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

Optimal. Leaf size=96 \[ -\frac{2 \sqrt{x} \sqrt{b x+c x^2} (4 b B-5 A c)}{15 c^2}+\frac{4 b \sqrt{b x+c x^2} (4 b B-5 A c)}{15 c^3 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x+c x^2}}{5 c} \]

[Out]

(4*b*(4*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(15*c^3*Sqrt[x]) - (2*(4*b*B - 5*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(15*c
^2) + (2*B*x^(3/2)*Sqrt[b*x + c*x^2])/(5*c)

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Rubi [A]  time = 0.0787564, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {794, 656, 648} \[ -\frac{2 \sqrt{x} \sqrt{b x+c x^2} (4 b B-5 A c)}{15 c^2}+\frac{4 b \sqrt{b x+c x^2} (4 b B-5 A c)}{15 c^3 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(4*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(15*c^3*Sqrt[x]) - (2*(4*b*B - 5*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(15*c
^2) + (2*B*x^(3/2)*Sqrt[b*x + c*x^2])/(5*c)

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

\begin{align*} \int \frac{x^{3/2} (A+B x)}{\sqrt{b x+c x^2}} \, dx &=\frac{2 B x^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{\left (2 \left (\frac{3}{2} (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right )\right ) \int \frac{x^{3/2}}{\sqrt{b x+c x^2}} \, dx}{5 c}\\ &=-\frac{2 (4 b B-5 A c) \sqrt{x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B x^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{(2 b (4 b B-5 A c)) \int \frac{\sqrt{x}}{\sqrt{b x+c x^2}} \, dx}{15 c^2}\\ &=\frac{4 b (4 b B-5 A c) \sqrt{b x+c x^2}}{15 c^3 \sqrt{x}}-\frac{2 (4 b B-5 A c) \sqrt{x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B x^{3/2} \sqrt{b x+c x^2}}{5 c}\\ \end{align*}

Mathematica [A]  time = 0.040393, size = 55, normalized size = 0.57 \[ \frac{2 \sqrt{x (b+c x)} \left (-2 b c (5 A+2 B x)+c^2 x (5 A+3 B x)+8 b^2 B\right )}{15 c^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(8*b^2*B - 2*b*c*(5*A + 2*B*x) + c^2*x*(5*A + 3*B*x)))/(15*c^3*Sqrt[x])

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Maple [A]  time = 0.004, size = 59, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -3\,B{c}^{2}{x}^{2}-5\,A{c}^{2}x+4\,Bbcx+10\,Abc-8\,{b}^{2}B \right ) }{15\,{c}^{3}}\sqrt{x}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(c*x+b)*(-3*B*c^2*x^2-5*A*c^2*x+4*B*b*c*x+10*A*b*c-8*B*b^2)*x^(1/2)/c^3/(c*x^2+b*x)^(1/2)

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Maxima [A]  time = 1.05934, size = 101, normalized size = 1.05 \begin{align*} \frac{2 \,{\left (c^{2} x^{2} - b c x - 2 \, b^{2}\right )} A}{3 \, \sqrt{c x + b} c^{2}} + \frac{2 \,{\left (3 \, c^{3} x^{3} - b c^{2} x^{2} + 4 \, b^{2} c x + 8 \, b^{3}\right )} B}{15 \, \sqrt{c x + b} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c^2*x^2 - b*c*x - 2*b^2)*A/(sqrt(c*x + b)*c^2) + 2/15*(3*c^3*x^3 - b*c^2*x^2 + 4*b^2*c*x + 8*b^3)*B/(sqrt
(c*x + b)*c^3)

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Fricas [A]  time = 1.64604, size = 131, normalized size = 1.36 \begin{align*} \frac{2 \,{\left (3 \, B c^{2} x^{2} + 8 \, B b^{2} - 10 \, A b c -{\left (4 \, B b c - 5 \, A c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{15 \, c^{3} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*B*c^2*x^2 + 8*B*b^2 - 10*A*b*c - (4*B*b*c - 5*A*c^2)*x)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17006, size = 112, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (3 \,{\left (c x + b\right )}^{\frac{5}{2}} B - 10 \,{\left (c x + b\right )}^{\frac{3}{2}} B b + 15 \, \sqrt{c x + b} B b^{2} + 5 \,{\left (c x + b\right )}^{\frac{3}{2}} A c - 15 \, \sqrt{c x + b} A b c\right )}}{15 \, c^{3}} - \frac{4 \,{\left (4 \, B b^{\frac{5}{2}} - 5 \, A b^{\frac{3}{2}} c\right )}}{15 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*(c*x + b)^(5/2)*B - 10*(c*x + b)^(3/2)*B*b + 15*sqrt(c*x + b)*B*b^2 + 5*(c*x + b)^(3/2)*A*c - 15*sqrt(
c*x + b)*A*b*c)/c^3 - 4/15*(4*B*b^(5/2) - 5*A*b^(3/2)*c)/c^3