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

Optimal. Leaf size=158 \[ \frac{x^{3/2} (A b-a B)}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\sqrt{x} (3 a B+A b)}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{3/2} b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-((A*b + 3*a*B)*Sqrt[x])/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b + 3*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2)*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0860282, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {770, 78, 47, 63, 205} \[ \frac{x^{3/2} (A b-a B)}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\sqrt{x} (3 a B+A b)}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{3/2} b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b + 3*a*B)*Sqrt[x])/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b + 3*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2)*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{x} (A+B x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((A b+3 a B) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{x}}{\left (a b+b^2 x\right )^2} \, dx}{4 a \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b+3 a B) \sqrt{x}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((A b+3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{8 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b+3 a B) \sqrt{x}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((A b+3 a B) \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b+b^2 x^2} \, dx,x,\sqrt{x}\right )}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b+3 a B) \sqrt{x}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b+3 a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{3/2} b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0561627, size = 105, normalized size = 0.66 \[ \frac{\sqrt{a} \sqrt{b} \sqrt{x} \left (-3 a^2 B-a b (A+5 B x)+A b^2 x\right )+(a+b x)^2 (3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{3/2} b^{5/2} (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-3*a^2*B + A*b^2*x - a*b*(A + 5*B*x)) + (A*b + 3*a*B)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sq
rt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2)*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.018, size = 194, normalized size = 1.2 \begin{align*}{\frac{bx+a}{4\,{b}^{2}a} \left ( A\sqrt{ab}{x}^{{\frac{3}{2}}}{b}^{2}+A\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){x}^{2}{b}^{3}-5\,B\sqrt{ab}{x}^{3/2}ab+3\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{2}a{b}^{2}+2\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) xa{b}^{2}+6\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) x{a}^{2}b-A\sqrt{ab}\sqrt{x}ab+A\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){a}^{2}b-3\,B\sqrt{ab}\sqrt{x}{a}^{2}+3\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){a}^{3} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(A*(a*b)^(1/2)*x^(3/2)*b^2+A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^2*b^3-5*B*(a*b)^(1/2)*x^(3/2)*a*b+3*B*arctan(
x^(1/2)*b/(a*b)^(1/2))*x^2*a*b^2+2*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x*a*b^2+6*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x
*a^2*b-A*(a*b)^(1/2)*x^(1/2)*a*b+A*arctan(x^(1/2)*b/(a*b)^(1/2))*a^2*b-3*B*(a*b)^(1/2)*x^(1/2)*a^2+3*B*arctan(
x^(1/2)*b/(a*b)^(1/2))*a^3)*(b*x+a)/(a*b)^(1/2)/a/b^2/((b*x+a)^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64653, size = 633, normalized size = 4.01 \begin{align*} \left [-\frac{{\left (3 \, B a^{3} + A a^{2} b +{\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \,{\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) + 2 \,{\left (3 \, B a^{3} b + A a^{2} b^{2} +{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt{x}}{8 \,{\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}, -\frac{{\left (3 \, B a^{3} + A a^{2} b +{\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \,{\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (3 \, B a^{3} b + A a^{2} b^{2} +{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt{x}}{4 \,{\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*s
qrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B*a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 +
2*a^3*b^4*x + a^4*b^3), -1/4*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(a
*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B*a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2
+ 2*a^3*b^4*x + a^4*b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18907, size = 132, normalized size = 0.84 \begin{align*} \frac{{\left (3 \, B a + A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a b^{2} \mathrm{sgn}\left (b x + a\right )} - \frac{5 \, B a b x^{\frac{3}{2}} - A b^{2} x^{\frac{3}{2}} + 3 \, B a^{2} \sqrt{x} + A a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a b^{2} \mathrm{sgn}\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2*sgn(b*x + a)) - 1/4*(5*B*a*b*x^(3/2) - A*b^2*x^
(3/2) + 3*B*a^2*sqrt(x) + A*a*b*sqrt(x))/((b*x + a)^2*a*b^2*sgn(b*x + a))