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

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

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Rubi [A]  time = 0.05, antiderivative size = 61, 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 = {794, 648} \begin {gather*} \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}} \end {gather*}

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 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 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 {(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*}

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Mathematica [A]  time = 0.03, size = 37, normalized size = 0.61 \begin {gather*} \frac {2 (x (b+c x))^{5/2} (7 A c-2 b B+5 B c x)}{35 c^2 x^{5/2}} \end {gather*}

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))

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IntegrateAlgebraic [A]  time = 0.51, size = 39, normalized size = 0.64 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{5/2} (7 A c-2 b B+5 B c x)}{35 c^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 76, normalized size = 1.25 \begin {gather*} \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 {gather*}

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))

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giac [B]  time = 0.19, size = 149, normalized size = 2.44 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.05, size = 39, normalized size = 0.64 \begin {gather*} \frac {2 \left (c x +b \right ) \left (5 B c x +7 A c -2 b B \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{35 c^{2} x^{\frac {3}{2}}} \end {gather*}

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)

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maxima [B]  time = 0.62, size = 129, normalized size = 2.11 \begin {gather*} \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 {gather*}

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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