^{2} \ln \relax (x )+6 B a \,b^{2} x^{3}-6 A \,a^{2} b x -2 B \,a^{3} x -A \,a^{3}\right )}{2 \left (b x +a \right )^{3} x^{2}} \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^3,x)

[Out]

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

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maxima [B]  time = 0.63, size = 351, normalized size = 1.76 \begin {gather*} 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{2} b \log \left (2 \, b^{2} x + 2 \, a b\right ) + 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{2} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2} x + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3} x}{2 \, a} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a b + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{2 \, a^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{2 \, a^{2} x^{2}} \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^3,x, algorithm="maxima")

[Out]

3*(-1)^(2*b^2*x + 2*a*b)*B*a^2*b*log(2*b^2*x + 2*a*b) + 3*(-1)^(2*b^2*x + 2*a*b)*A*a*b^2*log(2*b^2*x + 2*a*b)
- 3*(-1)^(2*a*b*x + 2*a^2)*B*a^2*b*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - 3*(-1)^(2*a*b*x + 2*a^2)*A*a*b^2*log(2
*a*b*x/abs(x) + 2*a^2/abs(x)) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^2*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*
A*b^3*x/a + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*b + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^2 + 1/2*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*A*b^2/a^2 - (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B/x - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*
b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^2)

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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 )}^{3/2}}{x^3} \,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^3,x)

[Out]

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

[Out]

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

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