3.2.98 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{\sqrt {x}} \, dx\) [198]

Optimal. Leaf size=61 \[ -\frac {2 (2 b B-5 A c) \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}} \]

[Out]

-2/15*(-5*A*c+2*B*b)*(c*x^2+b*x)^(3/2)/c^2/x^(3/2)+2/5*B*(c*x^2+b*x)^(3/2)/c/x^(1/2)

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Rubi [A]
time = 0.03, 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 = {808, 662} \begin {gather*} \frac {2 B \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {2 \left (b x+c x^2\right )^{3/2} (2 b B-5 A c)}{15 c^2 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b*B - 5*A*c)*(b*x + c*x^2)^(3/2))/(15*c^2*x^(3/2)) + (2*B*(b*x + c*x^2)^(3/2))/(5*c*Sqrt[x])

Rule 662

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 808

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) \sqrt {b x+c x^2}}{\sqrt {x}} \, dx &=\frac {2 B \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}+\frac {\left (2 \left (\frac {1}{2} (b B-A c)+\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{5 c}\\ &=-\frac {2 (2 b B-5 A c) \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}\\ \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))^{3/2} (-2 b B+5 A c+3 B c x)}{15 c^2 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(-2*b*B + 5*A*c + 3*B*c*x))/(15*c^2*x^(3/2))

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Maple [A]
time = 0.52, size = 37, normalized size = 0.61

method	result	size
default	\(\frac {2 \left (c x +b \right ) \left (3 B c x +5 A c -2 B b \right ) \sqrt {x \left (c x +b \right )}}{15 c^{2} \sqrt {x}}\)	\(37\)
gosper	\(\frac {2 \left (c x +b \right ) \left (3 B c x +5 A c -2 B b \right ) \sqrt {c \,x^{2}+b x}}{15 c^{2} \sqrt {x}}\)	\(39\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (3 B \,c^{2} x^{2}+5 A \,c^{2} x +b B x c +5 A b c -2 b^{2} B \right )}{15 \sqrt {x \left (c x +b \right )}\, c^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15*(c*x+b)*(3*B*c*x+5*A*c-2*B*b)*(x*(c*x+b))^(1/2)/c^2/x^(1/2)

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Maxima [A]
time = 0.28, size = 45, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (c x + b\right )}^{\frac {3}{2}} A}{3 \, c} + \frac {2 \, {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt {c x + b} B}{15 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(1/2),x, algorithm="maxima")

[Out]

2/3*(c*x + b)^(3/2)*A/c + 2/15*(3*c^2*x^2 + b*c*x - 2*b^2)*sqrt(c*x + b)*B/c^2

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Fricas [A]
time = 1.22, size = 53, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (3 \, B c^{2} x^{2} - 2 \, B b^{2} + 5 \, A b c + {\left (B b c + 5 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, 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)^(1/2)/x^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**(1/2),x)

[Out]

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

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Giac [A]
time = 2.94, size = 60, normalized size = 0.98 \begin {gather*} \frac {2}{15} \, 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}{3} \, A {\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)^(1/2)/x^(1/2),x, algorithm="giac")

[Out]

2/15*B*(2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 2/3*A*((c*x + b)^(3/2)/c - b^(3/2)/c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(1/2),x)

[Out]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(1/2), x)

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