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

Optimal. Leaf size=52 \[ \frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {656, 648} \begin {gather*} \frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/Sqrt[x],x]

[Out]

(-4*b*(b*x + c*x^2)^(5/2))/(35*c^2*x^(5/2)) + (2*(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 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]

Rubi steps

\begin {align*} \int \frac {\left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx &=\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {(2 b) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{7 c}\\ &=-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}+\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.60 \begin {gather*} \frac {2 (x (b+c x))^{5/2} (5 c x-2 b)}{35 c^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 54, normalized size = 1.04 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-2 b^3+b^2 c x+8 b c^2 x^2+5 c^3 x^3\right )}{35 c^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(-2*b^3 + b^2*c*x + 8*b*c^2*x^2 + 5*c^3*x^3))/(35*c^2*Sqrt[x])

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fricas [A]  time = 0.39, size = 48, normalized size = 0.92 \begin {gather*} \frac {2 \, {\left (5 \, c^{3} x^{3} + 8 \, b c^{2} x^{2} + b^{2} c x - 2 \, b^{3}\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((c*x^2+b*x)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*c^3*x^3 + 8*b*c^2*x^2 + b^2*c*x - 2*b^3)*sqrt(c*x^2 + b*x)/(c^2*sqrt(x))

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giac [B]  time = 0.19, size = 86, normalized size = 1.65 \begin {gather*} -\frac {2}{105} \, 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 {\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((c*x^2+b*x)^(3/2)/x^(1/2),x, algorithm="giac")

[Out]

-2/105*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*(
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 = 0.04, size = 33, normalized size = 0.63 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-5 c x +2 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((c*x^2+b*x)^(3/2)/x^(1/2),x)

[Out]

-2/35*(c*x+b)*(-5*c*x+2*b)*(c*x^2+b*x)^(3/2)/c^2/x^(3/2)

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maxima [A]  time = 1.47, size = 77, normalized size = 1.48 \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}}{105 \, c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)/(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}}{\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)/x^(1/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/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}}}{\sqrt {x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(1/2),x)

[Out]

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

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