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

Optimal. Leaf size=61 \[ \frac{2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac{2 \left (b x+c x^2\right )^{7/2} (2 b B-9 A c)}{63 c^2 x^{7/2}} \]

[Out]

(-2*(2*b*B - 9*A*c)*(b*x + c*x^2)^(7/2))/(63*c^2*x^(7/2)) + (2*B*(b*x + c*x^2)^(7/2))/(9*c*x^(5/2))

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Rubi [A]  time = 0.0486153, 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} \[ \frac{2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac{2 \left (b x+c x^2\right )^{7/2} (2 b B-9 A c)}{63 c^2 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b*B - 9*A*c)*(b*x + c*x^2)^(7/2))/(63*c^2*x^(7/2)) + (2*B*(b*x + c*x^2)^(7/2))/(9*c*x^(5/2))

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 )^{5/2}}{x^{5/2}} \, dx &=\frac{2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}+\frac{\left (2 \left (-\frac{5}{2} (-b B+A c)+\frac{7}{2} (-b B+2 A c)\right )\right ) \int \frac{\left (b x+c x^2\right )^{5/2}}{x^{5/2}} \, dx}{9 c}\\ &=-\frac{2 (2 b B-9 A c) \left (b x+c x^2\right )^{7/2}}{63 c^2 x^{7/2}}+\frac{2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0336292, size = 44, normalized size = 0.72 \[ \frac{2 (b+c x)^3 \sqrt{x (b+c x)} (9 A c-2 b B+7 B c x)}{63 c^2 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(-2*b*B + 9*A*c + 7*B*c*x))/(63*c^2*Sqrt[x])

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Maple [A]  time = 0.004, size = 39, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 7\,Bcx+9\,Ac-2\,bB \right ) }{63\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(c*x+b)*(7*B*c*x+9*A*c-2*B*b)*(c*x^2+b*x)^(5/2)/c^2/x^(5/2)

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Maxima [B]  time = 1.26518, size = 311, normalized size = 5.1 \begin{align*} \frac{2 \,{\left (35 \, b^{2} c x^{3} + 35 \, b^{3} x^{2} +{\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} x^{2} + 14 \,{\left (3 \, b c^{2} x^{3} + b^{2} c x^{2} - 2 \, b^{3} x\right )} x\right )} \sqrt{c x + b} A}{105 \, c x^{2}} + \frac{2 \,{\left ({\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} x^{3} + 6 \,{\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2} + 21 \,{\left (3 \, b^{2} c^{2} x^{4} + b^{3} c x^{3} - 2 \, b^{4} x^{2}\right )} x\right )} \sqrt{c x + b} B}{315 \, c^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(35*b^2*c*x^3 + 35*b^3*x^2 + (15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*x^2 + 14*(3*b*c^2*x^3 + b^2*
c*x^2 - 2*b^3*x)*x)*sqrt(c*x + b)*A/(c*x^2) + 2/315*((35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 1
6*b^4)*x^3 + 6*(15*b*c^3*x^4 + 3*b^2*c^2*x^3 - 4*b^3*c*x^2 + 8*b^4*x)*x^2 + 21*(3*b^2*c^2*x^4 + b^3*c*x^3 - 2*
b^4*x^2)*x)*sqrt(c*x + b)*B/(c^2*x^3)

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Fricas [B]  time = 1.55173, size = 224, normalized size = 3.67 \begin{align*} \frac{2 \,{\left (7 \, B c^{4} x^{4} - 2 \, B b^{4} + 9 \, A b^{3} c +{\left (19 \, B b c^{3} + 9 \, A c^{4}\right )} x^{3} + 3 \,{\left (5 \, B b^{2} c^{2} + 9 \, A b c^{3}\right )} x^{2} +{\left (B b^{3} c + 27 \, A b^{2} c^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{63 \, 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)^(5/2)/x^(5/2),x, algorithm="fricas")

[Out]

2/63*(7*B*c^4*x^4 - 2*B*b^4 + 9*A*b^3*c + (19*B*b*c^3 + 9*A*c^4)*x^3 + 3*(5*B*b^2*c^2 + 9*A*b*c^3)*x^2 + (B*b^
3*c + 27*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)/(c^2*sqrt(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17691, size = 365, normalized size = 5.98 \begin{align*} \frac{2}{315} \, B c^{2}{\left (\frac{16 \, b^{\frac{9}{2}}}{c^{4}} + \frac{35 \,{\left (c x + b\right )}^{\frac{9}{2}} - 135 \,{\left (c x + b\right )}^{\frac{7}{2}} b + 189 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{3}}{c^{4}}\right )} - \frac{4}{105} \, B 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}{105} \, A c^{2}{\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^{2}{\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{4}{15} \, A b 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^{2}{\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)^(5/2)/x^(5/2),x, algorithm="giac")

[Out]

2/315*B*c^2*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x
+ b)^(3/2)*b^3)/c^4) - 4/105*B*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) - 2/105*A*c^2*(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*(2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 4/15*A*b*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^2*((c*x + b)^(3/2)/c - b^(3/2)/c)