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

Optimal. Leaf size=133 \[ -\frac{16 b^2 \sqrt{b x+c x^2} (6 b B-7 A c)}{105 c^4 \sqrt{x}}-\frac{2 x^{3/2} \sqrt{b x+c x^2} (6 b B-7 A c)}{35 c^2}+\frac{8 b \sqrt{x} \sqrt{b x+c x^2} (6 b B-7 A c)}{105 c^3}+\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c} \]

[Out]

(-16*b^2*(6*b*B - 7*A*c)*Sqrt[b*x + c*x^2])/(105*c^4*Sqrt[x]) + (8*b*(6*b*B - 7*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2]
)/(105*c^3) - (2*(6*b*B - 7*A*c)*x^(3/2)*Sqrt[b*x + c*x^2])/(35*c^2) + (2*B*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c)

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Rubi [A]  time = 0.107513, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {794, 656, 648} \[ -\frac{16 b^2 \sqrt{b x+c x^2} (6 b B-7 A c)}{105 c^4 \sqrt{x}}-\frac{2 x^{3/2} \sqrt{b x+c x^2} (6 b B-7 A c)}{35 c^2}+\frac{8 b \sqrt{x} \sqrt{b x+c x^2} (6 b B-7 A c)}{105 c^3}+\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2*(6*b*B - 7*A*c)*Sqrt[b*x + c*x^2])/(105*c^4*Sqrt[x]) + (8*b*(6*b*B - 7*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2]
)/(105*c^3) - (2*(6*b*B - 7*A*c)*x^(3/2)*Sqrt[b*x + c*x^2])/(35*c^2) + (2*B*x^(5/2)*Sqrt[b*x + c*x^2])/(7*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^{5/2} (A+B x)}{\sqrt{b x+c x^2}} \, dx &=\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{\left (2 \left (\frac{5}{2} (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right )\right ) \int \frac{x^{5/2}}{\sqrt{b x+c x^2}} \, dx}{7 c}\\ &=-\frac{2 (6 b B-7 A c) x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{(4 b (6 b B-7 A c)) \int \frac{x^{3/2}}{\sqrt{b x+c x^2}} \, dx}{35 c^2}\\ &=\frac{8 b (6 b B-7 A c) \sqrt{x} \sqrt{b x+c x^2}}{105 c^3}-\frac{2 (6 b B-7 A c) x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c}-\frac{\left (8 b^2 (6 b B-7 A c)\right ) \int \frac{\sqrt{x}}{\sqrt{b x+c x^2}} \, dx}{105 c^3}\\ &=-\frac{16 b^2 (6 b B-7 A c) \sqrt{b x+c x^2}}{105 c^4 \sqrt{x}}+\frac{8 b (6 b B-7 A c) \sqrt{x} \sqrt{b x+c x^2}}{105 c^3}-\frac{2 (6 b B-7 A c) x^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B x^{5/2} \sqrt{b x+c x^2}}{7 c}\\ \end{align*}

Mathematica [A]  time = 0.0577603, size = 75, normalized size = 0.56 \[ \frac{2 \sqrt{x (b+c x)} \left (8 b^2 c (7 A+3 B x)-2 b c^2 x (14 A+9 B x)+3 c^3 x^2 (7 A+5 B x)-48 b^3 B\right )}{105 c^4 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x) + 3*c^3*x^2*(7*A + 5*B*x) - 2*b*c^2*x*(14*A + 9*B*x)))
/(105*c^4*Sqrt[x])

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Maple [A]  time = 0.007, size = 83, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 15\,B{c}^{3}{x}^{3}+21\,A{x}^{2}{c}^{3}-18\,B{x}^{2}b{c}^{2}-28\,Ab{c}^{2}x+24\,B{b}^{2}cx+56\,A{b}^{2}c-48\,{b}^{3}B \right ) }{105\,{c}^{4}}\sqrt{x}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(c*x+b)*(15*B*c^3*x^3+21*A*c^3*x^2-18*B*b*c^2*x^2-28*A*b*c^2*x+24*B*b^2*c*x+56*A*b^2*c-48*B*b^3)*x^(1/2)
/c^4/(c*x^2+b*x)^(1/2)

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Maxima [A]  time = 1.11578, size = 132, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (3 \, c^{3} x^{3} - b c^{2} x^{2} + 4 \, b^{2} c x + 8 \, b^{3}\right )} A}{15 \, \sqrt{c x + b} c^{3}} + \frac{2 \,{\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )} B}{35 \, \sqrt{c x + b} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*c^3*x^3 - b*c^2*x^2 + 4*b^2*c*x + 8*b^3)*A/(sqrt(c*x + b)*c^3) + 2/35*(5*c^4*x^4 - b*c^3*x^3 + 2*b^2*c
^2*x^2 - 8*b^3*c*x - 16*b^4)*B/(sqrt(c*x + b)*c^4)

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Fricas [A]  time = 1.62079, size = 186, normalized size = 1.4 \begin{align*} \frac{2 \,{\left (15 \, B c^{3} x^{3} - 48 \, B b^{3} + 56 \, A b^{2} c - 3 \,{\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{2} + 4 \,{\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{105 \, c^{4} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*c^3*x^3 - 48*B*b^3 + 56*A*b^2*c - 3*(6*B*b*c^2 - 7*A*c^3)*x^2 + 4*(6*B*b^2*c - 7*A*b*c^2)*x)*sqrt(
c*x^2 + b*x)/(c^4*sqrt(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{5}{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**(5/2)*(B*x+A)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**(5/2)*(A + B*x)/sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.14396, size = 149, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (15 \,{\left (c x + b\right )}^{\frac{7}{2}} B - 63 \,{\left (c x + b\right )}^{\frac{5}{2}} B b + 105 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{2} - 105 \, \sqrt{c x + b} B b^{3} + 21 \,{\left (c x + b\right )}^{\frac{5}{2}} A c - 70 \,{\left (c x + b\right )}^{\frac{3}{2}} A b c + 105 \, \sqrt{c x + b} A b^{2} c\right )}}{105 \, c^{4}} + \frac{16 \,{\left (6 \, B b^{\frac{7}{2}} - 7 \, A b^{\frac{5}{2}} c\right )}}{105 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*(c*x + b)^(7/2)*B - 63*(c*x + b)^(5/2)*B*b + 105*(c*x + b)^(3/2)*B*b^2 - 105*sqrt(c*x + b)*B*b^3 + 2
1*(c*x + b)^(5/2)*A*c - 70*(c*x + b)^(3/2)*A*b*c + 105*sqrt(c*x + b)*A*b^2*c)/c^4 + 16/105*(6*B*b^(7/2) - 7*A*
b^(5/2)*c)/c^4