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

Optimal. Leaf size=133 \[ -\frac {16 b^2 \sqrt {b x+c x^2} (6 b B-7 A c)}{105 c^4 \sqrt {x}}+\frac {8 b \sqrt {x} \sqrt {b x+c x^2} (6 b B-7 A c)}{105 c^3}-\frac {2 x^{3/2} \sqrt {b x+c x^2} (6 b B-7 A c)}{35 c^2}+\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c} \]

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Rubi [A]  time = 0.11, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {16 b^2 \sqrt {b x+c x^2} (6 b B-7 A c)}{105 c^4 \sqrt {x}}-\frac {2 x^{3/2} \sqrt {b x+c x^2} (6 b B-7 A c)}{35 c^2}+\frac {8 b \sqrt {x} \sqrt {b x+c x^2} (6 b B-7 A c)}{105 c^3}+\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2*(6*b*B - 7*A*c)*Sqrt[b*x + c*x^2])/(105*c^4*Sqrt[x]) + (8*b*(6*b*B - 7*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2]
)/(105*c^3) - (2*(6*b*B - 7*A*c)*x^(3/2)*Sqrt[b*x + c*x^2])/(35*c^2) + (2*B*x^(5/2)*Sqrt[b*x + c*x^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 \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx &=\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}+\frac {\left (2 \left (\frac {5}{2} (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right )\right ) \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx}{7 c}\\ &=-\frac {2 (6 b B-7 A c) x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}+\frac {(4 b (6 b B-7 A c)) \int \frac {x^{3/2}}{\sqrt {b x+c x^2}} \, dx}{35 c^2}\\ &=\frac {8 b (6 b B-7 A c) \sqrt {x} \sqrt {b x+c x^2}}{105 c^3}-\frac {2 (6 b B-7 A c) x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {\left (8 b^2 (6 b B-7 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x+c x^2}} \, dx}{105 c^3}\\ &=-\frac {16 b^2 (6 b B-7 A c) \sqrt {b x+c x^2}}{105 c^4 \sqrt {x}}+\frac {8 b (6 b B-7 A c) \sqrt {x} \sqrt {b x+c x^2}}{105 c^3}-\frac {2 (6 b B-7 A c) x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 75, normalized size = 0.56 \begin {gather*} \frac {2 \sqrt {x (b+c x)} \left (8 b^2 c (7 A+3 B x)-2 b c^2 x (14 A+9 B x)+3 c^3 x^2 (7 A+5 B x)-48 b^3 B\right )}{105 c^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x) + 3*c^3*x^2*(7*A + 5*B*x) - 2*b*c^2*x*(14*A + 9*B*x)))
/(105*c^4*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.12, size = 83, normalized size = 0.62 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (56 A b^2 c-28 A b c^2 x+21 A c^3 x^2-48 b^3 B+24 b^2 B c x-18 b B c^2 x^2+15 B c^3 x^3\right )}{105 c^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(-48*b^3*B + 56*A*b^2*c + 24*b^2*B*c*x - 28*A*b*c^2*x - 18*b*B*c^2*x^2 + 21*A*c^3*x^2 + 1
5*B*c^3*x^3))/(105*c^4*Sqrt[x])

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fricas [A]  time = 0.41, size = 79, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (15 \, B c^{3} x^{3} - 48 \, B b^{3} + 56 \, A b^{2} c - 3 \, {\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, c^{4} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*B*c^3*x^3 - 48*B*b^3 + 56*A*b^2*c - 3*(6*B*b*c^2 - 7*A*c^3)*x^2 + 4*(6*B*b^2*c - 7*A*b*c^2)*x)*sqrt(
c*x^2 + b*x)/(c^4*sqrt(x))

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giac [A]  time = 0.25, size = 108, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {c x + b}}{c^{4}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} B - 63 \, {\left (c x + b\right )}^{\frac {5}{2}} B b + 105 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} + 21 \, {\left (c x + b\right )}^{\frac {5}{2}} A c - 70 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c\right )}}{105 \, c^{4}} + \frac {16 \, {\left (6 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )}}{105 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-2*(B*b^3 - A*b^2*c)*sqrt(c*x + b)/c^4 + 2/105*(15*(c*x + b)^(7/2)*B - 63*(c*x + b)^(5/2)*B*b + 105*(c*x + b)^
(3/2)*B*b^2 + 21*(c*x + b)^(5/2)*A*c - 70*(c*x + b)^(3/2)*A*b*c)/c^4 + 16/105*(6*B*b^(7/2) - 7*A*b^(5/2)*c)/c^
4

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maple [A]  time = 0.05, size = 83, normalized size = 0.62 \begin {gather*} \frac {2 \left (c x +b \right ) \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-18 B b \,c^{2} x^{2}-28 A b \,c^{2} x +24 B \,b^{2} c x +56 A \,b^{2} c -48 b^{3} B \right ) \sqrt {x}}{105 \sqrt {c \,x^{2}+b x}\, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x)

[Out]

2/105*(c*x+b)*(15*B*c^3*x^3+21*A*c^3*x^2-18*B*b*c^2*x^2-28*A*b*c^2*x+24*B*b^2*c*x+56*A*b^2*c-48*B*b^3)*x^(1/2)
/c^4/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 0.64, size = 98, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (3 \, c^{3} x^{3} - b c^{2} x^{2} + 4 \, b^{2} c x + 8 \, b^{3}\right )} A}{15 \, \sqrt {c x + b} c^{3}} + \frac {2 \, {\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )} B}{35 \, \sqrt {c x + b} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*c^3*x^3 - b*c^2*x^2 + 4*b^2*c*x + 8*b^3)*A/(sqrt(c*x + b)*c^3) + 2/35*(5*c^4*x^4 - b*c^3*x^3 + 2*b^2*c
^2*x^2 - 8*b^3*c*x - 16*b^4)*B/(sqrt(c*x + b)*c^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(1/2),x)

[Out]

int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}} \left (A + B x\right )}{\sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x)**(1/2),x)

[Out]

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

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