3.2.97 \(\int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=96 \[ \frac {4 b \left (b x+c x^2\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^{3/2}}-\frac {2 \left (b x+c x^2\right )^{3/2} (4 b B-7 A c)}{35 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \]

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {794, 656, 648} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{3/2} (4 b B-7 A c)}{35 c^2 \sqrt {x}}+\frac {4 b \left (b x+c x^2\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^{3/2}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(4*b*B - 7*A*c)*(b*x + c*x^2)^(3/2))/(105*c^3*x^(3/2)) - (2*(4*b*B - 7*A*c)*(b*x + c*x^2)^(3/2))/(35*c^2*
Sqrt[x]) + (2*B*Sqrt[x]*(b*x + c*x^2)^(3/2))/(7*c)

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 656

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*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[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 \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx &=\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (2 \left (\frac {1}{2} (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right )\right ) \int \sqrt {x} \sqrt {b x+c x^2} \, dx}{7 c}\\ &=-\frac {2 (4 b B-7 A c) \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {(2 b (4 b B-7 A c)) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{35 c^2}\\ &=\frac {4 b (4 b B-7 A c) \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {2 (4 b B-7 A c) \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.58 \begin {gather*} \frac {2 (x (b+c x))^{3/2} \left (-2 b c (7 A+6 B x)+3 c^2 x (7 A+5 B x)+8 b^2 B\right )}{105 c^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(8*b^2*B + 3*c^2*x*(7*A + 5*B*x) - 2*b*c*(7*A + 6*B*x)))/(105*c^3*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.09, size = 59, normalized size = 0.61 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{3/2} \left (-14 A b c+21 A c^2 x+8 b^2 B-12 b B c x+15 B c^2 x^2\right )}{105 c^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x + c*x^2)^(3/2)*(8*b^2*B - 14*A*b*c - 12*b*B*c*x + 21*A*c^2*x + 15*B*c^2*x^2))/(105*c^3*x^(3/2))

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fricas [A]  time = 0.43, size = 78, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (15 \, B c^{3} x^{3} + 8 \, B b^{3} - 14 \, A b^{2} c + 3 \, {\left (B b c^{2} + 7 \, A c^{3}\right )} x^{2} - {\left (4 \, B b^{2} c - 7 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, c^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*c^3*x^3 + 8*B*b^3 - 14*A*b^2*c + 3*(B*b*c^2 + 7*A*c^3)*x^2 - (4*B*b^2*c - 7*A*b*c^2)*x)*sqrt(c*x^2
+ b*x)/(c^3*sqrt(x))

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giac [A]  time = 0.24, size = 86, normalized size = 0.90 \begin {gather*} -\frac {2}{105} \, B {\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} \, A {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*B*(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*A*(
2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2)

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maple [A]  time = 0.05, size = 59, normalized size = 0.61 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-15 B \,c^{2} x^{2}-21 A \,c^{2} x +12 B b c x +14 A b c -8 b^{2} B \right ) \sqrt {c \,x^{2}+b x}}{105 c^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(c*x+b)*(-15*B*c^2*x^2-21*A*c^2*x+12*B*b*c*x+14*A*b*c-8*B*b^2)*(c*x^2+b*x)^(1/2)/c^3/x^(1/2)

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maxima [A]  time = 0.61, size = 75, normalized size = 0.78 \begin {gather*} \frac {2 \, {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt {c x + b} A}{15 \, c^{2}} + \frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b} B}{105 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x}\,\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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