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

Optimal. Leaf size=96 \[ \frac {4 b \left (b x+c x^2\right )^{5/2} (4 b B-9 A c)}{315 c^3 x^{5/2}}-\frac {2 \left (b x+c x^2\right )^{5/2} (4 b B-9 A c)}{63 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}} \]

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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 )^{5/2} (4 b B-9 A c)}{63 c^2 x^{3/2}}+\frac {4 b \left (b x+c x^2\right )^{5/2} (4 b B-9 A c)}{315 c^3 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(4*b*B - 9*A*c)*(b*x + c*x^2)^(5/2))/(315*c^3*x^(5/2)) - (2*(4*b*B - 9*A*c)*(b*x + c*x^2)^(5/2))/(63*c^2*
x^(3/2)) + (2*B*(b*x + c*x^2)^(5/2))/(9*c*Sqrt[x])

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 \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx &=\frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}+\frac {\left (2 \left (\frac {1}{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}}{\sqrt {x}} \, dx}{9 c}\\ &=-\frac {2 (4 b B-9 A c) \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}+\frac {(2 b (4 b B-9 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{63 c^2}\\ &=\frac {4 b (4 b B-9 A c) \left (b x+c x^2\right )^{5/2}}{315 c^3 x^{5/2}}-\frac {2 (4 b B-9 A c) \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}\\ \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))^{5/2} \left (-2 b c (9 A+10 B x)+5 c^2 x (9 A+7 B x)+8 b^2 B\right )}{315 c^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(8*b^2*B + 5*c^2*x*(9*A + 7*B*x) - 2*b*c*(9*A + 10*B*x)))/(315*c^3*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.49, size = 59, normalized size = 0.61 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{5/2} \left (-18 A b c+45 A c^2 x+8 b^2 B-20 b B c x+35 B c^2 x^2\right )}{315 c^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x + c*x^2)^(5/2)*(8*b^2*B - 18*A*b*c - 20*b*B*c*x + 45*A*c^2*x + 35*B*c^2*x^2))/(315*c^3*x^(5/2))

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fricas [A]  time = 0.41, size = 102, normalized size = 1.06 \begin {gather*} \frac {2 \, {\left (35 \, B c^{4} x^{4} + 8 \, B b^{4} - 18 \, A b^{3} c + 5 \, {\left (10 \, B b c^{3} + 9 \, A c^{4}\right )} x^{3} + 3 \, {\left (B b^{2} c^{2} + 24 \, A b c^{3}\right )} x^{2} - {\left (4 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{3} \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^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*c^4*x^4 + 8*B*b^4 - 18*A*b^3*c + 5*(10*B*b*c^3 + 9*A*c^4)*x^3 + 3*(B*b^2*c^2 + 24*A*b*c^3)*x^2 - (
4*B*b^3*c - 9*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

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giac [B]  time = 0.21, size = 199, normalized size = 2.07 \begin {gather*} \frac {2}{315} \, B c {\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 {2}{105} \, B 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}{105} \, A 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} \, A 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 )} \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^(1/2),x, algorithm="giac")

[Out]

2/315*B*c*(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) - 2/105*B*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) - 2/105*A*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*A*b*(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 = 0.06, size = 59, normalized size = 0.61 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-35 B \,c^{2} x^{2}-45 A \,c^{2} x +20 B b c x +18 A b c -8 b^{2} B \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 c^{3} 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^(1/2),x)

[Out]

-2/315*(c*x+b)*(-35*B*c^2*x^2-45*A*c^2*x+20*B*b*c*x+18*A*b*c-8*B*b^2)*(c*x^2+b*x)^(3/2)/c^3/x^(3/2)

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maxima [B]  time = 0.63, size = 182, normalized size = 1.90 \begin {gather*} \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} A}{105 \, c^{2} 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} + 3 \, {\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}\right )} \sqrt {c x + b} B}{315 \, c^{3} x^{3}} \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^(1/2),x, algorithm="maxima")

[Out]

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)*A/(c^2*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 - 16*b^4)*x^3 + 3*(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)*sqrt(c*x + b)*B/(c^3*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\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)^(3/2)*(A + B*x))/x^(1/2),x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^(1/2), 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 )}{\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)**(3/2)/x**(1/2),x)

[Out]

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

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