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

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

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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])

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*}

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.57 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.10, size = 59, normalized size = 0.61 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-10 A b c+5 A c^2 x+8 b^2 B-4 b B c x+3 B c^2 x^2\right )}{15 c^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 55, normalized size = 0.57 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.18, size = 81, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (B b^{2} - A b c\right )} \sqrt {c x + b}}{c^{3}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} B - 10 \, {\left (c x + b\right )}^{\frac {3}{2}} B b + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A 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 {gather*}

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

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maple [A]  time = 0.06, size = 59, normalized size = 0.61 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-3 B \,c^{2} x^{2}-5 A \,c^{2} x +4 B b c x +10 A b c -8 b^{2} B \right ) \sqrt {x}}{15 \sqrt {c \,x^{2}+b x}\, c^{3}} \end {gather*}

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 = 0.57, size = 75, normalized size = 0.78 \begin {gather*} \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 {gather*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {x \left (b + c x\right )}}\, dx \end {gather*}

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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