3.9.11 \(\int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [811]

Optimal. Leaf size=190 \[ -\frac {2 a (A b-a B) \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{3/2} (A b-a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

2/3*(A*b-B*a)*x^(3/2)*(b*x+a)/b^2/((b*x+a)^2)^(1/2)+2/5*B*x^(5/2)*(b*x+a)/b/((b*x+a)^2)^(1/2)+2*a^(3/2)*(A*b-B
*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(7/2)/((b*x+a)^2)^(1/2)-2*a*(A*b-B*a)*(b*x+a)*x^(1/2)/b^3/((b*x+
a)^2)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {784, 81, 52, 65, 211} \begin {gather*} \frac {2 x^{3/2} (a+b x) (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a \sqrt {x} (a+b x) (A b-a B)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{3/2} (a+b x) (A b-a B) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 65

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 81

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 211

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

Rule 784

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 {x^{3/2} (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^{3/2} (A+B x)}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (\frac {5 A b^2}{2}-\frac {5 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2}}{a b+b^2 x} \, dx}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (2 a \left (\frac {5 A b^2}{2}-\frac {5 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{a b+b^2 x} \, dx}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a (A b-a B) \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 a^2 \left (\frac {5 A b^2}{2}-\frac {5 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}{5 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a (A b-a B) \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 a^2 \left (\frac {5 A b^2}{2}-\frac {5 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \text {Subst}\left (\int \frac {1}{a b+b^2 x^2} \, dx,x,\sqrt {x}\right )}{5 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a (A b-a B) \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{3/2} (A b-a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 100, normalized size = 0.53 \begin {gather*} \frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} \left (15 a^2 B-5 a b (3 A+B x)+b^2 x (5 A+3 B x)\right )-15 a^{3/2} (-A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 129, normalized size = 0.68

method	result	size
risch	\(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +5 B a b x +15 a b A -15 a^{2} B \right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{3} \left (b x +a \right )}+\frac {\left (\frac {2 a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) A}{b^{2} \sqrt {a b}}-\frac {2 a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) B}{b^{3} \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(128\)
default	\(\frac {2 \left (b x +a \right ) \left (3 B \sqrt {a b}\, x^{\frac {5}{2}} b^{2}+5 A \sqrt {a b}\, x^{\frac {3}{2}} b^{2}-5 B \sqrt {a b}\, x^{\frac {3}{2}} a b -15 A \sqrt {a b}\, \sqrt {x}\, a b +15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b +15 B \sqrt {a b}\, \sqrt {x}\, a^{2}-15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \sqrt {a b}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 147, normalized size = 0.77 \begin {gather*} \frac {2 \, {\left (3 \, B b^{2} x^{2} + 5 \, B a b x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (7 \, B a b - 5 \, A b^{2}\right )} x^{2} + 5 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{3} x + a b^{2}\right )}} - \frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {{\left (7 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{2} - A a b\right )} \sqrt {x}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*(3*B*b^2*x^2 + 5*B*a*b*x)*x^(3/2) + (3*(7*B*a*b - 5*A*b^2)*x^2 + 5*(5*B*a^2 - 3*A*a*b)*x)*sqrt(x))/(b^
3*x + a*b^2) - 2*(B*a^3 - A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/3*((7*B*a*b - 5*A*b^2)*x^(3
/2) - 6*(B*a^2 - A*a*b)*sqrt(x))/b^3

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Fricas [A]
time = 2.93, size = 180, normalized size = 0.95 \begin {gather*} \left [-\frac {15 \, {\left (B a^{2} - A a 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 (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*(B*a^2 - A*a*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(3*B*b^2*x^2 + 15*
B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x)*sqrt(x))/b^3, -2/15*(15*(B*a^2 - A*a*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqr
t(a/b)/a) - (3*B*b^2*x^2 + 15*B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x)*sqrt(x))/b^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {3}{2}} \left (A + B x\right )}{\sqrt {\left (a + b x\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.81, size = 133, normalized size = 0.70 \begin {gather*} -\frac {2 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{4} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 5 \, B a b^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 5 \, A b^{4} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a^{2} b^{2} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - 15 \, A a b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a^3*sgn(b*x + a) - A*a^2*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*B*b^4*x^(
5/2)*sgn(b*x + a) - 5*B*a*b^3*x^(3/2)*sgn(b*x + a) + 5*A*b^4*x^(3/2)*sgn(b*x + a) + 15*B*a^2*b^2*sqrt(x)*sgn(b
*x + a) - 15*A*a*b^3*sqrt(x)*sgn(b*x + a))/b^5

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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