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

Optimal. Leaf size=70 \[ \frac {2 \sqrt {b x+c x^2} (2 b B-A c)}{b c^2 \sqrt {x}}-\frac {2 x^{3/2} (b B-A c)}{b c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {788, 648} \begin {gather*} \frac {2 \sqrt {b x+c x^2} (2 b B-A c)}{b c^2 \sqrt {x}}-\frac {2 x^{3/2} (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.49 \begin {gather*} \frac {2 \sqrt {x} (-A c+2 b B+B c x)}{c^2 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.66, size = 43, normalized size = 0.61 \begin {gather*} \frac {2 \sqrt {b x+c x^2} (-A c+2 b B+B c x)}{c^2 \sqrt {x} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 45, normalized size = 0.64 \begin {gather*} \frac {2 \, {\left (B c x + 2 \, B b - A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{c^{3} x^{2} + b c^{2} 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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 51, normalized size = 0.73 \begin {gather*} \frac {2 \, \sqrt {c x + b} B}{c^{2}} + \frac {2 \, {\left (B b - A c\right )}}{\sqrt {c x + b} c^{2}} - \frac {2 \, {\left (2 \, B b - A c\right )}}{\sqrt {b} c^{2}} \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)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(c*x + b)*B/c^2 + 2*(B*b - A*c)/(sqrt(c*x + b)*c^2) - 2*(2*B*b - A*c)/(sqrt(b)*c^2)

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maple [A]  time = 0.05, size = 38, normalized size = 0.54 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-B c x +A c -2 b B \right ) x^{\frac {3}{2}}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}} \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)^(3/2),x)

[Out]

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

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

[Out]

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

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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 )}{{\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^(3/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x)

[Out]

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

[Out]

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

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