3.810 \(\int \frac{x^{5/2} (A+B x)}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=238 \[ \frac{2 x^{5/2} (a+b x) (A b-a B)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a x^{3/2} (a+b x) (A b-a B)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 a^2 \sqrt{x} (a+b x) (A b-a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a^{5/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(2*a^2*(A*b - a*B)*Sqrt[x]*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*a*(A*b - a*B)*x^(3/2)*(a + b*x)
)/(3*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*(A*b - a*B)*x^(5/2)*(a + b*x))/(5*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]) + (2*B*x^(7/2)*(a + b*x))/(7*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*a^(5/2)*(A*b - a*B)*(a + b*x)*ArcTan[(S
qrt[b]*Sqrt[x])/Sqrt[a]])/(b^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.108314, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {770, 80, 50, 63, 205} \[ \frac{2 x^{5/2} (a+b x) (A b-a B)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a x^{3/2} (a+b x) (A b-a B)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 a^2 \sqrt{x} (a+b x) (A b-a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a^{5/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(A*b - a*B)*Sqrt[x]*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*a*(A*b - a*B)*x^(3/2)*(a + b*x)
)/(3*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*(A*b - a*B)*x^(5/2)*(a + b*x))/(5*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]) + (2*B*x^(7/2)*(a + b*x))/(7*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*a^(5/2)*(A*b - a*B)*(a + b*x)*ArcTan[(S
qrt[b]*Sqrt[x])/Sqrt[a]])/(b^(9/2)*Sqrt[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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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{x^{5/2} (A+B x)}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^{5/2} (A+B x)}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (\frac{7 A b^2}{2}-\frac{7 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{x^{5/2}}{a b+b^2 x} \, dx}{7 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (2 a \left (\frac{7 A b^2}{2}-\frac{7 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{x^{3/2}}{a b+b^2 x} \, dx}{7 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 a (A b-a B) x^{3/2} (a+b x)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 a^2 \left (\frac{7 A b^2}{2}-\frac{7 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{x}}{a b+b^2 x} \, dx}{7 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 a^2 (A b-a B) \sqrt{x} (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a (A b-a B) x^{3/2} (a+b x)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (2 a^3 \left (\frac{7 A b^2}{2}-\frac{7 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{7 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 a^2 (A b-a B) \sqrt{x} (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a (A b-a B) x^{3/2} (a+b x)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (4 a^3 \left (\frac{7 A b^2}{2}-\frac{7 a b B}{2}\right ) \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 )}{7 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 a^2 (A b-a B) \sqrt{x} (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a (A b-a B) x^{3/2} (a+b x)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B x^{7/2} (a+b x)}{7 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 a^{5/2} (A b-a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0652006, size = 120, normalized size = 0.5 \[ \frac{2 (a+b x) \left (\sqrt{b} \sqrt{x} \left (35 a^2 b (3 A+B x)-105 a^3 B-7 a b^2 x (5 A+3 B x)+3 b^3 x^2 (7 A+5 B x)\right )+105 a^{5/2} (a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )\right )}{105 b^{9/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(Sqrt[b]*Sqrt[x]*(-105*a^3*B + 35*a^2*b*(3*A + B*x) - 7*a*b^2*x*(5*A + 3*B*x) + 3*b^3*x^2*(7*A +
5*B*x)) + 105*a^(5/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(105*b^(9/2)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.007, size = 163, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{105\,{b}^{4}} \left ( 15\,B\sqrt{ab}{x}^{7/2}{b}^{3}+21\,A\sqrt{ab}{x}^{5/2}{b}^{3}-21\,B\sqrt{ab}{x}^{5/2}a{b}^{2}-35\,A\sqrt{ab}{x}^{3/2}a{b}^{2}+35\,B\sqrt{ab}{x}^{3/2}{a}^{2}b+105\,A\sqrt{ab}\sqrt{x}{a}^{2}b-105\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){a}^{3}b-105\,B\sqrt{ab}\sqrt{x}{a}^{3}+105\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){a}^{4} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x)

[Out]

2/105*(b*x+a)*(15*B*(a*b)^(1/2)*x^(7/2)*b^3+21*A*(a*b)^(1/2)*x^(5/2)*b^3-21*B*(a*b)^(1/2)*x^(5/2)*a*b^2-35*A*(
a*b)^(1/2)*x^(3/2)*a*b^2+35*B*(a*b)^(1/2)*x^(3/2)*a^2*b+105*A*(a*b)^(1/2)*x^(1/2)*a^2*b-105*A*arctan(x^(1/2)*b
/(a*b)^(1/2))*a^3*b-105*B*(a*b)^(1/2)*x^(1/2)*a^3+105*B*arctan(x^(1/2)*b/(a*b)^(1/2))*a^4)/((b*x+a)^2)^(1/2)/b
^4/(a*b)^(1/2)

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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(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32508, size = 524, normalized size = 2.2 \begin{align*} \left [-\frac{105 \,{\left (B a^{3} - A a^{2} b\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) - 2 \,{\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \,{\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \,{\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt{x}}{105 \, b^{4}}, \frac{2 \,{\left (105 \,{\left (B a^{3} - A a^{2} b\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \,{\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \,{\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt{x}\right )}}{105 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/105*(105*(B*a^3 - A*a^2*b)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(15*B*b^3*x^3
- 105*B*a^3 + 105*A*a^2*b - 21*(B*a*b^2 - A*b^3)*x^2 + 35*(B*a^2*b - A*a*b^2)*x)*sqrt(x))/b^4, 2/105*(105*(B*a
^3 - A*a^2*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (15*B*b^3*x^3 - 105*B*a^3 + 105*A*a^2*b - 21*(B*a*b^2
- A*b^3)*x^2 + 35*(B*a^2*b - A*a*b^2)*x)*sqrt(x))/b^4]

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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(x**(5/2)*(B*x+A)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11492, size = 228, normalized size = 0.96 \begin{align*} \frac{2 \,{\left (B a^{4} \mathrm{sgn}\left (b x + a\right ) - A a^{3} b \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{4}} + \frac{2 \,{\left (15 \, B b^{6} x^{\frac{7}{2}} \mathrm{sgn}\left (b x + a\right ) - 21 \, B a b^{5} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right ) + 21 \, A b^{6} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right ) + 35 \, B a^{2} b^{4} x^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right ) - 35 \, A a b^{5} x^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right ) - 105 \, B a^{3} b^{3} \sqrt{x} \mathrm{sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{4} \sqrt{x} \mathrm{sgn}\left (b x + a\right )\right )}}{105 \, b^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2*(B*a^4*sgn(b*x + a) - A*a^3*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/105*(15*B*b^6*x^
(7/2)*sgn(b*x + a) - 21*B*a*b^5*x^(5/2)*sgn(b*x + a) + 21*A*b^6*x^(5/2)*sgn(b*x + a) + 35*B*a^2*b^4*x^(3/2)*sg
n(b*x + a) - 35*A*a*b^5*x^(3/2)*sgn(b*x + a) - 105*B*a^3*b^3*sqrt(x)*sgn(b*x + a) + 105*A*a^2*b^4*sqrt(x)*sgn(
b*x + a))/b^7