3.7.1 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=199 \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{2 x^2 (a+b x)}-\frac {3 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{x (a+b x)}+\frac {b^2 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}+\frac {b^3 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 199, 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)}{2 x^2 (a+b x)}-\frac {3 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{x (a+b x)}+\frac {b^2 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}+\frac {b^3 B x \sqrt {a^2+2 a b x+b^2 x^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^4,x]

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
2*x^2*(a + b*x)) - (3*a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x)) + (b^3*B*x*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(a + b*x) + (b^2*(A*b + 3*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^4} \, 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^4} \, 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^6 B+\frac {a^3 A b^3}{x^4}+\frac {a^2 b^3 (3 A b+a B)}{x^3}+\frac {3 a b^4 (A b+a B)}{x^2}+\frac {b^5 (A b+3 a 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}}{3 x^3 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^2 (A b+3 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 = 0.03, size = 88, normalized size = 0.44 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a^3 (2 A+3 B x)+9 a^2 b x (A+2 B x)-6 b^2 x^3 \log (x) (3 a B+A b)+18 a A b^2 x^2-6 b^3 B x^4\right )}{6 x^3 (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^4,x]

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(18*a*A*b^2*x^2 - 6*b^3*B*x^4 + 9*a^2*b*x*(A + 2*B*x) + a^3*(2*A + 3*B*x) - 6*b^2*(A*b
+ 3*a*B)*x^3*Log[x]))/(x^3*(a + b*x))

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IntegrateAlgebraic [B]  time = 2.04, size = 511, normalized size = 2.57 \begin {gather*} -\frac {1}{2} A \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {1}{2} A \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+A b^3 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )-\frac {3}{2} a \sqrt {b^2} b B \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {3}{2} a \sqrt {b^2} b B \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+3 a b^2 B \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^3 A b-3 a^3 b B x-9 a^2 A b^2 x-18 a^2 b^2 B x^2-18 a A b^3 x^2+3 a b^3 B x^3+6 b^4 B x^4\right )+\sqrt {b^2} \left (2 a^4 A+3 a^4 B x+11 a^3 A b x+21 a^3 b B x^2+27 a^2 A b^2 x^2+15 a^2 b^2 B x^3+18 a A b^3 x^3-9 a b^3 B x^4-6 b^4 B x^5\right )}{6 x^3 \left (a b+b^2 x\right )-6 \sqrt {b^2} x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^3*A*b - 9*a^2*A*b^2*x - 3*a^3*b*B*x - 18*a*A*b^3*x^2 - 18*a^2*b^2*B*x^2 +
3*a*b^3*B*x^3 + 6*b^4*B*x^4) + Sqrt[b^2]*(2*a^4*A + 11*a^3*A*b*x + 3*a^4*B*x + 27*a^2*A*b^2*x^2 + 21*a^3*b*B*
x^2 + 18*a*A*b^3*x^3 + 15*a^2*b^2*B*x^3 - 9*a*b^3*B*x^4 - 6*b^4*B*x^5))/(6*x^3*(a*b + b^2*x) - 6*Sqrt[b^2]*x^3
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + A*b^3*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] + 3*a*b^2*B
*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - (A*(b^2)^(3/2)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/2 - (3*a*b*Sqrt[b^2]*B*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (A*(b^
2)^(3/2)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (3*a*b*Sqrt[b^2]*B*Log[a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/2

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fricas [A]  time = 0.41, size = 75, normalized size = 0.38 \begin {gather*} \frac {6 \, B b^{3} x^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} \log \relax (x) - 2 \, A a^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

1/6*(6*B*b^3*x^4 + 6*(3*B*a*b^2 + A*b^3)*x^3*log(x) - 2*A*a^3 - 18*(B*a^2*b + A*a*b^2)*x^2 - 3*(B*a^3 + 3*A*a^
2*b)*x)/x^3

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giac [A]  time = 0.17, size = 118, normalized size = 0.59 \begin {gather*} B b^{3} x \mathrm {sgn}\left (b x + a\right ) + {\left (3 \, B a b^{2} \mathrm {sgn}\left (b x + a\right ) + A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) + A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\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}{6 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 96, normalized size = 0.48 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 A \,b^{3} x^{3} \ln \relax (x )+18 B a \,b^{2} x^{3} \ln \relax (x )+6 B \,b^{3} x^{4}-18 A a \,b^{2} x^{2}-18 B \,a^{2} b \,x^{2}-9 A \,a^{2} b x -3 B \,a^{3} x -2 A \,a^{3}\right )}{6 \left (b x +a \right )^{3} x^{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^4,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(6*A*ln(x)*x^3*b^3+18*B*ln(x)*x^3*a*b^2+6*B*b^3*x^4-18*A*a*b^2*x^2-18*B*a^2*b*x^2-9*A*a^
2*b*x-3*B*a^3*x-2*A*a^3)/(b*x+a)^3/x^3

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maxima [B]  time = 0.66, size = 443, normalized size = 2.23 \begin {gather*} 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) + \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3} x}{2 \, a} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4} x}{2 \, a^{2}} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{2 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{6 \, a^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{2 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{2 \, a^{2} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{3 \, a^{2} x^{3}} \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^4,x, algorithm="maxima")

[Out]

3*(-1)^(2*b^2*x + 2*a*b)*B*a*b^2*log(2*b^2*x + 2*a*b) + (-1)^(2*b^2*x + 2*a*b)*A*b^3*log(2*b^2*x + 2*a*b) - 3*
(-1)^(2*a*b*x + 2*a^2)*B*a*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - (-1)^(2*a*b*x + 2*a^2)*A*b^3*log(2*a*b*x/a
bs(x) + 2*a^2/abs(x)) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^3*x/a + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^4*
x/a^2 + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^3/a + 1/2*(b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*B*b^2/a^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/a^3 - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^
(3/2)*B*b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2/(a^2*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a
^2*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^3)

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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^4} \,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^4,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^4, 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^{4}}\, 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**4,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**4, x)

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