3.14.44 \(\int \frac {(a+b x)^5}{(a c+b c x)^{9/2}} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 (a c+b c x)^{3/2}}{3 b c^6} \]

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Rubi [A]  time = 0.00, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 32} \begin {gather*} \frac {2 (a c+b c x)^{3/2}}{3 b c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^(9/2),x]

[Out]

(2*(a*c + b*c*x)^(3/2))/(3*b*c^6)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+b x)^5}{(a c+b c x)^{9/2}} \, dx &=\frac {\int \sqrt {a c+b c x} \, dx}{c^5}\\ &=\frac {2 (a c+b c x)^{3/2}}{3 b c^6}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.18 \begin {gather*} \frac {2 (a+b x) \sqrt {c (a+b x)}}{3 b c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^(9/2),x]

[Out]

(2*(a + b*x)*Sqrt[c*(a + b*x)])/(3*b*c^5)

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IntegrateAlgebraic [A]  time = 0.06, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 (a c+b c x)^{3/2}}{3 b c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^5/(a*c + b*c*x)^(9/2),x]

[Out]

(2*(a*c + b*c*x)^(3/2))/(3*b*c^6)

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fricas [A]  time = 1.31, size = 23, normalized size = 1.05 \begin {gather*} \frac {2 \, \sqrt {b c x + a c} {\left (b x + a\right )}}{3 \, b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(9/2),x, algorithm="fricas")

[Out]

2/3*sqrt(b*c*x + a*c)*(b*x + a)/(b*c^5)

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giac [B]  time = 1.07, size = 54, normalized size = 2.45 \begin {gather*} \frac {2 \, {\left (3 \, \sqrt {b c x + a c} a - \frac {3 \, \sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}}{c}\right )}}{3 \, b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(9/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 1.05 \begin {gather*} \frac {2 \left (b x +a \right )^{6}}{3 \left (b c x +a c \right )^{\frac {9}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(b*c*x+a*c)^(9/2),x)

[Out]

2/3*(b*x+a)^6/b/(b*c*x+a*c)^(9/2)

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maxima [A]  time = 1.39, size = 18, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (b c x + a c\right )}^{\frac {3}{2}}}{3 \, b c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(9/2),x, algorithm="maxima")

[Out]

2/3*(b*c*x + a*c)^(3/2)/(b*c^6)

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mupad [B]  time = 0.03, size = 17, normalized size = 0.77 \begin {gather*} \frac {2\,{\left (c\,\left (a+b\,x\right )\right )}^{3/2}}{3\,b\,c^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^5/(a*c + b*c*x)^(9/2),x)

[Out]

(2*(c*(a + b*x))^(3/2))/(3*b*c^6)

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sympy [A]  time = 8.31, size = 53, normalized size = 2.41 \begin {gather*} \begin {cases} \frac {2 a \sqrt {a c + b c x}}{3 b c^{5}} + \frac {2 x \sqrt {a c + b c x}}{3 c^{5}} & \text {for}\: b \neq 0 \\\frac {a^{5} x}{\left (a c\right )^{\frac {9}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**(9/2),x)

[Out]

Piecewise((2*a*sqrt(a*c + b*c*x)/(3*b*c**5) + 2*x*sqrt(a*c + b*c*x)/(3*c**5), Ne(b, 0)), (a**5*x/(a*c)**(9/2),
True))

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