3.789 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^{3/2} (a+b x)}-\frac{2 a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{x} (a+b x)} \]

[Out]

(-2*a*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^(5/2)*(a + b*x)) - (2*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
3*x^(3/2)*(a + b*x)) - (2*b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x))

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Rubi [A]  time = 0.0430058, antiderivative size = 118, normalized size of antiderivative = 1., 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} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^{3/2} (a+b x)}-\frac{2 a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{x} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(7/2),x]

[Out]

(-2*a*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^(5/2)*(a + b*x)) - (2*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
3*x^(3/2)*(a + b*x)) - (2*b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x))

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]

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])

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, 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^{7/2}} \, 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^{7/2}}+\frac{b (A b+a B)}{x^{5/2}}+\frac{b^2 B}{x^{3/2}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac{2 (A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{x} (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0255423, size = 48, normalized size = 0.41 \[ -\frac{2 \sqrt{(a+b x)^2} (a (3 A+5 B x)+5 b x (A+3 B x))}{15 x^{5/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(5*b*x*(A + 3*B*x) + a*(3*A + 5*B*x)))/(15*x^(5/2)*(a + b*x))

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Maple [A]  time = 0.003, size = 44, normalized size = 0.4 \begin{align*} -{\frac{30\,Bb{x}^{2}+10\,Abx+10\,aBx+6\,aA}{15\,bx+15\,a}\sqrt{ \left ( bx+a \right ) ^{2}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x)

[Out]

-2/15*(15*B*b*x^2+5*A*b*x+5*B*a*x+3*A*a)*((b*x+a)^2)^(1/2)/x^(5/2)/(b*x+a)

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Maxima [A]  time = 1.05705, size = 46, normalized size = 0.39 \begin{align*} -\frac{2 \,{\left (3 \, b x^{2} + a x\right )} B}{3 \, x^{\frac{5}{2}}} - \frac{2 \,{\left (5 \, b x^{2} + 3 \, a x\right )} A}{15 \, x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

-2/3*(3*b*x^2 + a*x)*B/x^(5/2) - 2/15*(5*b*x^2 + 3*a*x)*A/x^(7/2)

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Fricas [A]  time = 1.48838, size = 73, normalized size = 0.62 \begin{align*} -\frac{2 \,{\left (15 \, B b x^{2} + 3 \, A a + 5 \,{\left (B a + A b\right )} x\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

-2/15*(15*B*b*x^2 + 3*A*a + 5*(B*a + A*b)*x)/x^(5/2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14543, size = 69, normalized size = 0.58 \begin{align*} -\frac{2 \,{\left (15 \, B b x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a x \mathrm{sgn}\left (b x + a\right ) + 5 \, A b x \mathrm{sgn}\left (b x + a\right ) + 3 \, A a \mathrm{sgn}\left (b x + a\right )\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*B*b*x^2*sgn(b*x + a) + 5*B*a*x*sgn(b*x + a) + 5*A*b*x*sgn(b*x + a) + 3*A*a*sgn(b*x + a))/x^(5/2)