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

Optimal. Leaf size=116 \[ \frac {2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

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Rubi [A]  time = 0.04, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 76} \begin {gather*} \frac {2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x)) + (2*(A*b + a*B)*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/(a + b*x) + (2*b*B*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(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) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{3/2}} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right ) (A+B x)}{x^{3/2}} \, dx}{a b+b^2 x}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a A b}{x^{3/2}}+\frac {b (A b+a B)}{\sqrt {x}}+b^2 B \sqrt {x}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 (A b+a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 47, normalized size = 0.41 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} (b x (3 A+B x)-3 a (A-B x))}{3 \sqrt {x} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.04, size = 48, normalized size = 0.41 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (-3 a A+3 a B x+3 A b x+b B x^2\right )}{3 \sqrt {x} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 26, normalized size = 0.22 \begin {gather*} \frac {2 \, {\left (B b x^{2} - 3 \, A a + 3 \, {\left (B a + A b\right )} x\right )}}{3 \, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 53, normalized size = 0.46 \begin {gather*} \frac {2}{3} \, B b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 2 \, A b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, A a \mathrm {sgn}\left (b x + a\right )}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(3/2),x, algorithm="giac")

[Out]

2/3*B*b*x^(3/2)*sgn(b*x + a) + 2*B*a*sqrt(x)*sgn(b*x + a) + 2*A*b*sqrt(x)*sgn(b*x + a) - 2*A*a*sgn(b*x + a)/sq
rt(x)

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maple [A]  time = 0.05, size = 44, normalized size = 0.38 \begin {gather*} -\frac {2 \left (-B b \,x^{2}-3 A b x -3 B a x +3 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (b x +a \right ) \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^(3/2),x)

[Out]

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

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maxima [A]  time = 0.62, size = 33, normalized size = 0.28 \begin {gather*} \frac {2 \, {\left (b x^{2} + 3 \, a x\right )} B}{3 \, \sqrt {x}} + \frac {2 \, {\left (b x^{2} - a x\right )} A}{x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.32, size = 53, normalized size = 0.46 \begin {gather*} \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^2}{3}-\frac {2\,A\,a}{b}+\frac {x\,\left (6\,A\,b+6\,B\,a\right )}{3\,b}\right )}{x^{3/2}+\frac {a\,\sqrt {x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^(3/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(3/2),x)

[Out]

Timed out

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