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

Optimal. Leaf size=98 \[ -\frac{\sqrt{x} \sqrt{a+b x} (2 A b-3 a B)}{a b^2}+\frac{(2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{5/2}}+\frac{2 x^{3/2} (A b-a B)}{a b \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0391032, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 50, 63, 217, 206} \[ -\frac{\sqrt{x} \sqrt{a+b x} (2 A b-3 a B)}{a b^2}+\frac{(2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{5/2}}+\frac{2 x^{3/2} (A b-a B)}{a b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0580714, size = 85, normalized size = 0.87 \[ \frac{\sqrt{b} \sqrt{x} (3 a B-2 A b+b B x)-\sqrt{a} \sqrt{\frac{b x}{a}+1} (3 a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*(-2*A*b + 3*a*B + b*B*x) - Sqrt[a]*(-2*A*b + 3*a*B)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x
])/Sqrt[a]])/(b^(5/2)*Sqrt[a + b*x])

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Maple [B]  time = 0.011, size = 201, normalized size = 2.1 \begin{align*}{\frac{1}{2} \left ( 2\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{b}^{2}-3\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) xab+2\,Bx{b}^{3/2}\sqrt{x \left ( bx+a \right ) }+2\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) ab-4\,A{b}^{3/2}\sqrt{x \left ( bx+a \right ) }-3\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}+6\,Ba\sqrt{b}\sqrt{x \left ( bx+a \right ) } \right ) \sqrt{x}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*b^2-3*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2
*b*x+a)/b^(1/2))*x*a*b+2*B*x*b^(3/2)*(x*(b*x+a))^(1/2)+2*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2
))*a*b-4*A*b^(3/2)*(x*(b*x+a))^(1/2)-3*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2+6*B*a*b^(1/
2)*(x*(b*x+a))^(1/2))*x^(1/2)/(x*(b*x+a))^(1/2)/b^(5/2)/(b*x+a)^(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((B*x+A)*x^(1/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.77868, size = 478, normalized size = 4.88 \begin{align*} \left [-\frac{{\left (3 \, B a^{2} - 2 \, A a b +{\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{2 \,{\left (b^{4} x + a b^{3}\right )}}, \frac{{\left (3 \, B a^{2} - 2 \, A a b +{\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{b^{4} x + a b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 12.6464, size = 122, normalized size = 1.24 \begin{align*} A \left (\frac{2 \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{3}{2}}} - \frac{2 \sqrt{x}}{\sqrt{a} b \sqrt{1 + \frac{b x}{a}}}\right ) + B \left (\frac{3 \sqrt{a} \sqrt{x}}{b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{5}{2}}} + \frac{x^{\frac{3}{2}}}{\sqrt{a} b \sqrt{1 + \frac{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*x+a)**(3/2),x)

[Out]

A*(2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(3/2) - 2*sqrt(x)/(sqrt(a)*b*sqrt(1 + b*x/a))) + B*(3*sqrt(a)*sqrt(x)/(
b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*b*sqrt(1 + b*x/a)))

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Giac [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)*x^(1/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

Timed out