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

Optimal. Leaf size=61 \[ \frac{2 B \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2} (2 b B-7 A c)}{35 c^2 x^{5/2}} \]

[Out]

(-2*(2*b*B - 7*A*c)*(b*x + c*x^2)^(5/2))/(35*c^2*x^(5/2)) + (2*B*(b*x + c*x^2)^(5/2))/(7*c*x^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0464435, antiderivative size = 61, normalized size of antiderivative = 1., 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 = {794, 648} \[ \frac{2 B \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2} (2 b B-7 A c)}{35 c^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b*B - 7*A*c)*(b*x + c*x^2)^(5/2))/(35*c^2*x^(5/2)) + (2*B*(b*x + c*x^2)^(5/2))/(7*c*x^(3/2))

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 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{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx &=\frac{2 B \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}+\frac{\left (2 \left (-\frac{3}{2} (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right )\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{7 c}\\ &=-\frac{2 (2 b B-7 A c) \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}+\frac{2 B \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0328249, size = 37, normalized size = 0.61 \[ \frac{2 (x (b+c x))^{5/2} (7 A c-2 b B+5 B c x)}{35 c^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(-2*b*B + 7*A*c + 5*B*c*x))/(35*c^2*x^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 39, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 5\,Bcx+7\,Ac-2\,bB \right ) }{35\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(c*x+b)*(5*B*c*x+7*A*c-2*B*b)*(c*x^2+b*x)^(3/2)/c^2/x^(3/2)

________________________________________________________________________________________

Maxima [B]  time = 1.13604, size = 174, normalized size = 2.85 \begin{align*} \frac{2 \,{\left (5 \, b c x^{2} + 5 \, b^{2} x +{\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} x\right )} \sqrt{c x + b} A}{15 \, c x} + \frac{2 \,{\left ({\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} x^{2} + 7 \,{\left (3 \, b c^{2} x^{3} + b^{2} c x^{2} - 2 \, b^{3} x\right )} x\right )} \sqrt{c x + b} B}{105 \, c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*b*c*x^2 + 5*b^2*x + (3*c^2*x^2 + b*c*x - 2*b^2)*x)*sqrt(c*x + b)*A/(c*x) + 2/105*((15*c^3*x^3 + 3*b*c^
2*x^2 - 4*b^2*c*x + 8*b^3)*x^2 + 7*(3*b*c^2*x^3 + b^2*c*x^2 - 2*b^3*x)*x)*sqrt(c*x + b)*B/(c^2*x^2)

________________________________________________________________________________________

Fricas [A]  time = 1.48886, size = 174, normalized size = 2.85 \begin{align*} \frac{2 \,{\left (5 \, B c^{3} x^{3} - 2 \, B b^{3} + 7 \, A b^{2} c +{\left (8 \, B b c^{2} + 7 \, A c^{3}\right )} x^{2} +{\left (B b^{2} c + 14 \, A b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{35 \, c^{2} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*B*c^3*x^3 - 2*B*b^3 + 7*A*b^2*c + (8*B*b*c^2 + 7*A*c^3)*x^2 + (B*b^2*c + 14*A*b*c^2)*x)*sqrt(c*x^2 + b
*x)/(c^2*sqrt(x))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.15869, size = 201, normalized size = 3.3 \begin{align*} -\frac{2}{105} \, B c{\left (\frac{8 \, b^{\frac{7}{2}}}{c^{3}} - \frac{15 \,{\left (c x + b\right )}^{\frac{7}{2}} - 42 \,{\left (c x + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{2}}{c^{3}}\right )} + \frac{2}{15} \, B b{\left (\frac{2 \, b^{\frac{5}{2}}}{c^{2}} + \frac{3 \,{\left (c x + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x + b\right )}^{\frac{3}{2}} b}{c^{2}}\right )} + \frac{2}{15} \, A c{\left (\frac{2 \, b^{\frac{5}{2}}}{c^{2}} + \frac{3 \,{\left (c x + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x + b\right )}^{\frac{3}{2}} b}{c^{2}}\right )} + \frac{2}{3} \, A b{\left (\frac{{\left (c x + b\right )}^{\frac{3}{2}}}{c} - \frac{b^{\frac{3}{2}}}{c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*B*c*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)/c^3) + 2/15*B
*b*(2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 2/15*A*c*(2*b^(5/2)/c^2 + (3*(c*x + b)^(5
/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 2/3*A*b*((c*x + b)^(3/2)/c - b^(3/2)/c)