3.3.35 \(\int \frac {x^{5/2} (A+B x)}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac {4 \sqrt {b x+c x^2} (4 b B-3 A c)}{3 c^3 \sqrt {x}}+\frac {2 \sqrt {x} \sqrt {b x+c x^2} (4 b B-3 A c)}{3 b c^2}-\frac {2 x^{5/2} (b B-A c)}{b c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 106, 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 = {788, 656, 648} \begin {gather*} \frac {2 \sqrt {x} \sqrt {b x+c x^2} (4 b B-3 A c)}{3 b c^2}-\frac {4 \sqrt {b x+c x^2} (4 b B-3 A c)}{3 c^3 \sqrt {x}}-\frac {2 x^{5/2} (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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 788

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.51 \begin {gather*} \frac {2 \sqrt {x} \left (b (6 A c-4 B c x)+c^2 x (3 A+B x)-8 b^2 B\right )}{3 c^3 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.66, size = 65, normalized size = 0.61 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (6 A b c+3 A c^2 x-8 b^2 B-4 b B c x+B c^2 x^2\right )}{3 c^3 \sqrt {x} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.41, size = 67, normalized size = 0.63 \begin {gather*} \frac {2 \, {\left (B c^{2} x^{2} - 8 \, B b^{2} + 6 \, A b c - {\left (4 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (c^{4} x^{2} + b c^{3} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/3*(B*c^2*x^2 - 8*B*b^2 + 6*A*b*c - (4*B*b*c - 3*A*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^4*x^2 + b*c^3*x)

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giac [A]  time = 0.19, size = 89, normalized size = 0.84 \begin {gather*} -\frac {2 \, {\left (B b^{2} - A b c\right )}}{\sqrt {c x + b} c^{3}} + \frac {4 \, {\left (4 \, B b^{2} - 3 \, A b c\right )}}{3 \, \sqrt {b} c^{3}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} B c^{6} - 6 \, \sqrt {c x + b} B b c^{6} + 3 \, \sqrt {c x + b} A c^{7}\right )}}{3 \, c^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-2*(B*b^2 - A*b*c)/(sqrt(c*x + b)*c^3) + 4/3*(4*B*b^2 - 3*A*b*c)/(sqrt(b)*c^3) + 2/3*((c*x + b)^(3/2)*B*c^6 -
6*sqrt(c*x + b)*B*b*c^6 + 3*sqrt(c*x + b)*A*c^7)/c^9

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maple [A]  time = 0.04, size = 58, normalized size = 0.55 \begin {gather*} \frac {2 \left (c x +b \right ) \left (B \,c^{2} x^{2}+3 A \,c^{2} x -4 B b c x +6 A b c -8 b^{2} B \right ) x^{\frac {3}{2}}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x)

[Out]

2/3*(c*x+b)*(B*c^2*x^2+3*A*c^2*x-4*B*b*c*x+6*A*b*c-8*B*b^2)*x^(3/2)/c^3/(c*x^2+b*x)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, {\left (B c x + B b\right )} \sqrt {c x + b} x^{2}}{3 \, {\left (c^{3} x^{2} + 2 \, b c^{2} x + b^{2} c\right )}} + \int \frac {{\left (3 \, A b c x^{2} - {\left (4 \, B b^{2} + {\left (4 \, B b c - 3 \, A c^{2}\right )} x\right )} x^{2}\right )} \sqrt {c x + b}}{3 \, {\left (c^{4} x^{4} + 3 \, b c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + b^{3} c x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/3*(B*c*x + B*b)*sqrt(c*x + b)*x^2/(c^3*x^2 + 2*b*c^2*x + b^2*c) + integrate(1/3*(3*A*b*c*x^2 - (4*B*b^2 + (4
*B*b*c - 3*A*c^2)*x)*x^2)*sqrt(c*x + b)/(c^4*x^4 + 3*b*c^3*x^3 + 3*b^2*c^2*x^2 + b^3*c*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x)

[Out]

int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**(5/2)*(A + B*x)/(x*(b + c*x))**(3/2), x)

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